[
    {
        "Q": "\nAt 4:30,Of what physical quantity or what measure does the Boyle's Law P.V=K gives or indicates us?Like momentum gives us a measure of the force to be applied on a body in motion to oppose its motion.",
        "A": "Look at the units: Pressure = force / area = N/m^2 Volume = m^3 P*V = N/m^2 * m^3 = N*m N*m is the unit of force*displacement. What is force*displacement?",
        "video_name": "x34OTtDE5q8",
        "timestamps": [
            270
        ],
        "3min_transcript": "telling you that the box is squeezed, I'm not telling you whether it did any work, or anything like that-- the same constant is going to be equal to the new pressure times the new volume, which is equal to P2 times V2. You could just have the general relationship: P1 times V1 is equal to P2 times V2, assuming that no work was done, and there was no exchange of energy from outside of the system. In most of these cases, when you see this on an exam, that is the case. The old pressure was 500 pascals times 50 meters cubed. One thing to keep in mind, because this equivalence is not equal, and we're not saying it has to equal some necessary absolute number-- for example, we don't know exactly what this K is, although we could figure it out right now-- as long as you're using one unit for you just have to use the same units. We could have done this same exact problem the exact same way, if instead of meters cubed, they said liters, as long as we had liters here. You just have to make sure you're using the same units on both sides. In this case, we have 500 pascals as the pressure, and the volume is 50 meters cubed. That's going to be equal to the new pressure, P2, times the new volume, 20 meters cubed. Let's see what we can do: we can divide both sides by 10, so we can take the 10 out of there, and we could divide both sides by 2, so that becomes a 250. We we get 250 times 5 is equal to P2, and so P2 is equal to 1250 pascals, and if we kept with the units, you would have seen that. pressure actually increased by 2 1/2, so that gels with what we talked about before. Let's add another variable into this mix-- let's talk about temperature. Like pressure, volume, work, and a lot of concepts that we talk about in physics, temperature is something that you probably are at least reasonably familiar with. How do you view temperature? A high temperature means something is hot, and a low temperature means something is cold, and I think that also gives you intuition that a higher temperature object has more energy. The sun has more energy than an ice cube-- I think that's"
    },
    {
        "Q": "At 19:05, what is the pinkish stuff on the white blood cell (first picture)? Thanks.\n",
        "A": "It is a phagocyte (a kind of white blood cells) which helps kill bacteria.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1145
        ],
        "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from."
    },
    {
        "Q": "at 11:11 sal says the RNA goes to DNA. How does that happen? Also, how does DNA go to RNA in the first place?\n",
        "A": "These are things you learn more about in Biology class. But basically, DNA and RNA are the instructions of the cell. DNA can be copied inside a cell and create more RNA or DNA molecules. RNA can go to DNA in a reverse method - in Sal s video he mentions reverse transcriptase.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            671
        ],
        "3min_transcript": "Everything I talk about, these are specific ways that a virus might work. But viruses really kind of explore-- well different types of viruses do almost every different combination you could imagine of replicating and coding for proteins and escaping from cells. Some of them just bud. And when they bud, they essentially, you can kind of imagine that they push against the cell wall, or the membrane. I shouldn't say cell wall. The cell's outer membrane. And then when they push against it, they take some of the membrane with them. Until eventually the cell will-- when this goes up enough, this'll pop together and it'll take some of the membrane with it. And you could imagine why that would be useful thing to have with you. Because now that you have this membrane, you kind of look like this cell. So when you want to go infect another cell like this, you're not going to necessarily look like a foreign particle. So it's a very useful way to look like something that And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride."
    },
    {
        "Q": "\nIn about 12:09 how can they change our genetic makeup",
        "A": "it can affect a lot of things..like for example the way are mouth is shaped our body or eye color...ect i hope this helped a little x)",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            729
        ],
        "3min_transcript": "And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of"
    },
    {
        "Q": "\nWhat's the red thing at 19:20?",
        "A": "I believe that is part of the infection of the HIV infection. It might be a rupture or infection of the white blood cell from the HIV infection.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1160
        ],
        "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from."
    },
    {
        "Q": "In 13:10 what is a provirus?\n",
        "A": "The genetic material of a virus as incorporated into, and able to replicate with, the genome of a host cell.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            790
        ],
        "3min_transcript": "codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of But if this isn't eerie enough, they estimate-- so if this infects a cell in my nose or in my arm, as this cell experiences mitosis, all of its offspring-- but its offspring are genetically identical-- are going to have this viral DNA. And that might be fine, but at least my children won't get it. You know, at least it won't become part of my species. But it doesn't have to just infect somatic cells, it could infect a germ cell. So it could go into a germ cell. And the germ cells, we've learned already, these are the ones that produce gametes. For men, that's sperm and for women it's eggs. But you could imagine, once you've infected a germ cell, once you become part of a germ cell's DNA, then I'm passing on that viral DNA to my son or my daughter."
    },
    {
        "Q": "\nat 18:38, he says something about a nuclear membrane? Please explain",
        "A": "Nuclear membrane is the membrane that envelops the DNA and nucleolus. In lysogenic cycle, the viral DNA becomes a part of the host s genome.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1118
        ],
        "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from."
    },
    {
        "Q": "That example with the plane at about 9:15 in the video- I don't exactly get it. I've been on a plane multiple times, and I always know if it's moving or not. Am I missing something here?\n",
        "A": "Once you are at cruising altitude, how do you know the plane is moving and the world is not? Maybe the world is moving and you are standing still.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            555
        ],
        "3min_transcript": "The grass is going to stop it from going. And even while it's in the air, it's going to slow down. It's not going to have a constant velocity. Because you have all of these air particles that are going to bump into it and exert force to slow it down. So what was really brilliant about these guys is that they could imagine a reality where you didn't have gravity, where you did not have air slowing things down. And they could imagine that in that reality, something would just keep persisting in its motion. And the reason why Galileo, frankly, was probably good at thinking about that is that he studied the orbits of planets. And he could, or at least he's probably theorized that, hey, maybe there's no air out there. And that maybe that's why these planets can just keep going round and round in orbit. And I should say their speed, because their direction is changing, but their speed never slows down, because there's nothing in the space to actually slow down those planets. Because on some level, it's super-duper obvious. But on a whole other level, it's completely not obvious, especially this moving uniformly straightforward. And just to make the point clear, if gravity disappeared, and you had no air, and you threw a ball, that ball literally would keep going in that direction forever, unless some other unbalanced force acted to stop it. And another way to think about it-- and this is an example that you might see in everyday life-- is, if I'm in an airplane that's going at a completely constant velocity and there's absolutely no turbulence in the airplane. So if I'm sitting in the airplane right over here. And it's going at a constant velocity, completely smooth, no turbulence. There's really no way for me to tell whether that airplane is moving without looking out the window. Let's assume that there's no windows in that airplane. It's going at a constant velocity. And let's say, I can't hear anything. So I can't even hear the engines. There's no way for me to sense that the plane is moving. Because from my frame of reference, it looks completely identical to if I was in that same plane that was resting on the ground. And that's another way to think about it. That it's actually very intuitive that they're similar states, moving at a constant velocity or being at rest. And you really can't tell whether you are one or the other."
    },
    {
        "Q": "\nat 6:04 after the 3rd resonance structure, why isn't the proton next to NO2 shift over to the positively charged C to make the 4th resonance structure and give positive charge on the Carbon that NO2 is attached to?",
        "A": "You can move only electrons to form resonance strictures. If you move the H, you are creating a tautomeric isomer.",
        "video_name": "rC165FcI4Yg",
        "timestamps": [
            364
        ],
        "3min_transcript": "And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge, nucleophile-electrophile. And those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 the formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring, and we have created our product. We have added in our nitro group."
    },
    {
        "Q": "\nAt 10:06, I'm confused as to how creating an ova is different through meiosis then creating sperm. How does the cell know if it is creating an ova or a sperm, and wouldn't only one type of sex cell be made inside each gender? Why would the cell create smaller non-functional ova cells that are not useful? What exactly is the difference between the process of making an ova and a sperm? I am very confused about almost the whole last two minutes of the video. Any help appreciated!",
        "A": "Meiosis is basically the same for males and females, both males and females undergo the process but males make sperm and females make eggs (ova). There is effectively just a second cell division without replicating the DNA so that the germ cells end up with one copy of each chromosome instead of two (as happens in mitosis). The egg cell divides dispropotionately so that the one big cell has enough resources to support itself once it is fertilized, plus you don t need a lot of eggs.",
        "video_name": "TX7-Kdn6lJQ",
        "timestamps": [
            606
        ],
        "3min_transcript": "and were then pulled in half, but not here. In meiosis, each chromosome lines up next to it's homologous pair partner that it's already swapped a few genes with. Now, the homologous pairs get pulled apart and migrate to either end of the cell and that's anaphase I. The final phase of the first round, telophase I rolls out in pretty much the same way as mitosis. The nuclear membrane reforms, the nucleoli form within them, the chromosomes fray out back into chromatin, a crease forms between the two new cells called cleavage and then the two new nuclei move apart from each other, the cells separate in a process called cytokinesis, literally again, cell movement and that is the end of round one. We now have two haploid cells, each with 23 double chromosomes that are new, unique combinations of the original chromosome pairs. In these new cells, the chromosomes are still duplicated and still connected at the centromeres. They still look like X's, but remember, the aim is to end up with four cells. Here, the process is exactly the same as mitosis, except that the aim here isn't to duplicate the double chromosomes, but instead to pull them apart into separate single strand chromosomes. Because of this, there's no DNA replication involved in prophase II. Instead, the DNA just clumps up again into chromosomes and the infrastructure for moving them, the microtubules are put back in place. In metaphase II, the chromosomes are moved into alignment into the middle of the cell and in anaphase II, the chromotids are pulled apart into separate single chromosomes. The chromosomes uncoil into chromatin, the crease form in cleavage and the final separation of cytokinesis then mark the end of telophase II. From one original cell with 46 chromosomes, we now have four new cells with 23 single chromosomes each. If these are sperm, all four of the resulting cells are the same size, but they each have slightly different genetic information and half will be for making girls and half will be for making boys, but if this is the egg making process, and the result is only one egg. To rewind a little, during telophase I, more of the inner goodness of a cell, the cytoplasm, the organelles heads into one of the cells that gets split off then to the other one. In telophase II, when it's time to split again, the same thing happens with more stuff going into one of the cells than the other. This big ol' fat remaining cell becomes the egg with more of the nutrients and cytoplasm and organelles that it will take to make a new embryo. The other three cells that were produced, the little ones, are called polar bodies and they're totally useless in people, though they are useful in plants. In plants, those polar bodies actually, also get fertilized too and they become the endosperm. That's the starchy, protein-ey stuff that we grind into wheat, or pop into popcorn and it's basically the nutrients that feed the plant embryo, the seed. And that's all there is to it. I know you were probably were really excited when I started talking about reproduction, but then I rambled on for a long time"
    },
    {
        "Q": "At 7:17, why are the X chromosomes shown to have crossed over if they can't cross over?\n",
        "A": "Hank explains that since the chromosomes are the same, they can cross over. The XY pair can t cross over because they are different chromosomes.",
        "video_name": "TX7-Kdn6lJQ",
        "timestamps": [
            437
        ],
        "3min_transcript": "Each double chromosome has two chromotids. Here in meiosis prophase I includes two additional and very important steps, crossover and homologous recombination. Remember, the point here, is to end up with four sex cells that each have just one single chromosome from each of the homologous pairs, but unlike in mitosis where all the copies end up the same, here, every copy is going to be different from the rest. Each double chromosome lines up next to it's homologue, so there's your mother's version lined up right next to your father's version of the same chromosome. Now, if you look, you'll see that these two double chromosomes, each with two chromotids, add up to four chromotids. Now watch, one chromotid from each X gets tangled up with the other X. That's crossover and while they're all tangled up, they trade sections of DNA. That's the recombination. The sections that they're trading are from the same location on each chromosome, so one is giving up its genetic code for, it's getting the other chromosomes genes for that trait. Now this is important, what just happened here, creating new gene combinations on a single chromosome. It's the whole point of reproducing this way. Life might be a lot less stressful if we could just clone ourselves, but then we'd also clone all our bad gene combinations and we wouldn't be able to change and adapt to our environment. Remember, that one of the pillars of natural selection is variation and this is a major source of that variation. What's more? Since all of the four chromotids have swapped some DNA segment at random, that means that all four chromotids are now different. Later on in the process, each chromotid will end up in a separate sex cell and that's why all eggs produced by the same woman have a slightly different genetic code. The same for sperm in men and that's why my brother John and I look different even though we're both made from the same two sets of DNA, because of the luck of the genetic draw that happens in recombination, I got this mane of luscious hair and John was stuck with his trash brown puff and don't forget But then, of course, there is that one pair of chromosomes that doesn't always go through the crossover or recombination and that's the 23rd pair and those are your sex chromosomes. If you're female, you have two matched, beautiful, fully capable chromosomes. They are your X chromosomes. Since they are the same, they can do the whole crossover and recombination thing, but if you, like me, are a male, you get one of those X chromosomes and another from your dad that's kind of ugly and short and runted and doesn't have a lot of genetic information on it. During prophase, the X wants nothing to do with the little Y because they're not homologous, so they don't match up and because the XY pairs on these chromosomes will split later into single chromotids, half of the four resulting sperm will be X, leading to female offspring and half will be Y, leading to male offspring. Now, what comes next is another kind of amazing feat of alignment. This is metaphase I and in mitosis you might recall that all of the chromosomes"
    },
    {
        "Q": "\nat 4:39 when Sal tells us the magnetic field direction is it always like that? is the magnetic field given which hasn't been drawn in the videos ?",
        "A": "the magnetic field doesn t have to be to the left, it depends on where the source of it is, he s just arbitrarily chosen it to come from the left for this example.. is that what you meant?",
        "video_name": "XMkUDyl1ZRo",
        "timestamps": [
            279
        ],
        "3min_transcript": "Now this is when we're looking at top on, where it's popping out of the screen, where it's above the screen. And now we've rotated 180 degrees and this side is on this side, right? Let me pick a suitable color. If this side was green. Now this side, we flip the whole thing over 180 degrees. And now something interesting happens. Remember, before we had this commutator and everything, if we just flipped it over, the current, because before when we didn't have the commutator, the current here was flowing down here, up here. And before the commutator, we had the current flowing down here and up here. And so we were switching directions. And so you would have had this thing that would never completely rotate. It would just keep flipping over, right? Which may be useful for, I don't know, if you wanted to flip things. But it's not useful as a motor. So what happens here? Now this side, all of a sudden instead of being connected to And this green side is now connected to this lead. So something interesting happens. Now the current on the left side is still flowing down, right, and the current on the right side is still flowing up. So we're back to this configuration except that this contraption has flipped over. The brown side is now on the left and the green side is now on the right. And what that allows is that those net torques are still going in that same rotational direction. Use your right hand rule. The current is flowing down here. So if your magnetic field is coming to the left, then the net force is going to be down there and it's going to be up there. And so we can continue indefinite, and we solve our other problem. That we will never keep twisting these wires here. So now using the commutator, we have essentially created an electric motor. And remember I drew that little thing, that could be like the axle. Maybe that turns the wheels or something. So if you have a constant magnetic field and you just by using this commutator which, as soon as you get to that when you go a little bit past vertical, a little bit past 90 degrees, it switches the direction of the current. So on the left hand side you always have the current coming down, and on the right hand side you always have the current going up. So that the net torque is always going to be pushing, is always going to be rotating this contraption down on the left hand side and up on the right hand side. Coming out of the page on the right hand side and then down And you could actually turn a wheel now. You could create an electric car. So that is the basics really of how electric motors are created. Well, there's another way you could have done it. You didn't have to use the commutator. One methodology you could have used is you could have had the magnetic field going until you get to this point, and then you turn off the magnetic field, right? And maybe you wait for this situation to go all the way"
    },
    {
        "Q": "At 7:50 when he's talking about fluid being pushed in the cochlea, does he just mean that there's a pressure wave in the fluid filling the cochlea? Or is liquid actually sloshing back and forth in it?\n",
        "A": "There really is fluid in there! It s called endolymph (in the middle part [scala media] where the hair cells project) and perilymph (in the scalae vestibuli and tympani).",
        "video_name": "6GB_kcdVMQo",
        "timestamps": [
            470
        ],
        "3min_transcript": "and the next thing that they hit is the eardrum, or tympanic membrane. So, the next thing that they hit is the eardrum. What the eardrum does is it actually starts to vibrate. As this pressurized sound wave hits the eardrum, the eardrum starts vibrating back and forth. When it's vibrating back and forth, it actually causes these little bones - there are three little bones here, one, two and three - it causes these three little bones to vibrate. The first bone is known as the malleus. The second bone is known as the incus, and the third little bone over here is known as the stapes. Let's just recap real quick. The sound waves come in, get funneled by the pinna into the external auditory meatus, otherwise known as the auditory canal, then hit the eardrum, otherwise known as the tympanic membrane, This vibration causes three little bones, known as the malleus, incus and stapes to vibrate back and forth accordingly. The next thing that happens is the stapes is attached to this oval window over here. It's known as the elliptical window, which I'm underlining here. It's also known as the oval window. This oval window starts to vibrate back and forth as well. The next thing that happens is there's actually fluid, so this structure that the oval window is attached to is known as the cochlea. This round structure right here is known as the cochlea. Inside the cochlea is a bunch of fluid. As the oval window gets pushed inside and outside of the cochlea by the stapes, it actually pushes the fluid. It causes the fluid to be pushed this way, and causes the fluid to go all the way around the cochlea. It keeps going all the way around the cochlea, until it reaches the tip of the cochlea. what does it do? The only thing it can do is go back. Now the fluid is gonna have to go back. Let's just follow this green line over here. The fluid moves back towards where it came, but it actually doesn't go back to the oval window. It actually goes to this other window known as the circular, or round, window. Let me just fix that, so it goes to this circular or round window. It causes the round window to get pushed out. This basically keeps happening, so the fluid moves all the way to the tip of the cochlea, all the way back out, and back and forth, and back and forth, until the energy of this sound wave - eventually the fluid stops moving - all that energy is dissipated. Meanwhile, hair cells inside the cochlea are being pushed back and forth, and that transmits an electrical impulse"
    },
    {
        "Q": "At 9:20, when we are try to figure out the direction of the induced current are we referring to the conventional current or the actual electron flow? This becomes a point of confusion in a number of my homework problems for my AP class, do physicists generally stick to conventional current for all problems having to do with inducing a current?\n",
        "A": "Current always refers to conventional current unless it says electron current.",
        "video_name": "9q-T8o1HUcw",
        "timestamps": [
            560
        ],
        "3min_transcript": "amperes, 2.5 amperes. So we now know the magnitude of the current that's going to be induced while we have this change in flux, remember this is going to happen while, over the course of those four seconds, as we have this rate of change of flux, this average rate of change of flux, which we'll assume is the actual rate of change of flux, we're assuming that it's changing at a constant rate and so while it is changing we were just able to figure out that it would induce a current of 2.5 amperes. Now the next question we should ask ourselves and this is where this little negative comes in, is a reminder for us to use Lenz's law is, well which direction is that current going to go in? Is it going to go in, let me pick two orientations, is it going to go in a, is it going to go in a, in a clockwise direction, is it going to go that way over the course of this change in flux or is it going to go in a counterclockwise direction, And to think about that we just have to use the right hand rule, take our right hand, point our thumb in the direction of the proposed direction of the current and so if we went with this one, our right hand, our right hand would look like this, I'm literally taking my left hand out and-- I mean my right hand out and I'm drawing it and I'm looking at it to think about what would happen, so that's my right hand so if I use the right hand if the current went in this direction then it would induce a magnetic field that went, that went like this and so if the current went in this direction the magnetic field it induces inside the surface would only reinforce the change in flux so it would only add to the flux so, and it's going in the same direction as the change in flux, which would just keep us, you know as we talked about in the Lenz's law video, that would turn into just this source of energy that comes out of nowhere and defies the law of conservation of energy so this absolutely not, is not going to be the direction is going to be in a clockwise one. So the current, the 2.5 ampere current is going to flow, is going to flow like that, and we're done! By thinking about our change in flux and how long it's taking us, we were able to figure out not only the magnitude of the current, we were able to figure out the orientation of the direction that it's actually going to flow in."
    },
    {
        "Q": "Hmm im not sure if the explanation at 9:20 is sound. Would there be a magnetic flux outside of the circle also, then the circuit could not \"countertact (lenz)\" in any way. If the B Field created on the inside counteracts it will strengthen in on the outside ?\n",
        "A": "We are only talking about the flux THROUGH the coil. The field lines (flux) are passing only inside the coil in this case. This is just a theoretical problem. You can t really do this irl",
        "video_name": "9q-T8o1HUcw",
        "timestamps": [
            560
        ],
        "3min_transcript": "amperes, 2.5 amperes. So we now know the magnitude of the current that's going to be induced while we have this change in flux, remember this is going to happen while, over the course of those four seconds, as we have this rate of change of flux, this average rate of change of flux, which we'll assume is the actual rate of change of flux, we're assuming that it's changing at a constant rate and so while it is changing we were just able to figure out that it would induce a current of 2.5 amperes. Now the next question we should ask ourselves and this is where this little negative comes in, is a reminder for us to use Lenz's law is, well which direction is that current going to go in? Is it going to go in, let me pick two orientations, is it going to go in a, is it going to go in a, in a clockwise direction, is it going to go that way over the course of this change in flux or is it going to go in a counterclockwise direction, And to think about that we just have to use the right hand rule, take our right hand, point our thumb in the direction of the proposed direction of the current and so if we went with this one, our right hand, our right hand would look like this, I'm literally taking my left hand out and-- I mean my right hand out and I'm drawing it and I'm looking at it to think about what would happen, so that's my right hand so if I use the right hand if the current went in this direction then it would induce a magnetic field that went, that went like this and so if the current went in this direction the magnetic field it induces inside the surface would only reinforce the change in flux so it would only add to the flux so, and it's going in the same direction as the change in flux, which would just keep us, you know as we talked about in the Lenz's law video, that would turn into just this source of energy that comes out of nowhere and defies the law of conservation of energy so this absolutely not, is not going to be the direction is going to be in a clockwise one. So the current, the 2.5 ampere current is going to flow, is going to flow like that, and we're done! By thinking about our change in flux and how long it's taking us, we were able to figure out not only the magnitude of the current, we were able to figure out the orientation of the direction that it's actually going to flow in."
    },
    {
        "Q": "at 11:05 can a system ever truly have zero internal energy. Wouldn't the system itself contain energy? This sounds like a black hole situation. Is it possible for the balloon to have an absence of molecules and atoms...no internal energy at all? Or is this just hypothetical to make the equation simpler?\n",
        "A": "At 11:05 this was referring to no change in energy, not no energy.",
        "video_name": "aOSlXuDO4UU",
        "timestamps": [
            665
        ],
        "3min_transcript": "W is work done. Now, the system did work to something else. It didn't have work done to it. So if this is work done to the system, and it did work, then this is going to be a minus 10. Minus 10 joules. And then you solve both sides of this. You add 10 to both sides and you get 10 is equal to Q, which is exactly what we got up here. But this can get confusing sometimes, because you're like, oh, is this heat that the system did? Is this heat that was added to the system or taken away? The convention tends to be that this is heat added. But then sometimes it's confusing. Is this the work done to the system or work done by? And that's why I like just doing it this way. If the system does work it loses energy. If the system has work done to it, it gains energy. So let's do another problem. other formula that you'll sometimes see. Delta U is equal to Q minus the work that the system does by the system-- Work done by the system. And in this case, once again, change in internal energy was 0. That is equal to heat added to the system, minus the work done. So minus-- I told you at the beginning of the problem that the system did 10 joules of work-- so minus the work done. Minus 10 joules. We get the same situation up here from two different formulas. And we got 10 is equal to Q. Either way, the heat added to the system is 10 joules. Let's do one more. Let's say that, I don't know, 5 joules of heat taken away And let's say that 1 joule of work done on system. So maybe we're compressing the balloon on the system. What is our change in internal energy? Let's just figure out our change. So the way I think of it is 5 joules of heat taken away from the system, that's going to reduce our internal energy by minus 5. And if 1 joule work is done onto the system, we're putting energy into it so that'll be plus 1. So minus 5, plus 1, is going to be minus 4. Or enter our change in internal energy is minus 4 joules. Now we could have done that a little bit more formally with the formula, change in internal energy is equal to heat added to the system, plus work done on the system."
    },
    {
        "Q": "\nAt 6:43 Sal said that change in interal energy=Q+W is the definition for internal energy can you give me a more precise definition for it?",
        "A": "Internal energy is all the energy contained in an object, including both kinetic energy and potential energy.",
        "video_name": "aOSlXuDO4UU",
        "timestamps": [
            403
        ],
        "3min_transcript": "transfer amount? No, wire transfer was how money was deposited or taken Likewise, it didn't have a checking deposit account, so I can't really-- it just seems weird to me to say change in wire transfer is $10, or change in check deposits is $20 or minus $20. Would you say, I made a $10 wire transfer and I paid $20 in checks. So I had a net change in my bank account of $10. Likewise, I say how much work was done to me or I did, which is essentially a deposit or withdrawal of energy. Or I could say much heat was given to me or how much heat was released, which is another way of depositing or withdrawing energy from my energy bank account. So that's why I like to stick to this. And I like to stay away from this. And just like I said, you can't say how much heat is in the system. So someone will say, oh, how much heat is in this? There's no heat state variable for that system. You have internal energy. The closest thing to heat, we'll talk about it in a future video, is enthalpy. Enthalpy is essentially a way of measuring how much heat is in a system, but we won't talk about that just now. And you can't say how much work is in a system. You can't say, oh I have x amount of work in a system. The system can do work or have work done to it, but there's not a certain amount of work, because that energy in the system could be all used for work, it could all be used for heat, it could all be used for a ton of different things. So you can't say those things. And that's why I don't like treating them like state variables, or state functions. So with that said, this is our definition. Let's do a couple of simple problems, just to give you intuition. And I really want to make you comfortable. My real goal is to make you comfortable with, when to know to use plus or minus on the work. And the best way to do it is not to memorize a formula, but just to kind of think about what's happening. So let's say that I have some system here and, I don't know, And let's say that I have no change in internal energy. Internal energy is 0. And for our purposes we can kind of view it as the kinetic energy of the particles haven't changed. And let's say by expanding a bit, by my balloon expanding a bit-- I did some work. I'll do this in more detail in the next video. So the system does 10 joules of work. My question is, how much heat was added or taken away from the system? So the way I can think about it-- you don't even have to write the formula down, or you can write it-- you could say, look, the internal energy, the amount of energy in the system hasn't changed. The system did 10 joules of work. So that's energy going out of the system. It did 10 joules. So 10 joules went out in the form of work."
    },
    {
        "Q": "\nAt 2:14 Sal says that the egg cell is also a gamete. Isn't he wrong because a gamete is haploid while the egg cell, as a result of the fertilization, is diploid ? He confirms what I say at 4:45.",
        "A": "The egg is not diploid. The egg and the sperm cell are gametes and haploid. The zygote is the result of the fertilization of the egg cell by the sperm cell. The zygote is diploid.",
        "video_name": "dNp7vErqlaA",
        "timestamps": [
            134,
            285
        ],
        "3min_transcript": "Voiceover: So let's talk a little bit about how we all came into being. What we see right over here, this is a picture of a sperm cell fusing with an egg cell. So that's a sperm cell and this is an egg cell or we could call this an ovum. And even this scene depicted right over here, this is the end of an epic competition because this sperm cell is one of one of two to three hundred million that is vying for this ovum. So there's two to three hundred million of these characters and they're all vying for this ovum and the one that you see that's about to fuse for it, this is the winner of this incredibly - remember two to three hundred, 200 million to 300 million sperm are trying to get here so this is a major victory and to some degree we should all feel pretty good about ourselves because we are all the by-product of that that won this race getting to our mother's ovum. So the sperm cell came from our father and the egg cell, this is all happening inside of our mothers, the egg cell is from our mother. Now, once this happens, let's talk a little bit about the terminology. So once these two fuse, or the process of them fusing, we call that fertilization. Fertilization. And it produces a cell that then differentiates into all of the cells of our body, so you can imagine that this is an important process. So let's make sure that we understand the different terminology, the different words for the different things that are acting in this process. So each of these sex cells, I guess we could say, So this right over here is a gamete and the ovum is a gamete, the egg cell is also a gamete. And as we'll see, each gamete has half the number of chromosomes as your body cells or most of the somatic cells of your body so outside of your sex cells that might be in your ovaries or your testes, depending on whether you're male or female, these have half the number so let's dig a little bit deeper into what I mean there. So let's just do a blow up of this sperm cell right over here, so a blow up of a sperm cell and I'm not going to draw it to scale, you see the sperm cell is much smaller than the egg cell but just to get a sense, so let me draw the nucleus"
    },
    {
        "Q": "At 4:40, is it ever possible for an ovum to carry a Y chromosome?\n",
        "A": "Adding on to what Mubashra Igbal said, because the female has two X chromosomes, it is only possible to get an X from the female. The X chromosome is required for life so even if they could give a Y and if the male gave a Y, the baby would not survive.",
        "video_name": "dNp7vErqlaA",
        "timestamps": [
            280
        ],
        "3min_transcript": "If we're talking about a human being, and I'm assuming you are a human being, so that might be of interest to you, this will have 23 chromosomes from your father so let's do them. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and for the 23rd one, that's going to be your sex-determining chromosome so if your father contributes an x, you are going to be female, if your father contributes a y, you are going to be male. So these are the chromosomes in the male gamete or I guess I should say the gamete that your father's contributing, the sperm. So this is a gamete right over here and that's going to fuse with the egg, the ovum that your mother is contributing and once again, I'm not drawing that to scale. So this is the egg, and let me draw it's nucleus. none of this is drawn to scale. And your mother is also going to contribute 23 chromosomes. So one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and then she will contribute an x chromosome for the sex determining so your sex determining chromosomes are going to be xy, you're going to be male, if this was xx, you're going to be female so this is also a gamete here. So a gamete is the general term for either a sperm or an egg. Now once these two things are fused, what do we have? Once they're fused, then we're going to have you could say a fertilized egg but we are going to call that a zygote so let me draw that. I'm going to do this in a new color, and I'm running out of space and I want this all to fit and so let me draw the nucleus of the zygote, I'm going to make the nucleus fairly large so that we can focus on the chromosomes in it, once again none of this is drawn to scale. So you're going to have the 23 chromosomes from your father, so let me do that. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23, and then the 23 chromosomes from your mother. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23"
    },
    {
        "Q": "At 7:06 The drawing shows that we have 46 chromosomes, but do we los 23 of those chromosomes when sperm connects with the ovum\n",
        "A": "Normal cells have 46 chromosomes or 23 pairs of chromosomes. Normal cell devision, mitosis, produces 2 cells with 46 chromosomes but during the creation of sex cells, meiosis, cells with 23 chromosomes are produced so that when the two sex cells come together they produce a cell with 46 chromosomes.",
        "video_name": "dNp7vErqlaA",
        "timestamps": [
            426
        ],
        "3min_transcript": "and so let me draw the nucleus of the zygote, I'm going to make the nucleus fairly large so that we can focus on the chromosomes in it, once again none of this is drawn to scale. So you're going to have the 23 chromosomes from your father, so let me do that. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23, and then the 23 chromosomes from your mother. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23 And as you might have notice, I've drawn them in pairs so you now have a total, let me make it clear, you have 23 chromosomes here, 23 chromosomes in the sperm, you have 23 chromosomes in the egg and now you have 46 chromosomes in the fertilized egg, 46 chromosomes, and now that we have a full contingent of chromosomes and then this cell can now keep replicating, keep splitting and differentiating into all of what makes you, you, we call this right over here, we call this a zygote. So one way to think about it, the gametes are the sex cells that have half the number of chromosomes and the zygote is the cell that's now ready to differentiate into an actual organism that has double the number that has 46 chromosomes, and you see that I've made them in pairs and these pairs, we call these homologous pairs and in each of these pairs, this is a pair of homologous chromosomes. So what does that mean? Well that means that in general, these two chromosomes, you got one from your father, one from your mother, they code for the same things, they code for the same proteins but there are different variants of how they code for those proteins, those traits that you have so gross oversimplification is, let's say that there is a gene on, that one from your father that helps code for hair color well there would be a similar, there would be another variant of that gene on the chromosome from your mother"
    },
    {
        "Q": "\nAt 2:10, Sal talks about how if you had a mono-North pole on top of a mono-South pole, the mono-North pole would go around and get under the mono-South pole. He does this to describe magnetic fields, but what I don't get is why the mono-North pole doesn't just go strait down-strait to the mono-South pole. I don't understand why the field goes around rather than strait down. Somebody please explain.",
        "A": "magnetic field lines actually pass through the object, they do not stop at the poles. some field lines are actually passing straight down.",
        "video_name": "NnlAI4ZiUrQ",
        "timestamps": [
            130
        ],
        "3min_transcript": "We know a little bit about magnets now. Let's see if we can study it further and learn a little bit about magnetic field and actually the effects that they have on moving charges. And that's actually really how we define magnetic field. So first of all, with any field it's good to have a way to visualize it. With the electrostatic fields we drew field lines. So let's try to do the same thing with magnetic fields. Let's say this is my bar magnet. This is the north pole and this is the south pole. Now the convention, when we're drawing magnetic field lines, is to always start at the north pole and go towards the south pole. And you can almost view it as the path that a magnetic north monopole would take. So if it starts here-- if a magnetic north monopole, even though as far as we know they don't exist in nature, although they theoretically could, but let's just say for the sake of argument that we do have a magnetic north monopole. If it started out here, it would want to run away from this north pole and would try to get to the south pole. something like this. If it started here, maybe its path would look something like this. Or if it started here, maybe its path would look something like this. I think you get the point. Another way to visualize it is instead of thinking about a magnetic north monopole and the path it would take, you could think of, well, what if I had a little compass here? Let me draw it in a different color. Let's say I put the compass here. That's not where I want to do it. Let's say I do it here. The compass pointer will actually be tangent to the field line. So the pointer could look something like this at this point. It would look something like this. And this would be the north pole of the pointer and this would be the south pole of the pointer. Or you could-- that's how north and south were defined. People had compasses, they said, oh, this is the north seeking pole, and it points in that direction. of the larger magnet. And that's where we got into that big confusing discussion of that the magnetic geographic north pole that we're used to is actually the south pole of the magnet that we call Earth. And you could view the last video on Introduction to Magnetism to get confused about that. But I think you see what I'm saying. North always seeks south the same way that positive seeks negative, and vice versa. And north runs away from north. And really the main conceptual difference-- although they are kind of very different properties-- although we will see later they actually end up being the same thing, that we have something called an electromagnetic force, once we start learning about Maxwell's equations and relativity and all that. But we don't have to worry about that right now. But in classical electricity and magnetism, they're kind of a different force. And the main difference-- although you know, these field lines, you can kind of view them as being similar-- is that magnetic forces always come in dipoles, soon. while you could have electrostatic forces that are monopoles."
    },
    {
        "Q": "\nIs the formula he used, the same as the formula for the Lorentz power? At 8:10 he defined the formula, and indicated the unit for B as Tesla. In my book, the formula for the Lorentz power is given as:\nF=B*I*L.\nF= Lorentz power\nB= magnetic induction\nI= electric current\nL= length of the magnetic field\n\nIt's a bit confusing which formula I should use. Can somebody explain it please?",
        "A": "I think you mean Lorentz force, not power. This is a specific application of the Lorentz force.",
        "video_name": "NnlAI4ZiUrQ",
        "timestamps": [
            490
        ],
        "3min_transcript": "I know I'm confusing you at this point, so let's play around with it and do some problems. But before that, let's figure out what the units of the magnetic field are. So we know that the cross product is the same thing as-- so let's say, what's the magnitude of the force? The magnitude of the force is equal to? Well, the magnitude of the charge-- this is just a scalar quantity, so it's still just the charge-- times the magnitude of the velocity times the magnitude of the field times the sine of the angle between them. This is the definition of a cross product and then we could put-- if we wanted the actual force vector, we can just multiply this times the vector we get using the We'll do that in a second. Anyway we're just focused on units. Sine of theta has no units so we can ignore it for this discussion. We're just trying to figure out the units of the magnetic field. So force is newtons-- so we could say newtons equals-- this is times the-- I don't know what we'll call this-- the B units. We'll call it unit sub B. So let's see. If we divide both sides by coulombs and meters per second, we get newtons per coulomb. And then if we divide by meters per second, that's the same thing as multiplying by seconds per meter. Equals the magnetic field units. So the magnetic field in SI terms, is defined as newton seconds per coulomb meter. And that might seem a little disjointed, and they've come up with a brilliant name. And it's named after a deserving fellow, and that's Nikolai Tesla. And so the one newton second per coulomb meter is And I'm actually running out of time in this video, because I want to do a whole problem here. But I just want you to sit and think about it for a second. Even though in life we're used to dealing with magnets as we have these magnets-- and they're fundamentally maybe different than what at least we imagine electricity to be-- but the magnitude or actually the units of magnetism is actually defined in terms of the effect that it would have on a moving charge. And that's why the unit-- one tesla, or a tesla-- is defined as a newton second per coulomb. So the electrostatic charge per coulomb meter. Well, I'll leave you now in this video. Maybe you can sit and ponder that. But it'll make a little bit more sense when we do some actual problems with some actual numbers in the next video. See"
    },
    {
        "Q": "\nAt 7:30, the generic acid, A, was used to deprotonate the H from the nitrogen group. Could we have used H2O to deprotonate, because that way it would become H3O+ in the end?",
        "A": "Yes, the generic base A: could have been a water molecule.",
        "video_name": "Gl7bQNm92fE",
        "timestamps": [
            450
        ],
        "3min_transcript": "So this intermediate is called a carbinolamine. So let me go ahead and write that. So this is called a carbinolamine. And for the next step we can protonate the O-H group. So we take our generic acid once again so H-A plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let's go ahead and show that. So let's get once again some more space in here. So if we protonate our carbinolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the O-H, we would form water as a good leaving group. So right in here, this would be a plus one formal charge So in here, lone pair of electrons on the oxygen, pick up this proton and so forming this bond right here and now you can see we have a good leaving group. So we have water as a good leaving group. So if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water. So we're going to minus H2O and let's go ahead and show our next structure here. We have our carbon double bonded to our nitrogen this time, our nitrogen is still bonded to our Y group. And our hydrogen over here. That gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form And this is an important structure, this is called an iminium ion, so let's go ahead and draw that. So an iminium ion and then we lose water of course. So minus water at this stage. So we're almost to our final product, we would just have to deprotonate our iminium ion. And so we can do that with our imine. Could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our Y group here. A lone pair of electrons on that nitrogen and then we have our alkyl groups and so let's show those electrons here in blue moving in here, off onto our nitrogen. And then once again, if our Y is a hydrogen an alkyl group we have formed an imine. Alright, so if that's formation of an imine,"
    },
    {
        "Q": "At 3:35, How can we go around the sphere if its a 2 dimensional object? We can only go back and forth.\n",
        "A": "The surface of the Earth is pretty much two dimensional... how do you travel on it?",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            215
        ],
        "3min_transcript": "And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space. I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere."
    },
    {
        "Q": "\nHow is a sphere a 2D object? There's not much more to say other than it was at 3:21 in the video, so there's my question.",
        "A": "A sphere is a 3D object but you can imagine its surface as a curved 2 dimensional space.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            201
        ],
        "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space."
    },
    {
        "Q": "\nat 3:10 you are drawing a sphere while explaining about two dimensional space. please explain me this concept.",
        "A": "The surface of the sphere has only two dimensions to it. You can think of those dimensions as latitude/longitude or right ascension/declination.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            190
        ],
        "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space."
    },
    {
        "Q": "\nAround 6:22 Sal mentions a \"toroid would fit the bill\" Could someone explain this shape to me. I dd look up the definition by the way. I'm looking for something a bit more intuitive.",
        "A": "A toroid is kind of like a donut shape. Its basically just a cylinder bend so thats its two ends connect.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            382
        ],
        "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite."
    },
    {
        "Q": "At 6:30 sal talks about a torroid. Meaning, please ?!?!?\n",
        "A": "A torroid is something that has the properties of a torus. A torus is something shaped like a donut.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            390
        ],
        "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite."
    },
    {
        "Q": "\nAt 6:22 Sal mentiones a \"toroid\" (not sure of the spelling). What is a toroid, what does it look like?",
        "A": "A toroid is a donut shape.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            382
        ],
        "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite."
    },
    {
        "Q": "At 00:41 Sal said that evertything is packed in together? i dont understand whats that everything? And where this everything came from?\n",
        "A": "We don t know where it came from. We just know that everything in the universe now was once much, much closer together before it was flung apart by the big bang.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            41
        ],
        "3min_transcript": "Right now, the prevailing theory of how the universe came about is commonly called the Big Bang theory. And really is just this idea that the universe started as kind of this infinitely small point, this infinitely small singularity. And then it just had a big bang or it just expanded from that state to the universe that we know right now. And when I first imagined this-- and I think if it's also a byproduct of how it's named-- Big Bang, you kind of imagine this type of explosion, that everything was infinitely packed in together and then it exploded. And then it exploded outward. And then as all of the matter exploded outward, it started to condense. And then you have these little galaxies and super clusters of galaxies. And they started to condense. And then within them, planets condensed and stars condensed. And then we have the type of universe that we have right now. has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that"
    },
    {
        "Q": "\n@2:31 sal says finite area, what does that mean?",
        "A": "No infinite or has a definable distance.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            151
        ],
        "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space."
    },
    {
        "Q": "\nAt 5:15 when he is dividing by .3639, is he dividing by the molar mass of Hg?",
        "A": "No, he is dividing by the number of moles of Hg.",
        "video_name": "NM0WycKCCDU",
        "timestamps": [
            315
        ],
        "3min_transcript": "so that's why I said roughly. So how many moles is this? This is going to be 73 over 200.59 of a mole. If a mole is 200.59 we have 73, this is the fraction of a mole that we have, moles of mercury. Remember, moles are just a number, Avogadro's number of something, but let's just figure out what it is. If we take 73 divided by 200.59 we get .36, I'll just say 0.364, and once again this is, so approximately 0.364. That's how many moles of mercury that we have. We can do the same thing for chlorine. This is going to be 27 over 35.453 moles of chlorine, 27 divided by 35.453 is equal to .76, I'll just say two. So 0.762 moles of chlorine. What's going to be the ratio of mercury to chlorine? Or I guess we could say since chlorine, there's more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. We can literally think of this as the ratio. This is a certain number of moles, Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines."
    },
    {
        "Q": "6:50 how is there more chlorine than mercury if there is .73% and 200 moles of Hg and only .27% and 36 moles of Cl?\n",
        "A": "Because if you do 73/200.5(Mercury s Atomic Mass), you get .3640897756 as the amount of moles of Mercury. If you do 25/35.453, you get .7051589428 as the amount of moles of Chlorine. So, .7051589428 is greater than .3640897756. therefore, there are more moles of Chlorine than there are moles of Mercury. Is that clear?",
        "video_name": "NM0WycKCCDU",
        "timestamps": [
            410
        ],
        "3min_transcript": "Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. very likely that you have mercury two chloride. The reason why it's called mercury two chloride is because, well I won't go into too much detail right over here but chlorine is highly electronegative, it's an oxidizing agent, it likes to take other people's electrons or hog other people electrons. In this case it's hogging, since each of the chlorine likes to hog one electron, this case two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury. When you lose electrons or when your electrons are being hogged you're being oxidized, so the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury. The ratio, we have two chlorines for every one mercury, roughly."
    },
    {
        "Q": "\nAt 5:42, it is said that the ratio of Chlorine to Mercury is 2:1, because 0.762/0.364 = ~2. and therefor stating that for each Mercury, we have 2 Chlorine.\nBut what about the ratio of Mercury to Chorine? Wouldn't that be 0.364/0.762= ~0.5, stating that for each Chlorine, we have 0.5 Mercury?",
        "A": "You cannot have half of an atom.",
        "video_name": "NM0WycKCCDU",
        "timestamps": [
            342,
            121
        ],
        "3min_transcript": "27 divided by 35.453 is equal to .76, I'll just say two. So 0.762 moles of chlorine. What's going to be the ratio of mercury to chlorine? Or I guess we could say since chlorine, there's more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. We can literally think of this as the ratio. This is a certain number of moles, Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. very likely that you have mercury two chloride. The reason why it's called mercury two chloride is because, well I won't go into too much detail right over here but chlorine is highly electronegative, it's an oxidizing agent, it likes to take other people's electrons or hog other people electrons. In this case it's hogging, since each of the chlorine likes to hog one electron, this case two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury. When you lose electrons or when your electrons are being hogged you're being oxidized, so the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury."
    },
    {
        "Q": "\nAt 1:28 something was said about shining a wavelength of light is specifically sensitive to the solute. my question is why is the spectrophotemeter set at certain wavelengths, why cant it just be set at any wavelength of our choice?",
        "A": "That is because chemicals only absorb very specific wavelengths of light. This is why we can use a spectrophotometer to measure the concentration of a specific chemical. If the wavelength of the light is wrong, then the light won t be absorbed.",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            88
        ],
        "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here."
    },
    {
        "Q": "at 1:18 why is #1 a lower concentration?\n",
        "A": "For example s sake",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            78
        ],
        "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here."
    },
    {
        "Q": "\nAt 2:24, if the same amount of light shine through the same beaker, why does the concentration of each beaker differ and why does one have a higher concentration from another? Somebody please explain.",
        "A": "The concentration is not dependent on the amount of light shining through. It would be similar to having to glasses of water that have different amounts of food coloring in them. The concentrations differ for example s sake. He is simply showing how transmittance differs in varying conditions.",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            144
        ],
        "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here."
    },
    {
        "Q": "\nAt 1:30, what do you mean by \"sigma\"?",
        "A": "sigma usually stands for sum",
        "video_name": "24vtg9Ehr0Q",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Instructor] If you're face to face with a sophisticated Newton's Second Law problem, you're gonna need a sophisticated understanding of Newton's Second Law. That's what I'm gonna try to provide you with here, so that no matter what scenario you're faced with you can apply this law in a correct way. Most people know Newton's Second Law is F equals MA, which is fine, it's a simple way to understand it, and it's fine for simple problems, if I had an asteroid for instance, of mass m, out in outer space so there's no air resistance or friction, and there was only one force on it, a force F, and that force pointed to the right, let's say that force was 50 newtons, well I could plug the 50 newtons into the force, I could plug the mass of the asteroid, let's just say it's 10 kilograms, into the mass, and I'd find the acceleration of the asteroid, in this case, 50 over 10 would give me five meters per second squared. But, what if we had extra forces on this asteroid? What if there was another force that pointed to the left, that was 30 newtons? So let's call, let's name these now, let's call this F1, this 50 newtons, let's say F2 was the magnitude of the 30 newton force, it points to the left, yes, that's the negative direction, but let's just say these forces here are just giving the magnitude of it and then the direction is specified by the direction of the arrow. Now what would I do? Well, to handle this we need to understand that the left hand side here isn't just force, it's the net force. Or, you can call it the sum of the forces. So to denote the net force, we often write this Greek letter sigma, and sigma is a mathematical symbol that represents the sum of whatever comes after it. So this is the sum of the forces. Because F comes afterward. If I had G it would be the sum of the G's, and if I had H it would be the sum of the H's. And this is a little confusing already. People hear sum of, phonetically, and they think oh, sometimes they're like oh, so, some of? Like a few of? No no no, we mean all of, all of the forces. You add up all of the forces, that will equal So in this case, we'd take this 50 newtons, I can take 50 newtons because it goes to the right, and, I mean we can call leftward positive if we really wanted to, if there was a good reason, but unless otherwise specified, we're gonna just choose rightward as positive and upward as positive, so this 50 has to be positive. And I can't now, by sum of, I add up the forces, but I have to add them up like vectors. This force here is a vector. Forces are vectors, and so I have to add them up as vectors. This is a vector equation. I can't just take 50 plus 30 to get the answer, because vectors that point to the left we're gonna consider negative, and vectors that point to the right we'll consider positive, and so I'll take 50 newtons minus 30 newtons. That's what's gonna be equal to the mass times the acceleration, so I could plug in 10 kilograms if I wanted to, multiply by A, and in this case I'd get 20 over 10 is two, meters per second squared. So you have to add these up like vectors, and if I had more forces it'd be just as easy to deal with,"
    },
    {
        "Q": "I don't get something at 10:00. It doesn't intuitively make sense to me that on one side the force is lesser and on the other its more, shouldn't force be conserved? I know if I look at the mathematical equations, it makes sense there, but how could the force be larger on one side, even though its being applied on the same amount of volume, and just a different surface area? Shouldn't it be smaller since its on a larger surface area?\n",
        "A": "In this system, and for that matter all systems, it s the work that is conserved. work being force x distance means that the larger piston will travel a shorter distance than the smaller piston with a greater amount of force. It took me several times reviewing this video to understand the conclusion that he was coming to. he probably should have put that in the end of the video but we are all only human.",
        "video_name": "lWDtFHDVqqk",
        "timestamps": [
            600
        ],
        "3min_transcript": "it's a very natural unit. Let's say my pressure in is 10 pascals, and let's say that my input area is 2 square meters. If I looked the surface of the water there it would be 2 square meters, and let's say that my output area is equal to 4 meters squared. What I'm saying is that I can push on a piston here, and that the water's going to push up with some piston here. First of all, I told you what my input pressure is-- what's my input force? Input pressure is equal to input force divided by input area, so 10 pascals is equal to my input force divided by I get input force is equal to 20 newtons. My question to you is what is the output force? How much force is the system going to push upwards at this end? We know that must if my input pressure was 10 pascals, my output pressure would also be 10 pascals. So I also have 10 pascals is equal to my out force over my out cross-sectional area. So I'll have a piston here, and it goes up like that. That's 4 meters, so I do 4 times 10, and so I get 40 newtons is equal to my output force. So what just happened here? I inputted-- so my input force is equal to 20 newtons, and my output force is equal to 40 newtons, so I just doubled my force, or essentially I had a mechanical advantage of 2. it's a hydraulic machine. Anyway, I've just run out of time. I'll see you in the next video."
    },
    {
        "Q": "At 4:31 , he said about external pressure is distributed through out the fluid, That means if we jump into the ocean, then we give external pressure for throughout the ocean.?\n",
        "A": "An EXTREMELY minuscule amount, but yes.",
        "video_name": "lWDtFHDVqqk",
        "timestamps": [
            271
        ],
        "3min_transcript": "is exerting on this piston. So that's the output pressure, P2. And what's area 2 times D2? The cross sectional area, times the height at which how much the water's being displaced upward, that is equal to volume 2. But what do we know about these two volumes? I went over it probably redundantly in the previous video-- those two volumes are equal, V1 is equal to V2, so we could just divide both sides by that equation. You get the pressure input is equal to the pressure output, so P1 is equal to P2. I did all of that just to show you that this isn't a new concept: this is just the conservation of energy. The only new thing I did is I divided-- we have this notion of the cross-sectional area, and we have this notion of This actually tells us-- and you can do this example in multiple situations, but I like to think of if we didn't have gravity first, because gravity tends to confuse things, but we'll introduce gravity in a video or two-- is that when you have any external pressure onto a liquid, onto an incompressible fluid, that pressure is distributed evenly throughout the fluid. That's what we essentially just proved just using the law of conservation of energy, and everything we know about work. What I just said is called Pascal's principle: if any external pressure is applied to a fluid, that pressure is distributed throughout the fluid equally. Another way to think about it-- we proved it with this little drawing here-- is, let's say that I have a tube, and at the end of the tube is a balloon. It's saying that if I increase-- say I have some piston here. This is stable, and I have water throughout this whole thing. Let me see if I can use that field function again-- oh no, there must have been a hole in my drawing. Let me just draw the water. I have water throughout this whole thing, and all Pascal's principle is telling us that if I were to apply some pressure here, that that net pressure, that extra pressure I'm applying, is going to compress this little bit. That extra compression is going to be distributed through the whole balloon. Let's say that this right here is rigid-- it's some kind of middle structure."
    },
    {
        "Q": "Why does Sal divide at 00:45\n",
        "A": "because the force is being applied over an area A, in this case A1 which is just pressure right? When we compress something, we apply a pressure.",
        "video_name": "lWDtFHDVqqk",
        "timestamps": [
            45
        ],
        "3min_transcript": "Welcome back. To just review what I was doing on the last video before I ran out of time, I said that conservation of energy tells us that the work I've put into the system or the energy that I've put into the system-- because they're really the same thing-- is equal to the work that I get out of the system, or the energy that I get out of the system. That means that the input work is equal to the output work, or that the input force times the input distance is equal to the output force times the output distance-- that's just the definition of work. Let me just rewrite this equation here. If I could just rewrite this exact equation, I could say-- the input force, and let me just divide it by this area. The input here-- I'm pressing down this piston that's pressing down on this area of water. So this input force-- times the input area. Let's call the input 1, and call the output 2 for simplicity. Let me do this in a good color-- brown is good color. I have another piston here, and there's going to be some outward force F2. The general notion is that I'm pushing on this water, the water can't be compressed, so the water's going to push up on this end. The input force times the input distance is going to be equal to the output force times the output distance right-- this is just the law of conservation of energy and everything we did with work, et cetera. I'm rewriting this equation, so if I take the input force and divide by the input area-- let me switch back to green-- then I multiply by the area, and then I just multiply times D1. You see what I did here-- I just multiplied and divided by A1, which you can do. You can multiply and divide by any number, and these two cancel out. It's equal to the same thing on the other side, which is over A2 times A2 times D2. Hopefully that makes sense. What's this quantity right here, this F1 divided by A1? Force divided by area, if you haven't been familiar with it already, and if you're just watching my videos there's no reason for you to be, is defined as pressure. Pressure is force in a given area, so this is pressure-- we'll call this the pressure that I'm inputting into the system. What's area 1 times distance 1? That's the area of the tube at this point, the cross-sectional area, times this distance. That's equal to this volume that I calculated in the previous video-- we could say that's the input volume, or V1. Pressure times V1 is equal to the output pressure-- force 2"
    },
    {
        "Q": "\nAt 6:02, how can two different orbitals have same sub shell.\nHow can 1st shell have s-sub shell and 2nd shell also have s-sub shell?",
        "A": "The are not the same subshell, they are the same TYPE of subshell. So, 1s is the s-subshell belonging to the first shell. 2s is the s-subshell belonging to the second shell.",
        "video_name": "KrXE_SzRoqw",
        "timestamps": [
            362
        ],
        "3min_transcript": "So the electron is most likely to be found somewhere in that sphere. Let's do the next shell. n is equal to two. If n is equal to two, what are the allowed values for l? l goes zero, one, and so on all the way up to n minus one. l is equal to zero. Then n minus one would be equal to one. So we have two possible values for l. l could be equal to zero, and l could be equal to one. Notice that the number of allowed values for l is equal to n. So for example, if n is equal to one, we have one allowed value. If n is equal to two, we have two allowed values. We've already talked about what l is equal to zero, what that means. l is equal to zero means an s orbital, shaped like a sphere. Now, in the second main energy level, or the second shell, we have another value for l. l is equal to one. l is equal to one means a p orbital. The shape of a p orbital is a little bit strange, so I'll attempt to sketch it in here. You might hear several different terms for this. Imagine this is a volume. This is a three-dimensional region in here. You could call these dumbbell shaped or bow-tie, whatever makes the most sense to you. This is the orbital, this is the region of space where the electron is most likely to be found if it's found in a p orbital here. Sometimes you'll hear these called sub-shells. If n is equal to two, if we call this a shell, then we would call these sub-shells. These are sub-shells here. Again, we're talking about orbitals. l is equal to zero is an s orbital. l is equal to one is a p orbital. Let's look at the next quantum number. This is the magnetic quantum number, symbolized my m sub l here. m sub l indicates the orientation of an orbital around the nucleus. This tells us the orientation of that orbital. The values for ml depend on l. ml is equal to any integral value that goes from negative l to positive l. That sounds a little bit confusing. Let's go ahead and do the example of l is equal to zero. l is equal to zero up here. Let's go ahead and write that down here. If l is equal to zero, what are the allowed values for ml? There's only one, right? The only possible value we could have here is zero."
    },
    {
        "Q": "\nat 8:50, from the three values of ml: -1 0 1, how can we know that the orientation of the 3 types p orbital should be on the three axis but not somewhere else like in between the axis? And one more thing is that the spin quantum number indicates the way electrons spin around themselves, so why can't we use -1 and 1 or any different number instead of -1/2 and 1/2?",
        "A": "It s complicated! The answer comes from a combination of quantum mechanics and the theory of special relativity.",
        "video_name": "KrXE_SzRoqw",
        "timestamps": [
            530
        ],
        "3min_transcript": "Let me use a different color here. If l is equal to zero, we know we're talking about an s orbital. When l is equal to zero, we're talking about an s orbital, which is shaped like a sphere. If you think about that, we have only one allowed value for the magnetic quantum number. That tells us the orientation, so there's only one orientation for that orbital around the nucleus. And that makes sense, because a sphere has only one possible orientation. If you think about this as being an xyz axis, (clears throat) excuse me, and if this is a sphere, there's only one way to orient that sphere in space. So that's the idea of the magnetic quantum number. Let's do the same thing for l is equal to one. Let's look at that now. If we're considering l is equal to one ... Let me use a different color here. l is equal to one. If l is equal to one, what are the allowed values for the magnetic quantum number? ml is equal to -- This goes from negative l to positive l, so any integral value from negative l to positive l. Negative l would be negative one, so let's go ahead and write this in here. We have negative one, zero, and positive one. So we have three possible values. When l is equal to one, we have three possible values for the magnetic quantum number, one, two, and three. The magnetic quantum number tells us the orientations, the possible orientations of the orbital or orbitals around the nucleus here. So we have three values for the magnetic quantum number. That means we get three different orientations. We already said that when l is equal to one, we're talking about a p orbital. A p orbital is shaped like a dumbbell here, so we have three possible orientations If we went ahead and mark these axes here, let's just say this is x axis, y axis, and the z axis here. We could put a dumbbell on the x axis like that. Again, imagine this as being a volume. This would be a p orbital. We call this a px orbital. It's a p orbital and it's on the x axis here. We have two more orientations. We could put, again, if this is x, this is y, and this is z, we could put a dumbbell here on the y axis. There's our second possible orientation. Finally, if this is x, this is y, and this is z, of course we could put a dumbbell on the z axis, like that. This would be a pz orbital. We could write a pz orbital here, and then this one right here would be a py orbital."
    },
    {
        "Q": "4:22 How does it hold the hot water, wouldn't it melt? Or is it not plastic? Also on 6:03 wouldn't the water come out?\n",
        "A": "Some plastics can handle tempatures well about the boiling poit of water.",
        "video_name": "XQTIKNXDAao",
        "timestamps": [
            262,
            363
        ],
        "3min_transcript": "Oh wow, it pops apart there. You can see there's a little bracket on the inside that the screw goes into and holds this steel band in place. That's what the handle is help on by. Then at the top, there's just a little lip that holds the top of the handle there in place. Just a little piece that folds over the glass and snaps on. So that's how that's held. This is made out of one molded part, this is made out of another, and this is made out of another. So that's how they made the handle, there's three different molded plastic parts there and the molds came together like this. You can tell that because you can feel the mold seam on the inside of the handle, there. It would make it easy to pull the mold out this way. They probably also had, it was probably a three part mold and there was a section that also came out in this direction. Then this is just another injection molded part that snaps onto this one, as we've seen. This is the part that holds the handle on. Very important part. I think they definitely paid the extra money for a stainless piece there because it's really important that that doesn't come loose and it probably gets fairly wet, so if it was made out of regular steel or another material it might rust and could potentially come apart. We wouldn't want hot coffee on us, now would we. Alright, so that's the coffee kettle. So, inside, here's our coffee maker. We know that hot water... We've got a container here and in this container, is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart."
    },
    {
        "Q": "\nat 6:09,he said \"The heat to get in\" why do we need heat to get in?",
        "A": "I think that was just a slip up as he fixed it with allowing air to get in its so that as it heats up inside it doesn t pressurize the air and and blow up.",
        "video_name": "XQTIKNXDAao",
        "timestamps": [
            369
        ],
        "3min_transcript": "is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart. There we go. Again, this is an injection molded part. This is a co-molded part, it looks like. Which means that there were two different materials molded together. Let's see if I can knock that screw out. Okay, it wants to stay, that's fine. This material here is, these feet are made out of a softer material and this is a polypropylene material. So it's a plastic, a low-cost plastic. So the mold comes together and they injection mold this material and then once this material has begun to harden, they injection mold the softer material, so it's co-molded or it's a dual molded part. You can see other parts are done like this, like sometimes you'll see toothbrushes that have soft saniprene and then the hard toothbrush and they're molded in one mold. It's a dual shot mold. In any case, so that's the bottom. heater chamber to vent out, I think a little heat in there. So here is the heater. This is where all the magic happens. The water comes down this tube, and it goes around this horse shoe shape, and then it comes up here. What causes the water to raise back up and go all the way up this tube. It goes all the way up the tube, here and then it comes out this apparatus here and then it goes and drains out of these holes right into here. There's just a little piece of plastic that causes this part to line up right over the top of this, when you close it. The water just comes up that and drains right out and into your coffee grounds and then makes your coffee. So the heater does two things. It causes the water to go down this pipe, expand out, and to drip into your coffee maker. But it also heats up the plate, this plate right here."
    },
    {
        "Q": "At 3:05, you say Gibbs Free Energy is only when there's constant pressure and TEMPERATURE. Gibbs Free Energy formula has temperature as a constant so how does that work?\n",
        "A": "It means that the Gibb s Free Energy value can change if temperature changes. That kinda makes sense, since a reaction that may not be spontaneous with low inputs of energy may become spontaneous if there is a greater input of energy into the system.",
        "video_name": "J2L-X2sUigs",
        "timestamps": [
            185
        ],
        "3min_transcript": "assumption if you're doing something in a beaker that's open to the air or if you're thinking about a lot of different biological systems. Now based on that logic what do you think this word means, endothermic. Well endothermic, therm same root and now your prefix is endo so this is a process that absorbs heat. Absorbs heat. Or if you're thinking of a constant pressure, you can say your enthalpy after the reaction is gonna be higher than the enthalpy before the reaction. So your delta H is going to be greater than zero. All right, fair enough. Now let's look at these two characters over here. Exergonic and endergonic so exergonic the root here is ergon and you might not be as familiar with that as you are with therm but you might have heard the word ergonomic. Say, hey that's a nice ergonomic desk. or it's a nice ergonomic chair. An ergon does indeed come from the Greek for work. And so exergonic is a reaction that releases work energy or at least that's what the word implies. Releases, let me do that in the same color. This is something that is going to release work energy. And endergonic, same logic, well that's gonna be something based on just the way the word is setup that absorbs work energy or uses work energy. Now one of our variables or properties that we can use to think about energy that can be used for work is Gibbs free energy and the formula for Gibbs free energy, if we're thinking about constant pressure and temperature, so let me write that down. So if we're talking about constant pressure for Gibbs free energy or you can even view this as a definition of Gibbs free energy. The change in Gibbs free energy, let me do this in another color. The change in Gibbs free energy is equal to our change in enthalpy minus, use in the different color. Minus our temperature times our change in entropy and if this looks completely foreign to you, I encourage you to watch the video on Gibbs free energy but the reason why this is related to energy for work is okay, look I have my, whether I'm absorbing or I'm releasing heat and I'm subtracting out entropy which is kind of the energy that is going to the disorder of the universe and what's left over is the energy that I can do for work."
    },
    {
        "Q": "\nFrom 7:45 on, a intermolecular Fischer esterification that forms lactone is shown. Is it possible that at the same time of forming lactone, an OH part of one molecule attack the COOH part of another molecule, and form a ester with two 5-carbon chain (and no ring) in it?",
        "A": "yes, you can also get the two 5 carbon chain product and you could essentially form a polymer from it because the OH portion of one can keep attacking the COOH on the other end of the molecule. Then theres also the possibility of a ring forming at any point in that process as well.",
        "video_name": "ynBuPEmcjp4",
        "timestamps": [
            465
        ],
        "3min_transcript": "And so our end result is to form our ester and water. Ok, so that's a little bit of a long mechanism. Let's take a look at some reactions to form esters using the Fischer esterification reactions. So, let's start with this molecule over here on the left. So this is salicylic acid, and if we add methanol, and we use sulfuric acid as our source of protons, we're going to form an ester. And this is one of those famous labs that's always done in undergraduate organic chemistry. So if we think about the mechanism, remember that this oxygen on our alcohol, and in this case it's methyl, are going to ag. So we're going to lose this OH on our carboxylic acid, and we're going to put this oxygen and this methyl group on in place. So let's go ahead and draw the product. So we would have our benzene ring right here, to our oxygen, and we would have the oxygen from the alcohol, from methanol, and then our methyl group like that, and then we still have our OH right here. The reason why this is one of those classic undergraduate labs, is this is wintergreen. So this is an incredible smell. It's always a lot of fun to do this in an undergraduate lab, because the lab smells great when you're done, so the synthesis of wintergreen. Alright, let's look at another Fischer esterification. This one is a little bit different. This one is an intramolecular Fischer esterification. So if we look at our starting molecule on the left. This time we have our carboxylic acid and our alcohol in the exact same molecule. And we have all these single bonds in here, which we know we can have some free rotation. So if we draw the molecule in a different conformation, so let's go ahead and do that, so we have our carboxylic acid up here, so let me use red for that, so we have one, we have carbon one, two, three, four, and five. So we have five carbons, so let's go ahead and draw them in, so there's carbon one, two, three, four, five, and then we have our OH. So let me go ahead and number those carbons. So this is carbon one, carbon two, carbon three, carbon four, and carbon five. And so in a different conformation, we can think about this oxygen attacking this carbonyl in the mechanism. So we know that we're going to lose this OH, we know we're gonna lose this Hydrogen, and so we can stick those together and think about our final product. So we're going to form an ester, but it's a different ester than what we've seen before. So we have our carbonyl right here, and then we have this oxygen."
    },
    {
        "Q": "\non the 4:40 timestamp he talks about coronary artery disease but my question about the arteries is that since it caries deoxygenated blood to the lungs it would make it blue from the deoxygenated blood but why does it all just look red in the real life photo",
        "A": "Deoxygenated blood isn t actually blue, it is dark red. Blue is used on diagrams to make them easier to read.",
        "video_name": "vYnreB1duro",
        "timestamps": [
            280
        ],
        "3min_transcript": "And the plaques, the material inside of them are lipids, so things like fat, cholesterol and also dead white blood cells, which is this kind of messy substance right over here. This is what we call a plaque. And the formation of these plaques that obstruct the actual blood vessel, that actually obstruct the artery. We call it.....make it clear you see that. This is kind of tube over here. Let me draw the blood So this formation of these plaques we call atherosclerosis. So you can imagine if you have these things build up, So it would be destructing the blood flow downstream right over there. In that general process we talk about the restriction of blood flow, that is ischemia that's happening. So ischemia is deprivation of blood flow and oxygen downstream from this right over there. That's what we call coronary artery disease, or heart disease. So this causes coronary artery disease, which is sometimes called heart disease. Coronary heart disease would be redundant, because coronary is already referring to the heart. This is also sometimes called heart disease. And so you can imagine if downstream the muscle tissue is when they are exerting themselves, they need more oxygen. The heart needs to pump a little harder. If downstream the cells are not getting all of the oxygen they need, you can imagine that the heart maybe not able to provide all of the functions, whoever's heart this is, that they needed to do. And when that happens that's called heart failure. So heart disease is one of the causes of heart failure. Now I want to be clear, heart failure does not mean that the heart is stopping. That the heart is stopped and the person is dead, it literally just means that the heart is failing to do what it should be doing. It's failing to provide the needs of that person. So it's not pumping hard enough or well enough to provide adequate function for that person."
    },
    {
        "Q": "Is it just my imagination, or are the absolute configurations of the alcohol and bromoalkane incorrect at 5:48? I thought the alcohol (with OH on a wedge and thus H on a dash) would be S and the bromoalkane (with Br on a dash and thus H on a wedge) would be R. :)\n",
        "A": "The video is correct, perhaps take a second look at it? O is #1 priority, the right carbon is #2 and the left carbon is #3. Going from 1->2->3 is clockwise so it s R.",
        "video_name": "KPh60w6McPI",
        "timestamps": [
            348
        ],
        "3min_transcript": "to give us a much better leaving group, and here we have an excellent leaving group. So that's one of the reasons for using tosylates here. So in the second step we are going to add sodium bromide and we are going to get again a SN-two type mechanism. So a nucleophile is going to attack our electrophile. So we can identify our electrophile. It's the carbon bonded to the oxygen. So the oxygen is withdrawing some electron density from this carbon, so this the electrophilic portion of the molecule. And then once again our bromide anion is going to function as a nucleophile. So here is our bromide, and I am going to highlight this lone pair of electrons right here in blue. And so our bromide anion attacks our electrophilic carbon and forms a bond. At the same time, these electrons come off onto the oxygen. So once again a concerted SN-two type mechanism. And this time we do have to worry about stereochemistry. So we have this wedge in here. is coming out at us in space. And so the bromide anion has to attack from the opposite side. And so if it is attacking from the opposite side when you draw the final product, you would have to show this is a dash. So the bromide had to attack from the opposite side which gives us inversion of configuration. Inversion of absolute configuration here. So when you assign your absolute configuration, for your starting alcohol, this would be R. So our chiral center would be right here. So this carbon is chiral and so this carbon is chiral for our products, and for our product we would form the S in the [an-te-um] right here. So we formed S in the [an-te-um] row and we started with a R in the [an-te-um] row. So a SN-two type mechanism inversion of configuration because a nucleophile has to attack from the opposite side. So lets look at one more example here. Lets do a SN-one mechanism using a tertiary alcohol. So lets do that. reacting with concentrated hydrochloric acid. So concentrated hydrochloric acid is going to function as an acid. Our alcohol is going to function as a base. And let me just highlight the fact that this is going to be a SN-one type mechanism because we have a tertiary alcohol. This carbon bonds to the OH bond to three other carbons. And so the alcohol functions as a base, and is protonated and we would form the chloride anion here. So lets go ahead and draw that in. So we would have the chloride anion, so negative one formal charge. And lets go ahead and put those electrons in blue like we did before. So these electrons in here, lets say that those are these electrons. So the chloride anion. We protonate the oxygen, so lets go ahead and draw that. So if we protonate the oxygen, now we have our tertbutyl group over here, protonate the oxygen so a plus one formal charge"
    },
    {
        "Q": "At 7:43, Sal integrated the entire equation. The right hand side of equation is zero. The integration of zero must be a constant as derivative of a constant is zero. But, instead after integrating the equation was written still equal to zero.\nCan anyone explain? It will be very helpful. Thanks.\n",
        "A": "I did not watch the video but probably the answer is that Sal was doing a definite integral rather than an indefinite integral. When you do a definite integral, you are integrating the same function and calculating its value at two different points, and then subtract the starting point from the ending point. When you do that subtraction, the constant cancels out, so you can just say it is zero.",
        "video_name": "ixRtSV3CXPA",
        "timestamps": [
            463
        ],
        "3min_transcript": "Well, on this term, the n's cancel out, the R cancels out. Over here, this nRT cancels out with this nRT. And what are we left with? We're left with 3/2-- we have this 1 over T left-- times 1 over T delta T plus 1 over V delta V is equal to-- well, zero divided by anything is just equal to 0. Now we're going to integrate over a bunch of really small delta T's and delta V's. So let me just change those to our calculus terminology. We're going to do an infinite sum over infinitesimally small changes in delta T and delta V. So I'll rewrite this as 3/2 1 over T dt plus 1 over V dv is small change in volume. This is a very, very, very, small change, an infinitesimally small change, in temperature. And now I want to do the total change in temperature. I want to integrate over the total change in temperature and the total change in volume. So let's do that. So I want to go from always temperature start to temperature finish. And this will be going from our volume start to volume finish. Fair enough. Let's do these integrals. This tends to show up a lot in thermodynamics, these antiderivatives. The antiderivative of 1 over T is natural log of T. So this is equal to 3/2 times the natural log of T. We're going to evaluate it at the final temperature and then the starting temperature, plus the natural log-- the plus the natural log of V, evaluated from our final velocity, and we're going to subtract out the starting velocity. This is just the calculus here. And this is going to be equal to 0. I mean, we could integrate both sides-- well, if every infinitesimal change is equal to the sum is equal to 0, the sum of all of the infinitesimal changes are also So this is still equal to 0. See what we can do here. So we could rewrite this green part as-- so it's 3/2 times the natural log of TF minus the natural log of TS, which is just, using our log properties, the natural log of TF over the natural log of TS. Right? When you evaluate, you get natural log of TF minus the natural log of TS."
    },
    {
        "Q": "\nMin 3:40 How does he know that delta U and delta T have a linear relationship with teach other? Of course they are related, but the nature of this relationship could be non-linear...",
        "A": "No it can t because U is internal energy and T is average energy of the molecules. If the average energy increases then U has to increase proportionally.",
        "video_name": "ixRtSV3CXPA",
        "timestamps": [
            220
        ],
        "3min_transcript": "small change in volume. So that's what we have there. Now, if it's adiabatic, we know that this is 0. And if that's 0, we can add P delta V to both sides of this equation, and we will get that-- this is only true if it were adiabatic-- that delta U, our change in internal energy, plus our pressure times our change in volume, is equal to 0. And let's see if we can do this somehow, we can do something with this equation to get to that result that I'm trying to get to. So a few videos ago, I proved to you that U, the internal energy at any point in time-- let me write it here. The internal energy at any point in time is equal to 3/2 times n times R times T. Which is also equal to 3/2 times PV. change on this side? Something must have changed. Well, 3/2 can't change. n can't change. We're not going to change the number of molecules we have. The universal gas constant can't change. So the temperature must change. You have delta U could be rewritten as delta-- let me do it in a different color-- delta U could be written as 3/2 n times R times our change in temperature. And that's why I keep saying in this-- especially when we're dealing with the situation where all of the internal energy is essentially kinetic energy-- that if you don't have a change in temperature, you're not going to have a change in internal energy. Likewise, if you don't have a change in internal energy, you're not going to have a change in temperature. So let me put this aside right here. I'm going to substitute it back there. But let's see if we can do something with this P here. Well, we'll just resort to our ideal gas equation. Because we're dealing with an ideal gas, we might as well. This should be emblazoned in your mind, at this point. So if we want to solve for P, we get O is equal to nRT over V. Fair enough. So let's put both of these things aside, and substitute them into this formula. So delta U is equal to this thing. So that means that 3/2 nR delta T plus P-- P is this thing-- plus nRT over V times delta V is equal to 0. Interesting. So what can we do further here? And I'll kind of tell you where I'm going with this. So that tells me, my change in internal energy over a very small delta T-- this tells me my work done by the system"
    },
    {
        "Q": "\nReferring to 7:04, how do you know that all of the base will react with the acid but not vice-versa?",
        "A": "I would assume that this is because we have much less base than acid. In other words, the base is our limiting reagent. There are 0.005 mol of -OH and 0.0100 mol of H3O+. All of the available H3O+ cannot react with the added -OH because there is much more (double actually) of it than there is of -OH available to react to form water. Hopefully this makes sense.",
        "video_name": "JoGQYSTlOKo",
        "timestamps": [
            424
        ],
        "3min_transcript": "that we have, and what's the volume. We have 10 milliliters, so in liters that would be, move our decimal place three, so that's .01. So this is equal to .01 for our liters. Solve for moles. So we would just multiply .5 by .01 and we would get our moles equal to .005. So that's how many moles of hydroxide ions we have. Now that we've found moles of both our acid and our base, we can think about the neutralization reaction that occurs. The base that we add is going to neutralize the acid that we had present. So we had hydronium ions present, and we added some base. So the acid donates a proton to the base. If OH minus picks up an H plus, then we get H2O. molecule of H2O, so we end up with two H2O over here. Or, if you prefer, instead of writing H3O plus, you could have just written H plus as your proton, and your proton reacts with hydroxide to give you water. So either way of representing the neutralization reaction is fine. So let's plug in our moles here. We know that we started with .01 moles of H3O plus, so let me write that down here. .01 moles of H3O plus, and we started with, and we added I should say, .005 moles of hydroxide. We added .005 moles of hydroxide ions. All of the hydroxide is going to react. It's going to neutralize the same amount of hydronium ions. So we're going to lose all of our hydroxide ions so we're left with zero moles of our base. If we're losing that much hydroxide, we're also going to lose that much hydronium, so that much is reacting with the hydroxide ions. We're losing the same amount of hydronium ions. So .01 minus .005 is of course equal to .005. That's how many moles of hydronium are left over. We've neutralized half of the hydronium ions present, and so we have another half left over. So one half of the acid has been neutralized, so one half of the acid is left. Let's think about the new concentration of hydronium ions in solution, so the concentration of hydronium is equal to moles over liters."
    },
    {
        "Q": "\nSal mentioned the f-block at 2:14 . Why don't universities ask questions about the f-block on the chemistry exams?",
        "A": "Sal is awesome",
        "video_name": "qkLzAXUP_K0",
        "timestamps": [
            134
        ],
        "3min_transcript": "Voiceover: What I want to do in this video is think about the electron configuration for an atom whose highest energy electron is not in the S block or the P block but the D block. And to help us think about that with the periodic table, I'm going to rearrange the periodic table a little bit. I'm gonna take Helium, which it makes sense that's it's right over here in the top right because it's a noble gas, it's inert like the other noble gases, has very similar properties. But for the purpose of electron configurations, because Helium's highest energy electron is in the S sub-shell it's electron configuration is essentially 1S2. I'm gonna put it over here and then that allows us to construct, essentially, an S block. So let me cut and paste that here. So what do I mean by an S block? Well that means that any of these, there are some exceptions, but the general rule is that energy electron is going to be in an S sub-shell. So this is the S block right over there. Now, for example, Potassium's highest energy electron is going to be in the 4S sub-shell. While Hydrogen's highest energy electron, of course, is in the 1S sub-shell. But it's going to be in an S sub-shell. Now that allows us to divide the rest of the periodic table to other blocks. This right over here is the P block, same idea. The general rule is that the highest energy electron in the electron configuration of these elements is going to be in a P sub-shell. So this is the P block right over there. And, as you might have guessed, this in the middle is the D block. the F block in a future video. This is the F block. And you can actually place it here and kind of push over the D and the P blocks to make space for this. But we'll do that in a future video. So this right over here is the D block. Now why is this helpful? Well, if something is in the D block we can say that it's highest energy electron and there's going to be some special cases, but we can say, in general, the highest energy electron is going to be sitting in a D sub-shell. So, for example, if we were to focus on Iron, it's highest energy electron is going to be in a D sub-shell. Now the D sub-shell starts to become interesting. For something like the S and P sub-shells, you can say look at Oxygen, and you can say \"OK, Oxygen's in the P block it's highest \"energy electron is going to be in a P sub-shell.\""
    },
    {
        "Q": "why is Andromeda then moving towards us?\nAlso was sal trying to explain relativity at 4:18\n",
        "A": "Andromeda is moving toward us because it is drawn in by the mutual gravitational attraction between the mass in the Milky Way and the mass in Andromeda.",
        "video_name": "1V9wVmO0Tfg",
        "timestamps": [
            258
        ],
        "3min_transcript": "of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points. They're all on the surface of this sphere. Obviously I'm only dealing with two dimensions right now and it's nearly impossible, or maybe impossible, to imagine a three dimensional surface of a four dimensional sphere. But the analogy holds. If this is a surface the balloon, or the surface of a bubble, if the bubble were to expand over a few billion years-- and once again, not drawn to scale. So now we have a bigger bubble here. This part of the surface is all going to expand. So once again, you have your magenta. You have your blue dot. You have your green dot right over here. And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. about what is the apparent velocity with which things And remember, we're going have to say, not only how far things are moving away, but we're going to say how far they are moving away from-- if the observer is us-- depending on how far they already are. So what we're going to do-- we could say is-- let me write this down. All objects moving away from each other. And the apparent relative velocity is proportional to distance."
    },
    {
        "Q": "Why have the points all moved away from each other during 0:41\n",
        "A": "Here s a good experiment. Draw a couple dots on a deflated balloon. When you blow it up, all the dots move away from each other, like Sal explained in 4:26.",
        "video_name": "1V9wVmO0Tfg",
        "timestamps": [
            41
        ],
        "3min_transcript": "Over many videos now, we've been talking about how every interstellar object is moving away from Earth. And we've also been talking about how the further something is away from Earth, the faster it's moving. What I want to do in this video is to put a little bit of numbers behind it, or even better conceptualize what we've been talking about. So one way to think about it is that, if at an early stage in the universe, I were to pick some points. So that's one point, another point, another point, another point. Let me just pick nine points so that I have a proper grid. So this at an early stage in the universe. If we fast forward a few billion years-- and I'm clearly not drawing it to scale-- all of these points have all moved away from each other. So this point is over here-- actually, let me draw another column, just to make it clear. So if we fast forward a few billion years, the universe has expanded. And so everything has moved away from everything. Let me make this point magenta. So this point, the magenta point is now here. This green point has now moved away from the magenta point. And now this blue point has now moved away from the magenta point in that direction. And we could keep going. This yellow point is maybe over here now. I think you get the general idea. And I'll just draw the other yellow points. So they've all moved away from each other. So there's no center here. Everything is just expanding away from things next to it. And what you can see here is not only did this thing expand away from this, but this thing expanded away from this even further. Because it had this expansion, plus this expansion. Or another way to think about it is, the apparent velocity with which something is expanding is going to be proportional to how far it is. Because every point in between is also expanding away. of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points."
    },
    {
        "Q": "At around 4:30 Sal demonstrates the universe be a sphere with small dots in it. Will the dots ever stop moving away from each other? Will the sphere ever stop growing?\n",
        "A": "We currently think the growth will not ever stop.",
        "video_name": "1V9wVmO0Tfg",
        "timestamps": [
            270
        ],
        "3min_transcript": "of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points. They're all on the surface of this sphere. Obviously I'm only dealing with two dimensions right now and it's nearly impossible, or maybe impossible, to imagine a three dimensional surface of a four dimensional sphere. But the analogy holds. If this is a surface the balloon, or the surface of a bubble, if the bubble were to expand over a few billion years-- and once again, not drawn to scale. So now we have a bigger bubble here. This part of the surface is all going to expand. So once again, you have your magenta. You have your blue dot. You have your green dot right over here. And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. about what is the apparent velocity with which things And remember, we're going have to say, not only how far things are moving away, but we're going to say how far they are moving away from-- if the observer is us-- depending on how far they already are. So what we're going to do-- we could say is-- let me write this down. All objects moving away from each other. And the apparent relative velocity is proportional to distance."
    },
    {
        "Q": "2:07. Why does the lone pair of the oxygen attack the carbon?\n",
        "A": "Because it is more nucleophilic than the carbon. This is covered earlier in the organic chemistry lectures.",
        "video_name": "bFj3HpdC4Uk",
        "timestamps": [
            127
        ],
        "3min_transcript": "In this video, we're going to look at the cleavage of alkenes using a reaction called ozonolysis. So over here on the left I have my generic alkene, and to that alkene we're going to add O3 in the first step, which is ozone. And in the second step, we're going to add DMS, which is dimethyl sulfide. And be careful because there are different regions you could add in the second step, so make sure to learn the one that your professor wants you to use. If you use it DMS, you're going to get to a mixture of aldehydes and or ketones for your product, depending on what is attached to your initial alkene. So let's start by looking at a dot structure for ozone. So over here on the left is a possible structure for ozone. And we can draw a resonant structure by taking these electrons and moving them in here to form a double bond between those two oxygens. And now it pushed these electrons in here off onto the oxygen on the left. So let's go ahead and draw the other resonant structure. So now I would have a double bond between my two oxygens on the right. The oxygen on the far right has two lone pairs The oxygen in the center still has a lone pair of electrons on it, and the oxygen on the far left now has three lone pairs of electrons on it. So the oxygen on the far left now has a negative 1 formal charge, and the oxygen on the top here still has a plus 1 formal charge. And so those are our two resonant structures. Remember that the actual molecule is a hybrid of these two resonant structures. So let's go ahead and pick one of those resonant structures. I'm just going to take the one on the right. So let me just go ahead and redraw the resonant structure on the right. And so we're going to do the one that has the negative charge on the oxygen on the far left. And this top oxygen is a plus 1 formal charge. And the oxygen on the right has no charge in this resonant structure. So let's go ahead and draw in our alkene. And so here is our alkene with unknown substituents at the moment. on the oxygen-- this lone pair of electrons here-- is going to attack this carbon which would push these pi electrons off. And those pi electrons are actually going go to this oxygen right here which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons. Like that. And so we had so many electrons moving, let's say if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those"
    },
    {
        "Q": "\nLa and Ac belongs to d block, 3:05 as unlike Lanthanides and Actinides they possess no electron in f orbital.",
        "A": "These two of the four elements whose position in the periodic table is disputed. You will see periodic tables with the disputed elements in either the d or f block. In the official periodic table maintained by IUPAC, La, Ac, Lu, and Lr all appear in the f block. But this is far from a settled matter amongst chemists.",
        "video_name": "L-0FkEPPdXE",
        "timestamps": [
            185
        ],
        "3min_transcript": "And remember, the D block, even though we're in the fourth period, we're going to actually fill in the three D subshell. We take the period minus one, so it's going to be three D, and to fill it up we have one, two, three, four, five, six, seven, eight, nine, 10, three D 10. Then you're back in the P block and since you're in the fourth period, it's gonna be four P, and once again to fill it up, you have six electrons. And then we go to the fifth period. Five S two and then we're going to go... The period is five but we're in the D block so we're going to go back to the four D subshell. Four D 10 and then we come back and Xenon perfectly fills out the five P subshell. Five P six. So now let's think about what the electron configuration of Neodymium is. So one thing that you're saying, \"Hey, \"what's going on with this F block right over here? \"Just kind of this island of elements.\" And it's really not an island. The reason that they put these colors here is because the F block actually belongs right here and if people had more width on pages to display these periodic tables, they would shove the D and the P blocks over to the right and stuff the F blocks right over here. And we actually did that and I actually like to look at it that way because it makes it clear it's not really an island and it actually kind of fits in the pattern of the periodic table. That is right over here. This right over here is our F block. So that is our... This right over here is our f block. the electron configuration for Neodymium. And I'll give you a hint. In the D block you go back and back fill one subshell below the one shell below the period that you're in. In the F block you go two, you go back two shells. So here, we're not filling out the six F subshell, not even the five F subshell, but the four F subshell. So given that, figure out how much more do we have to add. How much more do we have to go past Xenon in order to get its electron configuration. So I encourage you to pause the video and figure that out right now for Neodymium. So I'm assuming you've had a go at it. So Neodymium is going to have the electron configuration of Xenon so Xenon gets us right over."
    },
    {
        "Q": "At 4:10, Sal says that Xenon is the last noble gas with a lower atomic number than Neodymium. Why doesn't he just use Barium (56) instead.??\n\nThx in advance!!\nClarissa\n",
        "A": "Barium is not a noble gas so can be changed at any time. In order to start electron configuration with an element you have to start with a element that cannot be changed, so we always start with a noble gas.",
        "video_name": "L-0FkEPPdXE",
        "timestamps": [
            250
        ],
        "3min_transcript": "So now let's think about what the electron configuration of Neodymium is. So one thing that you're saying, \"Hey, \"what's going on with this F block right over here? \"Just kind of this island of elements.\" And it's really not an island. The reason that they put these colors here is because the F block actually belongs right here and if people had more width on pages to display these periodic tables, they would shove the D and the P blocks over to the right and stuff the F blocks right over here. And we actually did that and I actually like to look at it that way because it makes it clear it's not really an island and it actually kind of fits in the pattern of the periodic table. That is right over here. This right over here is our F block. So that is our... This right over here is our f block. the electron configuration for Neodymium. And I'll give you a hint. In the D block you go back and back fill one subshell below the one shell below the period that you're in. In the F block you go two, you go back two shells. So here, we're not filling out the six F subshell, not even the five F subshell, but the four F subshell. So given that, figure out how much more do we have to add. How much more do we have to go past Xenon in order to get its electron configuration. So I encourage you to pause the video and figure that out right now for Neodymium. So I'm assuming you've had a go at it. So Neodymium is going to have the electron configuration of Xenon so Xenon gets us right over. has a smaller atomic number than Neodymium. So we can start with that, which we already saw is fairly involved. And then to that we just add the incremental higher energy electrons. So now we're in the sixth period so we're gonna go six S two, that's that right over there. That's six S two and now we're filling, we're gonna start going into the F subshell but it's not gonna be the sixth shell, it's gonna go two. It's gonna be our period minus two so it's gonna be four. Let me use a color you can see. So this gonna be, we are four electrons into that subshell. One, two, three, four so it's gonna be four F four. So we're in the sixth period but since we're in the F block, we're gonna go back, fill it to the fourth shell. That's the fourth shell right over there. And we are one, two, three, four,"
    },
    {
        "Q": "0:47, How did you know it was a first order reaction?\n",
        "A": "He is saying, IF we know that the reaction is first order, this is how to do the math.",
        "video_name": "Bt0mz4mGddk",
        "timestamps": [
            47
        ],
        "3min_transcript": "- [Voiceover] Let's say we have a first order reaction where A turns into our products, and when time is equal to zero we have our initial concentration of A, and after some time T, we have the concentration of A at that time T, and let's go ahead and write out the rate for our reaction. In an early video, we said that we could express the rate of our reaction in terms of the disappearance of A, so we said that's the change in the concentration of A over the change in time, and we put a negative in sign in here to give us a positive value for the rate. We could also write out the rate law, so for the rate law, the rate is equal to the rate constant K times the concentration of A, and since this is a first order reaction, this would be to the first power, so we talked about this in an early video. We can set these equal to each other, right? Since they're both equal to the rates, we can say that the rate of disappearance of A, so the negative change in the concentration of A times the concentration of A to the first power. This is the average rate of reaction over here on the left, and so if we wanted to write this as the instantaneous rate, we need to think about calculus, all right? So the instantaneous rate would be at negative D A, right? D A, the negative rate of change of A with respect to time, so this would be D T, and this is equal to K A to the first power, and now we have a differential equation, and when you're solving a differential equation eventually you get to a function, and our function would be the concentration as a function of time, so we will eventually get there, but your first step for solving a differential equation is to separate your variables, so you need to put all the A's on one side, and we'll put the T on the opposite side, so we need to divide through by A, We divide both sides by A, and we get D A over the concentration of A here. We're going to multiply both sides by D T to get the T on the right side, so we would get K D T over here on the right, and I'll go ahead and put the negative sign on the right as well, so we just rearranged a few things to get us ready for integration, right? After you separate your variables you need to integrate, so we're going to integrate the left, and since K is a constant I can pull it out of my integral, and I can go like that, and let's see, what would be integrating from? Well, for time, let's go back up to here. All right, time we would be integrating from time is equal to zero to time is equal to T, and for concentration we'd be integrating from the initial concentration to the concentration at some time T, so let's go ahead and plug those in, so we'd be integrating from zero to T on the right side, and the left side we'd be integrating from"
    },
    {
        "Q": "\nAt 3:17, Sal states that the cell takes in nutrients from the environment in order to grow. What sort of nutrients does the cell take in? And doesn't this intake of nutrients affect the concentration of cytosol which in turn may affect the organelles?",
        "A": "The nutrition are what you eat. Everything you eat is broken down and divided up into carbohydrates, portein, fats, etc. which get sent off to the different parts of the body including the protein, which is sent to the cells. The proteins do affect the concentration of the cytosol, which does affect the organelles, but it will not affect it negatively. If you have any more questions or need more detail, comment below...",
        "video_name": "VXLSTd_dlKg",
        "timestamps": [
            197
        ],
        "3min_transcript": "But what I wanna focus on in this video is interphase. To do that, let's draw ourselves a cell. So let me copy and paste. So this right over here, actually let me, I did that just to save time. So let's say this is a cell, so green. I have it's nuclear membrane, or not nuclear membrane, I have its cell membrane. Inside of that, of course, you have all of the, all of the cytosol, and then this, in this orangeish color, I have the nuclear membrane that defines the nucleus. And then inside of that I have the DNA. And you might be used to seeing DNA all tightly bound, or chromosomes all tightly bound like that and like that or like this, this would be another chromosome right over here in magenta. But during interphase, the chromosomes aren't tightly bound like that so that they're easy to see For most of a cell's life, the chromosomes are completely unwound. They are in their chromatin form. So they are in their chromatin form. It's actually hard to see if you have just a simple microphone (laughing) a simple microscope. It's all unwound, you just have the proteins and the DNA, it's all tangled together. Now there's one other thing that I drew here. You might say, why am I drawing it when I haven't drawn most of the other organelles? But I'm drawing this thing, which is called a centrosome, 'cause it's going to be important for, it's going to be important for when we go into mitosis. Now, this drawing as well, you might say, wait, doesn't a cell, at least a human cell that has a diploid number of chromosomes, and once again, if we're not talking about sex cells, we're talking about just our somatic cells, doesn't it have to have 46 chromosomes? It looks like you only drew two. And it is true, I only drew two chromosomes that this is the cell of some organism that's much simpler, that it only has two chromosomes. So anyway, this is the new cell right over here. It is going to grow. So it is going to grow, it's going to take in nutrients from its environment, and it's going to grow as we would expect it to. So that's that right over there. And then let me give it its nucleus and its centrosome just like that. And this phase, this phase, where it is just growing from this new cell, this is, this phase right over here, is the G1 phase, the G1, actually I'm gonna do that in a different color since I'm already using that green so much. This is the G1 phase and so that might look something like this, different cells are going to do this for different periods of time, the G1 phase. But then you can imagine, well look, it's going to need to replicate some of the, or, it's gonna replicate the information inside of, or that's coded by the DNA"
    },
    {
        "Q": "\nI don't get what the G1 and G2 phases are. This is mentioned in 3:40.",
        "A": "Interphase: G1 phase -First phase cell enters if it\u00e2\u0080\u0099s going to divide / Normal cell functions, cell growth, and protein synthesis / Organelles replicate / May take anywhere from 8-12 hrs to months Interphase: G2 Phase -Protein synthesis / Preparation for division / Lasts 2-5 hours",
        "video_name": "VXLSTd_dlKg",
        "timestamps": [
            220
        ],
        "3min_transcript": "For most of a cell's life, the chromosomes are completely unwound. They are in their chromatin form. So they are in their chromatin form. It's actually hard to see if you have just a simple microphone (laughing) a simple microscope. It's all unwound, you just have the proteins and the DNA, it's all tangled together. Now there's one other thing that I drew here. You might say, why am I drawing it when I haven't drawn most of the other organelles? But I'm drawing this thing, which is called a centrosome, 'cause it's going to be important for, it's going to be important for when we go into mitosis. Now, this drawing as well, you might say, wait, doesn't a cell, at least a human cell that has a diploid number of chromosomes, and once again, if we're not talking about sex cells, we're talking about just our somatic cells, doesn't it have to have 46 chromosomes? It looks like you only drew two. And it is true, I only drew two chromosomes that this is the cell of some organism that's much simpler, that it only has two chromosomes. So anyway, this is the new cell right over here. It is going to grow. So it is going to grow, it's going to take in nutrients from its environment, and it's going to grow as we would expect it to. So that's that right over there. And then let me give it its nucleus and its centrosome just like that. And this phase, this phase, where it is just growing from this new cell, this is, this phase right over here, is the G1 phase, the G1, actually I'm gonna do that in a different color since I'm already using that green so much. This is the G1 phase and so that might look something like this, different cells are going to do this for different periods of time, the G1 phase. But then you can imagine, well look, it's going to need to replicate some of the, or, it's gonna replicate the information inside of, or that's coded by the DNA So let's depict that. So let me draw, let me draw the nucleus and the centrosome again. Let me draw that again. Let me draw the cellular membrane. This nice healthy growing cell. And now, its DNA is actually going to replicate. So instead of having one copy of its DNA, it's essentially going to go to two copies. But I wanna be very very careful now. So if I draw that magenta chromosome up here, so once again it's all unwound like that. When it replicates, it's going to create a copy of its DNA, and once again, I'm not doing justice for how much DNA, how much DNA there actually is. But it was one chromosome before, it was one chromosome when it was just like this, and it's still one chromosome, even though it's copied its genetic material."
    },
    {
        "Q": "\nAt 4:08, are you saying that over time, carbon will become nitrogen?",
        "A": "Carbon-14 will decay into Nitrogen-14 over time.",
        "video_name": "gqrh8wbPXVE",
        "timestamps": [
            248
        ],
        "3min_transcript": "so equal numbers of protons and neutrons turns out to be stable. So for this example, the helium-four nucleus is stable. Thinking about that, let's look at carbon-14 next. We have carbon-14, so let's get a little space right down here. So carbon-14 atomic number of six. Therefore, carbon has six protons in the nucleus. So there are six protons. Number of neutrons will be 14 minus six, so eight neutrons. So what's the neutron to proton ratio? So what's the N to Z ratio here? Well the N to Z ratio would be eight neutrons and six protons, and obviously that number is greater than one, so we have an unstable nucleus. The carbon-14 nucleus is unstable, it's radioactive, it's going to undergo spontaneous decay. It's going to try to get a better neutron to proton ratio. which represents the spontaneous decay of carbon-14. So here is our nuclear equation. And when you're writing nuclear equations, you're representing only the nuclei here, so for example, on the left side of my nuclear equation, I have carbon-14, we're talking about only the nucleus, so we're talking about six protons and eight neutrons in the nucleus. And so let's look and see what happens here. So carbon-14, the nucleus, the carbon-14 nucleus is actually going to give off an electron, and so that's pretty weird, and we'll talk about more why in the next video. It's a conversion that's governed by the weak nuclear force. But we know that an electron has a negative one charge, and so that's what we're talking about here. Here for carbon, we have six protons, let me go and write that, six protons here. let's write a negative one charge here for the electron. The carbon-14 nucleus is turning into the nucleus for nitrogen here. Let's look at what we have. Our atomic number is seven, so we have seven protons, let's go ahead and write that here. Seven protons, and 14 minus 7 gives us seven neutrons. So we look at the mass number here, so 14 minus seven gives us seven neutrons. And so that ratio, the ratio of neutrons to protons is seven over seven, which is equal to one. That implies that we have a stable nucleus here. That's the reason why carbon-14 undergoes radioactive decay. Let's look at more details about a nuclear equation, because that's really what I'm most concerned about here in this video. The number of nucleons is conserved. Let's use a different color here."
    },
    {
        "Q": "\nBy 2:02 can we say that Deuterium and Tritium are radioactive? What about elements like Boron and Chlorine? Are they also radioactive as they do not follow this rule?",
        "A": "Deuterium isn t radioactive, tritium is. What about boron and chlorine exactly? They both have two stable isotopes so I m not sure what you are getting at. If you mean because their N/Z isn t exactly equal to 1, that s a general rule only. There are exceptions to almost every rule in chemistry.",
        "video_name": "gqrh8wbPXVE",
        "timestamps": [
            122
        ],
        "3min_transcript": "- [Voiceover] In the last video, we talked about the helium nucleus, which contains two protons and two neutrons. Protons and neutrons in the nucleus are called nucleons, and so I'll use that term a few times in this video. Here's a picture of the nucleus, with two protons and two neutrons, and we know it's stable, even though we know like charges repel. And so these two protons are repelling each other, and that's the electrostatic force. So let me go ahead and write that here. The electrostatic force says like charges repel. We know that this nucleus is stable, so there must be something else holding the nucleus together, which we call the strong force. So the nuclear strong force is stronger than the electrostatic force. The strong force acts only over short distances though, but it does act between all nucleons. For example, a proton-proton interaction is the same as a proton-neutron interaction, which is the same as a neutron-neutron interaction. about the strong force. That's not really the point of this video. The point is that this nucleus is stable. And let's think about why. We have equal numbers of protons and neutrons, and so that's interesting. So let's think about the atomic number, which tells us the number of protons, which we represent by Z. And the number of neutrons we could say is capital N. So if we're concerned with the ratio, the ratio of neutrons to protons, so the N to Z ratio. In this example, we have two protons and two neutrons. So two neutrons over two protons is equal to one. We have N to Z ratio of one. It turns out that nuclei that have small numbers of protons, so if we're talking about Z is less than 20, they have stable nuclei when the N to Z ratio is equal to one. So when N over Z is equal to one, so equal numbers of protons and neutrons turns out to be stable. So for this example, the helium-four nucleus is stable. Thinking about that, let's look at carbon-14 next. We have carbon-14, so let's get a little space right down here. So carbon-14 atomic number of six. Therefore, carbon has six protons in the nucleus. So there are six protons. Number of neutrons will be 14 minus six, so eight neutrons. So what's the neutron to proton ratio? So what's the N to Z ratio here? Well the N to Z ratio would be eight neutrons and six protons, and obviously that number is greater than one, so we have an unstable nucleus. The carbon-14 nucleus is unstable, it's radioactive, it's going to undergo spontaneous decay. It's going to try to get a better neutron to proton ratio."
    },
    {
        "Q": "From 5:35 to 6:50, it seems more reasonable to me that water molecule should attach to secondary carbocation instead of tertiary carbocation because I think since secondary carbocation is more positive than the tertiary one, secondary carbocation would be more strongly attracted to a lone pair of electrons on the water molecule. Why is it not the case in the video?\n",
        "A": "Yes, both C1 and C2 have \u00ce\u00b4\u00e2\u0081\u00ba character, but C1 has more of the + charge. The cyclic bromonium ion is a resonance hybrid of two contributors: one with the + on C1, and the other with the + on C2. The second one is the minor contributor, so its C-Br bond is closer to an ordinary covalent bond. The major contributor is closer to having a tertiary carbocation at C1, so that is where the water attacks.",
        "video_name": "FaOOx6IZxV8",
        "timestamps": [
            335,
            410
        ],
        "3min_transcript": "can examine the stereochemistry a little bit more here. So if I start with an alkene, and to this alkene we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left. So Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right. So BR2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to once again look at this molecule a little bit from above, so looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine's going to form on top here. And so there's going to be two lone pairs of electrons It's going to have a plus 1 formal charge. So I'm going to say that my methyl group is now going down in space. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here is H2O, and I think about which carbon will the oxygen attack? So I have two options, right? This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the option is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation, or a partial carbocation in character. So you could think about it as being if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red. And if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character, and that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons here we kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. OK, so what would we have here?"
    },
    {
        "Q": "Please refer instants 00:34 & 02:25 seconds of video Fluids (Part 5)\nIn the example you delt with, assume there is water in the jar, upto height h in the jar without any objects. Now we know, the pressure is going to vary as per formula (density times the (g * h)).\nPlease educate me on how the pressure at each and every point of the jar is going to get effected, if a cube denser than the water is suspended (and immersed) in the water using a thread. Is the profile of pressure with respect to the\n",
        "A": "the density (or weight/mass) of the cube will not matter beyond causing it to sink because it will be off set by the tension in the string suspending it. putting it in the jar will be no different than augmenting the shape of the jar",
        "video_name": "vzID7ds600c",
        "timestamps": [
            34,
            145
        ],
        "3min_transcript": "Let's say we have a cup of water. Let me draw the cup. This is one side of the cup, this is the bottom of the cup, and this is the other side of the cup. Let me say that it's some liquid. It doesn't have to be water, but some arbitrary liquid. It could be water. That's the surface of it. We've already learned that the pressure at any point within this liquid is dependent on how deep we go into the liquid. One point I want to make before we move on, and I touched on this a little bit before, is that the pressure at some point isn't just acting downwards, or it isn't just acting in one direction. It's acting in all directions on that point. So although how far we go down determines how much pressure there is, the pressure is actually acting in all directions, including up. The reason why that makes sense is because I'm assuming that this is a static system, or that the fluids in this down here, and it's stationary. The fact that it's stationary tells us that the pressure in every direction must be equal. Let's think about a molecule of water. A molecule of water, let's say it's roughly a sphere. If the pressure were different in one direction or if the pressure down were greater than the pressure up, then the object would start accelerating downwards, because its surface area pointing upwards is the same as the surface area pointing downwards, so the force upwards would be more. It would start accelerating downwards. Even though the pressure is a function of how far down we go, at that point, the pressure is acting in every direction. Let's remember that, and now let's keep that in mind to learn a little bit about Archimedes' principle. Let's say I submerge a cube into this liquid, and let's What I want to do is I want to figure out if there's any force or what is the net force acting on this cube due to the water? Let's think about what the pressure on this cube is at different points. At the depths along the side of the cube, we know that the pressures are equal, because we know at this depth right here, the pressure is going to be the same as at that depth, and they're going to offset each other, and so these are going to be the same. But one thing we do know, just based on the fact that pressure is a function of depth, is that at this point the pressure is going to be higher-- I don't know how much higher-- than at this point, because this point is deeper into the water. Let's call this P1. Let's call that pressure on top, PT, and let's call this"
    },
    {
        "Q": "\n@2:03, why did Jay say that he \"has to\" make chlorine point up (axial)? Could't he make it point equatorial?",
        "A": "He randomly chose that first carbon to be #1, and the up one on that carbon happened to be axial. He could have chosen the next carbon along instead and then it would have been equatorial, but then the problem is this mechanism wouldn t work.",
        "video_name": "uCW6154hPkc",
        "timestamps": [
            123
        ],
        "3min_transcript": "- [Instructor] We're doing E two mechanism for a substituted cyclohexane. The key is to think about your chair conformations. For example, let's look at this substituted cyclohexane on the left here, and let's first try to solve this problem without looking at a chair conformation. So I would start by saying my chlorine is attached to my alpha carbon, and the carbons directly bonded to the alpha carbon would be the beta carbons, so there's a beta carbon on the right, which I call beta one, there's a beta carbon on the left which I will call beta two. I know that sodium methoxide will be my strong base. The ethoxide anion will take a proton from one of those beta carbons and a double bond would form between the alpha carbon and one of those beta carbons, and the chlorine would leave as the chloride anion, so a double bond forms between the alpha and the beta one carbon, so I'll draw in a double bond here, and the chlorine is now gone, but I would still have this methyl group going away from us in space, so CH3, so that's one possible product, and you might think oh I might also form a double bond between the alpha and the beta two carbon, and so my methyl group would just be on a straight line this time, and so you might think that could be one of the products, but when you actually think about chair conformations, you'll realize that you would not realize this product, so this product is not observed, so you would only get this one, and let's look at why. So first, we need to number our cyclohexane ring, and remember, when you number a cyclohexane ring for a chair conformation, you're not necessarily doing IUPAC nomenclature. You're just using whatever numbers you need to help you draw a proper chair conformation, so I'm going to start with the chlorine being number one, then I'm going to go around to the right, so two, three, carbon four, carbon five, and carbon six, so we're not naming this. We're just trying to get a chair conformation, so on my chair conformation down here, I always say this is carbon one, and so I need to have the chlorine, the chlorine up at carbon one, The chlorine has to be up axial, and so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in a Me for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons, so let me highlight our beta carbons here. I'll use red, so this would be, what I've marked is being beta one, so I have two hydrogens on that carbon, so I'll draw those in here, and then my other beta carbon which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video, so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation,"
    },
    {
        "Q": "\nAt 2:09, does anyone know more about Robert Hooke's microscope structure?",
        "A": "When Hooke viewed a thin cutting of cork using his self designed microscope, he discovered empty spaces contained by walls, and termed them cells. They looked like honeycombs. In latin, cell means tiny rooms. Thus, as Robert Hooke thought the compartments looked like small rooms. Thus, he called them cells.",
        "video_name": "1aJBToJrlvA",
        "timestamps": [
            129
        ],
        "3min_transcript": "Man: This is an animal. This is also an animal. Animal. Animal carcass. Animal. Animal carcass again. Animal. The thing that all of these other things have in common is that they're made out of the same basic building block, the animal cell. (music) Animals are made up of your run-of-the-mill eukaryotic cells. These are called eukaryotic because they have a true kernel in the Greek, a good nucleus. That contains the DNA and calls the shots for the rest of the cell, also containing a bunch of organelles. There's a bunch of different kinds of organelles and they all have very specific functions. All of this is surrounded by the cell membrane. Of course, plants are eukaryotic cells too, but theirs are set up a little bit differently. Of course, they have oranelles that allow them to make their own food, which is super nice. We don't have those. Also, their cell membrane is actually a cell wall that's made of cellulose. It's rigid which is why plants can't dance. we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella."
    },
    {
        "Q": "at 3:28 Hank says microtubules are called (9+2) structure. Can anyone tell me that why are they called so?\n",
        "A": "They are called the 9+2 structure because as he says at 3:24 there are made of 9 microtubules forming a ring around 2 microtubules.",
        "video_name": "1aJBToJrlvA",
        "timestamps": [
            208
        ],
        "3min_transcript": "we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here."
    },
    {
        "Q": "\nat 4:20 , aren't centrosomes also called centrioles",
        "A": "Centrosomes are made up of a pair of centrioles",
        "video_name": "1aJBToJrlvA",
        "timestamps": [
            260
        ],
        "3min_transcript": "Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here. is kind of wet and squishy; it's a bit of a swampland. Each eukaryotic cell is filled with a solution of water and nutrients called cytoplasm. Inside of the cytoplasm is a scaffolding called the cytoskeleton. It's basically just a bunch of protein strands that reinforce the cell. Centrosomes are a special part of this reinforcement. They assemble long microtubules out of proteins that act like steel girders that hold all the cities' building together. The cytoplasm provides the infrastructure necessary for all the organelles to do all of their awesome amazing business, with the notable exception of the nucleus, which has its own kind of cytoplasm called the nucleoplasm, which is a more luxurious, premium environment befitting the cell's beloved leader. First, let's talk about the cell's highway system. The endoplasmic reticulum, or just ER, are organelles that create a network of membranes that carry stuff around the cell. These membranes are phospolipid bilayers same as in the cell membrane. There are two types of ER. There's the rough and the smooth; fairly similar, but slightly different shapes, slightly different functions. The rough ER looks all bumpy because it has"
    },
    {
        "Q": "At 4:23, what are centrosomes, are they centrioles?\n",
        "A": "An organelle near the nucleus of a cell that contains the centrioles (in animal cells) and from which the spindle fibers develop in cell division.",
        "video_name": "1aJBToJrlvA",
        "timestamps": [
            263
        ],
        "3min_transcript": "Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here. is kind of wet and squishy; it's a bit of a swampland. Each eukaryotic cell is filled with a solution of water and nutrients called cytoplasm. Inside of the cytoplasm is a scaffolding called the cytoskeleton. It's basically just a bunch of protein strands that reinforce the cell. Centrosomes are a special part of this reinforcement. They assemble long microtubules out of proteins that act like steel girders that hold all the cities' building together. The cytoplasm provides the infrastructure necessary for all the organelles to do all of their awesome amazing business, with the notable exception of the nucleus, which has its own kind of cytoplasm called the nucleoplasm, which is a more luxurious, premium environment befitting the cell's beloved leader. First, let's talk about the cell's highway system. The endoplasmic reticulum, or just ER, are organelles that create a network of membranes that carry stuff around the cell. These membranes are phospolipid bilayers same as in the cell membrane. There are two types of ER. There's the rough and the smooth; fairly similar, but slightly different shapes, slightly different functions. The rough ER looks all bumpy because it has"
    },
    {
        "Q": "\nAt 13:14 how do you get 2p6? I thought each orbital only had 2 electrons, so how does this one have 6?",
        "A": "each orbital can hold two electrons. in the p orbital shell, it has three orbitals, each holding two electrons, therefore holding a maximum of six",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            794
        ],
        "3min_transcript": "The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus,"
    },
    {
        "Q": "7:05 (carbon) I'm confused about the P orbitals.. so in carbon the 1s2 and 2s2 are filled and then he goes to the P orbitals and why is it 2p2? I thought it filled only one of the first P orbital and then did one in the second so why isn't it 1p1 and 2p1 instead of 2p2?\n",
        "A": "The base number measures the energy level, or the row # of the periodic table. The p block starts on level 2, so there is no 1p. Carbon is also the 2nd element in the p block of that row, so the exponent is 2.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            425
        ],
        "3min_transcript": "And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2."
    },
    {
        "Q": "\nAt 6:10, he refers to the \"z\" direction as up and down, and the \"y\" direction as forward and backward, but isn't that reversed? I thought, that \"x\" was left-right, \"y\" up-down, and \"z\" forward-backward (representing 3 dimensions). Are these interchangeable, now? I'm only asking because it has been 18 years since I last took a math course.",
        "A": "They are just labels for axes. Up and down depends on how you are oriented when you are looking at something.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            370
        ],
        "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon."
    },
    {
        "Q": "\nAt 1:30,would those be snapshots of different atoms or the same one?",
        "A": "it s the same atom. It is just the electron, that is represented in different snapshots. When you overlay the different snapshots it becomes apparent, that the electron is more likely to be near the nucleus.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            90
        ],
        "3min_transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state."
    },
    {
        "Q": "I don't understand why the P orbital has Pz, Px and Py? 6:26\n",
        "A": "Because there s 3 p orbitals in a shell (that s why a total p subshell can hold 6 electrons). There s one p orbital aligned on each 3-dimensional axis.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            386
        ],
        "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon."
    },
    {
        "Q": "At 13:26 how does he get the configuration for Silicon? Can someone please summarize this whole video. I'm confused\n",
        "A": "That s a good explanation, but you explained how to get to Sulfur, not Silicon. So 3p2, rather than 3p4, because Silicon has 14 electrons.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            806
        ],
        "3min_transcript": "The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus,"
    },
    {
        "Q": "At 13:37, why does one write out the whole electron configuration, rather than just the last \"shell\"? In other words, wouldn't it be enough to write 3p^2 for silicon, since all the lower energy shells are always filled up?\n",
        "A": "When you get to heavier elements you ll find what you say about the lower energy orbitals always being full isn t always true.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            817
        ],
        "3min_transcript": "So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video."
    },
    {
        "Q": "\nat around 12:08. How does he know 3 electrons go in the p orbital. Is it because that is what is left over after filling the s orbitals?",
        "A": "Yes, when determining the electron configuration, you count up from the bottom (1s) orbital and add electrons one at a time until all are used up for that atom or ion. The electrons are first put in singly, then double up with an opposite spin. You cannot move up to the next higher level until the lower orbitals are filled. When the level or sub-shells has more than one orbital (as there are 3 p orbitals), then each is filled up with a single electron first before they are doubled up.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            728
        ],
        "3min_transcript": "Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2."
    },
    {
        "Q": "at 5:19, do the electrons fill the orbital from lowest energy orbital to highest energy orbital?\n",
        "A": "Yes, nature always wants to minimize energy.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            319
        ],
        "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves."
    },
    {
        "Q": "He stated at 12:10 that the electron configuration for Nitrogen is 1s^2 2s^2 2p^3. However, how can 2p have 3 electrons when each subshell can only have 2 electrons?\n",
        "A": "It is orbitals that can only have 2 electrons, not subshells. The p subshell contains 3 orbitals, so it can hold at most 6 electrons. The d subshell contains 5 orbitals, so it can hold at most 10 electrons. The f subshell contains 7 orbitals, so it can hold at most 14 electrons. And, of course, the s subshell has only 1 orbital so it can only hold 2 electrons.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            730
        ],
        "3min_transcript": "Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2."
    },
    {
        "Q": "\nat 1:27 he talks about the s, why are the orbitals called s,p,f,etc?",
        "A": "Those letters represent terms that are now outdated and no longer used. The terms were sharp, principal, diffuse, and one I can t remember.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            87
        ],
        "3min_transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state."
    },
    {
        "Q": "Hi\u00ef\u00bc\u008cI am confused about P orbital. I can understand s orbital and it seems like it follows orders of electrons numbers in each circle. (it seems like follow 2,8,8,18,18,32,32..). However, in the video 7:56, it suddenly jumps to P orbitals without explanations. What does P orbitals refer to\n",
        "A": "The orbitals each have a given amount of electrons, s can have 2, and then it must move up to p. The p orbital can have up to 6 electrons, and when you run out, you must move on to the d orbital, which can have 10 electrons. Lastly, you have the f orbital, and that orbital can have up to 14 electrons.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            476
        ],
        "3min_transcript": "that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different,"
    },
    {
        "Q": "\nAt 14:00, could someone explain why all the electron configurations look so strange?",
        "A": "They are based on complex math equations that calculate the likelihood of an electron being within the configurations with 90% accuracy.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            840
        ],
        "3min_transcript": "So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video."
    },
    {
        "Q": "At 5:13 Sal mentions that the water molecules on the skin surface are linked by hydrogen bonds. Would the cooling effect of a liquid be different if there was no hydrogen bond? I would assume that evaporation of other liquids without hydrogen bonds would take away less energy (=heat), because not having to break up a hydrogen bond would need less energy? Am I correct?\n",
        "A": "Yes, hydrogen bonds take energy to form, and actually take a larger amount of heat to disrupt a hydrogen bond putting hydrogen bonds. So in theory other liquids with the same evaporating temperatures would take less energy.",
        "video_name": "_eEONOJHnEs",
        "timestamps": [
            313
        ],
        "3min_transcript": "So average kinetic energy is going to go down. Or another way of saying it, is that your temperature is going to go down. Your temperature is going to go down because as these molecules turn into water vapor, they're going to be the highest kinetic energy, energy is transferred to them, and then they escape. And so what's left over is going to have a lower average kinetic energy. And you're saying, \"Well, how does that \"actually cool down my hand?\" Well, your hand is made up of molecules as well. So let's say this is the surface of your hand, those are the molecules, they have some average kinetic energy, they are kind of vibrating in place, especially if we're talking about they're... they're a solid. And so maybe I'll draw the more, you know, they're vibrating like this, they're bonded to each other in some way. I won't go into the details of what types of molecules these are, but then if you have your water molecules here, water molecules that are sitting on the surface, and I'm drawing this is kind of a cross-section. I'll draw them as blue molecules. So this is an H2O right over here. H2O. This is an H2O. And this is an H2O. And they have some hydrogen bonding, so there is some hydrogen bonding going on. Well, as the high kinetic energy water molecules escape, I'll say this one right over here escapes, and so the average kinetic energy of what's left over is lower, so then the temperature has gone down, and now your body molecules, the ones that are all warmed up, and because of whatever's going on inside of your body, well, those can now bump into, they can vibrate and bump into these water molecules and increase their kinetic energy more than the ones that have the most kinetic energy. Those might escape again. And so it's a--one way of thinking about it is that all that heat is being used to allow these individual water molecules to escape in order to vaporize. And so that heat is leaving your body, so it allows you to cool down. Cooling down happens by heat actually leaving. I wrote evaporative cool... That's how evaporative cooling... That's how evaporative cooling actually works."
    },
    {
        "Q": "\nAt the beginning of the video (around 0:38) he said that life is carbon based. What is that supposed to mean?",
        "A": "What Sal means at 0:38 is that all known lifeforms contain the element carbon. Of course, there are other elements that are found in a majority of life-forms as well. But Carbon is the most prominent. Scientists have yet to find a living organism that does not contain carbon in it s Chemical Make-Up. Therefore, as far as we can tell, all life is carbon-based.",
        "video_name": "JgYlogdtJDo",
        "timestamps": [
            38
        ],
        "3min_transcript": "You're probably already familiar with some forms of carbon. For example this graph I write over here, this is one form carbon takes. Very important when you're writing with a pencil, otherwise you would not see any writing. If you didn't have the graphite scraping onto your, scraping onto your paper, which you and your paper is also. It's not pure carbon, but it has a lot of carbon in it. This right over here is a raw diamond, another form the carbon can take under intense time or after a long period of time under intense pressure. But what you may or may not realize is that carbon is actually essential for life. In fact life as we know it is carbon-based, so carbon-based, based life. When we look for signs of life, at least life as we know on other planets,we are looking for signs of carbon-based life. And there might be other forms, other other elements that form the backbone of life. But carbon is the only one that we have been able to observe. Now why is carbon so valuable for life? Why does it form the backbone of the molecules that make life possible ? What all comes down to where it and its atomic number and how it tends to bond with things? So this is why Chemistry is important. So carbon we see over here has an atomic number six, which by definition means it has 6 protons. So far we draw its nucleus it would have 1,2 , 3, 4, 5, 6 protons. And the most common isotope of carbon on earth is carbon 12, which also has six neutrons. So let me draw that in this nucleus 1, 2, 3, 4, 5, 6 neutrons. And then neutral carbon is going to have six electrons. And so two of them are going to be in their inner most in the first shell .So that's two of them right over there. These are the inner shell. I guess you could say or the so that's the first two electrons ,and then you have four remaining in its outermost shell. And these four considered valence electrons, these are the electrons that the reacting. And if any of this seems unfamiliar to you,I encourage you to watch the videos on Khan Academy on things like valence electrons, but this is a little bit overview right over here. Carbon has four valence electrons. Valence electrons are the ones that do or that tend to do the reacting. And so I could if I wanted to simplify this drawng over here,I could say ,\"Okay carbon.\"And if I register as valence electrons which is a typical thing to do. I could say carbon has 1, 2, 3 ,4 valence electrons. Now you might remember the octet rule that that atoms tend to be more stable when they at least pretend, like they with that they're sharing or that they have eight electrons in there outermost shell. So carbon can do that by forming 4 covalent bonds. For example it could do that with hydrogen.This hydrogen over here has one valence electron. It actually has one electron. Now the hydrogen feels good. It feels like its sharing two-electrons filling its its first shell .Hydrogen is just trying to fill out the first"
    },
    {
        "Q": "At 13:10 isn't the height 6m?\n",
        "A": "No, if you think about it, if that ball has a radius of 2m. So when the ball is touching the ground, it s center of mass will actually still be 2m from the ground. It s true that the center of mass is initially 6m from the ground, but when the ball falls and touches the ground the center of mass is again still 2m from the ground. That means the height will be 4m. If you work the problem where the height is 6m, the ball would have to fall halfway through the floor for the center of mass to be at 0 height.",
        "video_name": "5eX5WnPDnvs",
        "timestamps": [
            790
        ],
        "3min_transcript": "and so, now it's looking much better. We just have one variable in here that we don't know, V of the center of mass. This I might be freaking you out, this is the moment of inertia, what do we do with that? With a moment of inertia of a cylinder, you often just have to look these up. The moment of inertia of a cylinder turns out to be 1/2 m, the mass of the cylinder, times the radius of the cylinder squared. So we can take this, plug that in for I, and what are we gonna get? If I just copy this, paste that again. If we substitute in for our I, our moment of inertia, and I'm gonna scoot this over just a little bit, our moment of inertia was 1/2 mr squared. So I'm gonna have 1/2, and this is in addition to this 1/2, so this 1/2 was already here. There's another 1/2, from the moment of inertia term, 1/2mr squared, but this r is the same as that r, so look it, I've got a, I've got a r squared and a one over r squared, these end up canceling, and this is really strange, it doesn't matter what the radius of the cylinder was, and here's something else that's weird, not only does the radius cancel, So no matter what the mass of the cylinder was, they will all get to the ground with the same center of mass speed. In other words, all yo-yo's of the same shape are gonna tie when they get to the ground as long as all else is equal when we're ignoring air resistance. No matter how big the yo-yo, or have massive or what the radius is, they should all tie at the ground with the same speed, which is kinda weird. So now, finally we can solve for the center of mass. We've got this right hand side. The left hand side is just gh, that's gonna equal, so we end up with 1/2, V of the center of mass squared, plus 1/4, V of the center of mass squared. That's just equal to 3/4 speed of the center of mass squared. If you take a half plus a fourth, you get 3/4. So if I solve this for the speed of the center of mass, I'm gonna get, if I multiply gh by four over three, and we take a square root, we're gonna get the square root of 4gh over 3, and so now, I can just plug in numbers. If I wanted to, I could just say that this is gonna equal times 9.8 meters per second squared, times four meters, that's where we started from, that was our height, divided by three, is gonna give us a speed of the center of mass of 7.23 meters per second. Now, here's something to keep in mind, other problems might look different from this, but the way you solve them might be identical. For instance, we could just take this whole solution here, I'm gonna copy that. Let's try a new problem, it's gonna be easy. It's not gonna take long. Let's say we take the same cylinder and we release it from rest at the top of an incline that's four meters tall and we let it roll without slipping to the bottom of the incline, and again, we ask the question, \"How fast is the center of mass of this cylinder \"gonna be going when it reaches the bottom of the incline?\" Well, it's the same problem. It looks different from the other problem, but conceptually and mathematically, it's the same calculation. This thing started off with potential energy, mgh, and it turned into conservation of energy says"
    },
    {
        "Q": "\nAt 1:48, why isn't the concentration of Cl- taken into consideration? Although it's an extremely weak base, I thought it would still make some difference in the pH.",
        "A": "HCl is a strong acid, so that means Cl- has to be an extremely weak base as you stated. Essentially the reverse reaction of Cl- + H+ -> HCl does not happen, so no need to account for it.",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            108
        ],
        "3min_transcript": "- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH?"
    },
    {
        "Q": "\nThis may seem trivial, but at 3:40, why is the hydroxide ion written with the charge on the left-hand side, instead of the right? I've seen it that way consistently in these chemistry videos, but never anywhere else.",
        "A": "It is preferable to put the charge on the atom that has the charge, so we should write \u00e2\u0081\u00bbOH or HO\u00e2\u0081\u00bb. However, many people still write the formula as OH\u00e2\u0081\u00bb. It s OK, as long as you remember that the O atom has the charge.",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            220
        ],
        "3min_transcript": "So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH? moles of sodium hydroxide, and our total volume is .50. So if we divide moles by liters, that will give us the concentration of sodium hydroxide. .005 divided by .50 is 0.01 molar. So that's our concentration of sodium hydroxide. And since sodium hydroxide is a strong base, that's also our concentration of hydroxide ions in solution. So this is our concentration of hydroxide ions, .01 molar. So we're adding a base and think about what that's going to react with in our buffer solution. So our buffer solution has NH three and NH four plus. The base is going to react with the acids. So hydroxide is going to react with NH four plus. Let's go ahead and write out the buffer reaction here. ammonium is going to react with hydroxide and this is going to go to completion here. So if NH four plus donates a proton to OH minus, OH minus turns into H 2 O. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia. Alright, let's think about our concentrations. So we just calculated that we have now .01 molar concentration of sodium hydroxide. For ammonium, that would be .20 molars. So 0.20 molar for our concentration. And for ammonia it was .24. So let's go ahead and write 0.24 over here. So if .01, if we have a concentration of hydroxide ions of .01 molar, all of that is going to react with the ammonium. So we're gonna lose all of this concentration here for hydroxide."
    },
    {
        "Q": "\nAt 1:00 how do you know if the Ka value is 5.6 x10^10 ?",
        "A": "So in this problem, it is just given to you. Even though it doesn t appear to be written as part of the original question, during exams you would be given a Kb value or a Ka value to work with. Also, ammonia s Kb value is 1.8 x 10^-5 (memorized from constant use). We know that (Ka x Kb = Kw) so all we have to do is take 1.0 x 10^-14 / 1.8 x 10^-5 (Kw / Kb) which gives us 5.6 x 10^-10 (Ka). Hope that helped!",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH?"
    },
    {
        "Q": "at 0:01 the picture are real or not\n",
        "A": "Yes, those pictures are real",
        "video_name": "N6IAzlugWw0",
        "timestamps": [
            1
        ],
        "3min_transcript": "Let's explore the scientific method. Which at first might seem a bit intimidating, but when we walk through it, you'll see that it's actually almost a common-sense way of looking at the world and making progress in our understanding of the world and feeling good about that progress of our understanding of the world. So, let's just use a tangible example here, and we'll walk through what we could consider the steps of the scientific method, and you'll see different steps articulated in different ways, but they all boil down to the same thing. You observe something about reality, and you say, well, let me try to come up with a reason for why that observation happens, and then you try to test that explanation. It's very important that you come up with explanations that you can test, and then you can see if they're true, and then based on whether they're true, you keep iterating. If it's not true, you come up with another explanation. If it is true, but it doesn't explain everything, well once again, you try to explain more of it. in I don't know, northern Canada or something, and let's say that you live near the beach, but there's also a pond near your house, and you notice that the pond, it tends to freeze over sooner in the Winter than the ocean does. It does that faster and even does it at higher temperatures than when the ocean seems to freeze over. So, you could view that as your observation. So, the first step is you're making an observation. Observation. In our particular case is that the pond freezes over at higher temperatures than the ocean does, and it freezes over sooner in the Winter. Well, the next question that you might wanna, or the next step you could view as a scientific method. It doesn't have to be this regimented, but this is a structured way of thinking about it. Well, ask yourself a question. Ask a question. Why does, so in this particular question, or in this particular scenario, why does the pond tend than the ocean does? Well, you then try to answer that question, and this is a key part of the scientific method, is what you do in this third step, is that you try to create an explanation, but what's key is that it is a testable explanation. So, you try to create a testable explanation. Testable explanation, and this is kind of the core, one of the core pillars of the scientific method, and this testable explanation is called your hypotypothesis. Your hypothesis. And so, in this particular case, a testable explanation could be that, well the ocean is made up of salt water, and this pond is fresh water, so your testable explanation could be salt water, salt water has lower freezing point."
    },
    {
        "Q": "at about 9:00, you show the chlorine in AlCl4 giving its' electron to the hydrogen, then the hydrogen giving it's electron back to the benzene ring so it can form a double bond and become aromatic again. You don't mention what happens to that chlorine atom and hydrogen after? do they then form HCl? thank you!\n",
        "A": "I just watched the next video, Friedel Crafts Acylation Addendum, that answers my question!",
        "video_name": "vFfriC55fFw",
        "timestamps": [
            540
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:40, what is the general formula for an exponential decay and wat does each letter in the formula represent?\n",
        "A": "The formula for an exponential decay is (1-r)^t. T is time, r is the rate or percent,",
        "video_name": "Hqzakjo_dYg",
        "timestamps": [
            280
        ],
        "3min_transcript": "It makes the calculator math a little bit easier. The same thing. So if I do 2 natural log, divided by 0.05, it is equal to 13.86. So when t is equal to 13.86. And I'm assuming that we're dealing with time in years. That tends to be the convention, although sometimes it could be something else and you'd always have to convert to years. But assuming that this original formula, where they gave this k value 0.05, that was with the assumption that t is in years, and I've just solved its half-life. I just solved that after 13.86 years, you can expect to have 1/2 of the substance left. We started with 100, we ended up with 50. I could have started with x and ended up with x over 2. Let's do one more of these problems, just so that we're Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught."
    },
    {
        "Q": "\n6:26 What is \"e\" for? why the alphabet becomes 2.71?",
        "A": "Sal has done a few videos on where e comes from. Look for the Introduction to compound interest and e video in the Precalculus playlist.",
        "video_name": "Hqzakjo_dYg",
        "timestamps": [
            386
        ],
        "3min_transcript": "Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught. equal to 500 grams. Or I could multiply both sides by e, and I have N sub naught is equal to 500e, which is about 2.71. So it's 500 times 2.71. I don't actually have e on this calculator or at least I don't see it. So we'll have 1,355 grams. So it's equal to 1,355 grams, or 1.355 kilograms. That's what I started with. So hopefully you see now. I mean, I think we've approached this pretty much at almost any direction that a chemistry test or teacher could throw the problem at you. But you really just need to remember this formula. And this applies to a lot of things. Later you'll learn, you know, when you do compound interest in finance, the k will just be a positive value, but it's essentially the same formula."
    },
    {
        "Q": "6:26 What is \"e\" for? why the alphabet becomes 2.71?\n",
        "A": "e is an important mathematical constant that is roughly equal to 2.71.",
        "video_name": "Hqzakjo_dYg",
        "timestamps": [
            386
        ],
        "3min_transcript": "Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught. equal to 500 grams. Or I could multiply both sides by e, and I have N sub naught is equal to 500e, which is about 2.71. So it's 500 times 2.71. I don't actually have e on this calculator or at least I don't see it. So we'll have 1,355 grams. So it's equal to 1,355 grams, or 1.355 kilograms. That's what I started with. So hopefully you see now. I mean, I think we've approached this pretty much at almost any direction that a chemistry test or teacher could throw the problem at you. But you really just need to remember this formula. And this applies to a lot of things. Later you'll learn, you know, when you do compound interest in finance, the k will just be a positive value, but it's essentially the same formula."
    },
    {
        "Q": "At 1:11 Sal says light travels fastest in a vacuum, I thought the speed of light was constant.\n",
        "A": "light slows down more in more densely packed things since the photon crashes into more particle in a time period. however, the speed of light in a vacuum is constant.",
        "video_name": "y55tzg_jW9I",
        "timestamps": [
            71
        ],
        "3min_transcript": "In the last couple of videos we talked about reflection. And that's just the idea of the light rays bouncing off of a surface. And if the surface is smooth, the incident angle is going to be the same thing as the reflected angle. We saw that before, and those angles are measured relative to a perpendicular. So that angle right there is going to be the same as that angle right there. That's essentially what we learned the last couple of videos. What we want to cover in this video is when the light actually doesn't just bounce off of a surface but starts going through a different medium. So in this situation, we will be dealing with refraction. Refraction. Refraction, you still have the light coming in to the interface between the two surfaces. So let's say--so that's the perpendicular right there, actually let me continue the perpendicular all the way down like that. And let's say we have the incident light ray coming in at some, at some angle theta 1, just like that...what will happen--and so let's say that this up here, this is a vacuum. In a vacuum. There's nothing there, no air, no water, no nothing, that's where the light travels the fastest. And let's say that this medium down here, I don't know, let's say it's water. Let's say that this is water. All of this. This was all water over here. This was all vacuum right up here. So what will happen, and actually, that's kind of an unrealistic-- well, just for the sake of argument, let's say we have water going right up against a vacuum. This isn't something you would normally just see in nature but let's just think about it a little bit. Normally, the water, since there's no pressure, it would evaporate and all the rest. But for the sake of argument, let's just say that this is a medium where light will travel slower. What you're going to have is is this ray is actually going to switch direction, it's actually going to bend. Instead of continuing to go in that same direction, it's going to bend a little bit. It's going to go down, in that direction is the refraction. That's the refraction angle. Refraction angle. Or angle of refraction. This is the incident angle, or angle of incidence, and this is the refraction angle. Once again, against that perpendicular. And before I give you the actual equation of how these two things relate and how they're related to the speed of light in these two media-- and just remember, once again, you're never going to have vacuum against water, the water would evaporate because there's no pressure on it and all of that type of thing. But just to--before I go into the math of actually how to figure out these angles relative to the velocities of light in the different media I want to give you an intuitive understanding of not why it bends, 'cause I'm not telling you actually how light works this is really more of an observed property and light, as we'll learn, as we do more and more videos about it, can get pretty confusing. Sometimes you want to treat it as a ray, sometimes you want to treat it as a wave, sometimes you want to treat it as a photon."
    },
    {
        "Q": "At 8:19, what does axial and equatorial mean?\n",
        "A": "It refers to which way the groups of a cycloalkane are pointing in the chair conformation. Axial = up or down Equatorial = sideways",
        "video_name": "FGq9-R6Yw18",
        "timestamps": [
            499
        ],
        "3min_transcript": "I've shown the sigma bonds here rotated a little bit differently. So all these sigma bonds in here have free rotation. So they're going to rotate to allow our necleophile to attack our electrophile a little bit easier. So I can think about lone pairs of electrons on our oxygen, and now it's a little bit easier to see the nucleophilic attack. It's going to attack right here at this carbon, and push these electrons in here off onto this oxygen. So let's go ahead and show the result of that. So now I have my oxygen, so it is bonded to this carbon now, and I've formed my ring. So let's follow those electrons, these electrons right here in magenta now form this bond right here. And I'm going to say that this oxygen right here in red is actually going to go up in the plane, so we'll come back to that in a minute. It had two lone pairs of electrons around it. It picked up another one, so that's where our negative one formal charge is. The oxygen in our ring still has the hydrogen bonded to it, and a lone pair of electrons that gets a plus one formal charge like that. And then I'm saying this hydrogen here is down, so this is our intermediate. And without showing all of the acid-base steps here, let's just think about what happens. We know that we next deprotonate, so a base comes along and takes this proton, and then these electrons in here move off onto our oxygen. And then we know after that we protonate our negative charge right here. So let's go ahead and draw one of the possible products. So we have our oxygen, that's part of our ring here, drawn in the chair conformation. We now have an OH right here, and then, so we have an OH equitorial as one of our possible products. And then two lone pairs of electrons on this oxygen. Hopefully you can see how this is one of the possible products and that here we drew it with like a flat plane, and here we've drawn it in more of a chair conformation. So the OH equatorial for this one. Let's go back to our original situation over here on the left, where the nucleophile attacked the carbonyl carbon. We know that the geometry at our carbonyl carbon is trigonal planar, so it's possible the nucelophile could attack from the opposite side and if that happened, then the oxygen would go down relative to the plane. So let's go ahead and show that now. We have another possibility here. So let's go ahead and draw it out. So we have our oxygen is now going to be part of our ring, so we've formed our ring. This time instead of putting"
    },
    {
        "Q": "\nat 3:08 Sal says that \"w\" (omega) is the angular velocity but souldn't it be the angular frequency",
        "A": "Omega is in rad/s. I do believe that angular frequency would be in rev/s. They are essentially the same thing just their units are different.",
        "video_name": "xoUppFlif04",
        "timestamps": [
            188
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAround 5:30 the Hank talks about the centrosomes moving away from the nucleas and leaving behind a trail of microtubles. But the animation shows the tubules growing from the centrosomes towards the chromosomes. I am confused on how the microtubles form. Anyone got a good tubular breakdown for meh?",
        "A": "To my understanding, the centrosomes are building the mitotic spindle/microtubules in the prophase, whilst they are moving.",
        "video_name": "X1bmedVziGw",
        "timestamps": [
            330
        ],
        "3min_transcript": "And part of what's really amazing about this whole process is that while we know what these stages are, we don't always know the underlying mechanisms that make all of them happen and this is part of science. Science isn't like all the stuff we know. It's how we're trying to figure all this stuff out. Consider job security if you ever want to be a biologist. There is a lot of stuff that future biologists have to still figure out and this is one of them. All right, let's get our clone on. So, most of their lives, cells hang out in this limbo period called interphase, which means they're in between episodes of mitosis, mostly growing and working and doing all the stuff that makes them useful to us. During interphase, the long strings of DNA are loosely coiled and messy, like that dust bunny of dog fur and laundry lint under your bed. That mess of DNA is called chromatin, but as the mitosis process begins to gear up lots of things start happening in the cell to get ready for the big division. One of the more important things that happens is that this little set of protein cylinders next to the nucleus, called the centrosome, duplicates itself. We're going to have to move a lot of stuff around by these centrosomes. The other thing that happens is that all of the DNA begins to replicate itself too, giving the cell two copies of every strand of DNA. To brush up on how DNA replicates itself like this check out this episode and then come on back. Now the cell enters the first phase, or the prophase, when that mess of chromatin condenses and coils up on itself to produce thick strands of DNA wrapped around proteins. Those, my friends, are your chromosomes. Instead of dust bunnies, the DNA is starting to look a little bit more like dread locks and the duplicates that have been made don't just float around freely. They stay attached to the original and together they look like little x's. These are called the chromatids and one copy is the left leg and arm of the x and the other copy is the right leg and arm. Where they meet in the middle is called the centromere. Just so you know, these x's are also called chromosomes. Sometimes double chromosomes or double stranded chromosomes and when the chromatids separate, they're considered individual chromosomes too. Now, while the chromosomes are forming by like, completely disintegrating and the centrosomes then peel away from the nucleus, start heading to the opposite ends of the cell. As they go, they leave behind a wide trail of protein ropes called microtubules running from one centrosome to the other. You might recall from our anatomy of the animal cell the microtubules help provide a kind of structure to the cell and this is exactly what they're doing here. Now we reach the metaphase, which literally means after phase and it's the longest phase of mitosis. It could take up to 20 minutes. During the metaphase, the chromosomes attach to those ropey microtubules right in the middle at their centromeres. The chromosomes then begin to be moved around and this seems to be being done by molecules called motor proteins. And while we don't know too much about how these motors work, we do know, for instance that there are two of them on each side of the centromere. These are called centromere associated protein E. So these motor proteins attached to the microtubule ropes basically serve to spool up the tubule slack. Now at the same time, another protein called dynein"
    },
    {
        "Q": "At 4:47 , is Hank saying that a double chromosome is still considered one chromosome? His use of the plural is a bit confusing.\n",
        "A": "A chromosome that has reproduced and is still attached at the centromere is still considered one chromosome.",
        "video_name": "X1bmedVziGw",
        "timestamps": [
            287
        ],
        "3min_transcript": "And part of what's really amazing about this whole process is that while we know what these stages are, we don't always know the underlying mechanisms that make all of them happen and this is part of science. Science isn't like all the stuff we know. It's how we're trying to figure all this stuff out. Consider job security if you ever want to be a biologist. There is a lot of stuff that future biologists have to still figure out and this is one of them. All right, let's get our clone on. So, most of their lives, cells hang out in this limbo period called interphase, which means they're in between episodes of mitosis, mostly growing and working and doing all the stuff that makes them useful to us. During interphase, the long strings of DNA are loosely coiled and messy, like that dust bunny of dog fur and laundry lint under your bed. That mess of DNA is called chromatin, but as the mitosis process begins to gear up lots of things start happening in the cell to get ready for the big division. One of the more important things that happens is that this little set of protein cylinders next to the nucleus, called the centrosome, duplicates itself. We're going to have to move a lot of stuff around by these centrosomes. The other thing that happens is that all of the DNA begins to replicate itself too, giving the cell two copies of every strand of DNA. To brush up on how DNA replicates itself like this check out this episode and then come on back. Now the cell enters the first phase, or the prophase, when that mess of chromatin condenses and coils up on itself to produce thick strands of DNA wrapped around proteins. Those, my friends, are your chromosomes. Instead of dust bunnies, the DNA is starting to look a little bit more like dread locks and the duplicates that have been made don't just float around freely. They stay attached to the original and together they look like little x's. These are called the chromatids and one copy is the left leg and arm of the x and the other copy is the right leg and arm. Where they meet in the middle is called the centromere. Just so you know, these x's are also called chromosomes. Sometimes double chromosomes or double stranded chromosomes and when the chromatids separate, they're considered individual chromosomes too. Now, while the chromosomes are forming by like, completely disintegrating and the centrosomes then peel away from the nucleus, start heading to the opposite ends of the cell. As they go, they leave behind a wide trail of protein ropes called microtubules running from one centrosome to the other. You might recall from our anatomy of the animal cell the microtubules help provide a kind of structure to the cell and this is exactly what they're doing here. Now we reach the metaphase, which literally means after phase and it's the longest phase of mitosis. It could take up to 20 minutes. During the metaphase, the chromosomes attach to those ropey microtubules right in the middle at their centromeres. The chromosomes then begin to be moved around and this seems to be being done by molecules called motor proteins. And while we don't know too much about how these motors work, we do know, for instance that there are two of them on each side of the centromere. These are called centromere associated protein E. So these motor proteins attached to the microtubule ropes basically serve to spool up the tubule slack. Now at the same time, another protein called dynein"
    },
    {
        "Q": "At 8:23, why use the volume of room rather than the volume of the container? Shouldn't the ideal gas law produced in the video be designated exclusively to the container of the water since the water is exclusively within the two liters, rather than the 42500 liters of room?\n",
        "A": "The 2.00 L is the volume of the liquid water in the open container. When the water evaporates, the water vapour expands to fill the whole room, so that\u00e2\u0080\u0099s the volume you use.",
        "video_name": "-QpkmwIoMaY",
        "timestamps": [
            503
        ],
        "3min_transcript": "Now, the hardest thing about this is just making sure you have your units right and you're using the right ideal gas constant for the right units, and we'll do that right here. So what I want to do, because the universal gas constant that I have is in terms of atmospheres, we need to figure out this vapor pressuree- this equilibrium pressure between vapor and liquid-- we need to write this down in terms of atmospheres. So let me write this down. So the vapor pressure is equal to 23.8 millimeters of mercury. And you can look it up at a table if you don't have this One atmosphere is equivalent to 760 millimeters of mercury. atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per"
    },
    {
        "Q": "\nAt 1:23, WHAT keeps water in its liquid state ?",
        "A": "The attractive forces among water molecules keep water in its liquid state.",
        "video_name": "-QpkmwIoMaY",
        "timestamps": [
            83
        ],
        "3min_transcript": "This exercises is from chapter 12 of the Kotz, Treichel in Townsend and Chemistry and Chemical Reactivity book, and I'm doing it with their permission. So they tell us you place 2 liters of water in an open container in your dormitory room. The room has a volume of 4.25 times 10 to the fourth liters. You seal the room and wait for the water to evaporate. Will all of the water evaporate at 25 degrees Celsius? And then they tell us at 25 degrees Celsius, the density of water is 0.997 grams per milliliter. And its vapor pressure is 23.8 millimeters of mercury. And this is actually the key clue to tell you how to solve And just as a bit of review, lets just think about what vapor pressure is. Let's say it's some temperature, and in this case we're dealing at 25 degrees Celsius. I have a bunch of water, and let me do that in a water color. I have a bunch of water molecules At 25 degrees Celsius, they're all bouncing around in every which way. And every now and then one of them is going to have enough kinetic energy to kind of escape the hydrogen bonds and all the things that keep liquid water in its liquid state, and it will escape. It'll go off in that direction, and then another one will. And this'll just keep happening. The water will naturally vaporize in a room. But at some point, enough of these molecules have vaporized over here that they're also bumping back into the water. And maybe some of them can be captured back into the liquid state. Now, the pressure at which this happens is the vapor pressure. As you can imagine, as more and more these water molecules vaporize and go into the gaseous state, more and more will also create pressure, downward pressure. surface of the water. And the pressure at which the liquid and the vapor states are in equilibrium is the vapor pressure. And they're telling us right now. It is 23.8 millimeters of mercury. Now, what we need to do to figure out this problem is say, OK, if we could figure out how many molecules need to evaporate, how many molecules of water need to evaporate to give us this vapor pressure, we can then use the density of water to figure out how many liters of water that is. So how do we figure out how many molecules-- let me write this down-- how many molecules of water need to evaporate to give us the vapor pressure of 23.8 millimeters of mercury?"
    },
    {
        "Q": "\nAt 10:50, the video notes that the two chair forms of cyclohexane are in equilibrium. So, does a single molecule spontaneously flip back and forth between the two forms? Or, must a single molecule, say, collide with another molecule for the flip to occur?",
        "A": "When cyclohexane is synthesized, it is in the two forms, but remember that equilibrium s can lean towards one side. The chair formation is a little bit more stable so the atoms rearrange (this is kind of your question, not much is known about how they do this, but it would help to understand hybridization) to be different than the boat conformation. Also some of the chair configuration can go back into the boat configuration.",
        "video_name": "YUEkOBvJSNg",
        "timestamps": [
            650
        ],
        "3min_transcript": "chair configurations are equally stable. And let me just touch on that a second. So you have, well, I don't have to-- actually, let me see. I won't copy and paste. I'll just redraw the other chair configuration for this guy. Actually let me just do it separately over here because I've made the colors here so confusing. Let me draw two, the same cyclohexane, but in two different chair configurations that it could be equilibrium in. So you could have this one, you could have this one, so this could be one chair configuration, and I'll draw it like this. And then the same hydrocarbon could be in-- or the same cyclohexane could be in equilibrium with the this other chair configuration that looks like this. Let me have a little more space here. So it looks like this. Let me do the pink. It goes up like that, like that. This pink guy goes like this. And then the blue guy is going to be just like this. So notice, in this situation this carbon appears kind of at the top of the chair, and this carbon is at the bottom, and then they've flipped. But these are equally stable configurations. But one way to think about is all of the axial guys on this carbon here turned into equatorial on this carbon and Let me show it to you. Let me just draw the hydrogens on this carbon. This carbon's hydrogens has an axial hydrogen, and has an equatorial hydrogen, whose bond would be parallel to that just like that. And this guy would have an equatorial hydrogen whose bond is parallel to actually both of these guys. And an axial hydrogen. hydrogens, but when this structure flips like that, what happens? Well, this hydrogen over here goes into this position, and this yellow hydrogen over here goes into this position. So over here, it was equatorial, and now it becomes axial. The same argument can be made over here. This equatorial hydrogen, when it flips-- when this whole blue part flips down-- now becomes axial. And this axial hydrogen, when you flip it down, becomes equatorial. And you can actually do that for all of the hydrogens. Over here you have an axial hydrogen. Once you flip it, you have an axial hydrogen, and then you have an equatorial hydrogen. When you flip it, these two equatorial hydrogens become axial."
    },
    {
        "Q": "Around 4:00, where is the boundary between non-polar and polar?\n",
        "A": "The boundary is approximately a difference of 1.7 between elements, but there are exceptions.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            240
        ],
        "3min_transcript": "Carbon is losing a little bit of negative charge. So carbon used to be neutral, but since it's losing a little bit of negative charge, this carbon will end up being partially positive, like that. So the carbon is partially positive. And the oxygen is partially negative. That's a polarized situation. You have a little bit of negative charge on one side, a little bit of positive charge on the other side. So let's say it's still a covalent bond, but it's a polarized covalent bond due to the differences in electronegativities between those two atoms. Let's do a few more examples here where we show the differences in electronegativity. So if I were thinking about a molecule that has two carbons in it, and I'm thinking about what happens to the electrons in red. Well, for this example, each carbon has the same value for electronegativity. So the carbon on the left has a value of 2.5. The carbon on the right has a value of 2.5. That's a difference in electronegativity of zero. aren't going to move towards one carbon or towards the other carbon. They're going to stay in the middle. They're going to be shared between those two atoms. So this is a covalent bond, and there's no polarity situation created here since there's no difference in electronegativity. So we call this a non-polar covalent bond. This is a non-polar covalent bond, like that. Let's do another example. Let's compare carbon to hydrogen. So if I had a molecule and I have a bond between carbon and hydrogen, and I want to know what happens to the electrons in red between the carbon and hydrogen. We've seen that. Carbon has an electronegativity value of 2.5. And we go up here to hydrogen, which has a value of 2.1. So that's a difference of 0.4. So there is the difference in electronegativity between those two atoms, but it's a very small difference. And so most textbooks would consider the bond between carbon and hydrogen Let's go ahead and put in the example we did above, where we compared the electronegativities of carbon and oxygen, like that. When we looked up the values, we saw that carbon had an electronegativity value of 2.5 and oxygen had a value of 3.5, for difference of 1. And that's enough to have a polar covalent bond. Right? This is a polar covalent bond between the carbon and the oxygen. So when we think about the electrons in red, electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge. And since electron density is moving away from the carbon, the carbon gets a partial positive charge. And so we can see that if your difference in electronegativity is 1, it's considered to be a polar covalent bond. And if your difference in electronegativity is 0.4, that's considered to be a non-polar covalent bond. So somewhere in between there must"
    },
    {
        "Q": "\n09:55, it is said that if the electronegativity is greater than 1.7, the compound is usually considered ionic. In the previous video (\"Electronegativity\"), it is said that the compound is considered ionic if the electronegativity is greater than 2.0.\nWhich one is more reliabe? How can I tell in an exam if a compound is ionic or covalent or nonpolar?",
        "A": "Different books use different cutoffs, though 1.7 is the most common, as Ryan mentioned. Ask you teacher which cutoff they want you to use for an exam.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            595
        ],
        "3min_transcript": "And we'll put in our electrons. And we know that this bond consists of two electrons, like that. Let's look at the differences in electronegativity between sodium and chlorine. All right. So I'm going to go back up here. I'm going to find sodium, which has a value of 0.9, and chlorine which has a value of 3. So 0.9 for sodium and 3 for chlorine. So sodium's value is 0.9. Chlorine's is 3. That's a large difference in electronegativity. That's a difference of 2.1. And so chlorine is much more electronegative than sodium. And it turns out, it's so much more electronegative that it's no longer going to share electrons with sodium. It's going to steal those electrons. So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons. So these two electrons in red-- let me go ahead and show them-- these two electrons in red here between the sodium and the chlorine, since chlorine is so much more in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine, like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron, so it ends up with a positive formal charge, like that. And so we know this is an ionic bond between these two ions. So this represents an ionic bond. So the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks we'll see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond. Lower than 1.7, in the polar covalent range. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved on to the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here, we're treating it like an ionic bond. Full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute."
    },
    {
        "Q": "\nAt about 4:30, in discussing microstates, Sal says that we can know the position and the momentum. I thought that, because of the uncertainty principle, we can never know both the position and the momentum of any given particle. Is he just simplifying things so that we understand the difference between microstates and macrostates, or am I missing something?\nThanks.",
        "A": "According to the Heisenberg Uncertainty Principle, you cant know both the location and momentum of a particle so you are correct. I believe it is a simplification.",
        "video_name": "5EU-y1VF7g4",
        "timestamps": [
            270
        ],
        "3min_transcript": "Now we know that that pressure is due to things like, you have a bunch of atoms bumping around. And let's say that this is a gas-- it's a balloon- it's going to be a gas. And we know that the pressure is actually caused-- and I've done several, I think I did the same video in both the chemistry and the physics playlist. I did them a year apart, so you can see if my thinking has evolved at all. But we know that the pressure's really due by the bumps of these particles as they bump into the walls and the side of the balloon. And we have so many particles at any given point of time, some of them are bumping into the wall the balloon, and that's what's essentially keeping the balloon pushed outward, giving it its pressure and its volume. We've talked about temperature, as essentially the average kinetic energy of these-- which is a function of these particles, which could be either the molecules of gas, or if it's an ideal gas, it could be just the atoms of the gas. Maybe it's atoms of helium or neon, or something like that. So for example, I could describe what's going on with I could say, hey, you know, there are-- I could just make up some numbers. The pressure is five newtons per meters squared, or some The units aren't what's important. In this video I really just want to make the differentiation between these two ways of describing what's going on. I could say the temperature is 300 kelvin. I could say that the volume is, I don't know, maybe it's one liter. And I've described a system, but I've described in on a macro level. Now I could get a lot more precise, especially now that we know that things like atoms and molecules exist. What I could do, is I could essentially label every one of contained in the balloon. And I could say, at exactly this moment in time, I could say at time equals 0, atom 1 has-- its momentum is equal to x, and its position, in three-dimensional coordinates, is x, y, and z. And then I could say, atom number 2-- its momentum-- I'm just using rho for momentum-- it's equal to y. And its position is a, b, c. And I could list every atom in this molecule. Obviously we're dealing with a huge number of atoms, on the order of 10 to the 20 something. So it's a massive list I would have to give you, but I could literally give you the state of every atom in this balloon."
    },
    {
        "Q": "At 3:47, don't chloroplasts have a double membrane instead of just 1?\n",
        "A": "Yes , the chloroplast technically has 2 membranes ; the outer and the inner one",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            227
        ],
        "3min_transcript": "a little bit rigid. So let's say that these are plant cells right here. Each of these squares, each of these quadrilaterals is a plant cell. And then in these plant cells you have these organelles called chloroplasts. Remember organelles are like organs of a cell. They are subunits, membrane-bound subunits of cells. And of course, these cells have nucleuses and DNA and all of the other things you normally associate with cells. But I'm not going to draw them here. I'm just going to draw the chloroplasts. And your average plant cell-- and there are other types of living organisms that perform photosynthesis, but we'll focus on plants. Because that's what we tend to associate it with. Each plant cell will contain 10 to 50 chloroplasts. I make them green on purpose because the chloroplasts contain chlorophyll. Which to our eyes, appear green. But remember, they're green because they reflect green wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma. chloroplast itself, you have these little stacks of these folded membranes, These little folded stacks. Let me see if I can do justice here. So maybe that's one, two, doing these stacks. Each of these membrane-bound-- you can almost view them as pancakes-- let me draw a couple more. Maybe we have some over here, just so you-- maybe you have some over here, maybe some over here. So each of these flattish looking pancakes right here, these are called thylakoids. So this right here is a thylakoid. That is a thylakoid. The thylakoid has a membrane. And this membrane is especially important. We're going to zoom in on that in a second. So it has a membrane, I'll color that in a little bit. The inside of the thylakoid, so the space, the fluid inside"
    },
    {
        "Q": "At 3:00, Sal said that the chloroplasts reflect green light, so does that mean that if \"green\" photons were to hit the chloroplast, photosynthesis wouldn't occur?\n",
        "A": "Well, green photons can t be absorbed because they are the wrong wavelength. The energy comes from photons that are absorbed. So the green photons can t provide energy for photosynthesis, and if you shone only green light on the chloroplast it would have no energy to photosynthesize. So, you are correct.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            180
        ],
        "3min_transcript": "It just doesn't need the photons from the sun. But let's focus first on the light-dependent reactions. The part that actually uses photons from the sun. Or actually, I guess, even photons from the heat lamp that you might have in your greenhouse. And uses those photons in conjunction with water to produce ATP and reduce NADP plus to NADPH. Remember, reduction is gaining electrons or hydrogen atoms. And it's the same thing, because when you gain a hydrogen atom, including its electron, since hydrogen is not too electronegative, you get to hog its electron. So this is both gaining a hydrogen and gaining electron. But let's study it a little bit more. So before we dig a little deeper, I think it's good to know a little bit about the anatomy of a plant. So let me draw some plant cells. a little bit rigid. So let's say that these are plant cells right here. Each of these squares, each of these quadrilaterals is a plant cell. And then in these plant cells you have these organelles called chloroplasts. Remember organelles are like organs of a cell. They are subunits, membrane-bound subunits of cells. And of course, these cells have nucleuses and DNA and all of the other things you normally associate with cells. But I'm not going to draw them here. I'm just going to draw the chloroplasts. And your average plant cell-- and there are other types of living organisms that perform photosynthesis, but we'll focus on plants. Because that's what we tend to associate it with. Each plant cell will contain 10 to 50 chloroplasts. I make them green on purpose because the chloroplasts contain chlorophyll. Which to our eyes, appear green. But remember, they're green because they reflect green wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma."
    },
    {
        "Q": "At 12:00, When H+ goes through the ATP synthase, it \"puts\" ADP and Pi to form ATP?\n",
        "A": "Well, assume the synthase is a wind turbine. Wind turbines generate electricity, but they cannot do so without any wind. Now, pretend that the flow of protons (H+) is that wind. Once the wind flows, the turbine is able to spin and generate ATP.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            720
        ],
        "3min_transcript": "the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top And it puts ADP and phosphate groups together. It puts ADP plus phosphate groups together to produce ATP. So that's the general, very high overview. And I'm going to go into more detail in a second. But this process that I just described is called photophosphorylation. Let me do it in a nice color. Why is it called that? Well, because we're using photons. That's the photo part. We're using light. We're using photons to excite electrons in chlorophyll. As those electrons get passed from one molecule, from one electron acceptor to another, they enter into lower and lower energy states. As they go into lower energy states, that's used to drive,"
    },
    {
        "Q": "At 5:30, he mentions that organelles were independent organisms that at some point evolved together (a process called \"endosymbiosis\"). This is true of mitochondria and chloroplasts, but I believe it is inaccurate to say of all organelles, correct?\n",
        "A": "this is possible because they both have DNA",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            330
        ],
        "3min_transcript": "wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma. chloroplast itself, you have these little stacks of these folded membranes, These little folded stacks. Let me see if I can do justice here. So maybe that's one, two, doing these stacks. Each of these membrane-bound-- you can almost view them as pancakes-- let me draw a couple more. Maybe we have some over here, just so you-- maybe you have some over here, maybe some over here. So each of these flattish looking pancakes right here, these are called thylakoids. So this right here is a thylakoid. That is a thylakoid. The thylakoid has a membrane. And this membrane is especially important. We're going to zoom in on that in a second. So it has a membrane, I'll color that in a little bit. The inside of the thylakoid, so the space, the fluid inside This light green color right there. That's called the thylakoid space or the thylakoid lumen. And just to get all of our terminology out of the way, a stack of several thylakoids just like that, that is called a grana. That's a stack of thylakoids. That is a grana. And this is an organelle. And evolutionary biologists, they believe that organelles were once independent organisms that then, essentially, teamed up with other organisms and started living inside of their cells. So there's actually, they have their own DNA. So mitochondria is another example of an organelle that people believe that one time mitochondria, or the ancestors of mitochondria, were independent organisms. That then teamed up with other cells and said, hey, if I produce your energy maybe you'll give me some food or whatnot. And so they started evolving together."
    },
    {
        "Q": "Did he mean hydrogen ions instead of hydrogen \"protons\"? (11:07)\n",
        "A": "A hydrogen ion (H+) is a proton. So, the terms are used interchangeably.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            667
        ],
        "3min_transcript": "But the general idea-- I'll tell you the general idea and then we'll go into the specifics-- of what happens during the light reaction, or the light dependent reaction, is you have some photons. Photons from the sun. They've traveled 93 million miles. so you have some photons that go here and they excite electrons in a chlorophyll molecule, in a chlorophyll A molecule. And actually in photosystem II-- well, I won't go into the details just yet-- but they excite a chlorophyll molecule so those electrons enter into a high energy state. Maybe I shouldn't draw it like that. They enter into a high energy state. And then as they go from molecule to molecule they keep going down in energy state. But as they go down in energy state, you have hydrogen the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top"
    },
    {
        "Q": "At 1:00 I don't why there is NH3. Is it solvent?\n",
        "A": "NH3 is the solvent. If you are talking about -NH2, that is from NaNH2 and is a very strong base.",
        "video_name": "HbDWBeRJboE",
        "timestamps": [
            60
        ],
        "3min_transcript": "Let's look at two ways to prepare alkynes from alkyl halides. So here I have an alkyl halide. So this is a dihalide, and my two halogens are attached to one carbon. We call this a geminal dihalide. So this is going to be a geminal dihalide reacting with a very strong base, sodium amide. So this is going to give us an E2 elimination reaction. So we're going to get an E2 elimination reaction, and this E2 elimination reaction is actually going to occur twice. And we're going to end up with an alkyne as our final product. So let's take a look at the mechanism of our double E2 elimination of a geminal dihalide. So let's start with our dihalide over here. And this time we're going to put in all of our lone pairs of electrons on our halogen, like that. So let me go ahead and put all of those in there, and then I have two hydrogens on this carbon. Sodium amide is a source of amide anions, which So a strong base means that a lone pair of electrons here on our nitrogen is going to take this proton. And these electrons, in here, are going to kick in to form a double bond at the same time these electrons kick off onto our halogen. So an E2 elimination mechanism. You can watch the previous videos on E2 elimination reactions for more details. So we're going to form ammonia as one of our products. And our other product is going to be carbon double-bonded to another carbon. And then we're going to still have our halogen down here. And over here, in the carbon on the right, we're still going to have a hydrogen, like that. So we're not quite to our alkyne yet. So we've done one E2 elimination reaction, and we're going to do one more. So we get another-- another amide anion comes along, and it's negatively charged. It's going to function as a base. It's going to take this proton this time. in here to form our triple bond. And these electrons are going to kick off onto our halogen, like that. So that is going to finally form our alkyne here. So you always have to have your base in excess, if you're trying to do this. Let's look at a very similar reaction, a double E2 elimination. This time the halogens are not on the same carbon. So let's go ahead and draw the general reaction for this. We have two carbons right here, and we have two halogens right here. And then hydrogen, and then hydrogen. This time we have two halogens on adjacent carbons. So this is called vicinal dihalides. So let's go ahead and write that. So this is vicinal, and the one we did before was geminal. So a vicinal dihalide will react in a very similar way if you add a strong base like sodium amide and you use ammonia for your solvent."
    },
    {
        "Q": "\nAt about 4:20, you mention that we cannot decipher the design of the rim of a wheel of a moving car because the parvo pathway has poor temporal resolution. I agree that this is true. However, then why can we not use the Magno Pathway to determine the design of the rim of the wheel since it has better temporal resolution?",
        "A": "Because the magno pathway is not good at detecting fine detail, like the the rims of the wheels. (I think this is the right answer).",
        "video_name": "0ugcw7wOZBg",
        "timestamps": [
            260
        ],
        "3min_transcript": "we also need to figure out, OK, what are the boundaries of the rose, so the boundaries of the stem, the boundaries of the leaf, the boundaries of the petals, from the background. And this is also really important, because not only do we need to distinguish the boundaries, but we also need to figure out, OK, what shape are the leaves, what shape are the petals. And these are all very important things that your brain ultimately is able to break down. So in order for us to figure out what the form of an object is, we use a very specialized pathway that exists in our brain, which is known as the parvo pathway. So the parvo pathway is responsible for figuring out what the shape of an object is. So another way to say this is that the parvo pathway is really good at spatial resolution. Let me write that down. So spatial resolution. is that it's really good at figuring out what the boundaries of an object are, what the little details that make up the object are. So if something isn't moving, such as when you're looking at a picture or when you're looking at a rose, you're able to break down and look at the little tiny details. So you're able to look at the little veins that make up the rose leaf. You're able to see all the little nuances of the rose. And that's because you're using the parvo pathway, which has a really high level of spatial resolution. One negative aspect of the parvo pathway is that it has really poor temporal resolution. And what I mean by this is that temporal resolution is motion. So if a rose is in motion, if I throw it across the room, I can't really use the parvo pathway to track the rose. The parvo pathway is used for stationary objects to acquire high levels of detail of the object. But as soon as that object starts moving, in the object. And you've probably noticed this. So when you're in a car, you're driving along, and there's a little Volkswagen Beetle driving right by you, it's going pretty slow. You can acquire a good number of details about the car, about the driver. But if you look down at the wheels, the wheels are spinning really, really quickly. And so if you try and look at the rims and figure out what design are the rims, it's really hard to figure that out. That's because the wheels are spinning so fast that it's really difficult to acquire any type of detail about the shape of the rims, about what they look like, and things like that. And finally, the parvo pathway also allows us to see things in color. So the parvo pathway not only allows us to acquire fine details about what we're looking at, but it also allows us to see in color. So if something is moving, we can't use the parvo pathway. But what we do use is the magno pathway. So we use the magno pathway in our brain. And the magno pathway is basically a set of specialized cells-- just"
    },
    {
        "Q": "At 8:10, you say that the force of gravity (Fg) that acts on A is equal to the force that the table acts on A (Ft; normal force) because of Newton's second law of motion. Is it also correct to say that these two forces are equal because of Newton's first law of motion?\n",
        "A": "If the object is not accelerating, we know that the net force must be zero. The first law is sort of just a special case of the second law.",
        "video_name": "VfpKzwrhmqQ",
        "timestamps": [
            490
        ],
        "3min_transcript": "So even if I came in all guns a blazing, Chuck Norris style, trying to dropkick some wall. That does not look like the correct form for a drop kick. But even if I came in, flying at this wall, as soon as I start to make contact with the wall, I'm gonna exert a force on the wall, and the wall has to exert a force back. So I'd exert a force on the wall to the right. And this would be the force on the wall, by my foot. There'd have to be an equal and opposite force instantly transmitted backwards, on my foot. So this would be the force on my foot, by the wall. This happens instantaneously, there is no delay. You can't kick this wall fast enough, for this other force to not be generated instantaneously. As soon as your foot starts to exert any force on the wall what so ever, the wall is gonna start exerting that same force back on your foot. So Newtons's Third Law is universal, but people still have trouble identifying these third law partner forces. So one of the best ways to do it, as soon as you list both objects, well to figure out where the partner force is, you can just reverse these labels. So I know over here, if one of my forces is the force on the wall by my foot, to find the partner force to this force, I can just reverse the labels and say it's gotta be the force on my foot, by the wall, which I drew over here. So this is a great way to identify the third law partner forces, cause it's not always obvious what force is the partner force. So to show you how this can be tricky, consider this example. Say we got the ground and a table. So this example drives people crazy for some reason. If I've got a box sitting on a table, we'll call it box A. Box A is gonna have forces exerted on it. One of those forces is gonna be the gravitational force. So the force of gravity is gonna pull straight down on box A, and if I were to ask you, what force is the third law partner force to this force of gravity, I'm willing to bet a lot of people might say, well there's an upwards force on box A, exerted by the table. And if this box A is just sitting here, not accelerating, these two forces are going to be equal and opposite. So it's even more tempting to say that these two forces are equal and opposite because of the third law, but that's not true. These two forces are equal and opposite because of the second law. The second law says if there's no acceleration, then the net force has to be zero, the forces have to cancel. And that's what's happening here. These forces are equal and opposite, they're canceling on box A. Which is a way to know that they are not third law partner forces, cause third law partner forces are always exerted on different objects. They can never cancel if they're third law partner forces. So what's going on over here? We've got two forces that are canceling, that are equal and opposite, but they're not third law partner forces, they're partner forces are somewhere else. I haven't drawn their partner forces yet. So let's try to figure out what they're partner forces are. So let's get rid of this, let's come back to here, let's slow it down to figure out what the partner force is, name the two objects interacting. So this force of gravity, I shouldn't be vague,"
    },
    {
        "Q": "\nAt 0:25, David says opposite and writes a negative. Is there such a thing as a negative force?",
        "A": "Yes, it s a force that is in the opposite direction of whatever direction you decided is positive.",
        "video_name": "VfpKzwrhmqQ",
        "timestamps": [
            25
        ],
        "3min_transcript": "- [Voiceover] We should talk a little more about Newtons's Third Law, because there are some deep misconceptions that many people have about this law. It seems simple, but it's not nearly as simple as you might think. So people often phrase it as, for every action there's an equal and opposite reaction. But that's just way too vague to be useful. So a version that's a little better, says that for every force, there's an equal and opposite force. So this is a little better. The equal sign means that these forces are equal in magnitude. And this negative sign means they're just different by the direction of the vector. So these are vectors, so this says that this pink vector F, has the opposite direction, but equal in magnitude to this green vector F. But to show you why this is still a little bit too vague, consider this, if this is all you knew about Newtons's Third Law, that for every force, there's an equal and opposite force, you might wonder, if you were clever, you might be like, wait a minute, if for every force F, right, there's got to be a force that's equal and opposite. that every force in the universe cancels? Shouldn't every force just cancel then, at that point? Doesn't that just mean that there's no acceleration that's even possible? Because if I go and exert a force F on something, if there's gonna be a force negative F, doesn't that mean that no matter what force I put forward, it's just gonna get cancelled? And the answer no, and the reason it's no is because these two forces are exerted on different objects. So you have to be careful. So the reason I say that this statement of Newtons's Third Law is still a little bit too vague, is because this is really on different objects. So if this is the force on object A, exerted by object B, then this force over here has to be the force on object B, exerted by object A. In other words, these forces down here are exerted on different objects. I'm gonna move this over to this side. I'm gonna move this over to here. Let's draw two different objects So if there was some object A, so I put some object A in here. Just wanna make sure there's an object A. Let's say this is object A, and it had this green force exerted on it, F. So this object right here is A. Well, there's gonna be another object, object B. We'll just make it another circle. So we'll make it look like this. So here's object B. And it's gonna have this pink force, F, negative F exerted on it. So I'm gonna call this object B. Now we're okay, now we know these forces can't cancel, and the reason these forces can't cancel, is cause they're on two different objects. But when you just say that Newtons's Third Law, is that every force has an equal and opposite force, it's not clear that it has to be on different objects. But it does have to be on different objects. So these Newtons force law pairs, often times is called force pairs, or Newton's third law partner forces, are always on different objects."
    },
    {
        "Q": "\nOn 3:54, why does Sal say that helium's configuration is 1s2, when helium is in the p block?",
        "A": "Helium is not part of the p block. It appears on the far right of the periodic table because it is a Nobel gas, so has similar reactivity to the other elements on the far right of the periodic table. However, helium has no electrons in p orbitals so does not count as part of the p block.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            234
        ],
        "3min_transcript": "well it's base configuration is the same as neon. It has a base configuration of neon. Neon is one S two, two S two, two P six, that's what this represents and then to get to sodium, you would then have three S one. How many Valence electrons does sodium have? Well its highest energy, furthest out electron or I say the electron that's in a non-stable shell. That's in a shell that hasn't been stabilized. It hasn't gotten to its fully stable state. There's only one electron in that situation right over here, the three S one electron. Sodium as well, you could depict like that. It only has one Valence electron that's the electron that could be swiped away from it or that somehow could be involved in a covalent bond somehow. Now let's do things with more, more Valence electrons but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus"
    },
    {
        "Q": "\nIn 5:19, Sal said Carbon has 2 + 2 valence electrons. Meaning the electrons from the s2 shell and the p2 shell. Is the s2 shell not filled? Why not use only the p2, which is the outermost shell? Help please.",
        "A": "2s and 2p are both subshells of shell number 2. So you have to consider both the s and p.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            319
        ],
        "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange."
    },
    {
        "Q": "Apologies if this has already been answered, but at 9:20 Sal tells us that from the n=4 S subshell, the element \"backfills\" electrons into the n=3 D subshell. Why does it do this, rather than going directly to 4d^6? Is this a special case, or does it happen for all elements in the D subshell and/or beyond? Thanks!\n",
        "A": "This happens for all elements in the d block and beyond. In the fourth level and beyond, the energy levels get all mixed up. For example, in cerium (Ce, element 58), the outermost electron configuration is 6s\u00c2\u00b2 4f\u00c2\u00b9 5d\u00c2\u00b9 6s2 4f1 5d1",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            560
        ],
        "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy."
    },
    {
        "Q": "\n\"They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances.\" Could someone explain to me how can copper and chromium only go 4s^1 not 4s^2 (they have 2 groups in s subshells in their period)? 10:15",
        "A": "You would expect Cu to be 4s\u00c2\u00b23d\u00e2\u0081\u00b4. But a half-filled d subshell is more stable than one with only 4 d electrons. It takes a little energy to promote a 4s electron to the 3d level, but you get back more than this by getting a 4s3d\u00e2\u0081\u00b5 configuration. In the same way, a filled d subshell is more stable than one with only 9 d electrons. It takes a little energy to promote a 4s electron to the 3d level, but you get back more than this by getting a 4s3d\u00c2\u00b9\u00e2\u0081\u00b0 configuration.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            615
        ],
        "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow."
    },
    {
        "Q": "at 10:16, why does the d group belong to the number 3 not four. Since it is in the fourth group why is it written at 3d6\n",
        "A": "Well, when you get to the D sub-level you have to take into account that the 4s orbital is filled BEFORE the 3d! The 5s orbital fills before the 4d, so, it even though the 5s seems like it would be farther out than the 4d, the 4d has more energy, so to speak.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            616
        ],
        "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow."
    },
    {
        "Q": "At 6:50, Sal talks of Methane having a logical structure of CH^4, with 8 valence electrons. How then, is methane so reactive and explosive? Hydrogen is flammable, yet saying that Methane is flammable only because it has Hydrogen atoms would be like saying water is flammable because it is H^2O.\n",
        "A": "A full octet of electrons doesn t mean it will completely unreactive to anything. It s not simply hydrogen atoms that are flammable, methane can be turned into more stable products (CO2 and H2O) when it is burned.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            410
        ],
        "3min_transcript": "How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways"
    },
    {
        "Q": "\nAt 5:14, he says \"that has not been completed yet\" what does he mean by that? Doesn't the 2s^2 mean that it's 2nd s orbital has been full so it has to move onto the p orbital? A little confused on Valence electrons still.",
        "A": "The 2s orbital is full, but those electrons are still part of the outermost shell. The 2s and the three 2p orbitals make up the second shell.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            314
        ],
        "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange."
    },
    {
        "Q": "At 7:15, Sal mentions covalent bonds between hydrogen and carbon. What are covalent bonds?\n",
        "A": "As opposed to ionic bonds, which donate and receive electrons to get charged, get full shells and stick together, covalent bonds share electrons to stick together and get full shells. Covalent bonds happen between elements with a smaller electronegativity difference than the amount required for an ionic bond, so generally nonmetals.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            435
        ],
        "3min_transcript": "How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways"
    },
    {
        "Q": "at 5:38, i don't really understand why 2s\u00c2\u00b22p\u00c2\u00b3......also, for school purpose, should i memories the whole table like Sal did and how to make it easier...someone help me\n",
        "A": "just see and by heart the name of elements minimum till 30 and some important like gold or zinc fe etc.. and to explain the first question its not possible in this box i would suggest you to go through all the previous topics of periodic table or read a source like your text and then go through this. it will explain everything as Mr khan always does:)",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            338
        ],
        "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair"
    },
    {
        "Q": "\nAt 9:32 Sal asks : \"what are the highest energy electrons ?\" and then at 9:36 he asks \"what are the furthest out ?\".\n\nHow do we determine which one are the furthest out and the one which have highest energy ? Is it a specificity of blocks ?",
        "A": "You can t determine it unless you have prior knowledge of it. But Sal is not correct here, because the 4s electrons are both the furtherest out and the highest energy for the transition metals.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            572,
            576
        ],
        "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow."
    },
    {
        "Q": "In 5:43, Why is it written as [He]2s2, 2p2 instead of 1s2, 2s2, 2p2? And why does Carbon only have four valence?\n",
        "A": "Because Helium s electronic configuration is 1s2, which is a complete electron shell, so 1s2 is replaced with [He] instead. For example, potassium(K) s electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1. Since the last complete electron shell is 1s2, 2s2, 2p6, 3s2, 3p6, which is Argon, it could be substituted into K s electronic configuration. Thus, it could be written as [Ar] 4s1. Carbon has four valency electrons and needs to gain or lose FOUR more electrons to obtain a complete electron shell.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            343
        ],
        "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair"
    },
    {
        "Q": "Why does Sal, at 7:39 state that any element in Carbon's group will have four valence electrons?\n",
        "A": "Because of the group it is in. All group 1 elements have one valence electron, group 2 elements have two, skip groups 3-12 for now (they are the exceptions to the rule), group 13 has three, group 14 has four (carbon being part of this group), and so on.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            459
        ],
        "3min_transcript": "Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base"
    },
    {
        "Q": "\n8:22 so? H2Te or Li20 could both be similar to H20? or even Li2Te? or did I just give a completely different molecule",
        "A": "Yeah H2Te and Li2O are similar to H2O and also to Li2Te. Slightly tricky since Hydrogen is NOT an alkali metal(group I metal) It s position is analogous.That s why Hydrogen s box is green. So, Hydrogen won t have the properties of Lithium. But Oxygen and Tellurium on the other hand are both Chalcogens and exhibit similar properties.So, H2Te is similar to H2O Li2O is similar to Li2Te",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            502
        ],
        "3min_transcript": "Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base"
    },
    {
        "Q": "At 1:56 How did we know that Carbon has sp3 hybridized orbitals and that it forms a tetrahedral shape?\n",
        "A": "Because each carbon has four other atoms directly attached to it. You have already learned that this requires sp3 hybridization and the sp3 bond angles are 109.5\u00c2\u00b0.",
        "video_name": "IkmM4CPnqF0",
        "timestamps": [
            116
        ],
        "3min_transcript": "In the video on sp3 hybridized orbitals, we went in pretty good detail about how a methane molecule looks. But just as a bit of a review, it's the tetrahedral shape. You have a carbon in the middle, and then you would have a hydrogen-- you can imagine I'm drawing it like this because this hydrogen is poking out of the page. Then maybe you have another hydrogen that's in the page. You have one above the carbon, and you have one that's behind the page. You could imagine it's like a tripod with a pole sticking out of the top of the tripod. Or if you were to imagine the shape another way, if you were to connect the hydrogens, you would have a four-sided pyramid with a triangle as each of the sides. So it would to look something like this. I'm trying my best to draw-- the pyramid would look something like this if you could see through it. So this would be one side, another side would be over here, and then the backside would be over here, and then transparent out front. So the fourth side would be the actual kind of thing that we're looking through when we look at this pyramid. It would be this front side right over here. So you can imagine it different ways, this was the case with methane. Now let's extend this into a slightly more complex molecule, and that's ethane. So the way we've been drawing it so far-- I guess the simplest way to draw ethane, is just like that. By implication this is ethane. By implication you have a carbon there, and a carbon there, and they'll each have three hydrogens bonded to it. And we've drawn it something like this. Three hydrogens bonded to each of these guys. But now we know that carbon has these sp3 hybridized orbitals, that it likes to form more of a tetrahedral shape when it bonds. So an ethane molecule would actually look more like this. So I'll do the carbons in orange. So if that's the carbon and that's the carbon. So you can imagine you have a carbon molecule here. I'll draw it as this little circle. And then if we have some perspective, so the carbon-carbon bond is going to look like that. And then you have another carbon molecule right over there. So that's that bond over here. And we want both the carbons, all of their bonds to be kind of in a tetrahedral shape. So then you could imagine this bond over here going like this. This bond going like that. And you have your hydrogen at the end. Let's make the green circles the hydrogens. So you have that hydrogen, and then-- or actually just the circles, we'll call them hydrogen-- and then you can imagine this one, maybe it's coming out of the page a little bit. That is that hydrogen. Let me label the hydrogens, actually. I'm doing it in all different colors so you can see what I'm And then this hydrogen is going right below it, maybe pointed back a little bit."
    },
    {
        "Q": "\nAround 2:50 Sal started talking about prokaryotes. I know about how organisms and species have evolved over time but how is this possible for single celled organisms to evolve exactly? Did these organisms gather in massses and shift into the next cellular organism? Or did they latch on to soil thus creating plants?",
        "A": "This isn t known at all, but we do know that they evolved somehow. It s kind of like the apes evolving into humans, it s just natural I guess.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            170
        ],
        "3min_transcript": "Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound that perform specific functions, like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus, where all of their DNA is. So this is the nuclear membrane. And then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet."
    },
    {
        "Q": "\nAt 9:08 Sal said that all life was happening in the ocean and the ocean was still iron rich. Doesn't too much iron kill you?",
        "A": "It wouldn t be enough to kill them, in fact it might help (hemoglobin?).",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            548
        ],
        "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria"
    },
    {
        "Q": "9:00 on the video there is suddenly an ocean. What happened? Where did it come from?\n",
        "A": "actually, when a meteriote formed, it had some ice on it. that ice melted and started the water cycle. this happened over thousands of years",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            540
        ],
        "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria"
    },
    {
        "Q": "\ni have another question.My teacher mentioned that nuclues in unicellular organisms help in reproduction.ie-the organism breaks itself into 2 and so,there is a new organism.The video mentioned that there were so many in number(i am talking about the procaryotes).So did many procaryotes come into being in a flash because it cant reproduce?The time could be approximately 3:00",
        "A": "Nothing comes into being in a flash.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            180
        ],
        "3min_transcript": "Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound that perform specific functions, like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus, where all of their DNA is. So this is the nuclear membrane. And then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet."
    },
    {
        "Q": "At 8:10, he mentions stromatalites, something about sediment and bacteria. What are they exactly?\n",
        "A": "Stromatolites are layered mounds, columns, and sedimentary rocks. They were formed upon layer by layer of cyanobacteria, a single celled organism that lives in many different habitats. I hope that helped you.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            490
        ],
        "3min_transcript": "oh, 4.6 billion years ago to 3.8 billion. Oh, that's just 800 million years. Remember-- and I'll talk about this. Grass has only existed for 50 million years. This is 800 million years. Humans and chimpanzees only diverged 5 million years ago. This is 800 million years we're talking about, from ancient Greece to now, we're only talking about 2,500 years. You multiply that times 1,000, you get 2 and 1/2 a million years. And this is 800 million years we're talking about. So these are extremely huge periods of time. And that's why we call them eons. Eons are 500 million to a billion years. Now, the dividing line between the Hadean Eon and the Archean Eon-- and it's kind of a fuzzy dividing line, but most people place it about 3.8 billion years ago. And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear."
    },
    {
        "Q": "at the end of the video< im kind of confused on how sal got the ratio 1:2. can you explain?\n",
        "A": "We discovered, in this video, that the bottle of molecular substance we were dealing with had 1.02 moles of Sulfur, 2.04 moles of Hydrogen, and 4.08 moles of Oxygen. The goal of the Empirical Formula is to represent the elements in simplest form. To simplify, we can divide each element by the least common factor (in this case, 1.02 happens to be the lcf) and we come up with 1, 2 and 4. We cannot reduce these any more.",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            62
        ],
        "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16."
    },
    {
        "Q": "My question is: In the case of Fe2O3 that are in the ratio of 70% iron and 30% oxygen.\nThe atomic weight of iron is 56 and I get 70/56 = 1.25 moles.\nThe atomic weight of oxygen is 16 and I get 30/16 = 1.87 moles.\nHow do I know that their ratio is 2:3?\nI know that 1.25 / 1.87 = 0.666666667, so I have to find which numbers divided among themselves make 0.666666667 and discover that is 2/3.\nThere is no way to know immediately without having to do a lot of divisions? Thank you.\n",
        "A": "Another way to write 0.6 repeating is 6/9 or 2/3. If you think of the fraction as the ratio of the number of Iron Atoms (numerator) to the number of Oxygen Atoms (denominator) then you get Fe2O3 pretty quickly. Conversely, if you were to multiply the numerator and denominator by 1000 you get 1250/1875 (you rounded 30/16 to 1.87, but it is 1.875). when you reduce that you get 2/3. As the numerator represented Fe, you have 2 atoms of that and the denominator represented O, you had 3 atoms of that",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            123
        ],
        "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16."
    },
    {
        "Q": "\nAt 0:07 why is hydrogen at 2.04 %",
        "A": "That is just the way the problem is set up.",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            7
        ],
        "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16."
    },
    {
        "Q": "\nAt 3:30, doesn't P1=V2 and P2=V1",
        "A": "No. Because its not P1/Vi=0 and P2/V2=0. If that was the case then you would have been correct. But as the temperature, no. of moles and constant does not change, both P1*V1 and P2*V2 are equal to nRT. Hope this helps :)",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            210
        ],
        "3min_transcript": "and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres. is equal to our new pressure times 3 liters. And if we divide both sides of the equation by 3, we get 3 liters cancel out, we're left with 9 atmospheres. And that should make sense. When you decrease the volume by 2/3 or when you make the volume 1/3 of your original volume, then your pressure increases by a factor of three. So this went by times 3, and this went by times 1/3. That's a useful thing to know in general. If temperature is held constant, then pressure times volume are going to be a constant. Now, you can take that even further."
    },
    {
        "Q": "\nAt 2:30, how can V1 and V2 be equal to each other when their volume is clearly decreasing?",
        "A": "The volume in each IS different V1=9L and V2=3Liters. When setting up the equation both are equal to nRT since he states these are constants. The next step he makes them equal to each other divides which eliminates 3 and the L leaving 9 atm.",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's do some more problems that involve the ideal gas equation. Let's say I have a gas in a container and the current pressure is 3 atmospheres. And let's say that the volume of the container is 9 liters. Now, what will the pressure become if my volume goes from 9 liters to 3 liters? So from the first ideal gas equation video, you can kind of have the intuition, that you have a bunch of -- and we're holding-- and this is important. We're holding the temperature constant, and that's an important thing to realize. So in our very original intuition behind the ideal gas equation we said, look, if we have a certain number of particles and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres."
    },
    {
        "Q": "At 3:30 in the video the answer is revealed to be 9 atm. When you divide the 3 over, why isn't the 9 divided as well, because should ((3atm x 9L)/(3L))= 3atm?\n",
        "A": "Sorry, but no. If you divide a sum by a number, then you divide each member of the sum. But if the numbers are all multiplied, you only divide one time.",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            210
        ],
        "3min_transcript": "and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres. is equal to our new pressure times 3 liters. And if we divide both sides of the equation by 3, we get 3 liters cancel out, we're left with 9 atmospheres. And that should make sense. When you decrease the volume by 2/3 or when you make the volume 1/3 of your original volume, then your pressure increases by a factor of three. So this went by times 3, and this went by times 1/3. That's a useful thing to know in general. If temperature is held constant, then pressure times volume are going to be a constant. Now, you can take that even further."
    },
    {
        "Q": "\nSo, before Sal changed the pressure in the initial problem at 6:10, the proportion was set up as (1atm x 2m^3)/300K = (2atm x 1m^3)/T. If the proportion was solved as such, wouldn't a temperature of 300K have been achieved even though the pressure of the second container was increased and the volume of the second container was decreased?",
        "A": "I think I understand the algebra behind the problem, I was just a little confused conceptually. If the pressure of a container is increased and the volume is decreased, wouldn t the result be a smaller, more pressurized space? And would this scenario not lead to a higher temperature?",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            370
        ],
        "3min_transcript": "the two things that we know don't change in the vast majority of exercises we do is the number of molecules we're dealing with, and obviously, R isn't going to change. So if we divide both sides of this by T, we get PV over T is equal to nR, or you could say it's equal to a constant. This is going to be a constant number for any system where we're not changing the number of molecules in the container. So if initially we start with pressure one, volume one, and some temperature one that's going to be equal to this constant. And if we change any of them, if we go back to pressure two, volume two, temperature two, they should still be equal to this constant, so they equal each other. So for example, let's say I start off with a pressure of 1 atmosphere. and I have a volume of-- I'll switch units here And let's say our temperature is 27 degrees Celsius. Well, and I just wrote Celsius because I want you to always remember you have to convert to Kelvin, so 27 degrees plus 273 will get us exactly to 300 Kelvin. Let's figure out what the new temperature is going to be. Let's say our new pressure is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been decreased by half and the pressure is doubled by half. Actually, no. Let me make the pressure even larger. Let me make the pressure into 5 atmospheres. Now we want to know what the second temperature is, and we set up our equation. And so we have 2/300 atmosphere meters cubed per Kelvin is equal to 5/T2, our new temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750 degrees Kelvin, which makes sense, right? We increased the pressure so much and we decreased the volume at the same time that the temperature just had to go up. Or if you thought of it the other way, maybe we increased the temperature and that's what drove the pressure to be so much higher, especially since we decreased the volume."
    },
    {
        "Q": "At 6:45 sal says that 2 over 300 = 5 over t2. I get that, but then he says that 1500=2t2. how did he get 1500 and how did he get 2t2?wouldn't it be t2?\n",
        "A": "If you have 2/300=5/t2, you cross multiply on both sides. So basically, first you multiply both sides by t2, which results in 2t2/300=5. Then you multiply both sides by 300, resulting in 2t2=5x300 which is 2t2=1500.",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            405
        ],
        "3min_transcript": "And let's say our temperature is 27 degrees Celsius. Well, and I just wrote Celsius because I want you to always remember you have to convert to Kelvin, so 27 degrees plus 273 will get us exactly to 300 Kelvin. Let's figure out what the new temperature is going to be. Let's say our new pressure is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been decreased by half and the pressure is doubled by half. Actually, no. Let me make the pressure even larger. Let me make the pressure into 5 atmospheres. Now we want to know what the second temperature is, and we set up our equation. And so we have 2/300 atmosphere meters cubed per Kelvin is equal to 5/T2, our new temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750 degrees Kelvin, which makes sense, right? We increased the pressure so much and we decreased the volume at the same time that the temperature just had to go up. Or if you thought of it the other way, maybe we increased the temperature and that's what drove the pressure to be so much higher, especially since we decreased the volume. is this pressure went up so much, it went up by factor of five, it went from 1 atmosphere to 5 atmospheres, because on one level we shrunk the volume by a factor of 1/2, so that should have doubled the pressure, so that should have gotten us to two atmospheres. And then we made the temperature a lot higher, so we were also bouncing into the container. We made the temperature 750 degrees Kelvin, so more than double the temperature, and then that's what got us to 5 atmospheres. Now, one other thing that you'll probably hear about is the notion of what happens at standard temperature and pressure. Let me delete all of the stuff over here. Standard temperature and pressure. Let me delete all this stuff that I don't need. Standard temperature and pressure. And I'm bringing it up because even though it's called standard temperature and pressure, and sometimes called STP,"
    },
    {
        "Q": "at 4:41 What is quantum mechanics ?\n",
        "A": "Quantum mechanics is basically what chemists and physicists use to describes how subatomic particles behave. You ll learn more about it as you learn more about chemistry :)",
        "video_name": "Rd4a1X3B61w",
        "timestamps": [
            281
        ],
        "3min_transcript": "that have all of these different properties. So when you think about chemistry, yes, it might visually look something like this. These are obviously much older pictures. But at its essence, it's how do we create models and understand the models that describe a lot of the complexity in the universe around us? And just to put chemistry in, I guess you could say, in context with some of the other sciences, many people would say at the purest level, you would have mathematics. That math, you're studying ideas, which could even be independent, you're seeing logical ideas that could be even independent of anything that you've ever observed or experienced. And a lot of folks that say if we ever communicate with another intelligent species that could be completely different than us, math might be that common language. Because even if we perceive the world differently math might be that common language. But on top of math, we start to say, well how is our reality actually structured? At the most basic level, what are the constituents of matter and what are the mathematical properties that describe how they react together? And then, or interact with each other? Then you go one level above that, you get to the topic of this video, which is chemistry. Which is very closely related to physics. When we talk about these chemical equations and we create these molecular structures, the interactions between these atoms, these are quantum mechanical interactions which we do not fully understand at the deepest level yet. But with chemistry, we can start to make use of the math and they physics to start to think about how all of these different building blocks can interact to explain all sorts of different phenomena. This chemical equation you see right here, This is hydrogen combusting with oxygen to produce a lot of energy. To produce energy. When we imagine combustion, we think of fire. But what even is fire at its most fundamental level? How do we get, why do we perceive this thing here? And chemistry is super important because on top of that, we build biology. We build biology. And as you'll see as you study all of these things, there's points where these things start to bleed together. But the biology in, say, a human being, or really in any species, it's based on molecular interactions. Interactions between molecules, between atoms, which, at the end of the day, is all about chemistry. As I speak, the only reason why I'm able to speak is because of really, hard to imagine the number of chemical interactions happening in me right now to create this soundness. To create this thing that thinks it exists that wants to make a video about how awesome"
    },
    {
        "Q": "at 2:30 the carbon is more electronegative than hydrogen has been said and it atract the shared pair of electrons\nbut why cant oxygen atom do this inspite o is more electronegative than c and h.\n",
        "A": "Oxygen is more electronegative than carbon and hydrogen. It does attract the bonding electrons significantly more than those two.",
        "video_name": "rhuYuerbhIE",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] In this video, we're going to find the oxidation state of carbon in several different molecules. In earlier video, we've already seen the definition for oxidation state, and also how to calculate it. So let's start with methane, and let's find the oxidation state of carbon and methane. One approach is more of a general chemistry approach where we know that hydrogen usually has an oxidation state of plus one, and we have four hydrogens for a total of plus four. The sum has to be equal to zero, so we know that carbon's oxidation state must be minus four immediately, since we only have one carbon here. So let's go ahead and verify that with our dot structure. So remember, when we're calculating the oxidation state using dot structures, we're thinking about bonding electrons, and we know that each bond consists of two electrons, so we need to put in the bonding electrons for all of our bonds. Next, we think about the oxidation state for carbon, and we start with a number or the number of valence electrons that carbon is supposed to have, and we know carbon is supposed to have four valence electrons, so from that number, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons carbon has in our drawing. But now we need to think about these covalent bonds as being ionic, and so the more electronegative atom is going to take all of the electrons in the bond, so we need to think about electronegativity differences, and we're comparing carbon to hydrogen. So which is more electronegative? We know that carbon is more electronegative than hydrogen, so the two electrons in this bond here, carbon is going to take both of them, so it's winner takes all. Carbon's going to hog those electrons in this bond. All right, same for this next carbon-hydrogen bond. Carbon is more electronegative, so it takes those electrons, and all the way around. by eight electrons right, let's count them up here, one, two, three, four, five, six, seven, eight, so four minus eight is equal to minus four, so we already knew that minus four was going to be the oxidation state for carbon. Let's move on to another molecule here, so C2H4, this is ethene, or ethylene. What's the oxidation state of carbon in this molecule? Well hydrogen should be plus one, and we have four of them for a total of plus four. So the total for carbon should be minus four, because that total has to sum to zero, but this time we have two carbons, so minus four divided by two gives us minus two, each carbon should have an oxidation state of minus two. And let's verify that, let's put in our bonding electrons,"
    },
    {
        "Q": "\nAt 3:58 Sal says a \"slightly negative pH\" i think what he means is a pH slightly below 7, but can you actually have a negative pH?",
        "A": "Yes, you can. You can easily have a 10 mol/L solution of a strong acid, so [H+] = 10 mol/L. pH = -log[H+] = -log10 = -1. When acids are very acidic, though, we must use measures of acidity other than pH. You have not yet learned about these, though.",
        "video_name": "BBIGR0RAMtY",
        "timestamps": [
            238
        ],
        "3min_transcript": "But since it was a strong acid, those conjugate bases don't do anything. They don't add anything to the pH. They're not really basic. The chlorine in hydrogen chloride, the chlorine ion, doesn't change the pH. So this is a strong acid. And this one, when we got to the equivalence point-- when we had used up all of the acid in a solution, and then we hit this in inflection point, where any OH we added was significantly increasing the pH-- when we hit that equivalence point, our pH was already basic. And that's because we had all of the conjugate base of the weak acid, which does make the solution more basic. So this is a weak acid. And in both of these situations, we were increasing the concentration of OH minus. Maybe by adding sodium hydroxide to the solution, a strong base. Now, In these situations, we start with a base, and we add a strong acid to it. Maybe whatever base. We're adding hydrogen chloride, something that will Here, we want to sop up the OH and bring its concentration down, until some point that we have sopped up all of the OH. All of the base is gone. Or most of it is gone. In this situation, we're in a completely neutral situation. So when we sopped up all of the base, we're completely neutral. No basic conjugate bases left. So this is a strong base. And here, the titration, we're increasing the hydrogen solution, or the hydrogen concentration, to sop up all the base. Same thing here. We're sopping up all of the base. We start over here. But over here, the inflection point happens right over here. So we've sopped up all of its base, but some of its conjugate acid is still left over, even after we've sopped up all of its base. So we end up with a slightly negative pH at the equivalence point. So this is a weak base. Let me actually draw that reaction for you. Maybe its A minus is in equilibrium-- that second equilibrium arrow is a little too wild for my blood-- is equilibrium with AH. It grabs hydrogen ions from the surrounding water. Everything is in an aqueous solution. So after you add hydrochloric acid to this-- remember, HcL disassociates completely into hydrogen ions plus chlorine anions. If you add hydrochloric acid to this, these things are going to just completely sop up these things. So we keep sopping up those things. Our concentration of OH goes down and down and down. And as we sop up this, our reaction goes in that direction because Le Chatelier's Principle. More and more of this is going to get formed into this and that. Until some point, we're out of that, and we have a ton of this left."
    },
    {
        "Q": "at 2:16 what does he mean by PH\n",
        "A": "I am struggling with my online Chemistry class and was wondering if someone could help me understand the group numbers. For example what are the group numbers for X and Y?",
        "video_name": "BBIGR0RAMtY",
        "timestamps": [
            136
        ],
        "3min_transcript": "I've drawn a bunch of titration curves here. So let's see if we can review everything we've learned to kind of have a more holistic understanding of interpreting these things. So the first thing to look at is which of these are the titration of acids versus bases? And everything I've done now is acids, but the logic for base titration is the exact same thing as acid. So for example, these are acid titrations. We start with low pH's. In all of these, this axis is pH. I should have drawn that ahead of time before I asked you the question, but I think you knew that already. So before we add any of the titrator or the reagent, in this reaction, we're starting with a low pH. So this is kind of our starting point. So we have a low pH there. We have a low pH there. So these are both clearly acids. Here, our starting point before we start titrating at all, it's a high pH. So both of these are bases. Let me write that down. These are clearly both bases. Now, we haven't covered bases. But it's the same exact idea. In an acid titration, you start with an acid and you add a strong base to it to sop up all of the acid until all of the acid is sopped up and you hit the equivalence point. You hit the point that all of the acid is sopped up. And now, as you add more and more strong base, you're making it superbasic. So in this acid, our equivalence point is over here. And in this acid, our equivalence point is over here. This is how much solution we had to add to sop up all of the acid. Right there. So given what we already know, which one's a strong acid, which one's a weak acid? Well, this one, when sopped up all of the acid, we have a completely neutral solution. So this must have been a strong acid. Everything has been converted to water in its natural state. pH of 7. And we might have had some neutral leftover conjugate But since it was a strong acid, those conjugate bases don't do anything. They don't add anything to the pH. They're not really basic. The chlorine in hydrogen chloride, the chlorine ion, doesn't change the pH. So this is a strong acid. And this one, when we got to the equivalence point-- when we had used up all of the acid in a solution, and then we hit this in inflection point, where any OH we added was significantly increasing the pH-- when we hit that equivalence point, our pH was already basic. And that's because we had all of the conjugate base of the weak acid, which does make the solution more basic. So this is a weak acid. And in both of these situations, we were increasing the concentration of OH minus. Maybe by adding sodium hydroxide to the solution, a strong base. Now, In these situations, we start with a base, and we add a strong acid to it. Maybe whatever base. We're adding hydrogen chloride, something that will"
    },
    {
        "Q": "1:21 Sal says that Sulphur's mass number is 32, but on the periodic table it says that the mass number is 32.07. Why?\n",
        "A": "We calculate the mass number of one specific sulphur atom, which has 16 neutrons. So, it s mass number is exactly 32. In this big wide world, there exist other isotopes of sulphur too, having different mass numbers. The average mass of all sulphur atoms is 32.07, as calculated by scientists.",
        "video_name": "koAFBScR41A",
        "timestamps": [
            81
        ],
        "3min_transcript": "- [Narrator] An isotope contains 16 protons, 18 electrons, and 16 neutrons. What is the identity of the isotope? And I encourage you to pause the video and see if you can figure it out and I'll give you a hint, you might want to use this periodic table here. All right, so I'm assuming you've had a go at it. So, an element is defined by the number of protons it has. So if someone tells you the number of protons, you should be able to look at a periodic table and figure out what element they are talking about. So, because it is 16 protons, well we can go right over here to the atomic number, what has 16 protons, well anything that has 16 protons by definition is going to be sulfur right over here. So I could write a big S. Now, the next thing we might want to think about is the mass number of this particular isotope. Remember, an isotope, all sulfur atoms are going numbers of neutrons. So, the sulfurs that have different number of neutrons, those would be different isotopes. So, this case we have 16 protons and we have 16 neutrons, so if you add the protons plus the neutrons together, you're going to get your mass number. So 16 plus 16 is 32. Now let's figure out if there's going to be any charge here. Well, the protons have a positive charge. The electrons have a negative charge. If you have an equal amount of protons and electrons, then you would have no charge. But in this case, we have a surplus of electrons. We have two more electrons than protons and since we have a surplus of the negative charged particles we, and we have two more, we're going to have a negative two charge and we write that as two minus. So this is actually an ion, it has a charge. So this is the isotope of sulfur that has a mass and it has two more electrons than protons which gives it this negative charge. Let's do another example where we go the other way. Where we are told, we are given some information about what isotope and really what ion we're dealing with because this has a negative charge and we need to figure out the protons, electrons, and neutrons. Well, the first thing that I would say is, well look, they tell us that this is fluorine. As soon as you know what element we're dealing with, you know what it's atomic number is when you look at the periodic table and you can figure out the number of protons. Remember, your atomic number is the number of protons and that's what defines the element. That's what makes this one fluorine. So let's go up to the, our periodic table and we see fluorine right over here has an atomic number of nine. That means any fluorine has nine protons."
    },
    {
        "Q": "\nAt 0:58, Sal mentions that every isotope of Sulfur will have 16 protons.\nAre there exceptions to this rule? Can an isotope have a number of protons different from its' Atomic Number?",
        "A": "No, because then it would not be an isotope of sulfur. Every sulfur atom in the universe has 16 protons. If it had say 15 protons then it would be phosphorus instead.",
        "video_name": "koAFBScR41A",
        "timestamps": [
            58
        ],
        "3min_transcript": "- [Narrator] An isotope contains 16 protons, 18 electrons, and 16 neutrons. What is the identity of the isotope? And I encourage you to pause the video and see if you can figure it out and I'll give you a hint, you might want to use this periodic table here. All right, so I'm assuming you've had a go at it. So, an element is defined by the number of protons it has. So if someone tells you the number of protons, you should be able to look at a periodic table and figure out what element they are talking about. So, because it is 16 protons, well we can go right over here to the atomic number, what has 16 protons, well anything that has 16 protons by definition is going to be sulfur right over here. So I could write a big S. Now, the next thing we might want to think about is the mass number of this particular isotope. Remember, an isotope, all sulfur atoms are going numbers of neutrons. So, the sulfurs that have different number of neutrons, those would be different isotopes. So, this case we have 16 protons and we have 16 neutrons, so if you add the protons plus the neutrons together, you're going to get your mass number. So 16 plus 16 is 32. Now let's figure out if there's going to be any charge here. Well, the protons have a positive charge. The electrons have a negative charge. If you have an equal amount of protons and electrons, then you would have no charge. But in this case, we have a surplus of electrons. We have two more electrons than protons and since we have a surplus of the negative charged particles we, and we have two more, we're going to have a negative two charge and we write that as two minus. So this is actually an ion, it has a charge. So this is the isotope of sulfur that has a mass and it has two more electrons than protons which gives it this negative charge. Let's do another example where we go the other way. Where we are told, we are given some information about what isotope and really what ion we're dealing with because this has a negative charge and we need to figure out the protons, electrons, and neutrons. Well, the first thing that I would say is, well look, they tell us that this is fluorine. As soon as you know what element we're dealing with, you know what it's atomic number is when you look at the periodic table and you can figure out the number of protons. Remember, your atomic number is the number of protons and that's what defines the element. That's what makes this one fluorine. So let's go up to the, our periodic table and we see fluorine right over here has an atomic number of nine. That means any fluorine has nine protons."
    },
    {
        "Q": "\nAt 0:56, Sal used the word reproduce. What does that mean? I forgot.",
        "A": "Reproduction is the process by which a living organism is able to create their own offsprings.",
        "video_name": "dQCsA2cCdvA",
        "timestamps": [
            56
        ],
        "3min_transcript": "- [Voiceover] I would like to welcome you to Biology at Khan Academy. And biology, as you might now, is the study of life. And I can't really imagine anything more interesting than the study of life. And when I say \"life,\" I'm not just talking about us, human beings. I'm talking about all animals. I'm talking about plants. I'm talking about bacteria. And it really is fascinating. How do we start off with inanimate molecules and atoms? You know, this right here is a molecule of DNA. How do we start with things like that, and we get the complexity of living things? And you might be saying, well, what makes something living? Well, living things convert energy from one form to another. They use that energy to grow. They use that energy to change. And I guess growth is a form of change. They use that energy to reproduce. And these are all, in and of themselves, How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism."
    },
    {
        "Q": "\nAt 5:41pmvideo Biology overview why is biology so important to us humans.",
        "A": "It s important because we need to know how life around us works. How the living matters that surround us came to be. It s good to have a basic knowledge down about living matter.",
        "video_name": "dQCsA2cCdvA",
        "timestamps": [
            341
        ],
        "3min_transcript": "They don't necessarily use energy and growth in the same way that we would associate with life. And then perhaps one of the biggest questions of all is how did life come about? And we will study that as we look at evolution and natural selection. So welcome to Khan Academy's Biology section. I think you're going to find it fascinating. You're going to realize that biology, in some way, is the most complex of the sciences. And in a lot of ways, the one that we understand the least. It's going to be built on top of chemistry, which in turn is built on top of physics, which in turn is built on top of mathematics. And biology is one of our Frankly, even in the last hundred years, we're just starting to scratch the surface of understanding it. But what's really exciting is where the field of biology is going. at a molecular level, we're going to start thinking about how can we even do things like engineer biology, or affect the world around us? It's going to raise all sorts of fascinating and deep and ethical questions. So, hopefully you enjoy this. Biology is one of the most, arguably, maybe the most fascinating subject of all. I don't want to offend the chemists and the physicists out there. I actually find those quite fascinating as well. But we're going to answer, or attempt to start to answer, some of the most fundamental questions of our existence."
    },
    {
        "Q": "At about 3:00, bacteria is between life and non-life, what would you call non-life?\n",
        "A": "Viruses are non-living.",
        "video_name": "dQCsA2cCdvA",
        "timestamps": [
            180
        ],
        "3min_transcript": "How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism. They don't necessarily use energy and growth in the same way that we would associate with life. And then perhaps one of the biggest questions of all is how did life come about? And we will study that as we look at evolution and natural selection. So welcome to Khan Academy's Biology section. I think you're going to find it fascinating. You're going to realize that biology, in some way, is the most complex of the sciences. And in a lot of ways, the one that we understand the least. It's going to be built on top of chemistry, which in turn is built on top of physics, which in turn is built on top of mathematics. And biology is one of our Frankly, even in the last hundred years, we're just starting to scratch the surface of understanding it. But what's really exciting is where the field of biology is going."
    },
    {
        "Q": "at 6:00\nthe alcohol used is connected to a tertiary carbon. Why does an Sn2 reaction occur and not an Sn1 ?\n\nThank you\n",
        "A": "The alcohol used is not connected to a tertiary carbon, it is connected to a secondary carbon. Tertiary carbons are required to be bonded to three other carbon atoms (do not confuse these with the number of bonds they make with atoms that aren t carbon). In the video, the alpha carbon is connected to two other carbons and one oxygen atom. This makes it a secondary carbon. Therefore, SN2 can occur.",
        "video_name": "j-rBgs_p-bg",
        "timestamps": [
            360
        ],
        "3min_transcript": "So 4 minus 1 gives us an oxidation state of plus 3. So once again, an increase in the oxidation state means oxidation. If you oxidize an aldehyde, you will get a carboxylic acid. Let's look at a secondary alcohol now. All right. So we'll go down here to our secondary alcohol. And once again, identify the alpha carbon-- the one attached to your OH group. We need to have at least one hydrogen on that alpha carbon. We have one right here. If we were to oxidize our secondary alcohol-- so we're going to oxidize our secondary alcohol. Once again, a simple way of doing is thinking-- my alpha carbon has one bond to oxygen. So I could increase that to two bonds, and that should be an oxidation reaction. In the process, I'm going to lose a bond it to my alpha hydrogen. So I'm now going to have two bonds of carbon to oxygen, with the hydrogen there. So that leaves my two alkyl groups. So now I have two alkyl groups. And of course, this would be a ketone functional group. If you oxidize a secondary alcohol, you're going to end up with a ketone. I can assign oxidation states. So once again, let's show that this really is an oxidation reaction here. And I go ahead and put in my electrons on my alpha carbon and think about electronegativity differences. Once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And carbon versus carbon, of course, is a tie again. Normally, four valence electrons. In this example, it's surrounded by four. 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I'm going to get this ketone over here So let's take a look at the oxidation state of the carbon that used to be is now our carbonyl carbon. Once again, we put in our electrons. And we think about electronegativity difference. Right? So oxygen is going to beat carbon. So we go like that. Carbon versus carbon is a tie. Carbon versus carbon is a tie. Once again, it's normally 4. Minus 2 this time around that carbon, giving us an oxidation state of plus 2. So to go from a secondary alcohol to a ketone, we see there's an increase in the oxidation state. So this is definitely an oxidation reaction. Let's look now at a tertiary alcohol. So here is my tertiary alcohol. And when I find my alpha carbon, I see that this time there are no hydrogen bonded to my alpha carbon. According to the mechanism-- which we'll see in a minute-- there's no way we can oxidize this tertiary alcohol under normal conditions anyway. If we attempted to oxidize this, we would say there's no reaction here"
    },
    {
        "Q": "How are you going to know which six month period to choose in order to have an isosceles triangle?At 1:49 he mentions finding the maximum parallax angle that the star makes during the year.Doesn't that require daily observation for a year?\n",
        "A": "isn t it just the 2 different 6 month periods? aren t there only 2?",
        "video_name": "6FP-hLuAlr4",
        "timestamps": [
            109
        ],
        "3min_transcript": "I got a comment on the video where we first introduced parallax, especially relative to stars. Essentially asking, how do we know that this angle and this angle is always the same? Or how do we know that we're always looking at an isosceles triangle, where this side is equal to this side? It worked out for this example that I drew right here. But what if the star was over here? What if the star was over here. Then if you just look at it this way. If you take at this point, the triangle is no longer, it's clearly no longer, an isosceles triangle. It looks more like a scalene triangle, I guess, where all of the sides are different. And so a lot of that trigonometry won't apply. Because we won't be able to assume that this is a right triangle over here. And what I want to make clear is that that is true. You would not be able to pick these two points during the year. These two points in our orbit six months apart, in order to do the same math that we did in the last video. In order to calculate this and still is pick two different points six months apart. So you want to do is if this is the sun, you want to pick two different points six months apart, where it does form an isosceles triangle. So if this is the distance from the sun to this other star right over here, you want to pick a point in Earth's orbit around the sun here. And then another point in the orbit six months later, which would put us right over here. And if you do that, then we are, now all of a sudden, we are looking at two right triangles, if we pick those periods correctly. And the best way to think about whether this is a perpendicular angle, is you're going to try to find the maximum parallax from center in each of these time periods. Here it's going to be maximally shifted in one direction. And then when you go to this six months later, it's going to be maximally shifted in the other direction. So to answer that question, the observation is right. all stars will not form an isosceles triangle with the sun and the earth. But you could pick other points in time around the year six months apart where any star will form an isosceles triangle. Hopefully you found that helpful."
    },
    {
        "Q": "\nAt 4:55, you mention prothrombin becoming thrombin, but this isn't mentioned again as you go through the cascade. Is prothrombin one of the Roman Numerals in the cascade or how does it fit in?",
        "A": "Yes, it is one of the numerals in the cascade. Prothrombin is Clotting Factor II, and Thrombin is IIa (activated factor 2). For whatever reason, most people use prothrombin/thrombin rather than calling it Factor 2 / Factor 2a.",
        "video_name": "FNVvQ788wzk",
        "timestamps": [
            295
        ],
        "3min_transcript": "how does your body know to convert fibrinogen into fibrin at the site of injury? The answer is that when you injure your endothelium here, you're going to expose your blood to new proteins. And maybe your actual endothelial cells will release some proteins because they're damaged. So basically, you have new proteins that weren't seen before and that are seen now. Those proteins will eventually cause fibrinogen to turn into fibrin. So while evolution was designing us, it could've said, let's use these little yellow guys to convert fibrinogen to fibrin. That might've worked, but it's actually not the most efficient way to do things. The reason is, imagine if you and a couple of friends have a huge amount of work to do. Let's say you need to convert a million fibrinogen to fibrin. Is the best way to do it, to actually sit down and crank it out? Or would it be more efficient to have each of you And ask those friends to each call five friends. And ask those friends to each call five friends. Well, obviously that would get the job done much faster assuming you had those friends. That's also what your body does. So actually, it doesn't use these yellow guys to convert fibrinogen to fibrin. There's another player which does that, and it's an important one, so we'll give it a little drawing like that, and it is called thrombin. Thrombin, just like fibrinogen, is activated from an inactive form, which we call prothrombin. The prothrombin has a little piece on the end that prevents it from working, so this is prothrombin. That piece is removed when you want to get to work. Well, actually, he's not either because the chain of amplification is much longer than that. To draw the actual amplification cascade, you just need to see it and practice. But there is an easier way to draw it than it usually is drawn, so we'll do that now. So let's say you were counting down from XII. Normally you would start at XII. You'd go to XI. You'd go to X, and then you'd go to IX. But let's say that you weren't very good at counting. You would start with XII. You'd go to XI. Then you'd make a little mistake - you'd go to IX. Then you'd realize you forgot X, so you'd go to X. Now it's good that you remembered X because X is a big deal, and he's going to help bring us thrombin. Turns out that thrombin is also known as II. We know very well that thrombin helps give us fibrin,"
    },
    {
        "Q": "if this sn2 mechanism only works in 1* or 2* alcohols, why is he starting with tert butanol? at 0:00?\n",
        "A": "It s not impossible for SN2 to happen to tertiary alcohols, it s just slow. There is a strong nucleophile and we have formed a good leaving group so it is possible.",
        "video_name": "LccmkSz-Y-w",
        "timestamps": [
            0
        ],
        "3min_transcript": "In this video we're going to see how to prepare alkyl halides from alcohols. And so if we start with this alcohol over here on the left, and we add SOCl2, which is called thionyl chloride, and pyridine to it. We're going to substitute a chlorine atom for the OH group. And this mechanism occurs via an SN2 type mechanism, which means that it's only going to work with primary or secondary alcohols. And you will get inversion of configuration if you have a chirality center present in your final product. So let's take a look at the mechanism. And we'll start with our alcohol. And so the oxygen is going to have to leave somehow. But by itself OH is not the best leaving group. And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group. And so if we draw the dot structure for thionyl chloride, we would have to sulfur double bonded to an oxygen here. And then the sulfur is also bonded to chlorine. So I'll go ahead and put in those lone pairs of electrons on the chlorines, like this. it turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also a bonded to a hydrogen. One lone pair of electrons formed that new bond, so one pair of electrons is left behind. Which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons. Picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorine. So we can go ahead and draw those in. And we can go ahead and draw that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons"
    },
    {
        "Q": "Why does the formation of an alkyl chloride (at 3:00) require a base , while the formation of an alkyl bromide (at 8:00) doesn't?\n",
        "A": "Pyridine is required in chlorination of alcohols by SOCl2 to nuetralise the HCl produced during the reaction. By the way, preparation of alkyl chloride by SOCl2 can also take place in abscence of pyridine via sNi mechanism{look up in wikipedia}.",
        "video_name": "LccmkSz-Y-w",
        "timestamps": [
            180,
            480
        ],
        "3min_transcript": "it turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also a bonded to a hydrogen. One lone pair of electrons formed that new bond, so one pair of electrons is left behind. Which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons. Picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorine. So we can go ahead and draw those in. And we can go ahead and draw that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that-- let's go ahead and get some more room here-- so what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur, and our chlorine, and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, right? Oxygen being more electronegative, it will be partially negative. And this carbon here be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism."
    },
    {
        "Q": "From 6:33 to 6:46, Sal talks about how light takes 8 minutes to hit the earth from the sun. At 6:42 he says that if the sun were to disappear, it would take 8 minutes for the people on Earth to know that light disappeared. Say the sun did disappear. What would happen in terms of gravity? Would we feel the absence of its gravitational effects right away? If not, why? And if so, wouldn't that mean the effects were faster than the speed of light?\n",
        "A": "gravity travels at the speed of light",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            393,
            406,
            402
        ],
        "3min_transcript": "than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers. it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the sun. And just to put that in perspective, if we want it in days, there's 24 hours per day. So this would be 6,250 days. Or, if we divided by 365, roughly 17 years. If you were to shoot a bullet straight at the sun it would take 17 years to get there, if it could maintain its velocity somehow. So this would take a bullet or a jet plane 17 years to get to the sun. Or another way to visualize it-- this sun right over here, on my screen it has about a five- or six-inch diameter. If I were to actually do it at scale, this little dot right here, which is the earth, this"
    },
    {
        "Q": "6:35 Sal says if the sun disappeared it would take 8 minutes for us to notice it. Would the sun's gravitational effects travel as quickly were it to somehow disappear? Does gravity propagate as quickly as light?\n",
        "A": "Yes the effect of gravity propagates through space at the speed of light.",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            395
        ],
        "3min_transcript": "than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers. it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the sun. And just to put that in perspective, if we want it in days, there's 24 hours per day. So this would be 6,250 days. Or, if we divided by 365, roughly 17 years. If you were to shoot a bullet straight at the sun it would take 17 years to get there, if it could maintain its velocity somehow. So this would take a bullet or a jet plane 17 years to get to the sun. Or another way to visualize it-- this sun right over here, on my screen it has about a five- or six-inch diameter. If I were to actually do it at scale, this little dot right here, which is the earth, this"
    },
    {
        "Q": "3:13 109 times the circumfrence. Is that 109 times bigger/wider/taller (Pretty much is the sun 109 x, y, z or all of the above then the earth)\n",
        "A": "Probably bigger.",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            193
        ],
        "3min_transcript": "depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth. if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the earth. Well, how long would it take to go around the sun? So if you were to get on a jet plane and try to go around the sun, or if you were to somehow ride a bullet and try to go around the sun-- do a complete circumnavigation of the sun-- it's going to take you 109 times as long as it would have taken you to do the earth. So it would be 100 times-- I could do 109, but just for approximate-- it's roughly 100 times the circumference of the earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is-- actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the earth times 40 hours."
    },
    {
        "Q": "At 1:42, how did we find the circumference of the Earth?\n",
        "A": "I think I read somewhere in the physics textbook that you can approximate the radius of Earth by using two point of surface to the center of earth and use basic phytagoras to solve it. And yeah, after finding the radius you can easily find the circumference since Earth is nearly round. So, it s possible to find it with math, especially with geometry.",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            102
        ],
        "3min_transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth."
    },
    {
        "Q": "\nabout 10:20, route 19.6h m^2/s^2 = route velocity(f)^2\nbut it turns out velocity(f) = - route 19.6h m/s. Where - < negative sign come from?",
        "A": "he explains it a about 8:55. It s because you took a square root (which has positive and negative solutions) and you want the one that goes in the downward direction, which you ve defined as negative",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            620
        ],
        "3min_transcript": "Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right"
    },
    {
        "Q": "\nAt 9:45 Sal says that we need to take the negaitve square root of 19.6*h m/s. I don\u00c2\u00b4t quite understand why he does this. Is it because our convention was that downward vectors have to have a negative algebraic sign? Thank you for your answers!",
        "A": "You can define direction however you want. If you make up positive, then down is negative, and you just have to stay consistent throughout the problem. So if he defined up as positive earlier in the problem, then down has to be negative.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            585
        ],
        "3min_transcript": "Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right"
    },
    {
        "Q": "Is the formula shown at 9:00 applicable for every free falling projectile dropped with zero initial velocity? Vf=(-)sqrt(19.6h)\n",
        "A": "Yes. As long as the air resistance is neglected and your g = 9.8m/s^2, you can use the kinematic equations for uniform motion. Here for calculating final velocity: v^2 - u^2 = 2*g*h if u = 0 and downward direction is taken as positive, then v = sqrt(2*9.8*h). Note that the final velocity doesn t depend on the mass and dimensions of the body. As long as your projectile can be approximated as a point mass particle, you are good to go.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            540
        ],
        "3min_transcript": "So let me write this over here. So this is negative 9.8. So we have 2 times negative 9.8-- let me just multiply that out. So that's negative 19.6 meters per second squared. And then what's our displacement going to be? What's the displacement over the course of dropping this rock off of this ledge or off of this roof? So you might be tempted to say that our displacement is h. But remember, these are vector quantities, so you want to make sure you get the direction right. From where the rock started to where it ends, what's it doing? It's going to go a distance of h, but it's going to go a distance of h downwards. And our convention is down is negative. So in this example, our displacement from when it leaves your hand to when it hits the ground, the displacement is going to be equal to negative h. It's going to travel a distance of h, but it's going to travel that distance downwards. Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something."
    },
    {
        "Q": "at 0:29 sal says cellular membrane is semi permeable. But actually it is selectively permeable. *why does he say so?*\n",
        "A": "Both semi permeable and selectively permeable refer to the same thing that the cellular membrane only allows certain substances (molecules or ions) to pass through it. Usually, the solvent molecules can pass through however certain solute molecules can not. That s why both terms can be used.",
        "video_name": "afWnU10ZNfg",
        "timestamps": [
            29
        ],
        "3min_transcript": "- [Voiceover] I have three different scenarios here of a cell being immersed in a solution, and the cell is this magenta circle, that's the cellular membrane. I have the water molecules depicted by these blue circles, and then, I have the solute inside of the solution, inside of the water solution that we depict with these yellow circles. I've clearly exaggerated the size of the water molecules and the solute particles relative to the size of the cell, but I did that so that we can visualize what's actually going on. We're going to assume that the cellular membrane, this phospholipid bilayer, is semipermeable, that it will allow water molecules to pass in and out, so a water molecule could go from the inside to the outside, or from the outside to the inside, but we're gonna assume that it does not allow the passage of the solute particles, so that's why it's semipermeable. It's permeable to certain things, or we could say, selectively permeable. Now, what do we think is going to happen? Well, the first thing that you might observe is we have a lower concentration of solute on the outside than we have on the inside, some water molecules moving in just the right direction to go from the outside to the inside, and you will also have some water molecules that might be in just the right place to go from the inside to the outside, but what's more likely to happen, and what's going to happen more over a certain period of time? The water molecules that are on the outside, and we talk about this in the osmosis video, they're going to be less obstructed by solute particles. If this one happens to be moving in that direction, well, it's gonna make its way to the membrane, and then, maybe get through the membrane, while something, maybe, if this water molecule was moving in this direction, well, gee, it's gonna be obstructed now, maybe this is bouncing back, and it's gonna ricochet off of it, so the water molecules on the inside are more obstructed. They're less likely to be able to fully interact with the membrane or move in the right direction. They're being obstructed by these solute particles. Even though you're going to have water molecules going back and forth, in a given period of time, you have a higher probability so you're going to have a net inflow. Net inflow of H2O, of water molecules. Now, a situation like this, where we're talking about a cell and it's in a solution that has a lower concentration of solute, it's important that we're talking about a solute that is not allowed to go to the membrane, the membrane is not permeable to that solute. We call this type of situation, this type of solution that the cell is immersed in, we call this a hypotonic solution. Hypotonic solution. Anytime we're talking about hypotonic, or as we'll see, isotonic and hypertonic, we're talking about relative concentrations of solute that cannot get through some type of a membrane. The word hypo, you might've seen it in other things. It's a prefix that means less of something, so in this case, we have a lower concentration of solute in the solution"
    },
    {
        "Q": "\nAt 4:49, how do you know which ones are Adenine?",
        "A": "He assigned random bases as Adenine (not accurate in the sense that those were actually Adenine) then assigned Thymine to complement Adenine.",
        "video_name": "AmOO4j0E408",
        "timestamps": [
            289
        ],
        "3min_transcript": "Well, the word deoxyribonucleic acid comes from the fact that this backbone is made up of a combination of sugar and phosphate. And the sugar that makes up the backbone is deoxyribose. So that's essentially the D in DNA. And then the phosphate group is acidic and that's now where you get the acid part of it. And nucleic is, hey this was found in nuclei of cells. It is nucleic acid. Deoxyribonucleic acid. It is actually mildly acidic all in total but for every acid it actually also has a base, and those bases form the rung of the ladders. And actually each rung is a pair of bases and as I said, that's where the information is actually stored. Well what am I talking about? Well let me talk about the four different bases that make up the rungs of a DNA molecule. So, you have adenine. And so for example, this part right over here. This section of that rung might be adenine. Maybe this right over here is adenine. This right over here. Remember, each of these rungs are made up by it's a pair of bases. And that might be adenine. Maybe this is adenine and I could stop there, I mean I'll do a little more adenine. Maybe that's adenine right over there. And adenine always pairs with the base thymine. So let me write that down. So adenine pairs with thymine. Thymine. So, if that's an adenine there then this is going to be a thymine. If this is an adenine then this is going to be a thymine. Or if I drew the thymine first, well say, okay it's gonna pair with the adenine. So this is going to be a thymine right over here. This is going to be a thymine. If I were to draw this, this would be a thymine right over here. Now the other two bases, you have cytosine which pairs with guanine So guanine and we're not gonna go into the molecular structure of these bases just yet, although these are good names to know because they show up a lot and they really form kind of the code, your genetic code. Guanine. Guanine pairs with cytosine. Guanine and cytosine. Cytosine. So actually if this is, let's say there's some cytosine there, let's say cytosine right over here. Maybe this is a cytosine, maybe this is cytosine, maybe this is cytosine, this is cytosine and maybe this is cytosine. Then it always pairs with the guanine. So, let's see, this is guanine then and this will be guanine. This is guanine, this is guanine. I actually didn't draw stuff here. This is guanine, I didn't say what these could be but these would be maybe the pairs of they could be adenine-thymine pairs and it could be adenine on either side"
    },
    {
        "Q": "at 6:59 what if your characteristics are nothing like you parents and does this mean we all come from two people and if so why are we all so different?\n",
        "A": "Characteristics don t always have to be like your parents, sometimes you find that people have eyes like their grandmother or their uncle s nose. Remember that mutations do occur, and the process of crossing over contributes to genetic variety. These characteristics are recessive. The idea that we all come 2 people is biblical and not necessarily scientific.",
        "video_name": "AmOO4j0E408",
        "timestamps": [
            419
        ],
        "3min_transcript": "So guanine and we're not gonna go into the molecular structure of these bases just yet, although these are good names to know because they show up a lot and they really form kind of the code, your genetic code. Guanine. Guanine pairs with cytosine. Guanine and cytosine. Cytosine. So actually if this is, let's say there's some cytosine there, let's say cytosine right over here. Maybe this is a cytosine, maybe this is cytosine, maybe this is cytosine, this is cytosine and maybe this is cytosine. Then it always pairs with the guanine. So, let's see, this is guanine then and this will be guanine. This is guanine, this is guanine. I actually didn't draw stuff here. This is guanine, I didn't say what these could be but these would be maybe the pairs of they could be adenine-thymine pairs and it could be adenine on either side and they could be made of guanine-cytosine pairs where the guanine or the cytosine is on the other side. Actually just to make it a little bit more complete let me just color in the rungs here as best as I can. So those are guanines so they're gonna pair with cytosine. Pair with cytosine, pair with cytosine. When you straw in this way you might start to see how this is essentially a code, the order of which the bases are... I guess the order in which we have these or the sequence of these bases essentially in code the information that make you, you, and you could be. Well how much of it is nature versus nurture and when people say nature, you know, it's literally genetic, and that's an ongoing debate, an ongoing debate but it does code for things like your hair color. When you see that your smile is similar to your parents it is because that information to a large degree is encoded genetically. It affects a lot of what makes you you and actually not even just within a species Humans have more genetic material in common with other humans than they do with say a plant. But all living creatures as we know them have genetic information. This is the basis by which they are passing down their actual traits. Now you might be saying well, how much genetic information does a human being have? And the number will either disappoint you or you might find it mind-boggling. The human genome and every species has a different number of base pairs to large degree correlated with how complex they are although not always. But the human genome has 6,000,000. Sorry, not 6,000,000, 6,000,000,000. 6,000,000 would be disappointing, even billion might be disappointing. 6,000,000,000 base pairs. 6,000,000,000. 6,000,000,000 base pairs. And when you have your full complement of chromosomes and this is in most of the cells in your body and outside of your sex cells,"
    },
    {
        "Q": "12:00 Why are Adenine pairing with Thymine and Guanine with Cytosine ??\n",
        "A": "It is a direct result of the molecular structure of the bases, A/T only have 2 side groups which can form hydrogen bonds whereas C/G have 3.",
        "video_name": "AmOO4j0E408",
        "timestamps": [
            720
        ],
        "3min_transcript": "kind of the scale of this thing. This is a very dense way to actually store information. But just to have an appreciation of and you might have seen it when I was coloring in on why the structure lends itself to being able to replicate the information or even to be able to translate or express the information. Let's think about if you were to take this ladder and you were to just kind of split all the base pairs. So, you just have 1/2 of them. So you essentially have half of the ladder. And so if you only have half of the ladder, you're able to construct the other half of the ladder. Let's take an example, let's say and I'll just use the first letter to abbreviate for each of these bases. Let's say you have some... So let's say this is one of the, this is the sugar phosphate backbone right over here. So this could be one of the sides. Let's say there's some adenine. Actually we do in the right color. So you got some adenine, adenine. Maybe some adenine right over here and maybe there's an adenine there. And maybe you have some thymine, thymine, and then you have some guanine, guanine, guanine. And then let's say you have some cytosine and you have some cytosine. So with just half of this ladder I guess you could say, you're able to construct the other half, and this is actually how DNA replicates. This ladder splits and then each of those two halves of that ladder are able to construct versions of the other half, or versions of the other half are able to constructed on top of that, on top of that half. So how does that happen? Well, it's based on how these bases pair. Adenine always pairs with thymine if we're talking about DNA. So if you have an A there, you're gonna have a T on this end, T on this end. T's right all over here, T right over there. If you have a T on that end you're gonna have an A right over there. A, A. If you have a G, a guanine on this side, you're gonna have a cytosine on the other side. And if you have a cytosine you're gonna have a guanine on the other side. Hopefully that gives you an appreciation of how DNA can replicate itself. And as we'll see also how this information can be translated to other forms of either related molecules but eventually to proteins. And just to kind of round out this video, to get a real visual sense what the DNA molecule looks like or I guess a different visual depiction from this. I found this animated gif that, you know, if you haven't fully digested what a double helix looks like, this is it. And you see here, you see your sugar phosphate bases here. You see kind of the sugars and phosphate, the sugars and the phosphates alternating along this backbone, and then the rungs of the ladder are these base pairs. So this is one of the bases, that's the corresponding, that's this corresponding, I guess you can say partner. And you can see that along all the way up and down in this molecule. Very exciting."
    },
    {
        "Q": "at 9:50 you said that the carbon dioxide will diffuse across and into the plasma, and that some of the carbon dioxide can make its way across the membrane into the red blood cell itself where it will be converted into carbonic acid, so does only some of the carbon dioxide get converted into carbonic acid and the rest remain in the plasma? because i thought that all the carbon dioxide gets converted into carbonic acid?\n",
        "A": "thank you, but does it travel in the blood as carbonic acid or as carbon dioxide?",
        "video_name": "LWtXthfG9_M",
        "timestamps": [
            590
        ],
        "3min_transcript": "the small capillaries, which a lot of people believe helps them release their contents and maybe some of the oxygen that they have in them. So you have a red blood cell that's coming in here. It's being squeezed through this capillary right here. It has a bunch of hemoglobin-- and when I say a bunch, you might as well know right now, each red blood cell has 270 million hemoglobin proteins. And if you total up the hemoglobin in the entire body, it's huge because we have 20 to 30 trillion red blood cells. And each of those 20 to 30 trillion red blood cells have 270 million hemoglobin proteins in them. So we have a lot of hemoglobin. So anyway, that was a little bit of a-- so actually, red blood cells make up roughly 25% of all of the We have about 100 trillion or a little bit more, give or take. I've never sat down and counted them. But anyway, we have 270 million hemoglobin particles or proteins in each red blood cell-- explains why the red blood cells had to shed their nucleuses to make space for all those hemoglobins. They're carrying oxygen. So right here we're dealing with-- this is an artery, right? It's coming from the heart. The red blood cell is going in that direction and then it's going to shed its oxygen and then it's going to become a vein. Now what's going to happen is you have this carbon dioxide. You have a high concentration of carbon dioxide in the It eventually, just by diffusion gradient, ends up-- let me do that same color-- ends up in the blood plasma just like that and some of it can make its way across the membrane into the actual red blood cell. In the red blood cell, you have this carbonic anhydrase essentially become carbonic acid, which then can release protons. Well, those protons, we just learned, can allosterically inhibit the uptake of oxygen by hemoglobin. So those protons start bonding to different parts and even the carbon dioxide that hasn't been reacted with-- that can also allosterically inhibit the hemoglobin. So it also bonds to other parts. And that changes the shape of the hemoglobin protein just enough that it can't hold onto its oxygens that well and it starts letting go. And just as we said we had cooperative binding, the more oxygens you have on, the better it is at accepting more-- the opposite happens. When you start letting go of oxygen, it becomes harder to retain the other ones. So then all of the oxygens let go. So this, at least in my mind, it's a brilliant, brilliant mechanism because the oxygen gets let go just where it needs to let go. It doesn't just say, I've left an artery and"
    },
    {
        "Q": "(2:35) Sal says that cooperative binding is when \"one binding makes other bindings more likely.\" What causes binding to stop in the first place?\n",
        "A": "So when one oxygen binds (perhaps in the lungs), that encourages additional oxygen to bind (up to 3 more) as hemoglobin can carry 4 O2 molecules. When the red blood cell carrying hemoglobin gets to other tissues outside the lung, the increased carbon dioxide diffuses into the red blood cell and through carbonic anhydrase and the production of H+, induces hemoglobin to let go of the oxygen - where the binding stops. The oxygen then diffuses out of the RBC to the tissues.",
        "video_name": "LWtXthfG9_M",
        "timestamps": [
            155
        ],
        "3min_transcript": "You have four heme groups and the globins are essentially describing the rest of it-- the protein structures, the four peptide chains Now, this heme group-- this is pretty interesting. It actually is a porphyrin structure. And if you watch the video on chlorophyll, you'd remember a porphyrin structure, but at the very center of it, in chlorophyll, we had a magnesium ion, but at the very center of hemoglobin, we have an iron ion and this is where the oxygen binds. So on this hemoglobin, you have four major binding sites for oxygen. You have right there, maybe right there, a little bit behind, right there, and right there. Now why is hemoglobin-- oxygen will bind very well here, but hemoglobin has a several properties that one, make it really good at binding oxygen and then also really good at dumping oxygen when it needs to dump oxygen. And this is just the principle that once it binds to one oxygen molecule-- let's say one oxygen molecule binds right there-- it changes the shape in such a way that the other sites are more likely to bind oxygen. So it just makes it-- one binding makes the other bindings more likely. Now you say, OK, that's fine. That makes it a very good oxygen acceptor, when it's traveling through the pulmonary capillaries and oxygen is diffusing from the alveoli. That makes it really good at picking up the oxygen, but how does it know when to dump the oxygen? This is an interesting question. this guy's running right now and so he's generating a lot of carbon dioxide right now in these capillaries and he needs a lot of oxygen in these capillaries surrounding his quadriceps. I need to deliver oxygen. It doesn't know it's in the quadraceps. How does the hemoglobin know to let go of the oxygen there? And that's a byproduct of what we call allosteric inhibition, which is a very fancy word, but the concept's actually pretty straightforward. When you talk about allosteric anything-- it's often using the context of enzymes-- you're talking about the idea that things bind to other parts. Allo means other. So you're binding to other parts of the protein or the enzyme-- and enzymes are just proteins-- and it affects the ability of the protein or the enzyme to do"
    },
    {
        "Q": "At 11:40 the carbon that takes the 1st H ends up having 5 bonds towards the end. How would that be correct? The last H that he adds in there shouldn't be there. It was a double bond that \"broke\" and an H is added, or did I misunderstand?\n",
        "A": "I can t see where any carbon has 5 bonds? Are you sure you counted correctly? Is it the front of back one that originally had the double bond? Note that there looks to be one very long bond from the front carbon to a hydrogen, but that s actually a bond between the front and back carbons, and then a bond between the back carbon and the hydrogen.",
        "video_name": "fSk1Crn3R2E",
        "timestamps": [
            700
        ],
        "3min_transcript": "off like that. And my double bond is going to go right here. And then this is going to be a methyl group. And then up here there are going to be two methyl groups, like that. So this is alpha-pinene, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pinene molecule and I wanted to hydrogenate it, I could use palladium and charcoal, palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat, like that. And so, when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond, think about this guy over here, think about the alpha-pinene as molecules like a spaceship, And the spaceship is approaching the docking station. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top, because of the steric hindrance of these methyl groups. Right? So this is the way that it approaches. In this part of the molecule, your alkene, is the flat part, right? So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction. And let's see if we can draw it here. And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below, right? So this hydrogen, let's say it adds right here. That's going to push this methyl group up. when that hydrogen adds right down here. And then this other hydrogen is going to add to the opposite side. And so we can show the addition of that hydrogen. So there's my syn addition of these two hydrogens. And there was something else in that carbon. It was another hydrogen. So another hydrogen got pushed up right here as well. So that is your only product, the only product of this reaction. The hydrogenation reaction is very sensitive to steric conditions."
    },
    {
        "Q": "\nDoes adsorb mean anything (0:34) or is it just a different pronunciation of absorb?",
        "A": "They are different words. Adsorb means to hold molecules of a gas as a thin film on the outside surface of a solid. Absorb means to suck up or soak up. For example, a sponge absorbs water.",
        "video_name": "fSk1Crn3R2E",
        "timestamps": [
            34
        ],
        "3min_transcript": "To hydrogenate an alkene, you need hydrogen gas and a metal catalyst, something like platinum or palladium or nickel. And there are many others, but these are the ones most commonly used. So what happens is those two hydrogens from the hydrogen gas are added across their double bond. And they're added on the same side of where the double bond used to be. So it's a syn addition. Let's take a look at why this is a syn addition of hydrogens. So we have our metal catalyst over here. So let's go ahead and draw our flat metal catalyst. And these metals are chosen, because they adsorb hydrogen really well, which means that if you bubble hydrogen gas through, the hydrogen is going to be adsorbed to the surface of that metal catalyst, like that. And then your alkene comes along. And your alkene is also flat, right? The portion of the molecule that contains the double bond, right? So these two carbons, this carbon and this carbon, these sp2 hybridized, which means that the stereo chemistry around those two carbons is going to be flat. So you have one thing that's flat approaching something else that's flat. So the only way those hydrogens can add are to add them on to the same side, right? So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, right? This would now be a wedge and then a dash, and then this would be a wedge and then a dash. You can see those two hydrogens have added on to the same side. So these two hydrogens or these two hydrogens for our syn addition. Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right. So we have to think about stereochemistry for this reaction for your products as well. So let's take a look at an actual reaction here. And let's see if we can follow along. So if this was my reaction, I want to hydrogenate this alkene. So I would add some hydrogen gas and I could choose whichever metal catalyst I wanted to. So I could add two hydrogens on the same side. So just like I did up there. So we would get now everything changes from sp2 hybridization So we have wedges and dashes to worry about. And usually you wouldn't see it drawn like this. That's too much work, quite frankly. It would be much easier just to say, well, all I have to do is take away the double bond, and there's my product. So for some of these reactions, they're very, very, very simple. Just take away the double bond and you'll end up with your alkene-like product. Let's take a look at oxidation states for this reaction. So I'm going to redraw this reaction. And this time I'm going to draw in my atoms. And I'm also going to draw in my electrons here in a second. So I'm just drawing out all the atoms here. So I have all these methyl groups to worry about. And then I have electrons in these bonds, right? Each one of these bonds consists of two electrons."
    },
    {
        "Q": "I'm confused.... at 6:30 where jay says that Carbon has been reduced as it gained an electron from Hydrogen, I thought C-H interactions were non-polar so they share electrons as oppose to carbon gaining one.... :(\n",
        "A": "In reality it does not mean that C-H is non-polar. Because of the small difference in electronegativities, the C\u00e2\u0088\u0092H bond is generally regarded as being non-polar. But theres still a difference in electronegativities. But it doesnt really matter in this case because concerning the oxidation state you simply look at the atom which is more electronegative, which gain more electron.",
        "video_name": "fSk1Crn3R2E",
        "timestamps": [
            390
        ],
        "3min_transcript": "we just drew around it for our dot structure. So that would be four. Each one of those carbons has four. So each of these carbons has an oxidation state of zero. Let's look at the product, and let's see if we can assign some oxidation states for the product. So our product over here on the right, we had a carbon and we had some methyl groups bonded to that carbon. We added on a hydrogen. And so each one of these carbons got a hydrogen added onto it. And let's go ahead and fill in our electrons in these bonds. So once again, each bond consists of two electrons, like that. And now we have a single bond between our carbons. And let's assign some oxidation states. So once again we know that the two carbons have the same electronegativity, right? So the tug of war for these two electrons right here, it's a tie. So it's a tie, it's a tie. Carbon is actually more electronegative than hydrogen. So in the war over the two electrons in the carbon-hydrogen bond, carbon wins, because it's a little bit more electronegative. So we're going to assign this extra electron here to carbon. And then again, carbon versus carbon. So that carbon gets that electron as well. Same thing down here, right? So it's a tie, it's a tie. Carbon beats hydrogen. And over here, it's a tie. So in the dot structure on the right, the oxidation states that the normal number of valence electrons would be four. From that we subtract the number of electrons in our picture here, which would be five electrons. Each one of these carbons has five electrons around it. So it gained electron. And it's a 4 minus 5 will give us a negative 1. So the oxidation states of these two carbons is negative 1. And we can look at our original oxidation states of being zero, went from zero to negative 1. That's a decrease in the oxidation state, right? A decrease in the oxidation state means a reduction. So the alkene is reduced by the addition of these two hydrogens. And you'll see other definitions for oxidation states. You'll see a gain in hydrogens is reduction. That's another definition that's often found in organic chemistry textbooks. And while that's true, to me it makes more sense to go ahead and assign your oxidation states and watch the oxidation states change as you add those hydrogens, as your molecule gains hydrogens. So this is a reduction. Let's look at the stereochemistry of the hydrogenation reaction. So let's do an example involving stereochemistry. So let's say your alkene-- let's do that ring again, it wasn't a very good one-- so let's say your alkene looked something like this. And you're going to react that with hydrogen and with platinum. All right, well, your first thought"
    },
    {
        "Q": "At 2:10, Sal says that speed is the magnitude of velocity but speed may not always be the magnitude of velocity (as in the case of average speed and average velocity). I'm still a little confused by what Sal meant here\n",
        "A": "THe magnitude of velocity is a speed It may or may not be the speed you are looking for in a particular problem",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            130
        ],
        "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope,"
    },
    {
        "Q": "\nAt 1:56, Sal said that if he'd written the word velocity rather than speed then the second statement would be true, but if we consider an unbalanced force on a body, so maybe the force will \"only\" change it's direction or the other way around, so i didn't understood that how can we say the unbalanced force will Always impact it's \"velocity\" (speed+direction) when it maybe changing it's direction \"only\" and not the speed or vise versa.\nI hope my question is clear.",
        "A": "If the force changes the direction of motion, then it changed the velocity, even if the speed is the same.",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            116
        ],
        "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope,"
    },
    {
        "Q": "@7:14 i m bit confuse here like centripetal force balances centrifugal force so here will the ice scater have balanced force............\nnd the speed example was really good\nwant to learn through such examples from daily life\n",
        "A": "The forces are not balanced, if they were the skater would have uniform motion and not be accelerating in a circle. Centripetal force is a real force which accelerates an object to travel in a circle. Centrifugal force is fictitious force which is felt from being in a reference frame which is rotating. The perceived centrifugal force is equal and opposite to the centripetal force.",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            434
        ],
        "3min_transcript": "It could be its speed, its direction, or both, but it doesn't have to be both. It could be just the speed or just the direction. So this is an incorrect statement. Now the third statement, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" This is absolutely true. And this is the example we gave. If I take an object, if I take my book and I try to slide it across the desk, the reason why it eventually comes to stop is because we have the unbalanced force of friction-- the grinding of the surface of the book with the grinding of the table. If I'm inside of a pool or even if there's absolutely no current in the pool, and if I were to try to push some type of object inside the water, it eventually comes to stop because of all of the resistance of the water itself. It's providing an unbalanced force in a direction opposite it's motion. That is what's slowing it down. why we don't see these things go on and on forever is that we have these frictions, these air resistants, or the friction with actual surfaces. And then the last statement, \"An unbalanced force on an object will always change the object's direction.\" Well, this one actually is maybe the most intuitive. We always have this situation. Let's say I have a block right over here, and it's traveling with some velocity in that direction-- five meters per second. If I apply an unbalanced force in that same direction-- so that's my force right over there. If I apply it in that same direction, I'm just going to accelerate it in that same direction. So I won't necessarily change it. Even if I were to act against it, I might decelerate it, but I won't necessarily change its direction. I could change its direction by doing something like this, but I don't necessarily. I'm not always necessarily changing the object's direction. So this is not true. An unbalanced force on an object will not always It can, like these circumstances, but not always. So \"always\" is what makes this very, very, very wrong."
    },
    {
        "Q": "at 1:30, sal says the is a minuscule amount of gravity in space, what causes gravity in space? is their any forces apart from the small about of gravity in space?\n",
        "A": "If you look at the equation for the gravitational acceleration A = G * M/r^2 you can see that A goes to 0 at an infinite distance so even deep space in not infinity far from any mass so there still be a small amount of gravity out there.",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            90
        ],
        "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope,"
    },
    {
        "Q": "At 6:46 in the video, Mr. Kahn mentions 'deceleration' but my science book tells me there is no such thing as \"deceleration.\" It says that there is acceleration in the direction of velocity or in the opposite direction of velocity, but in physics, it is always called acceleration. Which is accurate?\n",
        "A": "It s better to talk about negative acceleration but of course people do use the term deceleration and scientists do need to be able to talk to regular people. It s certainly not right to say there s no such thing as deceleration, because everyone knows there is. But that doesn t make the term or concept something we want to use when we are trying to do physics.",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            406
        ],
        "3min_transcript": "It could be its speed, its direction, or both, but it doesn't have to be both. It could be just the speed or just the direction. So this is an incorrect statement. Now the third statement, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" This is absolutely true. And this is the example we gave. If I take an object, if I take my book and I try to slide it across the desk, the reason why it eventually comes to stop is because we have the unbalanced force of friction-- the grinding of the surface of the book with the grinding of the table. If I'm inside of a pool or even if there's absolutely no current in the pool, and if I were to try to push some type of object inside the water, it eventually comes to stop because of all of the resistance of the water itself. It's providing an unbalanced force in a direction opposite it's motion. That is what's slowing it down. why we don't see these things go on and on forever is that we have these frictions, these air resistants, or the friction with actual surfaces. And then the last statement, \"An unbalanced force on an object will always change the object's direction.\" Well, this one actually is maybe the most intuitive. We always have this situation. Let's say I have a block right over here, and it's traveling with some velocity in that direction-- five meters per second. If I apply an unbalanced force in that same direction-- so that's my force right over there. If I apply it in that same direction, I'm just going to accelerate it in that same direction. So I won't necessarily change it. Even if I were to act against it, I might decelerate it, but I won't necessarily change its direction. I could change its direction by doing something like this, but I don't necessarily. I'm not always necessarily changing the object's direction. So this is not true. An unbalanced force on an object will not always It can, like these circumstances, but not always. So \"always\" is what makes this very, very, very wrong."
    },
    {
        "Q": "\nAt 1:01, If the net force on a body is zero, would it have any velocity or would it be a stationary object?",
        "A": "it can either be in motion or can also be in rest. force is not required to keep the object which is already in constant velocity in motion thats why they say a object in constant velocity to have a net force of 0",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            61
        ],
        "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope,"
    },
    {
        "Q": "At 10:45 Sal said that a hot spot could have created the Mid-Atlantic Ridge as well as the African Rift Valley. The Mid-Atlantic Ridge is much bigger than the African rift valley, so how could the same type of hot spot have created them both?\n",
        "A": "A hot spot cannot create a ridge, they create volcanic island chains, Hawaii and its islands are prime examples. The Mid-Atlantic Ridge was created by diverging tectonic plates, the magma which streamed up from the mantel solidified in the ocean and created the ridge.",
        "video_name": "FK1s1-OJ5BE",
        "timestamps": [
            645
        ],
        "3min_transcript": "You essentially have the Indian Ocean flowing into this rift that formed from this hot spot. And then if you fast forward a bunch so that finally the magma can kind of surface. So let's fast forward from even this point even more. So let's fast forward even more, and let's say now the land has been pushed a good bit apart. Now the hot spot has actually surfaced. Now the crust might look something like this. So it's been pushed apart a good bit at this point. Now we're talking about on the order of hundreds of thousands of years or tens of thousands of years. So the land, for example, the land that was here, this part of the land might now be out here. And this part of the land might now be out here. What's going to happen is that this hot spot is going to continue to fuel, and we're assuming everything's underwater at this point. Since this depression that was created is now so low the crust was stretched thin. The hot spot is essentially going to come out of underwater volcanoes and start creating what's now-- this body of water's gotten large enough that we can call it a mid-oceanic ridge. And so it'll actually start creating an actual ridge with volcanoes in the center. So that's why one, we see things like the Rift Valley in Africa, we see things like the Red Sea. And maybe even more importantly, that's why we see something like the mid-Atlantic rift in the middle of the Atlantic Ocean, where you have all of this depressed land that was essentially analogous to that Rift Valley but it's at a much later stage. And that's why it's able to collect water, because when the land was pushed out and stretched thin water was able to flow into it, going when this bread was baking and this part of the crust pushed outwards, you had this rift form, and then if there was some water on the bread, or if it was raining, or if it was connected to a body of water, water would've eventually flowed in here. And if that bread kept growing this rift would have kept growing, eventually to the size of the Atlantic Ocean in our theoretical bread. And so that's why you have this huge depressed area where the ocean can form, but in the middle of it you kind of have this you have this submersed you have this actual submersed mountain chain, this submersed chain of volcanoes, this submersed ridge where the land actually does go up a little bit because of all that magma flowing directly out of it. So hopefully that clears up a little bit. That was always confusing to me why you see uplifted land but then everything around the uplifted land is much lower, and why the whole thing is submersed as it's moving away. So hopefully that clears things up a little bit."
    },
    {
        "Q": "\nAround 7:40, Sal says the time taken for the momentum to change is 2x/v, bt isn't the actual time taken that tiny instant it JUST strikes the wall and reverses its velocity?",
        "A": "| _.->_| Starting |<-. __| Back at the same point, but not with the same velocity. | _.->_| Back at the same point with the same velocity. The distance traveled therefore must be twice the length. This distance over the velocity component in this direction will be the time required for the average pressure equation.",
        "video_name": "qSFY7GKhSRs",
        "timestamps": [
            460
        ],
        "3min_transcript": "Now, if I come in with a momentum of mv, and I ricochet off with a momentum of minus mv, what's my change in momentum? My change in momentum, off of that ricochet, is equal to-- well, it's the difference between these two, which is just 2mv. Now, that doesn't give me the force. I need to know the change in momentum per unit of time. So how often does this happen? How frequently? Well, it's going to happen every time we come here. We're going to hit this wall. Then the particle is going to have to travel here, bounce off of that wall, and then come back here and hit it again. So that's how frequently it's going to happen. So how long of an interval do we have to wait between the collisions? Well, the particle has to travel x going back. It's going to collide. It's going to have to travel x to the left. This distance is x. Let me do that in a different color. It's going to have to travel x to go back. Then it's going to have to travel x back. So it's going to have to travel 2x distance. And how long will it take it to travel 2x distance? Well, the time, delta T, is equal to, we know this. Distance is equal to rate times time. Or if we do distance divided by rate, we'll get the amount of time we took. This is just our basic motion formula. Our delta T, the distance we have to travel is back and forth. So it's 2 x's, divided by-- what's our rate? Well, our rate is our velocity. Divided by v. There you go. So this is our delta T right here. So our change in momentum per time is equal to 2 times our Because we ricocheted back with the same magnitude, but negative momentum. So that's our change in momentum. And then our change in time is this value over here. It's the total distance we have to travel between collisions of this wall, divided by our velocity. So it is, 2x divided by v, which is equal to 2mv times the reciprocal of this-- so this is just fraction math-- v over 2x. And what is this equal to? The 2's cancel out. So that is equal to mv squared, over x. Interesting. We're getting someplace interesting already. And if it doesn't seem too interesting, just hang on with me for a second. Now, this is the force being applied by one particle, is this-- force from one particle on this wall."
    },
    {
        "Q": "\nAround 6:31, why is the change in momentum a positive number? if P2 = -mv due to the velocity being in the opposite direction, then should change in momentum not be = P2 - P1, which is -2mv ?? Thanks for helping me out of my misery:-(",
        "A": "I have this exact same question, and no one seems to have answered it yet. The only explanation I can come up with is that the negative is just an indication of the direction, and when looking at the force that results from this pressure, a positive value should be used.",
        "video_name": "qSFY7GKhSRs",
        "timestamps": [
            391
        ],
        "3min_transcript": "And, of course, we know that this could be rewritten as this is equal to-- mass is a constant and shouldn't change for the physics we deal with-- so it's delta. We could put that inside of the change. So it's delta mv over change in time. And this is just change in momentum, right? So this is equal to change in momentum over change in time. So that's another way to write force. So what's the change in momentum going to be for this particle? Well, it's going to bump into this wall. In this direction, right now, it has some momentum. Its momentum is equal to mv. And it's going to bump into this wall, and then going to ricochet straight back. And what's its momentum going to be? Well, it's going to have the same mass and the same velocity. We'll assume it's a completely elastic collision. Nothing is lost to heat or whatever else. But the velocity is in the other direction. So the new momentum is going to be minus mv, because the Now, if I come in with a momentum of mv, and I ricochet off with a momentum of minus mv, what's my change in momentum? My change in momentum, off of that ricochet, is equal to-- well, it's the difference between these two, which is just 2mv. Now, that doesn't give me the force. I need to know the change in momentum per unit of time. So how often does this happen? How frequently? Well, it's going to happen every time we come here. We're going to hit this wall. Then the particle is going to have to travel here, bounce off of that wall, and then come back here and hit it again. So that's how frequently it's going to happen. So how long of an interval do we have to wait between the collisions? Well, the particle has to travel x going back. It's going to collide. It's going to have to travel x to the left. This distance is x. Let me do that in a different color. It's going to have to travel x to go back. Then it's going to have to travel x back. So it's going to have to travel 2x distance. And how long will it take it to travel 2x distance? Well, the time, delta T, is equal to, we know this. Distance is equal to rate times time. Or if we do distance divided by rate, we'll get the amount of time we took. This is just our basic motion formula. Our delta T, the distance we have to travel is back and forth. So it's 2 x's, divided by-- what's our rate? Well, our rate is our velocity. Divided by v. There you go. So this is our delta T right here. So our change in momentum per time is equal to 2 times our"
    },
    {
        "Q": "isnt it sn=3+2=5 in ~5:40?\n",
        "A": "No. Before the electrons move, there are 3 \u00cf\u0083 bonds and 1 lone pair, so SN = 3 + 1 = 4",
        "video_name": "kQCS1AhAnMI",
        "timestamps": [
            340
        ],
        "3min_transcript": "they can't participate in resonance; that lone pair of electrons in blue is localized to that nitrogen. And so, this is why you can think about an amide being different from an amine, in terms of functional group, and in terms of how they react and how they behave. If we look at another example, so this molecule right here, and we assume the lone pair of electrons on that nitrogen is localized to that nitrogen, let's go ahead and calculate the steric number. So the steric number'd be equal to sigma bonds, so that's one, two, and three; so three sigma bonds. Plus lone pairs of electrons, there's one lone pair of electrons on that nitrogen, so three plus one is four; so four hybrid orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone pair of electrons is not localized to that nitrogen; that lone pair of electrons because we have this pattern here. All right, so this pattern of a lone pair of electrons, in blue, next to a pi bond, which I will make magenta, and so we could draw a resonance structure. So I could take the electrons in blue, move them into here, too many bonds to this carbon, so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here, the nitrogen's bonded to a hydrogen, the electrons in blue moved in to form a pi bond, and the electrons in magenta, moved off, onto this carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. All right, so the steric number will be equal to number of sigma bonds: So there's one sigma bond, in our double-bond, one of them is a sigma, and one of them is a pi; so I'm saying that's our sigma bond. So our steric number is equal to three, plus the number of lone pairs of electrons, now zero. So that's a steric number of three, which implies three hybrid orbitals, which says SP two hybridization. And since we know that, that lone pair is de-localized, it's going to occupy a P orbital, and so therefore this nitrogen is SP two hybridized, because we know SP two hybridization has three SP two hybrid orbitals, and one P orbital. So that lone pair in blue is actually de-localized; it's occupying a P orbital, and so let's go ahead and draw that, down here, so let's say this is the nitrogen, and you're looking at it, at a bit of an angle. If that nitrogen is SP two hybridized, that nitrogen has a P orbital, so we can go ahead and draw in a P orbital, on that nitrogen. And so, the electrons in blue, since those electrons"
    },
    {
        "Q": "\n@0:45 is the structure on the left trigonal pyramidal because of the lone pair? is it still a resonance structure with the right given the different pyramidal/planar structures of Nitrogen?",
        "A": "If it is not conjugated, the N atom with a lone pair is trigonal pyramidal. If the N atom is conjugated, it is sp\u00c2\u00b2 hybridized and therefore trigonal planar.",
        "video_name": "kQCS1AhAnMI",
        "timestamps": [
            45
        ],
        "3min_transcript": "Voiceover: Let's look at the amide, or \"amid\" functional group, and let's start by assigning a steric number to this nitrogen. So, the steric number is equal to the number of sigma bonds; so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds. Plus number of lone pairs of electrons, so there's one lone pair of electrons on that nitrogen, so I'll go ahead and highlight them there. So three plus one gives us a total of four for the steric number, which means four hybrid orbitals, which implies SP three hybridization for that nitrogen, and from earlier videos, you know that SP three hybridization means, a trigonal, pyramidal, geometry for that nitrogen. And so, that's one way of looking at this function group, and that lone pair of electrons being localized to that nitrogen; however, now that we know resonance structures, we know that, that lone pair of electrons is not localized to that nitrogen; it's de-localized in resonance. So we could take that lone pair of electrons, force us to push some pi electrons off, onto this oxygen, so let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative one formal charge, and there'd be a double-bond between this carbon and this nitrogen, so let's go ahead and draw in everything. This nitrogen how has a plus one formal charge, and we can go ahead and complete our resonance bracket here. So let's follow those electrons along, the electrons in magenta, the lone pair of electrons moved in here to form our pi bond, and the pi electrons over here in blue, came off onto the oxygen, to give the oxygen a negative one formal charge. All right, let's now calculate a steric number for this nitrogen, in our second resonance structure. So, steric number is equal to number of sigma bonds: So here's a sigma bond, here's a sigma bond, and our and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen, zero lone pairs of electrons, so three plus zero gives us a steric number of three; that implies three hybrid orbitals, which means SP two hybridization, and a trigonal planar geometry around that nitrogen, so a planar geometry. And, here we've shown our electrons being de-localized, so the lone pair of electrons are de-localized due to resonance, and so, experimental studies have shown that the amide function group is planar, so these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen; they are actually de-localized. And so, that implies this nitrogen is SP two hybridized, and has a P orbital, and that allow that lone pair of electrons in magenta to be de-localized. And so, here's a situation where drawing"
    },
    {
        "Q": "\nHow can p orbital from N (4:44) create pi bond if it already has 2 electrons? Confused...",
        "A": "What makes you think that would stop it from being able to form a pi bond..? Something to keep in mind in aromatic molecules is that they technically are not double bonds at all, every atom in the ring is sharing the 6 pi electrons because that makes it aromatic, and being aromatic is VERY favourable energetically.",
        "video_name": "kQCS1AhAnMI",
        "timestamps": [
            284
        ],
        "3min_transcript": "actually happening: That lone pair is participating in resonance, which makes this nitrogen SP two hybridized, so it has a P orbital. All right, let's look at this example down here, and let's look first at this left side of the molecule, and so we can see our amide functional group, and if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually de-localized, so this lone pair is participating in resonance. And so, that affects the geometry, and how you think about the hybridization of this nitrogen, here. So, the electrons in magenta are de-localized because they participate in resonance, and if I think about, let's make this a different color here; let's make these electrons in here blue, so the electrons in blue they can't participate in resonance; that lone pair of electrons in blue is localized to that nitrogen. And so, this is why you can think about an amide being different from an amine, in terms of functional group, and in terms of how they react and how they behave. If we look at another example, so this molecule right here, and we assume the lone pair of electrons on that nitrogen is localized to that nitrogen, let's go ahead and calculate the steric number. So the steric number'd be equal to sigma bonds, so that's one, two, and three; so three sigma bonds. Plus lone pairs of electrons, there's one lone pair of electrons on that nitrogen, so three plus one is four; so four hybrid orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone pair of electrons is not localized to that nitrogen; that lone pair of electrons because we have this pattern here. All right, so this pattern of a lone pair of electrons, in blue, next to a pi bond, which I will make magenta, and so we could draw a resonance structure. So I could take the electrons in blue, move them into here, too many bonds to this carbon, so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here, the nitrogen's bonded to a hydrogen, the electrons in blue moved in to form a pi bond, and the electrons in magenta, moved off, onto this carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. All right, so the steric number will be equal to number of sigma bonds: So there's one sigma bond,"
    },
    {
        "Q": "12:25 a willing mother could be one going for an abortion? i mean....instead of killing the embryo,doctors could extract it and utilise the stem cells?? just saying....: )\n",
        "A": "Actually no, embryonic stem cells are derived from the surplus embryos not used or not suitable for use from in-vitro fertilization. While a religious person might still consider this immoral, it is not technically the same thing as an abortion.",
        "video_name": "-yCIMk1x0Pk",
        "timestamps": [
            745
        ],
        "3min_transcript": "That's another word that you might hear. Let me write that down, too: plasticity. And the word essentially comes from, you know, like a plastic can turn into anything else. When we say that something has plasticity, we're talking about its potential to turn into a lot of different things. So the theory is, and there's already some trials that seem to substantiate this, especially in some lower organisms, that, look, if you have some damage at some point in your body-- let me draw a nerve cell. Let me say I have a-- I won't go into the actual mechanics of a nerve cell, but let's say that we have some damage at some point on a nerve cell right there, and because of that, someone is paralyzed or there's some nerve dysfunction. We're dealing with multiple sclerosis or who knows what. The idea is, look, we have these cell here that could turn into anything, and we're just really understanding how It really has to look at its environment and say, hey, what are the guys around me doing, and maybe that's what helps dictate what it does. But the idea is you take these things that could turn to anything and you put them where the damage is, you layer them where the damage is, and then they can turn into the cell that they need to turn into. So in this case, they would turn into nerve cells. They would turn to nerve cells and repair the damage and maybe cure the paralysis for that individual. So it's a huge, exciting area of research, and you could even, in theory, grow new organs. If someone needs a kidney transplant or a heart transplant, maybe in the future, we could take a colony of these embryonic stem cells. Maybe we can put them in some type of other creature, or who knows what, and we can turn it into a replacement heart or a So there's a huge amount of excitement about what these can do. I mean, they could cure a lot of formerly uncurable diseases might otherwise die. But obviously, there's a debate here. And the debate all revolves around the issue of if you were to go in here and try to extract one of these cells, you're going to kill this embryo. You're going to kill this developing embryo, and that developing embryo had the potential to become a human being. It's a potential that obviously has to be in the right environment, and it has to have a willing mother and all of the rest, but it does have the potential. And so for those, especially, I think, in the pro-life camp, who say, hey, anything that has a potential to be a human being, that is life and it should not be killed. So people on that side of the camp, they're against the destroying of this embryo. I'm not making this video to take either side to that argument, but it's a potential to turn to a human being."
    },
    {
        "Q": "at 00:27 seconds, you say that only one sperm can get into the egg. If so, how would a couple get twins?\n",
        "A": "Sumangal is right concerning monozygotic or identical twins. In the case of fraternal (dizygotic) twins, the mother produces 2 egg cells that are then fertilized by 2 separate sperm. Each develops in its own amniotic sac with its own placenta (identical twins share a placenta but usually have separate sacs depending on when they divided) and is no more or less identical than any other pair of siblings born at different times.",
        "video_name": "-yCIMk1x0Pk",
        "timestamps": [
            27
        ],
        "3min_transcript": "Where we left off after the meiosis videos is that we had two gametes. We had a sperm and an egg. Let me draw the sperm. So you had the sperm and then you had an egg. Maybe I'll do the egg in a different color. That's the egg, and we all know how this story goes. The sperm fertilizes the egg. And a whole cascade of events start occurring. The walls of the egg then become impervious to other sperm so that only one sperm can get in, but that's not the focus of this video. The focus of this video is how this fertilized egg develops once it has become a zygote. So after it's fertilized, you remember from the meiosis videos that each of these were haploid, or that they had-- oh, I added an extra i there-- that they had half the contingency of the DNA. As soon as the sperm fertilizes this egg, now, all of a sudden, you have a diploid zygote. So now let me pick a nice color. So now you're going to have a diploid zygote that's going to have a 2N complement of the DNA material or kind of the full complement of what a normal cell in our human body would have. So this is diploid, and it's a zygote, which is just a fancy way of saying the fertilized egg. And it's now ready to essentially turn into an organism. So immediately after fertilization, this zygote starts experiencing cleavage. It's experiencing mitosis, that's the mechanism, but it doesn't increase a lot in size. So this one right here will then turn into-- it'll just split up via mitosis into two like that. And, of course, these are each 2N, and then those are going to split into four like that. as that first zygote, and it keeps splitting. And this mass of cells, we can start calling it, this right here, this is referred to as the morula. And actually, it comes from the word for mulberry because it looks like a mulberry. So actually, let me just kind of simplify things a little bit because we don't have to start here. So we start with a zygote. This is a fertilized egg. It just starts duplicating via mitosis, and you end up with a ball of cells. It's often going to be a power of two, because these cells, at least in the initial stages are all duplicating all at once, and then you have this morula. Now, once the morula gets to about 16 cells or so-- and we're talking about four or five days. This isn't an exact process-- they started differentiating a"
    },
    {
        "Q": "\nStarting at 6:34, why did he not use pythagorean theorem to describe V?",
        "A": "Good question. The reason he said it this way is because he was referring to vectors and not the magnitude (length) of the vectors. If you want the magnitude, then you are correct in saying that you would need the Pythagorean theorem.",
        "video_name": "2QjdcVTgTTA",
        "timestamps": [
            394
        ],
        "3min_transcript": "or sometimes it's called a caret character-- that tells us that it is a vector, but it is a unit vector. It has a magnitude of 1. And by definition, the vector j goes and has a magnitude of 1 in the positive y direction. So the y component of this vector, instead of saying it's 5 meters per second in the upwards direction or instead of saying that it's implicitly upwards because it's a vertical vector or it's a vertical component and it's positive, we can now be a little bit more specific about it. We can say that it is equal to 5 times j. Because you see, this magenta vector, it's going the exact same direction as j, it's just 5 times longer. I don't know if it's exactly 5 times. I'm just trying to estimate it right now. It's just 5 times longer. Now what's really cool about this is besides just being able to express the components as now able to do that-- which we did do, we're representing the components as explicit vectors-- we also know that the vector v is the sum of its components. If you start with this green vector right here and you add this vertical component right over here, you have head to tails. You get the blue vector. And so we can actually use the components to represent the vector itself. We don't always have to draw it like this. So we can write that vector v is equal to-- let me write it this way-- it's equal to its x component vector plus the y component vector. And we can write that, the x component vector is 5 square roots of 3 times i. And then it's going to be plus the y component, the vertical component, which is 5j, which is 5 times j. vector in two dimensions by some combination of i's and j's or some scaled up combinations of i's and j's. And if you want to go into three dimensions, and you often will, especially as the physics class moves on through the year, you can introduce a vector in the positive z direction, depending on how you want to do it. Although z is normally up and down. But whatever the next dimension is, you can define a vector k that goes into that third dimension. Here I'll do it in a kind of unconventional way. I'll make k go in that direction. Although the standard convention when you do it in three dimensions is that k is the up and down dimension. But this by itself is already pretty neat because we can now represent any vector through its components and it's also going to make the math much easier."
    },
    {
        "Q": "\nso are we sapouse to add up all the small areas to get the total work amount? 7:14\nthanks again",
        "A": "Yep, on 7:14 you do.",
        "video_name": "M5uOIy-JTmo",
        "timestamps": [
            434
        ],
        "3min_transcript": "This is after removing each of the pebbles, so that our pressure and volume macro states are always well defined. But in state 2, we now have a pressure low and volume is high. The volume is high, you can just see that, because we kept pushing the piston up slowly, slowly, trying to maintain ourselves in equilibrium so our macrostates are always defined. And our pressure is lower just because we could have the same number of particles, but they're just going to bump into the walls a little bit less, because they have a little bit more room to move around. And that's all fair and dandy. So this describes the path of our system as it transitioned or as it experienced this process, which was a quasi-static process. Everything was defined at every point. Now we said that the work done at any given point by the system is the pressure times the change in volume. Now, how does that relate to here? Change in volume is just a certain distance along this x-axis. This is a change in volume. We started off at this volume, and let's say when we removed one pebble we got to this volume. Now, we want to multiply that times our pressure. Since we did it over such a small increment, and we're so close to equilibrium, we could assume that our pressure's is roughly constant over that period of time. So we could say that this is the pressure over that period of time. And so how much work we did, it's this pressure over here, times this volume, which is the area of this rectangle right there. And for any of you all who've seen my calculus videos, this should start looking a little bit familiar. And then what about when we could take our next pebble? Well now our pressure is a little bit lower. This is our new pressure. Our pressure is a little bit lower. And we multiply that times our new change in volume-- times this change in volume-- and we have that increment of work. And if you keep doing that, the amount of work we do is essentially the area of all of these rectangles as we remove each pebble. And now you might say, especially those of you who haven't watched my calculus videos, gee, you know, this might be getting close, but the area of these rectangles isn't exactly the area of this curve. It's a little inexact. And what I would say is, well if you're worried about that, what you should do is use smaller increments of volume. And if you want to have smaller changes in volume along each step, what you do is you remove even smaller pebbles. And this goes back to trying to get to that ideal quasi-static process. So if you did that of, eventually the delta V's would get smaller and smaller and smaller, and the rectangles would get thinner and thinner and thinner. You'd have to do it over more and more steps. But eventually you'll get to a point, if you assume really small changes in our delta V."
    },
    {
        "Q": "\nAt 0:59, aren't 3 sulfur atoms called sulfate instead of sulfide?",
        "A": "No, the sulfate ion has the formula SO4^2-",
        "video_name": "vVTwzjvWySs",
        "timestamps": [
            59
        ],
        "3min_transcript": "- [Instructor] So we have the formula for an ionic compound here, and the goal of this video is what do we call this thing? It clearly involves some cobalt and some sulfur, but how do we name it? Well, the convention is, is the first element to be listed is going to be our cation, and if we look at cobalt over here, we see that it is a D-block element and D-block elements are tricky because you don't know exactly how it will ionize. So we know that this is going to be our cation, it's going to be our positive ion, but we don't know what the charge on each of those cobalt is actually going to be. So now let's look at the anion, let's look at the sulfur, or as an anion, the sulfide. So let me underline that. And on the periodic table, we see sulfur is out here that in its group, it would want to gain two electrons in order to have a complete outer shell. So the sulfide anion will look like this. So it will have sulfur when it ionizes will have a two minus charge, just like oxygen, just like everything else in this group. It would want to gain one, two electrons so that its outer shell looks like that of a normal gas, looks like that of argon. We can use this as a clue to figure out what must be the charge on the cobalts because we have three of the sulfides. Each of the sulfides has a two minus charge, and we have three of them, so that's going to give us a six minus charge all in. And then the cobalt, we have two of them. And so these two cobalt have to offset They have to have a six plus charge. Well that means that each of them need to have a three plus charge. If each of these have a three plus charge and you have two of them, then you're gonna have six plus on the positive side and you're gonna have six minus from the sulfides. And the reason why this is useful for us is now we can name this. We would call this ionic compound Cobalt III, cobalt and you would write it with Roman numerals here, Cobalt III Sulfide, Cobalt III Sulfide. Now I know what you might be thinking. Hey, when we looked at other ionic compounds, I didn't have to write the charge of the cation there and the reason why the convention is to do it here, and I don't have to write in upper case there, so let me rewrite it as"
    },
    {
        "Q": "On the last example at 5:00, what is the difference between 5.60x10^4 and 5.6x10^4\nthat trailing 0 seems pretty important and I don't understand why.\n",
        "A": "I understand it simply as how detailed you want your number to be. 5.6 could actually be 5.61 or 5.64, depends on how many decimals you want, and that means that 5.60 is more accurate.",
        "video_name": "eMl2z3ezlrQ",
        "timestamps": [
            300
        ],
        "3min_transcript": "your best assumption is probably that they just measured to the nearest thousand. That they didn't measure exactly the one and just happened to get exactly on 37,000. So if there's no decimal, let me write it this way-- it's ambiguous, which means that you're not sure what it means, it's not clear what it means. And you're probably safer assuming to not count it. If someone really does measure, if you were to really measure something to the exact one, then you should put a decimal at the end like that. And there is a notation for specifying. Let's say you do measure-- and let me do a different number. Let's say you do measure 56,000. And there is a notation for specifying that 6 definitely is the last significant digit. And sometimes you'll see a bar put over the 6, And that could be useful because maybe your last significant digit is this zero over here. Maybe you were able to measure to the hundreds with a reasonable level of precision. And so then you would write something like 56,000, but then you would put the bar above that zero, or the bar below that zero to say that that was the last significant digit. So if you saw something like this, you would say three significant digits. This isn't used so frequently. A better way to show that you've measured to three significant digits would be to write it in scientific notation. There's a whole video on that. But to write this in scientific notation, you could write this as 5.60 times 10 to the fourth power. Because if you multiply this times to the fourth, you would move this decimal over four spaces and get us to 56,000. So 5.60 times 10 to the fourth. on scientific notation. It will hopefully clarify things a little bit. But when you write a number in scientific notation, it makes it very clear about your precision and how many significant digits you're dealing with. So instead of doing this notation that's a little bit outdated-- I haven't seen it used much with these bars below or above the high significant digit, instead you could represent it with a decimal in scientific notation. And there it's very clear that you have three significant digits. So hopefully that helps you out. In the next couple of videos, we'll explore a little bit more why significant digits are important, especially when you do calculations with multiple measurements."
    },
    {
        "Q": "\nWhat does trans specify at 5:20",
        "A": "Trans is a type of isomerism. All it specifies is that it is a type of isomer. Just so you know - Isomers have the same molecular formulae but different structural formulae. Hope that helps.",
        "video_name": "z8h7QgevqjM",
        "timestamps": [
            320
        ],
        "3min_transcript": "This has a double bond right here. So it's hepten. If this was just an alkene, we would just called heptene, but we're not going to put this last e here, because this is the carboxylic acid. And to specify where that double bond is, we need to start numbering, and we start numbering at the carbonyl carbon. One, two, three, four, five, six, seven. So you could either name this 3 hepten, and I haven't finished it yet, I haven't put this final e over here. Or you could name it hept 3 ene, just like that. This is the more typical one that you would see, because it tells you we have a double bond, and it starts at the number three carbon, goes from the three to the four carbon. But this isn't just a regular alkene, this is a carboxylic acid. So instead of writing that final e, for an alkene, we write it as we have a carboxyl group right here, so this is 3 And we are done. Actually if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of a range like this. Let me draw it like this. They both have other hydrogens off there that we didn't draw, they're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons, just like this. This one has a hydrogen popping up like that; that one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. And then this one over here-- I'll do it in green-- has this other functional group, has these three carbons. We can call that R prime. And if you look at it this way, the functional groups are on opposite sides of the double bond. So if you wanted to, you could also call this trans 3 heptenoic acid. And this will specify that these guys are on opposite ends. But this is only if you're assuming that I drew it in the actual three dimensional configuration in some way. Anyway, hopefully you found that useful."
    },
    {
        "Q": "At 4:37, Why did we put the hydrogen molecules opposite to each other? Why not in the same direction?\n",
        "A": "NO. This is an equilibrium reaction. Therefore, the most stable molecules will form. Trans is much better then cis....steric repulsion (eclipsed atoms).",
        "video_name": "z8h7QgevqjM",
        "timestamps": [
            277
        ],
        "3min_transcript": "This has a double bond right here. So it's hepten. If this was just an alkene, we would just called heptene, but we're not going to put this last e here, because this is the carboxylic acid. And to specify where that double bond is, we need to start numbering, and we start numbering at the carbonyl carbon. One, two, three, four, five, six, seven. So you could either name this 3 hepten, and I haven't finished it yet, I haven't put this final e over here. Or you could name it hept 3 ene, just like that. This is the more typical one that you would see, because it tells you we have a double bond, and it starts at the number three carbon, goes from the three to the four carbon. But this isn't just a regular alkene, this is a carboxylic acid. So instead of writing that final e, for an alkene, we write it as we have a carboxyl group right here, so this is 3 And we are done. Actually if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of a range like this. Let me draw it like this. They both have other hydrogens off there that we didn't draw, they're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons, just like this. This one has a hydrogen popping up like that; that one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. And then this one over here-- I'll do it in green-- has this other functional group, has these three carbons. We can call that R prime. And if you look at it this way, the functional groups are on opposite sides of the double bond. So if you wanted to, you could also call this trans 3 heptenoic acid. And this will specify that these guys are on opposite ends. But this is only if you're assuming that I drew it in the actual three dimensional configuration in some way. Anyway, hopefully you found that useful."
    },
    {
        "Q": "\nAt 6:28 Sal writes ||a|| = 5. What does '|| ||' mean?",
        "A": "It means the magnitude, or length, of the vector. This is the hypotenuse length formed by the vector components.",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            388
        ],
        "3min_transcript": "this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end."
    },
    {
        "Q": "At 3:40, Sal mentions that the formula for vectors was a + b = c\nwhere 'a' and 'b' are the horizontal and vertical vectors. But when you add up vector 'a' (3) and 'b' (4) in this problem the don't equal 'c' (5).\n",
        "A": "and ply if it forms a right triangle else you use the formula (a^2+b^2+2ab cos{theta})^1/2 which will be equal to the sum of given two vectors",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            220
        ],
        "3min_transcript": "As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector \"vector X.\" Let's call this \"vector X.\" this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this"
    },
    {
        "Q": "\nat 3:00 cant we draw vector A from the tail of vector of vector B\nwill there be any difference",
        "A": "No. There will not be any difference. If you find the resultant vector with any head-tail arrangement, you ll still get the same resultant vector.",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            180
        ],
        "3min_transcript": "by the direction of the arrow. So it's going in that direction. Now let's say I have another vector. Let's call it vector B. Let's call it vector B. It looks like this. Now what I wanna do in this video is think about what happens when I add vector A to vector B. So there's a couple things to think about when you visually depict vectors. The important thing is, for example, for vector A, that you get the length right and you get the direction right. Where you actually draw it doesn't matter. So this could be vector A. This could also be vector A. Notice, it has the same length and it has the same direction. This is also vector A. I could draw vector A up here. It does not matter. I could draw vector A up there. I could draw vector B. I could draw vector B over here. It's still vector B. It still has the same magnitude and direction. Notice, we're not saying that its tail has to start at the same place that vector A's tail starts at. I could draw vector B over here. So I can always have the same vector but I can shift it around. As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector \"vector X.\" Let's call this \"vector X.\""
    },
    {
        "Q": "Starting after 6:00 with the new problem, why were the numbers used so specific?\n",
        "A": "Well, for an example, if you take out your calculator and take the sin of a random number, it ll most likely be a non-terminating number and will look like something just vomited random numbers out. I think he used such a specific angle on purpose to get his answer close to 3 in this problem. A nice, whole number is better than something like 3.58912385423168.",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            360
        ],
        "3min_transcript": "this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end."
    },
    {
        "Q": "\nAt 5:25, Can the same fusion process that causes Hydrogen to turn into Deuterium and then Helium continue past Helium and make all the elements given enough gravity?",
        "A": "Yes, the next few videos explain it pretty well.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            325
        ],
        "3min_transcript": "to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4."
    },
    {
        "Q": "\nIn 1:14 you have said that the atoms condense. But condensation happens only when temperature reduces. But you have said temperature increases. Aren't they contradicting?",
        "A": "Not condensation as the moisture that appears on cool objects. Condensing as in the atoms fill a smaller volume, getting denser. This builds up pressure, which builds temperature.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            74
        ],
        "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring"
    },
    {
        "Q": "At around 5:37, what is a nucleon? Sorry if I missed it in another video...\n",
        "A": "A nucleon is a particle that exists in the center of an atom, like a proton or neutron.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            337
        ],
        "3min_transcript": "results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite"
    },
    {
        "Q": "At 3:52, Sal says that one of the protons degrades into a neutron - how does that happen?\n",
        "A": "The instability essentially forces one of the constituent up quarks to decay into a down quark, resulting in a proton becoming a neutron.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            232
        ],
        "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling."
    },
    {
        "Q": "At 5:56, why it wouldn't form tritium from deuterium?\n",
        "A": "It can if a free neutron fuses with deuterium. But free neutrons aren t readily available at the early stages of stellar fusion, and tritium is unstable and will decay to Helium-3.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            356
        ],
        "3min_transcript": "results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite"
    },
    {
        "Q": "@ 0:18 sal is talking about gravity...well, what type of gravity is present in the stars...?\nand how these hydrogen atoms are attracted to each other due to gravity...?\n",
        "A": "There is only one type of gravity Hydrogen atoms are attracted to each other because they have mass All mass attracts other mass. That s what gravity is.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            18
        ],
        "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring"
    },
    {
        "Q": "\nAt 2:20, why don't the positivily charged nucleuses want to go any where near each other?",
        "A": "The positively charged nuclei don t want to go anywhere near each other because both of them a positively charged, and just like a magnet, positives don t attract each other.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring"
    },
    {
        "Q": "\nat 3:40 sal says when the fusion occurs, the atoms that fuse weigh a little bit less. One of the atoms turns into a neutron. Where does that extra weight go? Is it turned into energy or another substance? If it's energy it needs to have weight and that's not possible.",
        "A": "Well, that doesn t exactly happen, You must have got confused between Beta decay and fusion, where the proton emits a positron and turns into a neutron. But that isn t exactly the case over here. The fusion occurs when two hydrogen atoms fuse together, and no weight is lost either.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            220
        ],
        "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling."
    },
    {
        "Q": "\nat 0:45 why does cloud get dense",
        "A": "because of the forces of attraction acting over forces of repulsion in between the two nucleus and this works because the size is massive.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            45
        ],
        "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring"
    },
    {
        "Q": "\nAt 0:03, it is said that we consider that two things have the same charge. What if they are not of the same charge? Will like charges not repel each other in that case? And opposite charges attract each other?",
        "A": "Like charges always repel. Opposites always attract.",
        "video_name": "2GQTfpDE9DQ",
        "timestamps": [
            3
        ],
        "3min_transcript": "- [Voiceover] So we've already started to familiarize ourselves with the notion of charge. We've seen that if two things have the same charge, so they're either both positive, or they are both negative, then they are going to repel each other. So in either of these cases these things are going to repel each other. But if they have different charges, they are going to attract each other. So if I have a positive and I have a negative they are going to attract each other. This charge is a property of matter that we've started to observe. We've started to observe of how these different charges, this framework that we've created, how these things start to interact with each other. So these things are going to, these two things are going to attract each other. But the question is, what causes, how can we predict how strong the force of attraction or repulsion is going to be between charged particles? And this was a question people have noticed, I guess what you could call electrostatics, for a large swathe of recorded human history. But it wasn't until the 16 hundreds and especially the 17 hundreds, as something that they could manipulate and even start to predict in a kind of serious, mathematical, scientific way. And it wasn't until 1785, and there were many that came before Coulomb, but in 1785 Coulomb formally published what is known as Coulomb's law. And the purpose of Coulomb's law, Coulomb's law, is to predict what is going to be the force of the electrostatic force of attraction or repulsion between two forces. And so in Coulomb's law, what it states is is if I have two charges, so let me, let's say this charge right over here, and I'm gonna make it in white, because it could be positive or negative, but I'll just make it q one, it has some charge. And then I have in Coulombs. and then another charge q two right over here. Another charge, q two. And then I have the distance between them being r. So the distance between these two charges is going to be r. that the magnitude of the force, so it could be a repulsive force or it could be an attractive force, which would tell us the direction of the force between the two charges, but the magnitude of the force, which I'll just write it as F, the magnitude of the electrostatic force, I'll write this sub e here, this subscript e for electrostatic. Coulomb stated, well this is going to be, and he tested this, he didn't just kind of guess this. People actually were assuming that it had something to do with the products of the magnitude of the charges and that as the particles got further and further away the electrostatic force dissipated. But he was able to actually measure this and feel really good about stating this law. Saying that the magnitude of the electrostatic force is proportional, is proportional, to the product of the magnitudes of the charges. So I could write this as q one times q two, and I could take the absolute value of each, which is the same thing as just"
    },
    {
        "Q": "At 11:13, why does the final answer have two significant figures when the measurements only have 1 significant figure? I thought that, when multiplying or dividing significant figures, the result can only have as much significant figures as the measurement with the least amount of significant figures (which in this case is 1).\n",
        "A": "sal often makes mistakes with sig figs",
        "video_name": "2GQTfpDE9DQ",
        "timestamps": [
            673
        ],
        "3min_transcript": "so that negative is going to go away. All of that over, all of that over and we're in kind of the home stretch right over here, 0.5 meters squared. 0.5 meters squared. And so, let's just do a little bit of the math here. So first of all, let's look at the units. So we have Coulomb squared here, then we're going to have Coulombs times Coulombs there that's Coulombs squared divided by Coulombs squared that's going to cancel with that and that. You have meters squared here, and actually let me just write it out, so the numerator, in the numerator, we are going to have so if we just say nine times five times, when we take the absolute value, it's just going to be one. So nine times five is going to be, nine times five times negative... five times negative one is negative five, but the absolute value there, so it's just going to be five times nine. So it's going to be 45 times 10 to the nine, minus three, minus one. so that's going to be 10 to the fifth, 10 to the fifth, the Coulombs already cancelled out, and we're going to have Newton meter squared over, over 0.25 meters squared. These cancel. And so we are left with, well if you divide by 0.25, that's the same thing as dividing by 1/4, which is the same thing as multiplying by four. So if you multiply this times four, 45 times four is 160 plus 20 is equal to 180 times 10 to the fifth Newtons. And if we wanted to write it in scientific notation, well we could divide this by, we could divide this by 100 and then multiply this by 100 and so you could write this as 1.80 times one point... and actually I don't wanna make it look like I have more significant digits than I really have. 1.8 times times 10 to the seventh units, I just divided this by 100 and I multiplied this by 100. And we're done. This is the magnitude of the electrostatic force between those two particles. And it looks like it's fairly significant, and this is actually a good amount, and that's because this is actually a good amount of charge, a lot of charge. Especially at this distance right over here. And the next thing we have to think about, well if we want not just the magnitude, we also want the direction, well, they're different charges. So this is going to be an attractive force. This is going to be an attractive force on each of them acting at 1.8 times ten to the seventh Newtons. If they were the same charge, it would be a repulsive force, or they would repel each other with this force. But we're done."
    },
    {
        "Q": "In 11:04s why the [10^-4*10^-1] do not have to take off the minus sign? Isn't it a absolute value?\n",
        "A": "There is a modulus sign there so -tive ans will not be considered\u00f0\u009f\u0098\u008a",
        "video_name": "2GQTfpDE9DQ",
        "timestamps": [
            664
        ],
        "3min_transcript": "so that negative is going to go away. All of that over, all of that over and we're in kind of the home stretch right over here, 0.5 meters squared. 0.5 meters squared. And so, let's just do a little bit of the math here. So first of all, let's look at the units. So we have Coulomb squared here, then we're going to have Coulombs times Coulombs there that's Coulombs squared divided by Coulombs squared that's going to cancel with that and that. You have meters squared here, and actually let me just write it out, so the numerator, in the numerator, we are going to have so if we just say nine times five times, when we take the absolute value, it's just going to be one. So nine times five is going to be, nine times five times negative... five times negative one is negative five, but the absolute value there, so it's just going to be five times nine. So it's going to be 45 times 10 to the nine, minus three, minus one. so that's going to be 10 to the fifth, 10 to the fifth, the Coulombs already cancelled out, and we're going to have Newton meter squared over, over 0.25 meters squared. These cancel. And so we are left with, well if you divide by 0.25, that's the same thing as dividing by 1/4, which is the same thing as multiplying by four. So if you multiply this times four, 45 times four is 160 plus 20 is equal to 180 times 10 to the fifth Newtons. And if we wanted to write it in scientific notation, well we could divide this by, we could divide this by 100 and then multiply this by 100 and so you could write this as 1.80 times one point... and actually I don't wanna make it look like I have more significant digits than I really have. 1.8 times times 10 to the seventh units, I just divided this by 100 and I multiplied this by 100. And we're done. This is the magnitude of the electrostatic force between those two particles. And it looks like it's fairly significant, and this is actually a good amount, and that's because this is actually a good amount of charge, a lot of charge. Especially at this distance right over here. And the next thing we have to think about, well if we want not just the magnitude, we also want the direction, well, they're different charges. So this is going to be an attractive force. This is going to be an attractive force on each of them acting at 1.8 times ten to the seventh Newtons. If they were the same charge, it would be a repulsive force, or they would repel each other with this force. But we're done."
    },
    {
        "Q": "\nAt 8:28: The Coulomb's Law equation is used to calculate the electrostatic force (in Newtons) between two charges. But is there an independent way to measure this force (and thus verify that Coulomb's equation is correct)?",
        "A": "You could think about lot of ways of measuring this, you could set the two charges (known charge) at a known distance and use a spring to keep them in place, you could extrapolate the force through the hook law, by knowing the constant or the spring. Or,for not depending on hooke law you could put your guessed Force(given by the Coulomb\u00c2\u00b4s equation) in the way of weight, and see if a spring , identical to the ones in the beginning , elongated the same, if it does the formula is correct.",
        "video_name": "2GQTfpDE9DQ",
        "timestamps": [
            508
        ],
        "3min_transcript": "let's that this distance right here is 0.5 meters. So given that, let's figure out what the what the electrostatic force between these two are going to be. it's going to be an attractive force because they have different signs. And that was actually part of Coulomb's law. This is the magnitude of the force, if these have different signs, it's attractive, if they have the same sign then they are going to repel each other. And I know what you're saying, \"Well in order to actually calculate it, \"I need to know what K is.\" What is this electrostatic constant? What is this electrostatic constant going to actually be? And so you can measure that with a lot of precision, and we have kind of modern numbers on it, but the electrostatic constant, especially for the sake of this problem, I mean if we were to get really precise it's 8.987551, we could keep gone on and on times 10 to the ninth. But for the sake of our little example here, where we really only have one significant digit for each of these. it'll make the math a little bit easier, I won't have to get a calculator out, let's just say it's approximately nine times 10 to the ninth. Nine times 10 to the ninth. Nine times, actually let me make sure it says approximately, because I am approximating here, nine times 10 to the ninth. And what are the units going to be? Well in the numerator here, where I multiply Coulombs times Coulombs, I'm going to get Coulombs squared. This right over here is going to give me, that's gonna give me Coulombs squared. And this down over here is going to give me meters squared. This is going to give me meters squared. And what I want is to get rid of the Coulombs and the meters and end up with just the Newtons. And so the units here are actually, the units here are Newtons. Newton and then meters squared, and that cancels out with the meters squared in the denominator. Newton meter squared over Coulomb squared. Over, over Coulomb squared. Over, over Coulomb squared. So, these meter squared will cancel those. Those Coulomb squared in the denomin... over here will cancel with those, and you'll be just left with Newtons. But let's actually do that. Let's apply it to this example. I encourage you to pause the video and apply this information to Coulomb's law and figure out what the electrostatic force between these two particles is going to be. So I'm assuming you've had your go at it. So it is going to be, and this is really just applying the formula. It's going to be nine times 10 to the ninth, nine times 10 to the ninth, and I'll write the units here, Newtons meter squared over Coulomb squared. And then q one times q two, so this is going to be, let's see, this is going to be, actually let me just write it all out for this first this first time. So it's going to be times five times ten to the negative three Coulombs. Times, times negative one. Time ten to the negative one Coulombs"
    },
    {
        "Q": "is there any mistakes on this video?\nI guess after meiosis I, you duplicate one diploid cell to two diploid cell; so that for meiosis II, each of two diploid cell will be \"simplified\" to four haploid cell? (4:14)\n",
        "A": "Yes, two division events occur. You start with a replication of DNA, then divide the cells so you have 2 diploid cells, then you divide each of those cells to get 4 haploid cells.",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            254
        ],
        "3min_transcript": "that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again."
    },
    {
        "Q": "\nAt 4:22, it's noted that the cells at the end of meiosis II become gametes. Why can't the cells produced at the end of meiosis I become gametes, if they're also haploid cells? And if meiosis isn't a cycle, how can so many gametes be produced--are there just a ton of diploid cells undergoing meiosis?",
        "A": "Prior to a cell division of a diploid cell, chromosomes are duplicated and condensed. In meiosis I homologous chromosomes swap chunks of genes with each other (called crossing over), then they are randomly assorted into two cells. Meioses II is responsible for separating the sister chromatids of a chromosome (of the two cells formed in meiosis I) yielding two more cells that are also haploid, for a total of four haploid cells.",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            262
        ],
        "3min_transcript": "that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again."
    },
    {
        "Q": "\naround 2:40, sal says that each of the two cells after the first division has 23 chromosomes. I thought that during S1 the parent cell duplicates its chromosomes to 46 pairs, 92 total, while still being 2n, and then the homologous pairs split and the two cells then each has 46 chromosomes 2n. Then after meiosis 2 each of the four daughter cells has 23 chromosomes. Is this correct?",
        "A": "The chromosomes are still considered 1 because they are connected. The sister chromatids are just split apart.",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            160
        ],
        "3min_transcript": "are just like this cell was, it can go through interphase again. It grows and it can replicate its DNA and centrosomes and grow some more then each of these can go through mitosis again. And this is actually how most of the cells in your body grow. This is how you turn from a single cell organism into you, or for the most part, into you. So that is mitosis. It's a cycle. After each of these things go through mitosis, they can then go through the entire cell cycle again. Let me write this a little bit neater. Mitosis, that s was a little bit hard to read. Now what happens in meiosis? What happens in meiosis? I'll do that over here. In meiosis, something slightly different happens and it happens in two phases. You will start with a cell that has a diploid number of chromosomes. that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number."
    },
    {
        "Q": "at 0:02 sal sayes arbitrary amino acids. what does that mean?\n",
        "A": "Any of a class of organic compounds that contains at least one amino group, \u00e2\u0080\u0093NH 2, and one carboxyl group, \u00e2\u0080\u0093COOH: the alpha-amino acids, RCH(NH 2)COOH, are the building blocks from which proteins are constructed. Your welcome!",
        "video_name": "nv2kfBFkv4s",
        "timestamps": [
            2
        ],
        "3min_transcript": "- [Voiceover] So I've got two arbitrary amino acids here. We recognize the telltale signs of an amino acid. We have an amino group right over here that gives us the amino and amino acid. We have a carboxyl group right over here. This is the acid part of an amino acid. And in between we have a carbon, and we call that the alpha carbon. And that alpha carbon is gonna be bonded to a hydrogen and some type of a side chain, and we're just gonna call this side chain R1, and then we're gonna call this side chain R2. And what we're gonna concern ourselves with in this video is, how do you take two amino acids and form a peptide out of them? And just as a reminder, a peptide is nothing more... than a chain of amino acids. And so, how do you take these two amino acids and form a dipeptide like this? That would be the smallest possible peptide, but then you could keep adding amino acids and form polypeptides. And a very high-level overview of this reaction is that this nitrogen uses its lone pair to form a bond with this carbonyl carbon right over here. So this lone pair goes to this carbonyl carbon, forms a bond, and then this hydrogen, this hydrogen, and this oxygen could be used net net to form a water molecule... that's let go from both of these amino acids. So this reaction, you end up with the nitrogen being attached to this carbon, and a release of a water molecule. And because you have the release of this water molecule, this type of reaction, and we've seen it many other times with other types of molecules, we call this a condensation reaction, or a dehydration synthesis. So condensation... or dehydration synthesis. We saw this type of reaction when we were putting glucoses together, when we were forming carbohydrates. Dehydration synthesis. But whenever I see a reaction like this, it's somewhat satisfying to just be able to do the counting and say, \"All right, this is gonna bond \"with that, we see the bond right over there, \"and I'm gonna let go of an oxygen and two hydrogens, \"which net net equals H2O, equals a water molecule.\" But how can we actually imagine this happening? Can we push the electrons around? Can we do a little bit of high-level organic chemistry to think about how this happens? And that's what I wanna do here. I'm not gonna do a formal reaction mechanism, but really get a sense of what's going on. Well, nitrogen, as we said, has got this lone pair, it's electronegative. And this carbon right over here, it's attached to two oxygens, oxygens are more electronegative. The oxygens might hog those electrons."
    },
    {
        "Q": "\nAt about 13:35 when he says it's intuitive for Heat to be represented as Q, because heat does not start with Q, was he being sarcastic, or am I missing some reason as to why Q is heat?",
        "A": "Sal is being sarcastic about Q being an intuitive representation of heat, since Q is completely unrelated to the word heat. However, he is serious about the use of Q to represent heat, as an accepted method.",
        "video_name": "Xb05CaG7TsQ",
        "timestamps": [
            815
        ],
        "3min_transcript": "We're going to assume we're dealing with an ideal gas. And even better, we're going to assume we're dealing with a monoatomic ideal gas. And maybe this is just helium, or neon. One of the ideal gases. They don't want to bond with each other. They don't form molecules with each other. Let's just assume that they're not. They're just individual atoms. And in that case, the internal energy, we really can simplify to it being the kinetic energy, if we ignore all of these other things. But it's important to realize, internal energy is everything. It's all of the energy inside of a system. If you said, what's the energy of the system? Its internal energy. So the first law of thermodynamics says that energy cannot be created or destroyed, only transformed. So let's say that internal energy is changing. So I have this system, and someone tells me, look, the So delta U, that's just a capital delta that says, what is the change an internal energy? It's saying, look, if your internal energy is changing, your system is either having something done to it, or it's doing something to someone else. Some energy is being transferred to it or away from it. So, how do we write that? Well the first law of thermodynamics, or even the definition of internal energy, says that a change in internal energy is equal to heat added to the system-- and once again a very intuitive letter for heat, because heat does not start with Q, but the convention is to use Q for heat. The letter h is reserved for enthalpy, which is a very, very, very similar concept to heat. We'll talk about that maybe in the next video. It's equal to the heat added to the system, minus the work done by the system. Sometimes it's written like this. Sometimes it's written that the change in internal energy is equal to the heat added to the system, plus the work done on the system. And this might be very confusing, but you should just always-- and we'll really kind of look at this 100 different And actually this is a capital U. Let me make sure that I write that as a capital U. But we're going to do it 100 different ways. But if you think about it, if I'm doing work I lose energy. I've transferred the energy to someone else. So this is doing work. Likewise, if someone is giving me heat that is increasing my energy, at least to me these are reasonably intuitive Now if you see this, you say, OK, if my energy is going up, if this is a positive thing, I either have to have this go up, or work is being done to me."
    },
    {
        "Q": "At 2:50 why does he round the mass of the elements? Isn't it more accurate to keep it unrounded?\n",
        "A": "He does that to make the examples easier to follow. In real calculations you cannot do that. It is standard practice to use 4 significant digits for your atomic masses, unless your problem requires using more.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            170
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n2:15 , 4:40 , 8:42 ! Why is the Charge on each of the individual Capacitors in series is SAME as the charge on the Equivalent Capacitor ?\nFull video is on this Idea!!\nBut, Why ?",
        "A": "Picture each conductor with one end at a capacitor plate, and the other end at another plate. Since this conductor needs to be with total charge 0, then if the plate at the right has charge +Q, the other must have charge -Q. Thus, the charge of each pair of plates is the same.",
        "video_name": "-MaD9Ycy3a4",
        "timestamps": [
            135,
            280,
            522
        ],
        "3min_transcript": "Having to deal with a single capacitor hooked up to a battery isn't all that difficult, but when you have multiple capacitors, people typically get much, much more confused. There's all kinds of different ways to hook up multiple capacitors. But if capacitors are connected one after the other in this way, we call them capacitors hooked up in series. So say you were taking a test, and on the test it asked you to find the charge on the leftmost capacitor. What some people might try to do is this. Since capacitance is the charge divided by the voltage, they might plug in the capacitance of the leftmost capacitor, which is 4 farads, plug in the voltage of the battery, which is 9 volts. Solving for the charge, they'd get that the leftmost capacitor stores 36 coulombs, which is totally the wrong answer. To try and figure out why and to figure out how to properly deal with this type of scenario, let's look at what's actually going on in this example. will start to flow from the right side of capacitor 3, which makes a negative charge get deposited on the left side of capacitor 1. This makes a negative charge flow from the right side of capacitor 1 on to the left side of capacitor 2. And that makes a negative charge flow from the right side of capacitor 2 on to the left side of capacitor 3. Charges will continue doing this. And it's important to note something here. Because of the way the charging process works, all of the capacitors here must have the same amount of charge stored on them. It's got to be that way. Looking at how these capacitors charge up, there's just nowhere else for the charge to go but on to the next capacitor in the line. This is actually good news. This means that for capacitors in series, the charge stored on every capacitor is going to be the same. So if you find the charge on one of the capacitors, you've found the charge on all of the capacitors. is going to be? Well, there's a trick we can use when dealing with situations like this. We can imagine replacing our three capacitors with just a single equivalent capacitor. If we choose the right value for this single capacitor, then it will store the same amount of charge as each of the three capacitors in series will. The reason this is useful is because we know how to deal with a single capacitor. We call this imaginary single capacitor that's replacing multiple capacitors the \"equivalent capacitor.\" It's called the equivalent capacitor because its effect on the circuit is, well, equivalent to the sum total effect that the individual capacitors have on the circuit. And it turns out that there's a handy formula that lets you determine the equivalent capacitance. The formula to find the equivalent capacitance of capacitors hooked up in series looks like this."
    },
    {
        "Q": "\nAt 6:32, why is the charge stored on each of the individual capacitors equal to the charge stored on the equivalent capacitor? Why aren't the charges divided between the four- like each one has 192/4 C of charge?",
        "A": "NO, remember that the Capacitance unit is F, not C, So basically you messed up, you should NOT sum like this, they have the same amount of Charge NOT Capacitance. So you add (1/48F) + (1/16F) + (1/96F) + (1/32F) = 0.125F, Then taking the reciprocal you get 8F which is the equivalent of CAPACITANCE. When you try to find the Voltage you do this ( 192/48 ) + ( 192/16 ) + ( 192/96 ) + ( 192/32 ) = 24v which is the same voltage of the battery. I Hope that helped!",
        "video_name": "-MaD9Ycy3a4",
        "timestamps": [
            392
        ],
        "3min_transcript": "This time, let's say you had four capacitors hooked up in series to a 24-volt battery. The arrangement of these capacitors looks a little different from the last example, but all of these capacitors are still in series because they're hooked up one right after the other. In other words, the charge has no choice but to flow directly from one capacitor straight to the next capacitor. So these capacitors are still considered to be in series. Let's try to figure out the charge that's going to be stored on the 16-farad capacitor. We'll use the same process as before. First we imagine replacing the four capacitors with a single equivalent capacitor. We'll use the formula to find the equivalent capacitance of capacitors in series. Plugging in our values, we find that 1 over the equivalent capacitance is going to equal 0.125. Be careful. We still have to take 1 over this value to get that the equivalent capacitance for this circuit Now that we know the equivalent capacitance, we can use the formula capacitance equals charge per voltage. We can plug in the value of the equivalent capacitance, 8 farads. And since we have a single capacitor now, the voltage across that capacitor is going to be the same as the voltage of the battery, which is 24 volts. So we find that our imaginary equivalent capacitor would store a charge of 192 coulombs. This means that the charge on each of the individual capacitors is also going to be 192 coulombs. And this gives us our answer, that the charge on the 16-farad capacitor is going to be 192 coulombs. In fact, we can go even further. Now that we know the charge on each capacitor, we can solve for the voltage that's going to exist across each of the individual capacitors. We'll again use the fact that capacitance is the charge per voltage. If we plug in the values for capacitor one, we'll plug in a capacitance of 32 farads. So we can solve for the voltage across capacitor 1, and we get 6 volts. If we were to do the same calculation for each of the other three capacitors, always being careful that we use their particular values, we'll get that the voltages across the capacitors are 2 volts across the 96-farad capacitor, 12 volts across the 16-fard capacitor, and 4 volts across the 48-farad capacitor. Now, the real reason I had us go through this is because I wanted to show you something neat. If you add up the voltages that exist across each of the capacitors, you'll get 24 volts, the same as the value of the battery. This is no coincidence. If you add up the voltages across the components in any single-loop circuit like this, the sum of the voltages is always going to equal the voltage of the battery. And this principle will actually let"
    },
    {
        "Q": "\nAt 6:19 why does Sal not put 3d10 instead of 4s2 and then 3d8? Does it make a difference which way you do it?",
        "A": "Yes, it makes a difference. 3d\u00c2\u00b9\u00e2\u0081\u00b0 means 10 electrons in the 3d orbitals. 4s\u00c2\u00b2 3d\u00e2\u0081\u00b8 means two electrons in a 4s orbital and 8 electrons in the 3d orbitals. Since the 4s orbital is lower in energy than the 3d orbitals, the 4s orbital gets filled up first. 3d\u00c2\u00b9\u00e2\u0081\u00b0 would represent an excited state of the atom.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            379
        ],
        "3min_transcript": "Well we're not going to get to f. But you could write f and g and h and keep going. What's going to happen is you're going to fill this one first, then you're going to fill this one, then that one, then this one, then this one. Let me actually draw it. So what you do is, these are the shells that exist, period. These are the shells that exist, in green. What I'm drawing now isn't the order that you fill them. This is just, they exist. So there is a 3d subshell. There's not a 3f subshell. There is a 4f subshell. Let me draw a line here, just so it becomes a little bit neater. And the way you fill them is you make these diagonals. So first you fill this s shell like that, then you fill this one like that. Then you do this diagonal down like that. Then you do this diagonal down like that. And then this diagonal down like that. six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition"
    },
    {
        "Q": "what does he mean by \"go back to fill the previous shell, so subtract 1\" at 2:27\n",
        "A": "when the shells are filled, there are certain spots, where are still no electrons, the d and f blocks fill up those spots(because thats closer to the nucleus).",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            147
        ],
        "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that"
    },
    {
        "Q": "For the element Zr, discussion starting around 7:50, energy level 5s (5s2) is filled before level 4d (4d2). Will the electrons \"peel\" in the opposite order that they're \"put on\"? or will 5s electrons \"peel\" before 4d electrons? for this element, which would be considered the valence electrons? Thanks.\n",
        "A": "The electrons fill in that order, but when they are used in bonding ( peeling? ), the electrons with the highest quantum number (period 5 before period 4 electrons) are used first. So the 5s2 will be used before the 4d electrons.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            470
        ],
        "3min_transcript": "Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10."
    },
    {
        "Q": "at 2:07 why do we fill 4s orbital 1st than 3d ?\nand if we do so why do we consider 4s as the valence shell and not 3d??\n",
        "A": "We put them in that order because that is the order they actually fill up in. The 4s is very much valence. A partially filled 3d is counted as valence electrons by some scientists but not by others. The reason for the disagreement has to do with the way that d electrons get involved in chemical reactions -- it is not as straightforward as what happens with the outermost s and p orbitals.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            127
        ],
        "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that"
    },
    {
        "Q": "How did Sal at 2:00 know the third shell was not full? Why does the formula 2(n)^2 determine the shells max capacity?\n",
        "A": "Sal was using the relative energy levels of the different orbitals to explain the order in which they are filled. The empty 4s orbital is of slightly lower energy than the empty 3d orbitals so is filled first. The formula you quote works simply because of the algebraic relationship between n and the magnetic quantum number.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            120
        ],
        "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that"
    },
    {
        "Q": "\nAt 8:01 Sal says you have to minus 1 from the period for d. Why is this?",
        "A": "It s just how it goes. The 3d orbitals happen to be about the same energy as the 4s orbital. Similar story for the 4d and 5s etc.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            481
        ],
        "3min_transcript": "Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10."
    },
    {
        "Q": "At 2:30, Sal says that in the d-block, the period is subtracted by 1. Why is this the case and how was it calculated?\n",
        "A": "Because the 4s2 orbital electrons are in a higher energy state than the 3d8 orbital electrons, so they are further from the nucleus. So instead of 4d8, it s 3d8 (-1, as in lower energy). The 4s2 orbital only fills with electrons before the 3d8 orbital",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that"
    },
    {
        "Q": "\nIs there any difference between \"3-phosphoglycerate\" (3:17), and Glycerate 3-phosphate (which is what I have in my textbook)?",
        "A": "no they are theoretically the same thing with just a slightly different structure (isomerization)",
        "video_name": "DnNqe8o0ehc",
        "timestamps": [
            197
        ],
        "3min_transcript": "but because it's a mouthful, the shorthand notation is R-U-B-P. Sometimes people might say Roo-B-P, or I guess you could even say Rube-P somehow, but each of these six Rube-P, or RuBPs, can then react with a carbon dioxide. So if I have six RuBPs, well, they're gonna react with six carbon dioxides, and so one way to think about it is, it's fixing the carbon in that carbon dioxide. It's taking this carbon that's part of this gaseous carbon dioxide, and fixing it as part of an organic molecule. Now, you might be tempted to say, well, it's gonna create six carbon molecules, but then those will immediately become 12 three carbon molecules. And notice, and it's important to keep doing this. Pause the video if you need to. You can make sure that the carbons are all accounted for. Well, we have six times five, so that's 30 carbons right over here, and here we have six times one carbon, so that's six carbons right over here. So if we wanna account for all of our carbons, we should have 36 carbons right over here, and we do. We have 12 three carbon molecules. This three carbon molecules, when we go into some detail here in the video on the Calvin cycle, it's called three phosphoglycerate, but that's not what the focus is on this video. The focus of this video is the enzyme that actually does the fixing of the carbon along with the RuBP. And that enzyme, that character, the character with the quirks that we're going to talk about, the shorthand, its name, you could call it ribulose one, five, bisphosphate oxygenase-carboxylase, but that's even more of a mouthful than RuBP, so people call it the nice friendly name rubisco, rubisco for short. But you can learn a lot about what rubisco does and you can even learn a little bit about its quirk that we're about to talk about. So it obviously involves ribulose one, five bisphosphate, and it does indeed involve that, and then you see oxygenase, dash, carboxylase. Well, the carboxylase is what tells us that it can deal with the carbon dioxide right over here. The carbon dioxide can be one of the substrates in a reaction with the ribulose one, five, bisphosphate. And so that's exactly what it's doing in this reaction. In a normal Calvin cycle, it's acting as a carboxylase. It is fixing that carbon. It's making it part of, if you view, you know, if you view that carbon-- Actually, I won't do it that way because here we have 12 as many. But it's taking these carbon molecules, and it's fixing them into organic molecules, some of which can eventually be used to create glucose. And that's what happens in a typical Calvin cycle."
    },
    {
        "Q": "At 4:41\nhow can three different orientation of the same sub shell exist at the same time?\nIn other words how can the sub shell be aligned at the x-axis, the y-axis and z-axis at the same time.\n",
        "A": "Because there are exactly three p orbitals per shell (except the first)",
        "video_name": "qLU0X154wlE",
        "timestamps": [
            281
        ],
        "3min_transcript": "that would be two minus one, that would be one. So we go from zero and then we go to one and then we have to stop. So we have only two allowed values for the angular momentum quantum number. So if you have n equal to two, you get two allowed values here. We already talked about what l is equal to zero means, l is equal to zero refers to an s orbital. And there's one s orbital. So in the second energy level, there's another s orbital. This is different from the s orbital in the first energy level that we just talked about. So there's another s orbital here. It too is shaped liked a sphere. What I drew here is misleading. I drew this as being a little bit smaller then the one before. Remember, when n is equal to two, you're further away. You're electron is on average further away from your nucleus here, l is equal to one so if l is equal to one, what is the allowed values for the magnetic quantum number? So remember, the magnetic quantum number So negative l would be negative one, and then we include zero and then we go to positive one. So there are three possible values for the magnetic quantum number. One, two, three, the magnetic quantum number told us the orientation so there are three different orientations. And we talked about l is equal to one is referring to a p orbital which is shaped like a dumbbell. So we have three different orientations, we have three different p orbitals in the second energy level. One of them goes along the x-axis, one of them the y and one of them the z. So we talked about this in the previous video. So a total of three p orbitals here. So how many orbitals are there in the second energy level? Well we have one s orbital and three p orbitals. So one plus three gives us four. We could have done this math, n squared, so two is n squared which gives us four. Alright, let's do electrons now. So four, let's go back to the s orbital here. fit a maximum of two electrons in one orbital. For the p orbitals, we have three p orbitals. If each p orbital's holding a maximum of two electrons, three times two gives us six, so we have a total of eight electrons in the second energy level. So eight electrons and we can get that from two n squared again right? Because if n is equal to two, square that and you get four, multiply that by two and you get eight. Alright, let's go to the third energy level or the third shell here. So when n is equal to three what are the allowed values for the angular momentum quantum number l? So remember l goes from zero all the way up to n minus one. So l goes from zero all the way to n minus one. So l is equal to zero, l is equal to one and l is equal to two because three minus one is equal to two, so if we have n is equal to three,"
    },
    {
        "Q": "\n@ 1:07 Sal says that when silk is rubbed with glass for long, one of the objects might discharge on touching another object ? What does that mean ?",
        "A": "That means that the charges are transferred to the earth through that object.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            67
        ],
        "3min_transcript": "- I'm guessing that you've had the experience of rubbing a balloon against your hair and then when you take the balloon away from your hair, your hair sticks up. And if you haven't had that experience, you might think about trying to lead a more rich and fun life, but I'm guessing most of you all have done that. And you had a sense that it had something to do with the balloon or your hair, somehow exchanging charge or now one is going to be more positive or negative than the other, and so now they are somehow attracted. And if you were thinking of those things, you are generally right. What you just experienced after you rubbed the balloon on your head, and then your hair is now attracted to the balloon, that's actually called the triboelectric effect, let me write that down, tribo, triboelectric, electric effect. And human beings have been observing this for a long long time, and it wasn't necessarily with balloons at birthday parties or whatever, it's with other things, they rub a silk cloth on a piece of attraction, or they might see that if they do that enough, one of the objects might discharge when it touches another object. People have observed things like lightning, where it looks like there's some type of a buildup and some type of a potential and then all of a sudden it discharges and you have this lightning and then this thunder blast sound that happens too. So this is something that humans have observed for a long long time, and scientists or people with a, I guess you could say a scientific mind have been trying to understand it for a long long time, and trying to come up with a framework for what exactly is happening. Well lucky for us, we now have a framework for it that explains it quite well. And that framework for what is going on with that triboelectric effect, is a framework around charge. Is a framework that we now have around charge. And this tells us, this way of looking at the world, says look, there's some things that just have a property called charge. Some things have a positive charge, and it's somewhat of an arbitrary name, we just happen to call it positive. And some things have what we say is an opposite charge, or a negative charge, a negative charge. We could have called this the magenta charge, and this the green charge, we could have called this the hippopotamus charge and this the ostrich charge. And we could have said that hippopotami, I believe plural for hippopotamus, they're always attracted to ostriches, but they always repel other hippopotami, and likewise. The like charges repel or like hippo... You get the general idea. But I'll stick to the words that people are used to using. And so if we say something has a charge, say a positive charge, and something else has a negative charge, then in our framework that we're setting up, these two things are going to attract. So opposite charges are going to attract, while like charges are going to repel. So if you have a positive charge, and you have a positive charge,"
    },
    {
        "Q": "\nAt 6:35... when the hair losing electrons , does that affect on the structure of the hair because the atoms of the hair lost electrons ?",
        "A": "great question! My answer is this. I dont think it would; with these reasons a) the electrons come from surface of the hair not from inside. So it is less likely to effect structure. b) although the effect is strong , the actual number of electrons exchanged is relatively small. c) your hair is possibly being charged up many times a day as objects (including the wind) rub against it and no harm seems to come of it. ok??",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            395
        ],
        "3min_transcript": "We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon. of transfer of electrons, that's exactly what's happening. And so when you think that way, it's like ok, we are scientists, this is a nice model, we can start to think about what's happening here. This model actually explains a whole ton of behavior that we've observed in the universe, including things like, lightning and whatever else, you know the static shock that you get when you might touch a doorknob after rubbing your shoes along the carpet. But we like to start, we like to quantify things, so we can start seeing how much they repel or how much they attract each other. And so the fundamental unit of charge, or one of the fundamental units of charge, or I guess you could say the elementary unit of charge is defined in terms of the charge of a proton or an electron. So the fundamental, or I guess you could say the elementary unit of charge is denoted by the letter e, and this is the charge of a proton, this is e for elementary, charge of proton."
    },
    {
        "Q": "\nAt 2:57 why did Sal said like charges are going to repel (It is just a convention). Why can't they attract each other.",
        "A": "They just don t. It s like asking why does gravity pull mass together instead of repel them. That s just not how it works. Like charges repel, and unlike charges attract. Why they do that, we don t know.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            177
        ],
        "3min_transcript": "of attraction, or they might see that if they do that enough, one of the objects might discharge when it touches another object. People have observed things like lightning, where it looks like there's some type of a buildup and some type of a potential and then all of a sudden it discharges and you have this lightning and then this thunder blast sound that happens too. So this is something that humans have observed for a long long time, and scientists or people with a, I guess you could say a scientific mind have been trying to understand it for a long long time, and trying to come up with a framework for what exactly is happening. Well lucky for us, we now have a framework for it that explains it quite well. And that framework for what is going on with that triboelectric effect, is a framework around charge. Is a framework that we now have around charge. And this tells us, this way of looking at the world, says look, there's some things that just have a property called charge. Some things have a positive charge, and it's somewhat of an arbitrary name, we just happen to call it positive. And some things have what we say is an opposite charge, or a negative charge, a negative charge. We could have called this the magenta charge, and this the green charge, we could have called this the hippopotamus charge and this the ostrich charge. And we could have said that hippopotami, I believe plural for hippopotamus, they're always attracted to ostriches, but they always repel other hippopotami, and likewise. The like charges repel or like hippo... You get the general idea. But I'll stick to the words that people are used to using. And so if we say something has a charge, say a positive charge, and something else has a negative charge, then in our framework that we're setting up, these two things are going to attract. So opposite charges are going to attract, while like charges are going to repel. So if you have a positive charge, and you have a positive charge, are going to accelerate away from each other. And that's not just true for positive positive, that's also true for negative and negative, these two things are going to repel because they are like charges. Now it's very interesting to think about this because we are so used to thinking in terms of charge, even you know if, especially in kind of the world of electricity you have the positive and negative terminal. You think of charging up your phone or whatever else. That it seems like, we completely, charge is just something that is fundamental about the universe, and that's true to some, that's true, but you'd have to appreciate that these are arbitrary words and they're really just to describe a property that we have observed in the world. And if you go down to the atomic level, we can get to a fundamental level of where the charge is happening. But once again, these are really models for our brain to describe, these are frameworks and models for our brain to be able to predict and describe"
    },
    {
        "Q": "How does top part of the hair have positive charge and bottom part have negative charge and doesn't the hair build up tension at 6:51?\n",
        "A": "By rubbing the balloon electrons move from your hair to the balloon. The top of the hair becomes positively charged because they are attracted to the negative charge of the balloon. The negative ions still in the hair are repelled from the negative charge of the balloon. Yes, it will a little bit!",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            411
        ],
        "3min_transcript": "We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon. of transfer of electrons, that's exactly what's happening. And so when you think that way, it's like ok, we are scientists, this is a nice model, we can start to think about what's happening here. This model actually explains a whole ton of behavior that we've observed in the universe, including things like, lightning and whatever else, you know the static shock that you get when you might touch a doorknob after rubbing your shoes along the carpet. But we like to start, we like to quantify things, so we can start seeing how much they repel or how much they attract each other. And so the fundamental unit of charge, or one of the fundamental units of charge, or I guess you could say the elementary unit of charge is defined in terms of the charge of a proton or an electron. So the fundamental, or I guess you could say the elementary unit of charge is denoted by the letter e, and this is the charge of a proton, this is e for elementary, charge of proton."
    },
    {
        "Q": "At 6:38, why we got different masses for different points? Aren't they the same? because they are all on the same ball.\n",
        "A": "Great question. If you sliced up the baseball into lots of little chunks, then, if you were careful, you might be able to make every chunk have the same mass. But dividing a sphere into many, many little pieces that all have the same volume (and thus mass if the density is constant) is very difficult. So it would probably be the case that each small mass element would have a different mass.",
        "video_name": "o7_zmuBweHI",
        "timestamps": [
            398
        ],
        "3min_transcript": "times the angular velocity, or the angular speed gives you the regular speed. This formula is really handy, so we're gonna replace V with R omega, and this is gonna give us R omega and you still have to square it and at this point you're probably thinking like this is even worse, what do we do this for. Well watch, if we add this is up I'll have one half M. I'm gonna get an R squared and an omega squared, and the reason this is better is that even though every point on this baseball has a different speed V, they all have the same angular speed omega, that was what was good about these angular quantities is that they're the same for every point on the baseball no matter how far away you are from the axis, and since they're the same for every point I can bring that out of the summation so I can rewrite this summation and bring everything that's constant for all of the masses out of the summation so I can write this as one half times the summation of M times R squared and end that quantity, because it's the same for each term. I'm basically factoring this out of all of these terms in the summation, it's like up here, all of these have a one half. You could imagine factoring out a one half and just writing this whole quantity as one half times M one V one squared plus M two V two squared and so on. That's what I'm doing down here for the one half and for the omega squared, so that's what was good about replacing V with R omega. The omega's the same for all of them, you can bring that out. You might still be concerned, you might be like, we're still stuck with the M in here cause you've got different Ms at different points. We're stuck with all these R squareds in here, all these points at the baseball are different Rs, they're all different points from the axis, different distances from the axis, we can't bring those out so now what do we do, well if you're clever you recognize this term. This summation term is nothing but the total moment of inertia of the object. Remember that the moment of inertia of an object, so the moment of inertia of a point mass is M R squared and the moment of inertia of a bunch of point masses is the sum of all the M R squareds and that's what we've got right here, this is just the moment of inertia of this baseball or whatever the object is, it doesn't even have to be of a particular shape, we're gonna add all the M R squareds, that's always going to be the total moment of inertia. So what we've found is that the K rotational is equal to one half times this quantity, which is I, the moment of inertia, times omega squared and that's the formula we got up here just by guessing. But it actually works and this is why it works, because you always get this quantity down here, which is one half I omega squared, no matter what the shape of the object is. So what this is telling you, what this quantity gives us is the total rotational kinetic energy of all the points on that mass about the center of the mass but here's what it doesn't give you. This term right here does not include"
    },
    {
        "Q": "dose that dash preforms away must be three lines? or it can be random dashes?\nat 2:40\n",
        "A": "It doesn t have to be three lines, just dash it :-)",
        "video_name": "ZAgQH2azx3w",
        "timestamps": [
            160
        ],
        "3min_transcript": "tetrahedral geometry around that carbon. And if you look at that carbon on the picture here, you can see that this bond and this bond are in the same plane. So if you had a flat sheet of paper, you could say those bonds are in the same plane. So a line represents a bond in the plane of the paper, let me go ahead and draw that, so this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper. Next, let's look at what else is connected to the carbon in magenta. Well, obviously, there is an OH, so let me go ahead and circle that. There is an OH we can see there is an OH here, and the OH, the OH in our picture, is coming out at us in space, so hopefully you can visualize that this bond in here is coming towards you in space, which is why this oxygen, this red oxygen atom, looks so big. So this is coming towards you, so let me go ahead and draw a wedge in here, and a wedge means that the bond is in front of your paper, so this means the OH is coming out at you in space, let me draw in the OH like that. Now let's look at what else is connected to that carbon in magenta, we know there's a hydrogen. We didn't draw it over here, but we know there's a hydrogen connected to that carbon, and we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space, so this bond is going away from us in space, or into the paper, or the bond is behind the paper. And we represent that with a dash, so I'm going to draw a dash here, showing that this hydrogen is going away from us. So we're imagining our flat sheet of paper and the OH coming out at us, and that hydrogen going away from us. All right, next, let's look at the carbon on the left here, so this carbon in blue, so that's this carbon, and I'll say that's this carbon over here on the left. we can see that this bond, and this bond are in the same plane, so let's go ahead and draw in the carbon, so the carbon that I just put in is the carbon in blue, and this hydrogen over here on the left, right, this bond is in the same plane, so I'm going to draw a line representing the bond is in the plane of the paper, and so we have a hydrogen right here. What about the other two hydrogens? Well, let me highlight those, so this hydrogen, hopefully you can see that this is coming out at us in space. So we represent that with a wedge, so we draw a wedge right here, and then we draw in a hydrogen, so the bond is in front of the paper, the bond is coming towards us in space. And there's another hydrogen bonded to the carbon in blue, and my thumb here is blocking it a little bit but hopefully you can see that's going away from us in space, so this hydrogen is going away from us in space. So the carbon in blue, is also SP3 hybridized."
    },
    {
        "Q": "at around 3:30 he calls carbon-12 an isotope, but i thought that isotopes were defined as atoms that have more neutrons than protons in their nucleus. carbon-12 has 6 protons and 6 neutrons in the nucleus, why is it an isotope?\n",
        "A": "All atoms are isotopes. Isotope simply refers to the specific number of nucleons for a given atom. It doesn t matter if the number of neutrons is more, less or equal to the number of protons, or if there are no neutrons, the atom is still a specific isotope.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            210
        ],
        "3min_transcript": "Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then,"
    },
    {
        "Q": "\nAt 3:29 in this video Sal says that an Atomic Mass Unit is a very, very small fraction of a gram; but in an earlier video he said that an Atomic Mass Unit was a very, very small fraction of a kilogram. So what exactly is an Atomic Mass Unit a very small fraction of?",
        "A": "A gram is a small fraction of a kilogram (1/1000) So both of those statements you mentioned are true, aren t they?",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            209
        ],
        "3min_transcript": "Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then,"
    },
    {
        "Q": "At 3:35 Sal references carbon-12 as being an isotope. Why is carbon-12 considered an isotope? The number of protons matches the number of neutrons in the nucleus.\n",
        "A": "Every atom is an isotope. It has nothing at all to do with the number of protons matching the number of neutrons. There is nothing special at all about having equal protons and neutrons. There are quite a few elements whose most common isotopes happen to have that, but if you look at the periodic table you will see that as the elements get heavier you tend to have more neutrons than protons.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            215
        ],
        "3min_transcript": "Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then, to figure out what the atomic mass, or the mass of any other atom. And you might say, \"Oh, why didn't they just do a hydrogen, \"and just say that's one atomic mass unit, and all that,\" and actually, they had started there. They had been there at an earlier stage, but for a whole set of reasons, carbon-12 is kind of being the benchmark, as having 12 atomic mass units, is what people went with. Now, what is atomic weight, then? Atomic weight. Let me write this in a different color. I'll do it in blue. Atomic weight. So if you draw the same analogy that we did up here, you might say, \"Okay, this must be a ... \"This must be a force. \"It should maybe, you know, \"an atomic weight unit would be a small fraction of, \"very small fraction of a unit.\" But it turns out in chemistry, when we talk about atomic weight, we're still measuring in atomic mass units."
    },
    {
        "Q": "I thought atomic weight is determined using the relative abundances of only stable isotopes of elements. At 7:16, Mr. Khan says that the weighted average of C-12 and C-14 is used to obtain the atomic weight of carbon. Is this a mistake? The only stable isotopes of C are C-12 and C-13. C-14 is radioactive.\n",
        "A": "Atomic weight is determined using the relative abundance of the element as it occurs in nature. Highly unstable isotopes will only be found in very small amounts in nature, because they will have decayed. C-14 has a half-life over 5000 years and it is constantly produced in the atmosphere, so a small amount of the natural carbon is C14",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            436
        ],
        "3min_transcript": "Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is, the most common isotope of carbon ... Remember, when we're saying \"isotopes,\" we're saying the same element, we have the same number of protons, but we have different number of neutrons. The most common isotope on Earth is carbon-12, but there's also some carbon-14. If you were to take a weighted average, as found on the Earth, of all the carbon-12 and all of the carbon-14, the weighted average of the atomic masses is the atomic weight. And the atomic weight of carbon ... And you'll see this on a periodic table. In fact, I have one right over here. Notice, the six protons, this is what defines it to be carbon. But then they write 12.011, which is the weighted average of the masses of all of the carbons. Now, it's very close to 12, as opposed to being closer to 14, because most of the carbon on Earth is carbon-12. We could write this down. This is the atomic weight. This is the atomic weight of carbon on Earth. This is 12.011."
    },
    {
        "Q": "\nat 6:28 you explained that carbon 12 has 6/ neutrons and protons each. so doesnt it make a balanced atom but you are saying its an isotope??",
        "A": "All atoms are isotopes. There is no such concept of a balanced atom as far as protons and neutrons are concerned. There are stable isotopes, but stability can apply to multiple isotopes, and the ratio of protons to neutrons varies as the atomic size increases.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            388
        ],
        "3min_transcript": "to figure out what the atomic mass, or the mass of any other atom. And you might say, \"Oh, why didn't they just do a hydrogen, \"and just say that's one atomic mass unit, and all that,\" and actually, they had started there. They had been there at an earlier stage, but for a whole set of reasons, carbon-12 is kind of being the benchmark, as having 12 atomic mass units, is what people went with. Now, what is atomic weight, then? Atomic weight. Let me write this in a different color. I'll do it in blue. Atomic weight. So if you draw the same analogy that we did up here, you might say, \"Okay, this must be a ... \"This must be a force. \"It should maybe, you know, \"an atomic weight unit would be a small fraction of, \"very small fraction of a unit.\" But it turns out in chemistry, when we talk about atomic weight, we're still measuring in atomic mass units. Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is,"
    },
    {
        "Q": "At 0:55, he says it's confusing when you go to Europe, where they use the metric system to give you their weight. Why is it different if you were say in the USA where they use pounds? Are pounds a form of mass or weight??\n",
        "A": "I suppose it s because Sal is based in the USA so therefore he sees weight from the US pound perspective.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            55
        ],
        "3min_transcript": "Let's have a little bit of a primer on weight and mass, especially if we start talking about atomic weight and atomic mass. If we're sitting in a physics class, weight and mass mean something very, very ... well, they mean different things. It might be a discovery, or a new learning, for some of you, because in everyday life, when we say something's mass, we think, \"Well, the more mass it has, the more weight it has.\" Or, if we think something has more or less weight, we think, \"Okay, that relates to its mass.\" But in physics class, we see that these actually represent two different ideas, albeit related ideas. Mass is a notion of how much of something there is, or you could say, how hard is it to accelerate or decelerate it. Or you could view it as a measure of an object's inertia. We typically, it, kind of a human scale, might measure mass in terms of grams or kilograms. What's confusing is, if you go to Europe, and you ask someone their weight, they'll often give you their weight in terms of kilogram, Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ..."
    },
    {
        "Q": "\nat 2:58 , can the (CH3)2CHOH also be (CH3)2 CH2O?",
        "A": "The -O- is connected to the -C-. One -H- is connected to the -C- , while the other -H- is connected to the -O-. In short : No.",
        "video_name": "XEPdMvZqCHQ",
        "timestamps": [
            178
        ],
        "3min_transcript": "And the carbon in magenta is bonded to three other hydrogens. So we could represent that as a CH three. So I could write CH three here, and the carbon in red is this one and the carbon in magenta is this one. On the left side, the carbon in red is bonded to another carbon in blue and the carbon in blue is bonded to three hydrogens, so there's another CH three on the left side, so let me draw that in, so we have a CH three on the left and the carbon in blue is directly bonded to the carbon in red. So this is called a partially condensed structure so this is a partially condensed, partially condensed structure. We haven't shown all of the bonds here but this structure has the same information as the Lewis structure on the left. it's just a different way to represent that molecule. We could keep going. We could go for a fully condensed structure. So let's do that. Focus in on the carbon in red. So this one right here. So let me draw in that carbon over here. So that's that carbon. That carbon is bonded to two CH three groups. There's a CH three group on the right, so there's the CH three group on the right. And there's a CH three group on the left. So I could write CH three and then I could write a two here which indicates there are two CH three groups bonded to directly bonded to the carbon in red. What else is bonded to the carbon in red? There's a hydrogen, so I'll put that in. So the carbon is bonded to a hydrogen. The carbon is also bonded to an OH, so I'll write in here an OH. This is the fully condensed version, so this is completely condensed and notice there are no bonds shown. you have to infer you have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure, so we'll start with a condensed and then we'll do partially condensed structure, and then we'll go to a full Lewis structure. Just to get some more practice here. So I'll draw in a condensed one, so we have CH three, three and then COCH three. All right let's turn that into a partially condensed structure. So this carbon in red right here we're gonna start with that carbon, so I'll start drawing in that carbon right here. What is bonded to that carbon? Well, we have CH three groups and we have three of them. So there are three CH three groups directly bonded to that carbon. So let me draw them in. So here's one CH three group here is another CH three group, and then finally here's the third CH three group."
    },
    {
        "Q": "At 5:28, Jay says that the lone pairs for Oxygen can form a pi bond between Oxygen and Carbon. Why can't they form pi electrons inside the Benzene ring?\n",
        "A": "If the electrons in the C=O pi bond (from the oxygen lone pair) were pushed into the ring, that would leave oxygen with a +2 charge and an incomplete octet. This would be very unstable so does not happen.",
        "video_name": "i9rfWOAEplk",
        "timestamps": [
            328
        ],
        "3min_transcript": "we need to show the nitro group adding onto the ortho position. So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here. I'm showing the nitro group adding onto the ortho position. And remember there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. We can show some resonance structures. So we can show some resonance stabilization of this cation here. So I could show these pi electrons moving over to here. And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. So we get a +1 formal charge on this carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro group in the ortho position. And now we show the movement of those pi electrons over to here. So let me go ahead and highlight those. These electrons in red move down to here. I took a bond away from this carbon. So that carbon is the one that gets a plus 1 formal charge now. Since the oxygen is right next to this carbon-- the oxygen has a lone pair of electrons. And so that lone pair of electrons can give us yet another resonance structure. So these electrons could move into here to draw a fourth resonance structure. So the presence of that methoxy substituent with the lone pair of electrons on that oxygen allows you to draw a fourth resonance structure. So this will give this oxygen a +1 formal charge. We have these pi electrons over here. We have our nitro group, once again, in the ortho position. And let me just go ahead and show the movement of those electrons. So these electrons-- I'll make them green-- these electrons right here are going"
    },
    {
        "Q": "At 2:40, would not natural selection cause organisms that were attracted to useless traits to drop out of the gene pool? Organisms attracted to useful traits would survive and sexual selection would no longer be a factor as being attracted to useful traits would actually make the organism fitter, thereby making it natural, and not sexual, selection.\n",
        "A": "No. Other ones with okay mutations wouldn t drop, so the gene pool gets rid of REALLY useless organisms.",
        "video_name": "tzqZsPjHFVQ",
        "timestamps": [
            160
        ],
        "3min_transcript": "it's the study of how populations of a species change genetically overtime leading to species evolving. Let's start up by defining what a population is. It's simply a group of individuals of a species that can interbreed. Because we have a whole bunch of fancy genetic testing gadgets and because unlike Darwin we know a whole lot about heredity. We can now study the genetic change in populations over just a couple of generations. This is really exciting and really fun because it's basically like scientific instant gratification. I can now observe evolution happening within my lifetime. You know, just cross that off the bucket list. Now, part of population genetics or pop-gen and now we've got fancy abbreviations for everything now, involves the study of factors that cause changes and what's called allele frequency. Which is just how often certain alleles turn up within a population. Those changes are at the heart of how and why evolution happens. There are several factors that change allele frequency Just like Fast and Furious movies, there are five of them. Unlike Fast and Furious movies, they're actually very, very important and are the basic reason why all complex life on earth exists. The main selective pressure is simply natural selection itself, Darwin's sweet little baby which he spent a lot of his career defending from haters. Obviously we know these natural selection makes the alleles that make animals the strongest and most virile and least likely to die more frequent in the population. Most selective pressures are environmental ones like food supply, predators or parasites. At the population level, one of the most important evolutionary forces is sexual selection. Population genetics gets its special attention particularly when it comes to what's called non-random mating which is a lifestyle that I encourage in all of my students, do not mate randomly. Sexual selection is the idea that certain individuals will be more attractive mates than others because of specific traits. This means they'll be chosen to have more sex and therefore offspring. The pop-gen spend on things if that sexual selection There are specific traits that are preferred even though they may not make the animals technically more fit for survival. Sexual selection changes a genetic make up of a population because the alleles of the most successful maters are going to show up more often in the gene-pool. Maters are going to mate. Another important factor here, and another thing that Darwin wished he had understood is mutation. Sometimes when eggs and sperm are formed through meiosis, a mistake happens in the copying process of DNA, that errors in the DNA could result in the death or deformation of offspring. But not all mutations are harmful. Sometimes these mistakes can create new alleles that benefit the individual by making it better at finding food or avoiding predators or finding a mate. These good errors and the alleles they made are then passed to the next generation and into the population. Fourth, we have genetic drift which is when an alleles frequency changes due to random chance. A chance that's greater if the population is small. Those happens much more quickly if the population"
    },
    {
        "Q": "\nIs the cyclical form of ribose a furanose ? And where does the Oxygen attached to Carbon 2 goes ? I guess the Hydrogen attached to Carbon 2 in deoxyribose form is not the same from the O-H. 08:56. So, where does the Hydrogen comes from? Thanks.",
        "A": "I think it is because the phosphate groups are negative and hydrogen is a positive ion, so it attracts it.",
        "video_name": "L677-Fl0joY",
        "timestamps": [
            536
        ],
        "3min_transcript": "that hydrogen proton right over there and this green bond that gets formed between the four prime carbon and or between the oxygen that's attached to the four prime carbon and the one prime carbon, that's this. That's this bond right over here. This oxygen is that oxygen right there. Notice, this oxygen is bound to the four prime carbon and now it's also bound to the one prime carbon. It was also attached to a hydrogen. It was also attached to a hydrogen so that hydrogen is there but then that can get nabbed up by another passing water molecule to become hydronium so it can get lost. It grabs up a hydrogen proton right over here and so it can lose a hydrogen proton right there. It's not adding or losing in that net. You form this cyclical form and the cyclical form right over here is very close to what we see in a DNA molecule. It's actually what we would see in an RNA molecule, in a ribonucleic acid. when we say deoxyribonucleic acid. Well, you can start with you have a ribose here but if we got rid of one of the oxygen groups and in particular one of... Well, actually if we just got rid of one of the oxygens we replace a hydroxyl with just a hydrogen, well then you're gonna have deoxyribose and you see that over here. This five-member ring, you have four carbons right over here. it looks just like this. The hydrogens are implicit to the carbons, we've seen this multiple time. The carbons are at where these lines intersect or I guess at the edges or maybe and also where these lines end right over there. But you see this does not have an... This molecule if we compare these two molecules, if we compare these two molecules over here, we see that this guy has an OH, and this guy implicitly just has... This has an OH and an H. This guy implicitly has just two hydrogens over here. He's missing an oxygen. This is deoxyribose. Deoxyribose doesn't have this oxygen. It does not have the oxygen on the two prime carbon. So this if you get rid of that, this is deoxyribose. So let me circle that. This thing right over here, this thing right over here, that is deoxyribose. Deoxy or it's based on deoxyribose I guess before it bonded to these other constituents. You could consider this deoxyribose. That's where the deoxyribo comes from and then the last piece of it, the last piece of it is this chunk right over here. These we call nitrogenous bases. Nitrogenous. Nitrogenous. Nitrogenous bases. You could see we have different types of nitrogenous bases. This is a nitrogenous base. This right over here is a different nitrogenous base. This right over here is another different nitrogenous base."
    },
    {
        "Q": "At around 0:26 in the video, why does Sal do \"deoxy\" and \"ribo\" in two different colors? Isn't deoxyribose one word and one type of sugar?\n",
        "A": "he is separating the roots of the word. deoxy means lack of oxygen and ribo, I think means with sugar, and you know acid. So, therefore, it is a lack of acid, with sugar and acid.",
        "video_name": "L677-Fl0joY",
        "timestamps": [
            26
        ],
        "3min_transcript": "- [Voiceover] We already have an overview video of DNA and I encourage you to watch that first. What I want to do in this video is dig a little bit deeper. Actually get into the molecular structure of DNA. This is a starting point. Let's just remind ourselves what DNA stands for. I'm gonna write the different parts of the word in different colors. It stands for deoxy. Deoxyribonucleic. Ribonucleic. Ribonucleic acid. Ribonucleic acid. So I'm just gonna put this on the side and now let's actually look at the molecular structure and how it relates to this actual name, deoxyribonucleic acid. DNA is just a junction for nucleic acid and it's the term nucleic that comes from the fact that it's found in the nucleus. It's found in the nucleus of eukaryotes. That's where the nucleic comes from and we'll talk about in a second why it's called an acid but I'll wait on that. Now each DNA molecule is made up of a chain What we call nucleotides. It's made up of nucleo, nucleo, nucleotides. What does a nucleotide look like? Well, what I have right over here is I have two strands, I've zoomed two strands of DNA or I've zoomed in two strands of DNA. You could view this side right over here as one of the, I guess you can say the backbones of one side of the ladder. This is the other side of the ladder and then each of these bridges, and I will talk about what molecules these are. These are kind of the rungs of the ladder. A nucleotide, let me separate off the nucleotide. A nucleotide would... What I am cordoning off, what I am cordoning off right over here could be considered, could be considered a nucleotide. That's one nucleotide and then it's connected to another. Another nucleotide right over here. On the right hand side we have a nucleotide, we have a nucleotide right over there and then, actually I want to do it, let me do it slightly different. We have a nucleotide right over here on the right side and then right below that we have another. We have another nucleotide. We have another nucleotide. Depicted here, we essentially have four nucleotides. These two are on this left side of the ladder, these two are on the right side of the ladder. Now let's think about the different pieces of that nucleotide. The one thing that might jump out at you is we have these phosphate groups. This is a phosphate group right over here. This is a phosphate group right over here. Each of these nucleotides have a phosphate group. This is a phosphate group over here and this is a phosphate group over here."
    },
    {
        "Q": "\nAt 11:34, can we classify Adenine and Guanine as Purines and Cytosine and Thymine as Pyrimidines",
        "A": "Yes, that s how they are classified.",
        "video_name": "L677-Fl0joY",
        "timestamps": [
            694
        ],
        "3min_transcript": "this one has two rings. This one over here has two rings and we have different names for these nitrogenous bases. The ones with two rings, the general categorization we call them purines. Nitrogenous bases if you have two rings, if you have two rings we call them purines. That's a general classification term. Let me make sure, purines. If you have one ring. Anyway, I'll just write this way. One ring. One ring, we call these pyrimidines. Pyrimidine. Pyrimidines. We call these pyrimidines. These particular, these two on the right, these two purines, this one up here this is adenine, and we talk about how they pair in the overview video on DNA. This one right over here is adenine, this nitrogenous base. This one over here is guanine. That is guanine. And then over here, over here, which makes it a pyrimidine, this is thymine. This right over here is thymine. This is thymine and then last but not least if we're talking about DNA, when we go into RNA, we're also gonna talk about uracil. But when we talk about DNA this one over here is cytosine. Cytosine. You could see the way it's structured. The thymine is attracted to adenine. It bonds with adenine and cytosine bonds with guanine. How are they bonding? Well, the way that these nitrogenous bases form the rungs of the ladder, how they want they're drawn to each other, this is our good old friend hydrogen bonds. This all comes out of the fact, that nitrogen is quite electronegative. When nitrogen is bound to a hydrogen you're going to have a partially negative charge at the nitrogen. Let me do this in green. You're going to have a partial negative charge at the nitrogen and a partially positive charge at the hydrogen. as being electronegative so it has a partial negative charge. The partial negative charge of this oxygen is going to be attracted to the partial positive charge of this hydrogen, and so you're going to have a hydrogen bond. That's then going to happen between this hydrogen which is going... Its electrons are being hogged by this nitrogen and this nitrogen with who, which itself hogs electrons. That forms a hydrogen bond. And then down here you have a hydrogen that has a partially positive charge because its electrons are being hogged. And then you have this oxygen with a partially negative charge, they're going to be attracted to each other. That's a hydrogen bond. Same thing between this nitrogen and that hydrogen, and same thing between this oxygen and that hydrogen. That's why cytosine and guanine pair up and that's why thymine and adenine pair up, and we talk about that as well in the overview video of DNA."
    },
    {
        "Q": "at 0:50 he dose not go deeply into what reactants are\n",
        "A": "Reactants are the things that react together in a reaction. They are on the left side of a chemical equation.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            50
        ],
        "3min_transcript": "- [Voiceover] Let's talk a little bit about chemical reactions. And chemical reactions are a very big deal. Without chemical reactions, you or I would not exist. In your body right now, there are countless chemical reactions going on every second. Without chemical reactions, we would have no life, we would not even have the universe as we know it. So what are chemical reactions. Well, they're any time that you have bonds being formed or broken between atoms or molecules. So what are we talking about there? Well this is maybe one of the most fundamental chemical reactions. Once again if one never occurred, we'd be in trouble, we would not have, we would not have any water. But let's think about what it is actually describing. So over here on the left-hand side we have the reactants. Let me write that down. So here we have the reactants. These are the molecules that are going to reaction. And then we have an arrow that moves us to the product. So let me do that in a different color. or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens."
    },
    {
        "Q": "At 6:35 what are the reactants and products in reversible reaction?\n",
        "A": "Typically, you name the ones on the left the reactants and the ones on the right the products, although they are technically both reactants and products. See also which chemicals you start with, and these will be your reactants. Alternatively, those being reduced in concentration to suit the equilibrium are the reactants.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            395
        ],
        "3min_transcript": "And just to get an appreciation of how much energy this produces, let me just show you this picture right over here. That's the space shuttle and this, this big tank right over here, let me... This big tank contains a bunch of liquid oxygen and hydrogen. And to create this incredible amount of energy, it actually just... You mix the two together with a little bit of energy and then you produce a ton of energy that makes the rocket, that makes the space shuttle. Well, space shuttle's been discontinued now, but back when they did it, to make it get it's necessary, it's necessary velocity. Now let's talk about the idea. So, you know, this reaction, strongly goes in this, in the direction of going to water. But it can actually go the other way, but it's very, very hard for it to go the other way. So in general we would consider this to be an irreversible reaction, even though it is. You know irreversible sounds like, It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side,"
    },
    {
        "Q": "\nat 3:30 Sal said energy is provided but from where does it come from?",
        "A": "Humans e.g. a spark of fire (Heat) can overcome the activation energy to kickstart the reaction and allow it to occur spontaneously from that point forward",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            210
        ],
        "3min_transcript": "or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds"
    },
    {
        "Q": "At 2:37 he calls the water molecules, but later calls them compounds. What is the difference between a molecule and a compound, and why is he able to call water both names?\n",
        "A": "A molecule made of more than one element is a compound. Water is a compound of hydrogen and oxygen.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            157
        ],
        "3min_transcript": "or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds"
    },
    {
        "Q": "\nAround 8:30, he mentions that carbolic acid is good for your body. But how does it help our bodies?",
        "A": "For digestion",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            510
        ],
        "3min_transcript": "It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side, are gonna, they're gonna be more likely to interact, Or if you provide more of this, it might go in the other direction because these might more likely react with their surroundings or disassociate in some way. Now just to get a sense of, you know, it's nice to kind of, you know, are these just some random letters that I wrote here? Carbonic acid is actually an incredibly important molecule, or we could call it a compound because it's made up of two or, two or more elements, in living systems and in fact, you know, even in the environment. And even when you go out to get some fast food. When you have carbonated drinks, it has carbonic acid in it that disassociates into carbon dioxide and that carbon dioxide is what you see bubbling up. Carbonic acid is incredibly important in how your body deals with excess carbon dioxide in its bloodstream. Carbonic acid is involved in the oceans taking up carbon dioxide from the atmosphere. So when you're studying chemistry, especially in the context of biology, these aren't just,"
    },
    {
        "Q": "what happens if at 8:15, you didn't have the valves? what would happen to you blood\n",
        "A": "If you didn t have the valves, your blood might flow backwards. Your blood might not get the oxygen it needs from the respiratory system. Also, the muscles in your body might not receive the oxygen it needs to expand and contract if the oxygenated blood is flowing backwards and away from the muscle tissue that needs oxygen to move. This will also cause the pressure in the circulatory system to be unstable. Hank talks about this at 8:16.",
        "video_name": "L1qpKn2hNF0",
        "timestamps": [
            495
        ],
        "3min_transcript": "I'm going to explain. We're used to talking about the heart as the head honcho of the circulatory system, and yeah, you would be in serious trouble if you didn't have a heart. But the heart's job is to basically power the circulatory system, move the blood all around your body, and get it back to the lungs so that it can pick up more oxygen and get rid of the CO2. As a result, the circulatory system of mammals essentially makes a figure 8. Oxygenated blood is pumped from the heart to the rest of the body, and then when it makes its way back to the heart again, it's then pumped on a shorter circuit to the lungs to pick up more oxygen and unload CO2 before it goes back to the heart and starts the whole cycle over again. So even though the heart does all the heavy lifting in the circulatory system, the lungs are the home base for the red blood cells, the postal workers that carry the oxygen and the CO2. Now the way that your circulatory system moves the blood around is pretty nifty. Remember when I was talking about air moving from high pressure to low pressure? Well, so does blood. A four-chambered heart, which is just one big honking beast of muscle, has very high pressure. In fact, the reason it seems like the heart is situated a little bit to the left of center is because the left ventricle is so freaking enormous and muscle-y. It has to be that way in order to keep the pressure high enough that the oxygenated blood will get out of there. From the left ventricle, the blood moves through the aorta, a giant tube, and then through the arteries and blood vessels that carry the blood away from the heart to the rest of the body. Arteries are muscular and thick-walled to maintain high pressure as the blood travels along. As arteries branch off to go to different places, they form smaller arterioles, and finally, the very little capillary beds, which, through their huge surface area, facilitate the delivery of oxygen to all of the cells in the body that need it. The capillary beds are also where the blood picks up CO2. From there, the blood keeps moving down the pressure gradient through a series of veins. These do the opposite of what the arteries did. Instead of splitting off from each other, Little ones flow together to make bigger and bigger veins to carry the deoxygenated blood back to the heart. The big difference between most veins and most arteries is that instead of being thick-walled and squeezy, veins have thinner walls and have valves that keep the blood from flowing backwards, which would be bad. This is necessary because the pressure in the circulatory system keeps dropping lower and lower until the blood flows in to two major veins. The first is the inferior vena cava, which runs pretty much down the center of the body and handles blood coming from the lower part of your body. The second is the superior vena cava, which sits on top of the heart and collects the blood from the upper body. Together, they run into the right atrium of the heart, which is the point of the lowest pressure in the circulatory system. All this deoxygenated blood is now back in the heart, and it needs to sop up some more oxygen. So, it flows into the right ventricle and then into the pulmonary artery. Now arteries, remember, flow away from the heart, even though, in this case, it contains deoxygenated blood. And pulmonary means \"of the lungs,\""
    },
    {
        "Q": "\nwhat does the sub exponent thing do? (at 1:24)",
        "A": "The subscripts on a and b allow you to have lots of different coefficients without using up all the letters of the alphabet. All the cosine terms have coefficients named a_sub_something and all the sine terms use b_sub_something. This way of naming is handy for talking about the terms in order: the 0th term, the 1st term, etc.",
        "video_name": "UKHBWzoOKsY",
        "timestamps": [
            84
        ],
        "3min_transcript": "- [Voiceover] So I have the graph of y is equal to f of t here, our horizontal axis is in terms of time, in terms of seconds. And this type of function is often described as a square wave, and we see that it is a periodic function, that it completes one cycle every two pi seconds. And so we could say its period is equal to two pi, if we wanna put the units we could say two pi, two pi seconds per cycle, we could write it like that. We could also just write s for seconds. And its frequency is gonna be one over that. So we could write its frequency is equal to one over two pi cycles per second, cycles per second, it could also be described as hertz. And what we're gonna explore in this video, is can we take a periodic function like this and represent it as an infinite sum of sines and cosines of different periods or different frequencies? So to write that out a little bit more clearly, and write it as the sum of sines and cosines? So can we write it, so it's going to be sum, let's say baseline constant, that'll shift it up or down, and as we'll see, that's going to be based on the average value of the function over one period. So a sub zero, and then, let's start adding some periodic functions here. And so let's take a sub one times cosine of t. Now, why am I starting with cosine of t? And I could also add a sine of t, so plus b sub one, of sine of t. Why am I starting with cosine of t and sine of t? Well, if our original function has a period of two pi, and I just set up this one so it does have a period of two pi, well it would make sense that it would involve some functions that have periods of two pi. If a one is much larger than b one, well it says, okay, this has a lot more of cosine of t in it, than it has of sine of t in it. And that by itself isn't going to describe this function, because we know what this would look like. This would look like a very clean sinusoid, not like a square wave. And so what we're gonna do is we're gonna add sinusoids of frequencies that are multiples of these frequencies. So let's add a sub two, so another waiting coefficient, times cosine of two t. This has a frequency of one over two pi, this has twice the frequency, this has a frequency of one over pi. And then a sub three times cosine of three t. And I'm gonna keep going on and on and on forever. And I'm gonna do the same thing with the sines. So plus b two sine of two t plus b three sine of three t."
    },
    {
        "Q": "\nat 7:55 the second molecule is determined to be di-substituted. How come the methyl group on the second carbon is not counted in to be tri-substituted?",
        "A": "It is the number of carbon atoms directly attached to the double-bonded carbons that determines the degree of substitution. C-1 has no C atoms attached to it. C-2 has two C atoms attached to it ( a methyl and an ethyl group). Total attached atoms = 0 + 2 = 2. So the alkene is disubstituted.",
        "video_name": "MDh_5n0OO2M",
        "timestamps": [
            475
        ],
        "3min_transcript": "So, let me use red for this. If we think about the degree of substitution for the alkene on the right, by drawing my hydrogen right here, it makes it a little bit easier to see we have three alkyl groups, so this one, this one, and this one. So this would be a trisubstituted alkene. So the one on the right is a trisubstituted alkene, and the one on the left, so this one right here, would be a disubstituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon, and the carbon on the right has two alkyl groups bonded to it. So this one is a disubstituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a disubstituted product, and a trisubstituted. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. the double bond formed in this region of the molecule, and for the trisubstituted product, the double bond formed in this region. The trisubstituted product is the major product, and it's also the more stable alkene. So remember, from the video in alkene stability, the more substituted your alkene is, the more stable it is, so this product is more stable, and that's why we form more of it. And the more stable products or the more substituted product is called the Zaitsev product. So we say that this E1 reaction is regioselective because it has a preference to form the more stable product, the more substituted product, which we call the Zaitsev product."
    },
    {
        "Q": "\nIn the equation shown at 1:52, why is H2O formed and not H3O+? How would I know which to write as a product?",
        "A": "Acid + base -> salt + water This is something you should memorise",
        "video_name": "aj34f2Bg9Vw",
        "timestamps": [
            112
        ],
        "3min_transcript": "- [Voiceover] Let's do another titration problem, and once again, our goal is to find the concentration of an acidic solution. So we have 20.0 milliliters of HCl, and this time, instead of using sodium hydroxide, we're going to use barium hydroxide, and it takes 27.4 milliliters of a 0.0154 molar solution of barium hydroxide to completely neutralize the acid that's present. All right, so let's start with what we know. We know the concentration of barium hydroxide. It's 0.0154 molar, and we also know that molarity is equal to moles over liters. All right, so we have 0.0154 as equal to, let's make moles X, over liters. 27.4 milliliters is 0.0274 liters, right? So that's 0.0274 liters. the moles of barium hydroxide. So let's get out the calculator here, and let's do that. So let's get some room over here. So we take 0.0154 and we multiply that by 0.0274, and that gives us, this will be 4.22 times 10 to the negative fourth, right? So that's equal to 0.000422 moles of barium hydroxide. All right, next, let's write the neutralization reaction. So we have barium hydroxide reacts with HCl. So barium hydroxide plus HCl gives us, for our products, we have H plus and OH minus, so that's H20. And then our other product, this is barium two plus, right? So we have BA2 plus and CL minus 1, so you could cross those over. So BACl2, right? So BACl2, barium chloride, as our other product here. All right, next we need to balance our equation, right? We need to balance the neutralization reaction here. So let's start by looking at the chlorines. So over here on the left, we have one chlorine. On the right, we have two. So we need to put a two right here, and now we have two chlorines on both sides. Next, let's look at hydrogens. So on the left side, we have two hydrogens here, and then we have two over here. So we have four hydrogens on the left. On the right, we have only two hydrogens. So we need to put a two here for this coefficient to give us four hydrogens on the right. So now we have four, and we should be balanced, right? Everything else should be balanced. Let's look at the mole ratio for barium hydroxide to HCl."
    },
    {
        "Q": "At 0:32 he said lesser and great apes. What is the difference between lesser and greater apes?\n",
        "A": "The great apes are chimpanzees, gorillas, humans, and orangutans. The lesser apes are gibbons.",
        "video_name": "oFGkYA_diDA",
        "timestamps": [
            32
        ],
        "3min_transcript": "In the first video on evolution, I drew something that I called an ape, and then I drew a tail on it. And what I want to do in this video is clarify that that was absolutely wrong. Apes - Apes have no tails. And it's actually one of the main distinguishing characteristics of apes. There's obviously other primates that also have no tails, but apes definitely have no tails. And just to clarify, there's kind of two families within apes and their common names are \"the lesser apes\" and \"the great apes.\" And the lesser apes - the lesser apes are things like gibbons, And the great apes are things like chimpanzees and gorillas, and, me, or human beings! So these right here, these are the great apes. These are the great apes, And clearly this great ape right here did not have a great sense of style in 1994. And really didn't feel the need to have a haircut!"
    },
    {
        "Q": "\nAt 4:00sec why did you take the force(10n) as kg.m/s2",
        "A": "That s what a Newton is. It is 1 kg*m/s^2. It s the force you need to apply to 1 kg to get it to accelerate at 1 m/s^2.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            240
        ],
        "3min_transcript": "going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration."
    },
    {
        "Q": "AT 3:23 HOW CAN WE WRITE F=M^2*A^1/2\n",
        "A": "We can t, that was just an example of a more complicated equation that, in another universe, could have been the relationship between mass, force, and acceleration. He mentioned it just to point out how simple the real equation, F=ma, was.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            203
        ],
        "3min_transcript": "-actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared"
    },
    {
        "Q": "\nAt 1:58 what does netforce mean?",
        "A": "net force means total force. remember that forces can cancel each other out as well if opposing forces act upon an object. for example, if your bank account is overdrawn by $100 and you deposit $100, your net amount will be $0 (-100+100=0), forces behave in the same way depending on their magnitude (size) and direction (which direction they act on an object).",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            118
        ],
        "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff"
    },
    {
        "Q": "\nIn many Khan academy videos, Sal does what I belive is called a dimensional analasis (5:10 in this video), and while I understand how this works, I cannot find anything on this site about where to use them, and the rules for doing so. Help?",
        "A": "The rules of it are just the rules of algebra. You should use it pretty much all the time when you are doing physics problems. It provides a good check on your answers. If the units don t come out right, you did something wrong.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            310
        ],
        "3min_transcript": "so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration. divide both sides by 2 kilograms So let's divide the left by 2 kilograms let's divide the right by 2 kilograms that cancels out. The 10 and the 2-- 10 divided by 2 is 5 and then you have kilograms cancelling kilograms. Your left hand side you get 5 metres per second squared and then that's equal to your acceleration. Now just for fun, what happens if I double that force? Well then I have 20Newtons--I'll actually work it out-- 20 kilograms.metres per second squared is equal to --I'll actually color-code this-- 2 kilograms times the acceleration"
    },
    {
        "Q": "At 4:00 why are seconds squared?\nAnd what does Seconds^2 mean? Other then Second * Seconds.\n",
        "A": "F = ma. Force is mass times acceleration. Acceleration is change in velocity over time. Velocity is distance over time. So acceleration is change in distance over time over time, or distance over time squared.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            240
        ],
        "3min_transcript": "going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration."
    },
    {
        "Q": "At 1:15, the second law of motion is stated. How is this equation derived?\n",
        "A": "It s not derived, it s an empirical observation.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            75
        ],
        "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff"
    },
    {
        "Q": "\nat 14:12 , to get the value of T2 cant we just assume that mg=T1+T2 i.e. 30N=60N+T2?\nThat would be T2=-30N..... would that be wrong?",
        "A": "Yes that would be wrong because you have to separate forces and tensions acting on the x-axis from those acting on the y-axis. That has to be done because forces and tensions acting on the x-axis have no effect on those acting on the y-axis (at least as long as they act on the same point of the object). T2 is horizontal so it acts on the x-axis, mg is vertical so it acts on the y-axis and T1 actually acts on both the x-axis and the y-axis which is why David had to break it into its components.",
        "video_name": "aHlOp5nYs28",
        "timestamps": [
            852
        ],
        "3min_transcript": "I'll get T one times sine theta, and then I divide both sides by sine theta. I'll end up with T one equals T one in the y direction divided by sine theta. I know T one in the y direction. That was 30 degree, or sorry, not 30 degrees. That was 30 Newtons. So I've got 30 Newtons, that's my force, upward. This vertical component right here had to be 30 Newtons 'cause it had to balance gravity, divided by sine of the angle. But what is this angle? We know this angle's 30. And you could probably convince yourself, if I draw a triangle this way, let's try to figure out, we wanna figure out what this angle is right here 'cause that's what this angle is here. So if this is 30 and that's 90, then this has to be 60. And if that's 60, and this is 90, then this has to be 30. So this angle is 30 degrees right here. So that's 30 degrees. So this angle right here, So when I'm taking my sine, I'm taking my sine of 30 degrees. And I get 30 Newtons divided by sine of 30, and sine of 30 is 1/2. So .5, so I get that this is 60 Newtons. And that might seem crazy. You might be like, \"Wait a minute. \"T one is 60 Newtons? \"60 Newtons? \"The weight of this chalkboard is only 30 Newtons. \"How in the world can the tension in this rope \"be 60 Newtons?\" I mean, if we just hung it by a single string, if we just hung this chalkboard by a single string over the center of mass, you'd just get a tension of 30 Newtons. How can this be 60 Newtons? And the reason is, this part's gotta be 30 Newtons. We know that 'cause it has to balance gravity. But that's only part of the total tension. So if the total tension, if part of the total tension is 30, all of the tension's gotta be more than 30. And in this case, it's 60 Newtons. 'cause it's at an angle. So this component has to equal gravity, and this total amount has to be bigger than that so that its component is equal to gravity. Right, how do we figure out T two? Well, you don't invent a new strategy. We keep going, we're just gonna say that the acceleration in the horizontal direction is the net force in the horizontal direction divided by the mass, so we still stick with Newton's Second Law even when we wanna find this other force. This force is horizontal, so it makes sense that we're gonna use Newton's Second Law for the horizontal direction. Again, if this chalkboard is not accelerating, the acceleration is zero, so I'll draw a line here to keep my calculations separate. Equals net force in the x direction. Okay, now I'm gonna have T one in the x direction. So this is gonna be T one in the x. So I'll have T one in the x direction. That's positive 'cause it points right, and I'm gonna consider rightward positive. Minus T two, all of T two. I don't have to break T two up. T two points completely in the horizontal direction."
    },
    {
        "Q": "\nAt 11:31 mvr = Iw. What is the Iw stand for?",
        "A": "moment of Inertia*angular velocity",
        "video_name": "viLpmZtQYzE",
        "timestamps": [
            691
        ],
        "3min_transcript": "if r is zero, there's gonna be no torque exerted by that axis. And if there's no torque exerted externally, there's no change in angular momentum of the system. So this system of ball and rod is gonna have no external torque on it. That means the angular momentum has to stay the same. This is a classic conservation of angular momentum problems. So we're gonna say that L initial, the initial angular momentum, has to equal the final angular momentum. And we'll just say, for our entire system, what had angular momentum initially? Well, it was this mass. So this mass had the angular momentum. And how do we find that? Remember it's m times v times R, and the r is that distance of closest approach. So we're gonna use this here for the whole four meters as this R. Yes, you can consider this hypotenuse R and a sine of the angle, but that's harder than it needs to be. You can find angular momentum, mvR, that's gonna equal the final angular momentum. So since this ball comes to rest, and it's only the bar that has angular momentum afterward, we only have to worry about the angular momentum of the bar on the final side. And to find the angular momentum of an extended object, a rigid object, you can use I omega. And this would let us solve for what is the final angular velocity of this rod after the collision. So that's what we wanna figure out. What is the final angular velocity of the rod after the collision? Now we can figure it out. We know the mass the of the ball, m. We know the speed of the ball initially. We know the R, line of closest approach. That's four meters. What's the moment of inertia here? Well, it's just gonna be 1/3 m L squared. Let me clean this up a little bit. Let me take this. I'll just copy that. Put that right down over here, and we could say that the moment of inertia of a mass of a rod, it's rotating around its end, is always gonna be 1/3 m L squared. So 1/3 times the mass of the rod, times the length of the rod squared, because this ball's line of closest approach was jus equal to the entire length of the rod, since it struck it at the very end, and then times omega. So we can solve this for omega now. We can say that omega. I'm gonna bring this down around here, so we go some room. Omega final of the rod is just gonna be, what? It's gonna be mass of the ball times the initial speed of the ball, times the line of closet approach. And then I'm gonna divide by 1/3 the mass of the rod times the length of the rod. I can just call that R, it's the same variable, length of the rod squared, So I can cancel off one of these Rs, and then I can plug in numbers if I wanted to actually get a number. I could say that the final angular velocity of this rod is gonna be five kilograms, that was the mass of the ball, times eight meters per second, that was the initial speed of the ball, and then I'm gonna divide by 1/3 of the mass of the rod was 10 kilograms, and then the length of the rod, which is this line of closest approach,"
    },
    {
        "Q": "At 10:10, David says that there is no external torque exerted on the system. But, isn't the mass exerting torque on the system by hitting the moment arm?\n",
        "A": "The system includes the mass, so the mass is not external to the system, although it is external to the rod. This is why the angular momentum of the rod can change without violating conservation of momentum, but the combined angular momentum of rod plus mass cannot.",
        "video_name": "viLpmZtQYzE",
        "timestamps": [
            610
        ],
        "3min_transcript": "is always just equal to that, so you could make your life easy. Just imagine, when this ball comes in, at what point is it closest to the axis? That would be this point. And then how far is it when it is closest? That gives you this R value. You can take mvR. That gives you the angular momentum of this point mass. It tells you the total amount of angular momentum that thing could transfer to something else if it lost all of its angular momentum. That's how much angular momentum something, like that rod, could get. So let's try an example. Let's do this example. It's actually a classic, a ball hitting a rod. Man, I'm telling you, physics teachers and professors, You should know how to do this. Let's get you prepared here. So let's say this ball comes in. It hits a rod, right? And so the ball is gonna come in. Ball is gonna hit a rod, and let's put some numbers on this thing, so we can actually solve this example. let's say the ball had a mass of five kilograms. It was going eight meters per second, hits the end of the rod, and the rod is 10 kilograms, four meters long. Let's assume this rod has uniform density, evenly throughout it, and it can rotate around the end. So when the ball gets in here, strikes the end of the rod, the rod is gonna rotate around its axis. And let's make another assumption. Let's assume when this ball does hit the rod, the ball stops. So after hitting the rod, the ball has stopped, and the rod moves on with all the angular momentum that the ball had. That will just make it a little easier. We'll talk about what to do if that doesn't happen. It's not that much harder. Let's just say that's the case initially or so. I'll move the ball back over to here. How do we solve this problem? Well, we're gonna try to use conservation of angular momentum. We're gonna say that even though there's an axis his exerting a force, the force that that axis is gonna exert on our system is gonna exert zero torque, because the r value. Torque is equal to r F sine theta. And if the r is zero, if r is zero, there's gonna be no torque exerted by that axis. And if there's no torque exerted externally, there's no change in angular momentum of the system. So this system of ball and rod is gonna have no external torque on it. That means the angular momentum has to stay the same. This is a classic conservation of angular momentum problems. So we're gonna say that L initial, the initial angular momentum, has to equal the final angular momentum. And we'll just say, for our entire system, what had angular momentum initially? Well, it was this mass. So this mass had the angular momentum. And how do we find that? Remember it's m times v times R, and the r is that distance of closest approach. So we're gonna use this here for the whole four meters as this R. Yes, you can consider this hypotenuse R and a sine of the angle, but that's harder than it needs to be. You can find angular momentum, mvR, that's gonna equal the final angular momentum."
    },
    {
        "Q": "\nAt around 09:00 , the second step of the Sn1 reaction, I was just wondering if instead of a new metanol, an Iodide could come and take the proton away? Isn't that even more electronegative than the metanol? And if so, which of the two happens more often? Thank you",
        "A": "Yes, an iodide ion could certainly remove the proton. but you must remember that the solvent is methanol, so there are many more methanol molecules than iodide ions present in the solution. Also, the cation is probably solvated by a shell of methanol molecules. The probability of the H being attacked by a methanol molecule is therefore much greater than the probability of being attacked by an iodide ion.",
        "video_name": "MtwvLru62Qw",
        "timestamps": [
            540
        ],
        "3min_transcript": "We're going to get substituted with the weak base, and the weak base here is actually the methanol. The weak base here is the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons and one of them, it's a weak base. It was willing to give an electron. It has a partial negative charge over here because oxygen is electronegative, but it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation, and that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this. That's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen. That is our oxygen. Here's that other pair of electrons on that oxygen, and it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron, so now it becomes-- it now is positive. So now you might have another methanol. You might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it It would give the electron to the hydrogen proton, really. then that becomes neutral. So in the final step, it'll all look like this. We have that over here. The methanol that had originally bonded has lost its hydrogen, so it looks like this. We just have the oxygen and the CH3 there. It is now neutral because it gained an electron when that hydrogen proton was nabbed. So if you wanted to draw it, it has actually those two extra electrons, just like that. And if you want to draw this last methanol, it's now a positive cation, so it looks like this. So it's OH, CH3, H, and then it has unbonded pair right there, and now this has a positive charge. So that was the Sn1 reaction. Now, the other reaction that's going to occur is the E1. Once again, our first step-- nope, I"
    },
    {
        "Q": "\nAt 13:20 Sal showns that the change in entropy after the cycle is (2Q_f)/T. The left side of the equation is a subtraction, not an addition like in the last video. In the last video, he showed that the change in entropy is equal to Q_1/T_1 + Q_2/T_2. What's going on here? These are clearly not the same. This doesn't make any sense.",
        "A": "In this video Sal talks about irreversible system. That s where (2Q_f)/T comes from: it s a heat generated by friction. About the signs of Q: in the previous video Sal puts a plus sign in front of the Q, but actual value of Q is negative (he actually mentioned that we ll see its actually negative somewhere in the video). In this video Sal assumes the value of Q to be positive (as he mentions at about 10:20, so he puts a negative in front.",
        "video_name": "PFcGiMLwjeY",
        "timestamps": [
            800
        ],
        "3min_transcript": "This is still positive. In either direction, when we move upwards or downwards, the system is generating friction. Now, we always said, we went all the way here, we went all the way back. So the sum of these has to be equal to 0, because this is a state variable. So if the sum of all of this has to be equal to 0, let's So this gets us to Qa minus Qr. So the heat accepted minus the heat released. The W's cancel out. Plus-- let me see right here-- plus 2 times the heat of friction in either direction. All of that has to be equal to 0. Let's see. What we can do is, we can rewrite this as the heat accepted minus the heat released is equal to minus 2 times the amount of heat generated from friction. And then if we just switch these around, we'll get the just wanted to get all positive numbers-- 2 times the heat of friction. Now why did I do all of this? Because I wanted to do an experiment with an irreversible system, and this was a very simple experiment with an irreversible system. Now, we said that delta S, which a long time ago I defined as Q divided by T-- and in this video, I said it had to be reversible. And I wanted to show you right now that what if I didn't make the constraint that this has to be reversible? Because if this doesn't have to be reversible, and I just use this definition right here, you'll see that your delta S here would be-- you just divide everything by T-- because our temperature was constant the entire time, we were just on a reservoir-- you'll see that this is going to be your delta s. This is your total change in the, I guess you could say, So this is, let me say, this is the heat added to the system. Let me do it this way. Heat added to the system, divided by the temperature at which it was added. Which is a positive number. Even though we got to the exact same place on this date diagram. So in an reversible system, this wouldn't be a valid state variable. So it's only a valid state variable if it's reversible. Now, does that mean that you can only talk about entropy for reversible reactions? No. You can talk about entropy for anything. But what you do is-- and this is another important point. So let's say that I have some irreversible reaction that goes from here to here. And I want to figure out its change in entropy."
    },
    {
        "Q": "\nAt 1:29, it's said that heat is released (Q2), and right after he says that the change is adiabatic, and no heat is being transferred to and from the system. Isn't that contradictory?",
        "A": "He was pointing to the horizontal paths when talking about heat transfer, but was pointing to the vertical paths when talking about the adiabatic paths.",
        "video_name": "PFcGiMLwjeY",
        "timestamps": [
            89
        ],
        "3min_transcript": "In the video where I first introduced the concept of entropy, I just tried something out. I defined my change in entropy as being equal to the heat added to a system, divided by the temperature at which it was added to the system. And then I tested to see if this was a valid state variable. And when I did that, I looked at the Carnot cycle. And this is a bit of a review. Never hurts to review. Let me draw the PV diagram here. We saw that we start at this state here, and then we proceed isothermically. We removed little pebbles off the piston. So we increased the volume and lowered the pressure. Then we proceed adiabatically, where we isolated things and we moved like that. That was adiabatically. Then at this other isotherm, we added the pebbles back. And then we isolated the system again. got back to our original state. And I did a couple of videos where I show that if you take the heat added here-- so this is all being done at some high temperature, T1. This is being done at some low temperature, T2. There's some heat being added here, Q1, and that there's some heat being released here, Q2. And since these are adiabatic, there's no transfer of heat to and from the system. And when I looked at this, and when I looked at the Carnot cycle, and I used this definition of entropy, I saw that the total change in S, when I go from this point all the way around and got back, the change in S, was equal to Q1 over T1 plus Q2 over T2. And then I actually showed you that this was equal to 0, which is exactly the result that I wanted to see. Because in order for this to be a state variable, in order how I got there. It should only be dependent on my state variables. So even if I go on some crazy path, at the end of the day, it should get back to 0. But I did something, I guess, a little bit-- what I did wasn't a proof that this is always a valid state variable. It was only a proof that it's a valid state variable if we look at the Carnot cycle. But it turns out that it was only valid because the Carnot cycle was reversible. And this is a subtle but super important point, and I really should've clarified this on the first video. I guess I was too caught up showing the proof of the Carnot cycle to put the reversibility there. And before I even show you why it has to be reversable, let me just review what reversibility means. Now, we know that in order to even define a path here, the system has to be pretty close to equilibrium the whole time. That's the whole reason why throughout these videos, I've"
    },
    {
        "Q": "at 6:27, how 2sp2 and why not 2sp3?\n",
        "A": "sp3 is in the case of a single bond sp2 is in the case of the double bond sp is in the case of a triple bond",
        "video_name": "lJX8DxoPRfk",
        "timestamps": [
            387
        ],
        "3min_transcript": ""
    },
    {
        "Q": "From 11:20 to 12:06, why do pi bonds pull the nuclei of the two carbons in ethylene closer together? Wouldn't the electrons be repelling each other?\n",
        "A": "pi bonds occurs when there is a parallel bond between two p orbitals.We know that an orbital is nothing but a probability cloud representing possible positions of electrons.Now, because in the pi bonds the two p orbitals overlap, this would mean that in that particular region there is a higher possibility of finding an electron.More electrons (-ve charge) would attract the nuclei of the two atoms more strongly, thus bringing them closer.This is also the reason of why these molecules are smaller. Hope this helped...!",
        "video_name": "lJX8DxoPRfk",
        "timestamps": [
            680,
            726
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:57, how do you find the probability, and if you use probability would the answer be accurate?\n",
        "A": "You don t! You measure the concentrations experimentally. Sal s use of probabilities here was just a way of trying to get you to intuitively understand why the expression for the equilibrium constant looks the way it does.",
        "video_name": "ONBJo7dXJm8",
        "timestamps": [
            417
        ],
        "3min_transcript": "what's the probability of having five heads? Well you would multiply the probability of one head five times. So the forward reaction probability is going to be the concentration of V to the a power, and, of course, that's not enough to have the reaction happen. You also need to have b of the X molecules there. So you have the concentration of X to the b power. And I want to make sure you understand this. My claim is that this is approximation -- or actually it's a pretty good way of calculating-- the probability. So let me write it this way. The rate is equal to some constant that takes into account the temperature and the molecular configurations times the probability of having a V molecules and b X molecules And the best way to approximate that is with their concentration. Obviously, the higher the concentration, the higher the moles per liter, the more likely you're going to find that many of molecules in kind of that little small space that you care about, and the temperature and the configuration are going to matter more. But if you use the concentration as the probability of a -- let me switch colors. If the probability of having a V molecule in some volume -- if we assume that the solution is homogeneous, that the V molecules are roughly evenly distributed, it's going to be -- this isn't even an approximation. It's going to be the concentration of the V molecules If we want the probability of a, where a is a number, it could be five V molecules, a V's in some volume, it's the probability of finding this a times. So it's going to be equal to -- and this is just from the probability concepts that we learned in the whole probability playlist. So if you want to have five heads in a row, it's 1/2 to the fifth power. If you want to have V molecules there, five of them at the same time in some volume, or a of them, it's going to be V to the a power times the volume. If you also care about the probability so you want all of that, so a V's and b X's in some volume,"
    },
    {
        "Q": "\nIf at around 11:20, Sal divided by [Y]^c [Z]^d, wouldn't that change the answer? But then again a constant divided by a constant is another constant. So couldnt the equilibrium constant be the reciprocal of what it actually is?",
        "A": "If he had divided by [Y]^c[Z]^d instead, he d end up with the reciprocal, yes. This, however, gives the equilibrium constant for the reverse direction of the reaction, i.e, the products reacting to form the reactants. As a general rule, the equlibrium constant for the reverse reaction is equal to the reciprocal of the equlibrium constant of the forward reaction.",
        "video_name": "ONBJo7dXJm8",
        "timestamps": [
            680
        ],
        "3min_transcript": "-- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other? -- the forward rate is this. That's our forward constant, which took into account a whole bunch of temperature and molecular structure and all of that-- times the concentration of our V molecule to the a power. You can kind of view that as what's the probability of finding in a certain volume -- and that certain volume can be factored into that K factor as well-- but what's the probability of finding V things, a V molecules in some volume. And it's the concentration of V to the a power times concentration of X to the b power -- that's the forward reaction-- and that has to equal the reverse reactions. So K-minus times the concentration of Y to the c power times the concentration of Z to the d power. Now, if we divide both sides by -- let me erase more space. Nope, not with that. All right. So let's divide both sides by K-minus so you get K-plus over K-minus is equal to that, is equal to Y to the c times Z to the d. All of that over that-- V to the a times the concentration of X to the b. Let me put this in magenta just so you know that this was this K-minus right here. And then, these are just two arbitrary constants, so we could just replace them and call them the equilibrium constant. And we're there where we need to be. We're at the formula for the equilibrium constant. Now, I know this was really hand wavy, but I want you to at least get the sense that this doesn't come from out of the blue, and there is -- at least I think there is-- there's an intuition here. These are really calculating the probabilities of finding -- this is the forward reaction rate probabilities"
    },
    {
        "Q": "\nAt 9:00, what does the Keq stand for?",
        "A": "The Equilibrium Constant K",
        "video_name": "ONBJo7dXJm8",
        "timestamps": [
            540
        ],
        "3min_transcript": "If we want the probability of a, where a is a number, it could be five V molecules, a V's in some volume, it's the probability of finding this a times. So it's going to be equal to -- and this is just from the probability concepts that we learned in the whole probability playlist. So if you want to have five heads in a row, it's 1/2 to the fifth power. If you want to have V molecules there, five of them at the same time in some volume, or a of them, it's going to be V to the a power times the volume. If you also care about the probability so you want all of that, so a V's and b X's in some volume, So it's going to be equal to the concentration of V to the a power times the concentration of X to the b power times the volume. So the probability of finding the right number of V particles and X particles in the right place in some volume is going to be proportional to exactly this. And we're saying that the reaction rate, the forward reaction rate, is also proportional to this thing. So that's where we get the forward reaction rate. So the rate forward is equal to the concentration of our V molecules to the a power times the concentration of our X molecules to the b power. Now, if we want to find the reverse rate, so this is the rate forward. If we want to find the rate of the reverse reaction, -- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other?"
    },
    {
        "Q": "@ 5:00 sal says that the concentration is a rough approximation of the probability. But what if Concentration is greater than one? we know that probability is always less than or equal to one. Then how can concentration be an approximation for probability in that case?\n",
        "A": "If you think of concentration as the number of molecules of the substance of interest divided by the number of all other molecules in the mixture then concentration will have the same limits as probability, that is, between 0 and 1.",
        "video_name": "ONBJo7dXJm8",
        "timestamps": [
            300
        ],
        "3min_transcript": "and in the right place and kind of close enough in order for the reaction to happen. So the reaction is really going to be driven by, if you think about it, the probability of finding a V molecules and b molecules all within close enough confines that they can actually react. So you could say that the rate is going to be driven by -- maybe it's going to be proportional. Let's say it's just equal to-- let's say some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent just on them being there. You have to have worry about their kinetic energies. You have to worry about their shape, because some shapes are going to be more conducive to reaction than others. So let's just let that be taken into account with a K. And we're talking about the forward reaction, right? So in order for the forward reaction to happen, let's call that K plus for the forward reaction. We have to have So what's the probability of having a molecules of X? Or what's a rough approximation of the probability? Well, the concentration. Let's think about this a second. When we write the concentration of the molecule V, which I think when I did this was the blue one right here, what is that given in? That is given in moles per liter. Moles is just a number, so this tells us, look, in any given volume, roughly how many of the molecules do you expect to find? That's what concentration is. So if I wanted to figure out the probability of finding a of these molecules, because that's how many I need, I need to multiply this by itself a times, because I need a of them. The probability of having just one molecule in just some small fraction, you would just use the concentration once. But you're going to use it a times, because you want a of those molecules there, right? what's the probability of having five heads? Well you would multiply the probability of one head five times. So the forward reaction probability is going to be the concentration of V to the a power, and, of course, that's not enough to have the reaction happen. You also need to have b of the X molecules there. So you have the concentration of X to the b power. And I want to make sure you understand this. My claim is that this is approximation -- or actually it's a pretty good way of calculating-- the probability. So let me write it this way. The rate is equal to some constant that takes into account the temperature and the molecular configurations times the probability of having a V molecules and b X molecules"
    },
    {
        "Q": "\nat 4:40,sal says when sun is just setting,why not observe it as sun is just rising; like in the previous case?",
        "A": "If you looked at the star as the sun was rising exactly 6 months later, while on the other side of the sun, you would be facing the wrong direction. At 5:21, Sal says that at sunset 6 months later straight up is the same direction .",
        "video_name": "ETzUpoqZIHY",
        "timestamps": [
            280
        ],
        "3min_transcript": "up is going to be at some angle to the left of straight up. It's going to be right over there. And obviously the star won't be that big relative to your entire field of vision, but you get the idea. Maybe I'll draw it a little bit smaller, just like that. So there's going to be some angle here. And this angle, whatever it is, let's just call it Theta, that's going to be the same angle as this. And when I talk about the angle, I'm talking about if you measure from one side of the horizon to the other side of the horizon, you're essentially looking halfway around the earth. That would be 180 degrees. So you could literally measure what this angle is right over here. Now let's say we waited six months. What's going to happen? Six months, we're going to be on this side of the sun. We're assuming that our distance is relatively constant at one astronomical unit. Remember, the earth is rotating like this. So if we wait, right at sunset, right when the last glimpse of the sun has just gone away-- because you can remember, right now, the sun is illuminating this side of the earth. The sun is going to be illuminating that side of the earth. So if we're sitting right at the equator right over there, right when the sun is just setting, we look straight up. Let me do that in the same color. We look straight up. So six months later when we look straight up, where is the star relative to straight up? Well now the star will be to the right. It'll be in the direction. So if this is our field of vision six months later, now the sun is setting all the way to the right, on the right horizon. And if we look straight up, this star now is going to be to the right of straight up. Well, it looks like relative to straight up-- and we're looking at the exact kind of position of the earth. We're making sure that we're picking times of year and times of day where straight up is the same direction. We're looking in the same direction of the universe. It looks like the position of that star has actually shifted. And let's say that this is the middle of summer, and that this is the middle of winter. Doesn't have to be. It could be any other two points in time six months apart. Then when we look at this star in the summer, it's going to be over here. Summer, it's going to be right over there. And when we look at the star in the winter, it is going to be over here. And, in general, for any star, especially stars that are in the same plane as the solar system, you can find two points in the year where that star is at a kind of a maximum distance from center."
    },
    {
        "Q": "\nAt 3:46, the switching on and off is called pulse-width modulation, right?",
        "A": "That is correct.",
        "video_name": "a16uKH2K7gM",
        "timestamps": [
            226
        ],
        "3min_transcript": "And that's an AC wave, sign wave. And we want the power to flow in a straight line, which is DC. So the AC, basically if you look at it it means the current's flowing in one direction, then it switches and flows in the other. So it's going like this. So we want to convert it to DC. So those four diodes, these four guys right here, convert it to DC by basically flipping the wave over like this. And then the power only flows in one direction. Which is a key aspect there. Let's look at some of the other components. We have this resistor right here. This resistor helps the board in current sensing. Then we have these two capacitors here. They are basically part of an electromagnetic interference filter. This is the transformer. The transformer is a high frequency transformer. In order for a transformer to function, there has to be a change in the voltage. The way we make change in the voltage is we do what's called pulse width modulation. The voltage is going to be coming out like this. And then it's creating a square wave, what's called a square wave. This is direct current, DC. But it's being... So this is 100% power and this is zero... So it's being switched on and off very fast. That turning on and off allows for the induction to happen, which occurs inside of this transformer here. The transformer steps the voltage down from 120 volts to both 12 volts and five volts. That pulse width modulation, or that square wave This is the IC chip, that controls the PWM or pulse width modulation. You can see there are a number of resistors here. They're to protect the circuitry on the IC chip most likely. We've got some diodes as well, those control the flow of current. The opto-isolator is used to sample output voltage and regulate it during different, basically different load conditions. And this guy right here, it looks like a transistor but it is actually a shunt regulator, which basically functions like a zener diode, or a variable zener diode. A zener diode allows current to flow in one direction to a certain point and then above a certain point, it allows current to flow in both directions. So, this sort of functions like a variable zener diode. And I wanted to say thank you to Toom-Too-Pro who provided some awesome feedback on the first video,"
    },
    {
        "Q": "@7:30 ish, if it was a high frequency, why would we hear? I thought we couldn't hear high frequencies and only certain animals could or is that just a myth?\n",
        "A": "He is saying that the high frequency affects other electronic devices, such as radios, it would have the same effect as when you are getting a phone call or message when you are near the radio.",
        "video_name": "a16uKH2K7gM",
        "timestamps": [
            450
        ],
        "3min_transcript": "This is the IC chip, that controls the PWM or pulse width modulation. You can see there are a number of resistors here. They're to protect the circuitry on the IC chip most likely. We've got some diodes as well, those control the flow of current. The opto-isolator is used to sample output voltage and regulate it during different, basically different load conditions. And this guy right here, it looks like a transistor but it is actually a shunt regulator, which basically functions like a zener diode, or a variable zener diode. A zener diode allows current to flow in one direction to a certain point and then above a certain point, it allows current to flow in both directions. So, this sort of functions like a variable zener diode. And I wanted to say thank you to Toom-Too-Pro who provided some awesome feedback on the first video, or switch-mode power supplies can work."
    },
    {
        "Q": "At 1:13 how does a radio uses electricity and how much ?\n",
        "A": "it uses 5watt-hours, or after 200 days of such use",
        "video_name": "a16uKH2K7gM",
        "timestamps": [
            73
        ],
        "3min_transcript": "- [Instructor] So this is our switch mode power supply for the DVD player. Now, the switch mode power supply operates at a high frequency. Linear power supplies operate at lower frequencies. The key differences are that switch mode power supplies tend to be smaller in form factor. They produce more electromagnetic interference, so they have to have a lot more filtering, but they also operate with much more efficiency. They can change the amount of power they supply and they therefore are able to... They don't use near as much power as a linear power supply. So those are the main differences there. Now, if you look right here, you can see this is where the power comes in from the house. There's a fuse here and this fuse is to protect the house from any failure on the board. So if there's a failure, the fuse will fail. This safety capacitor is also designed to fail in the case of a short. And it's got all these markings on it for regulation. It's basically meant to fail, and protect the house, again, if there's a problem. The induction coils here, and this capacitor here, they're basically used to filter out noise from the power supply. The power supply, since it operates at a high frequency produces noise. And these prevent the noise from going back through to the house and causing other problems in the line. Now, if you look here, you can see these four diodes. Diodes prevent current flow in one direction. So they act like an electrical valve. So, the four diodes are set up there to act as a bridge rectifier. And they take the current that comes in, which is DC current or AC current, and they convert it to DC current. So I can show you really quickly right here. And that's an AC wave, sign wave. And we want the power to flow in a straight line, which is DC. So the AC, basically if you look at it it means the current's flowing in one direction, then it switches and flows in the other. So it's going like this. So we want to convert it to DC. So those four diodes, these four guys right here, convert it to DC by basically flipping the wave over like this. And then the power only flows in one direction. Which is a key aspect there. Let's look at some of the other components. We have this resistor right here. This resistor helps the board in current sensing. Then we have these two capacitors here. They are basically part of an electromagnetic interference filter."
    },
    {
        "Q": "\nAt 0:24, the hydrogen atom is represented by the Bohr model but isn't the Bohr model incorrect because electrons are actually not orbiting the nucleus in a circle and actually can behave as a wave as well? Then why is this model used?",
        "A": "Because the Bohr model will help you explain basic things in Chemistry. You could use a quantum mechanical model of an atom or even think of electrons as cloud charges instead of having a particle-wave duality --- but why would you if a simpler model can suffice? Models are (almost) always wrong, not just in Chemistry either, and that s OK because they are supposed to help you understand and a Bohr model will do this.",
        "video_name": "AznXSVx2xX0",
        "timestamps": [
            24
        ],
        "3min_transcript": "- We've been talking about the Bohr model for the hydrogen atom, and we know the hydrogen atom has one positive charge in the nucleus, so here's our positively charged nucleus of the hydrogen atom and a negatively charged electron. If you're going by the Bohr model, the negatively charged electron is orbiting the nucleus at a certain distance. So, here I put the negatively charged electron a distance of r1, and so this electron is in the lowest energy level, the ground state. This is the first energy level, e1. We saw in the previous video that if you apply the right amount of energy, you can promote that electron. The electron can jump up to a higher energy level. If we add the right amount of energy, this electron can jump up to a higher energy level. So now this electron is a distance of r3, so we're talking about the third energy level here. This is the process of absorption. The electron absorbs energy and jumps up to a higher energy level. This is only temporary though, It's eventually going to fall back down to the ground state. Let's go ahead and put that on the diagram on the right. Here's our electron, it's at the third energy level. It's eventually going to fall back down to the ground state, the first energy level. Here's the electron going back to the first energy level here. When it does that, it's going to emit a photon. It's going to emit light. When the electron drops from a higher energy level to a lower energy level, it emits light. This is the process of emission. I could represent that photon here. This is how you usually see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is those two energy levels. We have energy with the third energy level and the first energy level. The difference between those... So, the energy of the third energy level minus the energy of the first energy level. That's equal to the energy of the photon. This is equal to the energy of that photon here. We know the energy of a photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant, this is Planck's constant. Nu is the frequency. We want to think about wavelength. We need to relate the frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light, lambda is the wavelength, and nu is the frequency."
    },
    {
        "Q": "5:25 what is brass?\n",
        "A": "Brass is basically an alloy of copper and zinc. If you do not know what an alloy is, it is just a metal formed by mixing two or more other metals to give it a special property. Steel for example is an alloy as well and its purpose is as you might know, is to make it super hard! Hope this was informative",
        "video_name": "qLMsZKx_a8s",
        "timestamps": [
            325
        ],
        "3min_transcript": "So it's really a handy way to do it. It used to be you'd have to look it up or just memorize what the different color band codes meant. But this particular resistor, it's got a green band. So we'll put it on green. There we go. And it's got a-- it looks like a navy blue band, and a gold band, green-- oh, wait. Actually, there's a black one. Sorry-- and then the gold one. There we go. So this is a 56 ohm resistor. And that's the amount of resistance that that resistor provides. And the switch right here is just a momentary switch. All right. It's not a momentary switch. It's a continuous switch. So that means when you push it down, it stays down. So the light will stay on after you push it. And the circuit is extremely simple. Basically, you've got the power from the batteries. It comes in through the loop here and the switch basically opens and closes and stops the power flow. Or when you push down on it, it closes and allows the power to flow in the continuous loop there. And so that's what's inside of a tap light. Let's take the batteries out really quick so. These are double A's. And now, it looks like this back panel here was injection molded. And you can see the ejector pins there. Those are the pins that push it out of the mold. And so it looks like it was injection molded. And I would say-- it doesn't have the plastic designation marking on it, but I would guess that it's probably either polypropylene or ABS plastic. are probably made out of-- I initially thought that they were made out of steel. But let's take a look. We've got some magnets here, so we'll take one of our magnets and-- I don't think they are. So they're probably brass contacts, because the magnets are not attracted to them. So it's not a ferrous metal. So then we've got this loop here. And this plastic loop prevents the positive terminal on the battery from slipping below the contacts. So it stays in constant contact and keeps everything together. And you can see there's another one on this side. And that's pretty much it. Oh, and there's also a feature right here so that if you have a screw or a nail on your wall, you can put the tap light in and just hang it like that. But that's pretty much the tap light."
    },
    {
        "Q": "at 9:08, when you say force, are you talking about the net force?\n",
        "A": "yes, as a body is constantly acted upon by a number of forces at any instant, it is rather easier to calculate net fore than the individual forces",
        "video_name": "Mz2nDXElcoM",
        "timestamps": [
            548
        ],
        "3min_transcript": "There will be no acceleration because this wedge is here. So the wedge is exerting a force that completely counteracts the force, the perpendicular component of gravity. You might guess what it's called. So the wedge is exerting a force, just like that, that's going to be 98 newtons upward. The wedge is going to be exerting a force that is 49 square roots of 3, because this right here is 49 square roots of 3 newtons into. And so this is 49 square roots of 3 newtons out of the surface, out of the surface. And this is the normal force. It is the force perpendicular to the surface that essentially, you could kind of view as the contact force that the, in this case, that the surface is exerting We're not talking about accelerating straight towards the center of the earth. We're talking about accelerating in that direction. We broke up the force into kind of the perpendicular direction and the parallel direction. So you have this counteracting normal force. And that's why you don't have the block plummeting or accelerating into the plane. Now what other forces do we have? Well, we have the force that's parallel to the surface. And if we assume that there's no friction-- and I can assume that there's no friction in this video because we are assuming that it is ice on ice-- what is going to happen? There's no counteracting force to this 49 newtons. 49 newtons parallel downwards, I should say parallel downwards, to the surface of the plane. So what's going to happen? Well, it's going to accelerate in that direction. You have force is equal to mass times acceleration. Or you divide both sides by mass, you get force over mass is equal to acceleration. Over here, our force is 49 newtons in that direction, parallel downwards to the surface of the plane. And so if you divide both by mass, if you divide both of these by mass. So that's the same thing as dividing it by 10 kilograms, dividing by 10 kilograms, that will give you acceleration. That will give you our acceleration. So acceleration is 49 newtons divided by 10 kilograms in that direction, in this direction right over there. And 49 divided by 10 is 4.9, and then newtons divided by kilograms is meters per second squared. So then you get your acceleration. Your acceleration is going to be 4.9 meters per second squared."
    },
    {
        "Q": "what is happening at 4:50?\n",
        "A": "At 4:50 Sal was finishing talking about the supposed formation of the Earth, and started talking about the supposed extinction of the dinosaurs.",
        "video_name": "DRtLXagrMHw",
        "timestamps": [
            290
        ],
        "3min_transcript": "Actually even on a worldwide basis, it was the first secular democracy based on a kind of a constitutional democracy that showed up on the planet. They said we don't want the king of England anymore. And this was about 234 years ago. And I always remembered because I was born almost on the 200th anniversary. So you just have to add my age to 200. So this is 234 years ago. So these are all events or periods of time that we've heard about and we've talked about. And people throw around these type of years. But what I want to do in this video is relate it to time scales that we can comprehend. So instead of the Big Bang occurring 13.7 billion years ago, let's pretend like it occurred 10 years ago. Because most of us, especially if you're over the age of 10, can kind of understand what 10 years is. It's a very, very long period of time. But something that's well within our lifetimes, well within our experience. Bang occurred 13.7 billion years ago, let's pretend like it occurred 10 years ago. And if we pretend that it occurred 10 years ago, let's think about how many years, or minutes, or hours ago each of these events would have occurred. So if Big Bang, which is really 13.7 billion years, if it really had occurred 10 years ago, and we scaled everything down, if we had scaled everything down, then the Earth would have been created about 3.3 years ago. So this would have been 3.3 years ago. So there's nothing kind of amazing about this. This is a significant fraction of the age of the universe. So not that mind blowing just yet. But if we go all the way to when the dinosaurs were extinct, the last land dinosaurs, now the 65 million years-- and this will give you an appreciation of the difference between million then the dinosaurs would have been extinct 17 days ago. Not even a month ago, the dinosaurs would have been extinct. So if the universe was created when I was just graduated-- well, I'm in my '30s now, so when I was 24-- just last month, the dinosaurs would have gone extinct. And it gets even crazier. 17 days ago, the dinosaurs would have extinct. Australopithecus afarensis would have walked on the Earth 19 hours ago, yesterday. 19 hours ago, she would have been walking around And modern humans wouldn't have shown up until 80 minutes ago, 80 minutes, a little over an hour. There wasn't even a modern human. Then the universe was 10 years, it didn't take until just very recently, the last hour, for us to see someone that looks something like us, looks and thinks something like us."
    },
    {
        "Q": "\n2:10 How can we know at what time humans had evolved to the point that they were \"modern humans\" who appeared and thought in similar ways to the way we did today? How can we really project an estimate for something like that? Evolutions is really a gradual process, and people say that we are still evolving now.",
        "A": "Highly sophisticated, beautiful paintings of animals found in caves in Europe have been dated back 40,000 years. That says to me that modern humans have been around for at least 40,000 years.",
        "video_name": "DRtLXagrMHw",
        "timestamps": [
            130
        ],
        "3min_transcript": "What I've done here is I've copied and pasted a bunch of pictures that signify events in our history, when you think about history on a grander scale, that most of us have some relation to or we kind of have heard it talked about a little bit. And the whole point of this is to try to understand, or try to begin to understand, how long 13.7 billion years is. So just to start off, I have here-- this is the best depiction I could find where it didn't have copyrights. This is from NASA-- of the Big Bang. And I've talked about it several times. The Big Bang occurred 13.7 billion years ago. And then if we go a little bit forward, actually a lot forward, we get to the formation of our actual solar system and the Earth. This is kind of the protoplanetary disk or a depiction of a protoplanetary disk forming around our young Sun. And so this right here is 4.5 billion years ago. Now this over here-- once again, these aren't pictures of them. These are just depictions because no one was there This is what we think the asteroid that killed the dinosaurs looked like when it was impacting Earth. And it killed the dinosaurs 65 million years ago. So until then, we had land dinosaurs. And then this, as far as the current theories go, got rid of them. Now, we'll fast forward a little bit more. At about 3 million years ago-- let me do this in a color that you can see-- about 3 million, so three million years ago, our ancestors look like this. This is Australopithecus afarensis. This is I think a depiction of-- this is Lucy. I believe the theory is that all of us have some DNA from her. But this was 3 million years ago. And you fast forward some more and you actually that looked and thought like you and me. This is 200,000 years ago. That's right over here. Obviously, this drawing was done much later. But this is a depiction of a modern human, so 200,000 years ago. And then you fast forward even more. And I don't want to keep picking on Jesus. I did that with him getting on the jet liner. And I genuinely don't mean any offense to anyone. I just keep picking Jesus because frankly our calendar is kind of-- he's a good person that most people know about, 2,000 years ago. And so when we associate kind of a lot of modern history occurring after his birth. So this right here is obviously a painting of the birth of Jesus. And this is 2,000 years ago. And then this might be a little bit American-centric. But the Declaration of Independence, it"
    },
    {
        "Q": "\nAt 6:40,how is velocity 19.6m/s if time is zero ?",
        "A": "You can start the stopwatch whenever you want, right?",
        "video_name": "T0zpF_j7Mvo",
        "timestamps": [
            400
        ],
        "3min_transcript": "so times zero, what is our velocity? well if we use this expression right here time zero or delta t equal zero this expression right here is gonna be zero, and it's just going to be initial velocity, in the last video we gave our initial velocity is going to be 19.6m/s, so it is going to be 19.6m/s I will plot that over here, time zero, it's going to be 19.6 m/s what is our initial displacement at time zero? Our change in time zero, so you look at this up here delta t is zero so this expression here is going to be zero so we haven't done any displacement yet when no time has gone by So we have done no displacement, we are right over there what is now our velocity? well our initial velocity right here is 19.6m/s that was a given, and our acceleration is negative 9.8m/s*s so it's negative right over there and you multiply that by delat t in every situation, so in this situation we are gonna multiply it by 1 delta t is 1, so you have 19.6 minus 9.8 that gives exactly 9.8m/s and the unit we got cause we multiply here is second so its give us meters per second 19.6m/s minus 9.8m/s one of these seconds go away multiply by second give you 9.8m/s so after 1s our velocity is now half of what it was before so we are now going 9.8m/s, let me draw a line here Let me rewrite this displacement formular here with all the infomation So we know that displacement is going to be equal to initial veloctiy which is 19.6m/s now I won't write the units here just for the sake of space times change in time, times our, use the same color to see what is what times our change in time, plus one half let me be clear one half times negative 9.8m/s*s so one half times a is going to be I rewrite this right over here cause this is gonna be negative 9.8m/s*s times one half, so this is going to be negative 4.9 All I did is one half times negative 9.8 over here It is important that is why the vector quantity is start to matter because if you put a positive here you wouldn't have the obeject slowing"
    },
    {
        "Q": "At 14:06 in the video, Sal says the velocity is decreasing at a constant pace. Isn't it really decreasing gradually, and then increasing gradually in the opposite direction?\n",
        "A": "isnt that the same thing? Think about the acceleration. What is its value throughout?",
        "video_name": "T0zpF_j7Mvo",
        "timestamps": [
            846
        ],
        "3min_transcript": "we are talking about delat t, our change in time is 3s so that's square So times 9 and that gives us 14.7 meters, so 14.7 meters So after 3s we are 14.7m again, the same position with 1s but the difference is now we are moving downard over here we were moving upwards and finally what happen after 4s? Or what's our velocity? Let me just get the calculator out or you might be figure this out in your head Our velocity is going to be 19.6 - 9.8 times 4s just minus 19.6 m/s So we are going a magnitude of velocity which is the same initially threw the ball except now it's going at opposite direction and what is our displacement? get the calculator out So we have out displacement is 19.6 times 4, 4s has gone by minus 4.9 times 4 square which is 16 times, which is equal to zero! the displacement here is zero! We are back on the ground So if you plot the displacement, you will actually got a parabola a downward opening parabola that looks something like this I best draw it relatively neatly So my check to do it better than that Dotted line, dotted line is always easier to adjust than its streamed so if you plot displacement verses time it look something like this it's velocity just downward sloping line, and the acceleration is constant that velocity the whole time is decreasing at constant rate and that make sense because that's the rate which the velocity increases and decreases as the acceleration, and the acceleration base on our convention is downward so that why it's decreasing, we have a negative slope here we have a negative slope of negative 9.8 m/s*s and just to think about what's happening for this ball I know this video is getting long as it goes I'm gonna draw the vectors for velocity So I'm gonna do that in orange, or maybe I will do that in blue So velocity in blue, so right when we start, it has a positive velocity of 19.6m/s, so I will draw a big vector like this 19.6m/s that's velocity, but after 1s is 9.8m/s so half of that so then its maybe would look something like this"
    },
    {
        "Q": "Around near 2:40,if both the vectors point toward the centre of circle i,e gravity,what will be the force on the Wooden floor then?\n",
        "A": "Newton s Third law says that if the Wooden floor is exerting a force on the ball then the floor is exerting the same amount of force on the ball.",
        "video_name": "Xpgsg-fY4DY",
        "timestamps": [
            160
        ],
        "3min_transcript": "the force between the two surfaces? This is what we'd have to know in order to figure out if our structure is strong enough to contain this bowling ball as it goes around in a circle. And it's also a classic centripetal force problem, so let's do this. What do we do first? We should always draw a force diagram. If we're looking for a force, you draw a force diagram. So what are the forces on this ball? You're gonna have a force of gravity downward, and the magnitude of the force of gravity is always given by M times G, where G represents the magnitude of the acceleration due to gravity. And we're gonna have a normal force as well. Now which way does this normal force point? A common misconception, people wanna say that that normal force points up because in a lot of other situations, the normal force points up. If you're just standing on the ground over here, the normal force on you is upward because it keeps you from falling through the ground, but that's not what this loop structure's doing up here. The loop structure isn't keeping you up. The loop structure's keeping you from flying out of the loop have to point downward. So this is weird for a lot of people to think about, but because the surface is above this ball, the surface pushes down. Surfaces can only push. If the surface is below you, the surface has to push up. If the surface was to the side of you, the surface would have to push right. And if the surface was to the right of you, the surface would have to push left. Normal forces in other words, always push. So the force on the ball from the track is gonna be downward but vice versa. The force on the track from the ball is gonna be upward. So if this ball were going a little too fast and this were made out of wood, you might see this thing splinter because there's too much force pushing on the track this way. But if we're analyzing the ball, the force on the ball from the track is downward. And after you draw a force diagram, the next step is usually, if you wanna find a force, to use Newton's Second Law. And to keep the calculation simple, we typically use Newton's Second Law for a single dimension at at time, i.e. vertical, horizontal, centripetal. because the normal force is pointing toward the center of the circular path and the normal force is the force we wanna find, we're gonna use Newton's Second Law for the centripetal direction and remember centripetal is just a fancy word for pointing toward the center of the circle. So, let's do it. Let's write down that the centripetal acceleration should equal the net centripetal force divided by the mass that's going in the circle. So if we choose this, we know that the centripetal acceleration can always be re-written as the speed squared divided by the radius of the circular path that the object is taking, and this should equal the net centripetal force divided by the mass of the object that's going in the circle and you gotta remember how we deal with signs here because we put a positive sign over here because we have a positive sign for our centripetal acceleration and our centripetal acceleration points toward the center of the circle always. Then in toward the center of the circle is going to be our positive direction,"
    },
    {
        "Q": "\nat 11:44 when Hank is describing the electron configuration chain I got lost what exactly is happening and where is the ATP being produced?",
        "A": "ATP is produced in the mitochondria inside the cell.",
        "video_name": "CIyAs0bxeoI",
        "timestamps": [
            704
        ],
        "3min_transcript": "that are related to B vitamins. Derivatives of Niacin and Riboflavin, which you might have seen in the vitamin aisle. These B vitamins are good at holding onto high energy electrons and keeping that energy until it can get released later in the electron transport chain. In fact, they're so good at it, that they show up in a lot of those high-energy vitamin powders that the kids are taking these days. NAD+s and FADs are like batteries, big awkward batteries that pick up hydrogen and energized electrons from each pyruvate, which in effect charges them up. The addition of hydrogen turns them into NADH and FADH2, respectively. Each pyruvate yields three NADHs and one FADH2 per cycle, and since each glucose has been broken down into two pyruvates that means each glucose molecule can produce six NADHs and two FADH2s. The main purpose of the Krebs Cycle is to make these powerhouses And now comes the time when your saying, \"Sweet pyruvate sandwiches, Hank, \"aren't we supposed to be making ATP here? \"Let's make it happen, Capt'n! What's the holdup?\" Well friends, your patience is finally paying off because when it comes to ATPs, the electron transport chain is the real moneymaker. In a very efficient cell, it can net a whopping 34 ATPs. So, remember all those NADHs and FADH2s we made in the Krebs Cycle? Well, their electrons are going to provide the energy that will work as a pump along a chain of channel proteins across the inner membrane of the mitochondria where the Krebs Cycle occurred. These proteins will swap these electrons to send hydrogen protons from inside the very center of the mitochondria, across its inner membrane to the outer compartment of the mitochondria. But once they're out, the protons want to get back to the other side of the inner membrane, because there's a lot of other protons out there and as we've learned, nature always tends to seek a nice, peaceful balance So all of these anxious protons are allowed back in through a special protein called ATP synthase. And the energy of this proton flow drives this crazy spinning mechanism that squeezes some ADP and some phosphates together to form ATP. So, the electrons from the 10 NADHs that come out of the Krebs Cycle, have just enough energy to produce roughly three ATPs each. And we can't forget our friends the FADH2s. We have two of them and they make two ATPs each. And voila! That is how animal cells, the world over, make ATP through cellular respiration. Now just to check, let's reset our ATP counter and do the math for a single glucose molecule once again. We made two ATPs for each pyruvate during glycolysis. We made two during the Krebs Cycle, and then during the electron transport chain we made about 34. And that is just for one molecule of glucose. Imagine how much your body makes and uses every single day."
    },
    {
        "Q": "\nat 3:57 Hank is saying that usually 29-30 molecules of ATP produced for 1 glucose, but if it was a best scenario it would be 38 ATPs, so where do the rest ATPs are going, why there is not always 38?",
        "A": "People are still studying this topic.",
        "video_name": "CIyAs0bxeoI",
        "timestamps": [
            237
        ],
        "3min_transcript": "Same goes with energy, in order to be able to use it, our cells need energy to be transferred into adenosine triphosphate to be able to grow, move, create electrical impulses in our nerves and brains, everything. A while back, for instance, we talked about how cells use ATP to transport some kinds of materials in and out of its membranes. To jog your memory about that, you can watch that episode right here. Now before we see how ATP is actually put together, let's look at how cells can cash in on the energy that's stashed in there. Well, adenosine triphosphate is made up of a nitrogenous base called adenine with a sugar called ribose and three phosphate groups attached to it. Now one thing you need to know about these three phosphate groups, is that they are super uncomfortable sitting together in a row like that, like three kids on a bus who hate each other all sharing the same seat. So because the phosphate groups are such terrible company for each other, ATP is able to do this nifty trick where it shoots one of the phosphate groups off the end of the seat, creating ADP, or adenosine diphosphate, because now, And this reaction when the third jerk kid is kicked off the seat, energy is released. And since there are a lot of water molecules just floating around nearby, an OH pairing, that's called a hydroxide, from one of the H2Os comes over and takes the place of that third phosphate group, and everybody is much happier. By the way, when you use water to break down a compound like this, it's called hydrolysis, \"hydro\" from water and \"lysis\" from the Greek word \"for separate\". So now that you know how ATP is spent, let's see how it is minted, nice and new, by cellular respiration. Like I said, it all starts with oxygen and glucose. In fact, textbooks make a point of saying that through cellular respiration, one molecule of glucose can yield a bit of heat and 38 molecules of ATP. Now, it's worth noting that this number is kind of a best-case scenario. Usually it's more like 29 or 30 ATPs, but whatever. People are still studying this stuff, so let's stick with that number, 38. Now, cellular respiration isn't something that just happens all at once. Glucose is transformed into ATPs over three separate stages. the Krebs Cycle, and the electron transport chain. Traditionally, these stages are described as coming one after the other but really everything in the cell, is kind of happening all at the same time. But let's start with the first step, glycolysis, or the breaking down of the glucose. Glucose, of course, is a sugar, you know this because it's got an \"ose\" at the end of it. And glycolysis is just the breaking up of glucose's six-carbon ring into two three-carbon molecules called pyruvic acids or pyruvate molecules. Now in order to explain how exactly glycolysis works, I'd need about an hour of your time, and a giant cast of finger puppets each playing a different enzyme, and though it would pain me to do it, I would have to use words like phosphoglucoisomerase, but a simple way of explaining it, is like this, if you wanna make some money, you gotta spend some money. Glycolysis needs the investment of two ATPs in order to work and in the end, it generates four ATPs, for a net profit if you will of two ATPs."
    },
    {
        "Q": "\nHow do you know when to do a hydride shift? Couldn't you of just moved the Cl where the positive charge is in the second hexane? 6:28",
        "A": "You want the most stable carbocation - meaning you want that positive charge to be on a carbon connected to as many other carbons as possible, or to be resonance stabilized. That hydride shift happens to move the positive charge onto the most stable carbocation (from the secondary to the tertiary carbon).... Tertiary > Secondary > Primary being the order of preference there.",
        "video_name": "iEKA0jUstPs",
        "timestamps": [
            388
        ],
        "3min_transcript": "So let's go ahead and write that. See if we can spell Markovnikov. The halogen adds the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement. So let's start out with this as our alkene and react that with hydrochloric acid once again. First steps-- pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? Right? We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course, we can. There's a hydrogen attached to this carbon. And we saw-- in our earlier video on carbocations and rearrangements-- we could get a hydride shift here. All right. So the proton and these two electrons here are hydride anion. to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well, now our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now, this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion, negatively charged nucleophile file. So a nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up"
    },
    {
        "Q": "At 12:00 carbon at 1 is categorized as tertiary because it is connected to 3 other carbons (implies connection to ring (connection beyond zigzag) is not taken into account. However, later at 12:42 carbon 1 is said to be connected to 3 other carbons and hence tertiary while according to earlier logic the connection to the ring should not be counted and so carbon #1 is not tertiary it is secondary. What am I missing?\n",
        "A": "That carbon 1 at 12:00 is quaternary, but he seems to be talking about where the prefixes might come from not the actual classification of those carbons.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            720,
            762
        ],
        "3min_transcript": "And I'm going to once again take off one of the hydrogens. So I'll make it an R double prime group. And now if I wanted to classify my central carbon, now this is connected to 1, 2, 3 other carbons, so it is said to be tertiary. So that is a tertiary carbon like that. And finally, I have one more example, of course. I take off the last hydrogen. So now I have R, R prime, R double prime, and R triple prime. So what is the classification of this carbon now connected to four other carbons? So it is said to be quaternary. So that is a quaternary carbon right here. All right, so if I'm trying to think about where some of these common names come from I can see, oh, well, right here I have-- well, that would be secondary. So S-E-C for my prefix. So let's go back up here, and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary, so I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons, so it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again I can say, well, that would make sense, because if I look at this carbon, it's bonded to three other carbons, right? And once again I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons, so it is said to be a secondary. This carbon right here is bonded to three other carbons so it is tertiary. This carbon is bonded to one other carbon so it is primary. This carbon is bonded to three other carbons, so it is tertiary. And all of the carbons on the ring right here are bonded to two other carbons, so they are all said to be secondary. So that's a very important skill to develop classifying your carbons."
    },
    {
        "Q": "\nAt 1:05, why is the longest carbon chain consisting of only 2 carbons? Why cant it be 3 carbons?",
        "A": "You number the continuously-attached carbons of the side-chain from the carbon that is attached to the main chain. If you start at C-1, you can go out only one carbon with your pencil. To get to the third carbon without removing your pencil, you would have to re-trace your steps back to C-1, and that is not permitted.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            65
        ],
        "3min_transcript": "So how do we name this molecule? Well, we start with the longest carbon chain. So there are seven carbons in my longest carbon chain. So I would call this heptane. And I number it to give the substituent the lowest number possible. So in this example, it doesn't really matter if I start from the left or from the right. In both examples, you would end up with a 4 for your substituent there. Now, this substituent looks different from ones we've seen before. There are three carbons in it, but those carbons are not in a straight-chain alkyl group. So if I look at it, right there are three carbons, but they're not going in a straight chain. They're branching of branching here. So this is kind of weird. How do we name this substituent? Well, down here, I have the same substituent, and I'm going to draw this little zigzag line to indicate that that substituent is coming off of some straight-chain alkane. And when you're naming a complex substituent like this, you actually use the same rules that you would use for a straight-chain alkane. which in this case is only two carbons. So that would be an ethyl group coming off of my carbon chain. So I'm going to go ahead and name that as an ethyl group. I'm going to go ahead number it to give my branching group there the lowest number possible. So I go 1 and 2. So what is my substituent coming off of my ethyl group? Well, that's a methyl group coming off of carbon 1. So I name it as 1-methylethyl. OK, so now, that complex substituent is named as 1-methylethyl. So I could go ahead and put that into my name. So coming off of carbon 4, I have 1-methylethyl. And I'm going to put that in parentheses. And all of that is coming off of carbon 4 for my molecule. way of naming that molecule. So if your naming your complex substituent as 1-methylethyl, that's the official IUPAC way, but there are also common names for these complex substituents. So the common name for 1-methylethyl is isopropyl. So isopropyl is the common name. And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well. So you could have said, oh, this is 4-isopropylheptane, and you would have been absolutely correct. So that's yet another IUPAC name. So iso means same, and it probably comes from the fact that you have these two methyl groups giving you this Y shape that are the same. So that's one complex substituent, one that has three carbons on it. Let's look at a bunch of complex substituents"
    },
    {
        "Q": "At 12:48, isn't that carbon bonded to 4 carbons\n",
        "A": "no it bonded to 3 carbons and the extra one available bond is attached to hydrogen , and this is why its called Primary",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            768
        ],
        "3min_transcript": "well, that would be secondary. So S-E-C for my prefix. So let's go back up here, and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary, so I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons, so it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again I can say, well, that would make sense, because if I look at this carbon, it's bonded to three other carbons, right? And once again I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons, so it is said to be a secondary. This carbon right here is bonded to three other carbons so it is tertiary. This carbon is bonded to one other carbon so it is primary. This carbon is bonded to three other carbons, so it is tertiary. And all of the carbons on the ring right here are bonded to two other carbons, so they are all said to be secondary. So that's a very important skill to develop classifying your carbons. with different functional groups."
    },
    {
        "Q": "\nAt about 8:45 in the video i just wanted to verify that FADH is being created and not FADH2. Is the 2 supposed to be out in front of the FADH?",
        "A": "It s easy to get these confused, because two things are created: NADH and FADH2. The 2 should appear as a subscript (as in, there are 2 hydrogens). In addition, plants don t use NAD+/NADH, they use NADP+/NADPH. Just remember, at the end of the day, these molecules are very similar in their function, as they all contribute to the electron transport chain (next video).",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            525
        ],
        "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's"
    },
    {
        "Q": "At 8:32 , Sal says that FAD is oxidised to FADH2 when before he mentioned that the NAD+ molecules are reduced to NADH2. This seems contradictory to me since they both involve the gain of hydrogen, therefore wouldn't that mean that both the NAD+ and FAD are reduced? If so then how do my notes say that the CAC includes oxidations which generate high-energy electrons that will be used to power the synthesis of ATP ?\n",
        "A": "Oh I think I get it now,,, they are produced by reduction making them high electron carriers and they are then oxidized in the ETC for synthesis of ATP. Thanks :)",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            512
        ],
        "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's"
    },
    {
        "Q": "\nat 08:37 sal says FADis oxideised......but FAD is gaining proton...so as per the rule(OIL RIG) reduction is gaining proton so FAD will be reduced not oxidised.",
        "A": "Oxidation and reduction involve the lose or gain or electrons, not protons. What Sal is saying is that in the Krebs cycle, FAD is first reduced and gains electrons becoming FADH2, which it later looses (is oxidized) for the electron transport chain.",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            517
        ],
        "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's"
    },
    {
        "Q": "\nAt 5:00 you say the force to move the particle towards the charge needs to be equal and opposite to the force being pushed out...but shouldnt it be greater? Shouldnt the force needed to move the particle towards the charge be greater since equal and opposite forces would just cancel each other out?",
        "A": "Equal and opposite force pairs do not cancel out because they don t act on the same object. The moon is attracted to earth with the same force that the earth attracts the moon. If you draw those forces, you will see that only one of them is acting on the moon, and one of them is acting on the earth, so there s no canceling going on. This is a common hurdle people face in understanding Newton s 3rd law.",
        "video_name": "CqsYCIjSm9A",
        "timestamps": [
            300
        ],
        "3min_transcript": "Because the field is pushing it outward. It takes work to push it inward. So let's say we want to push it in. Let's say it's at 10 meters. Let's say that this distance right here-- let me draw a radial line-- let's say that this distance right here is 10 meters, and I want to push this particle in 5 meters, so it eventually gets right here. This is where I'm eventually going to get it so then it's going to be 5 meters away. So how much work does it take to move it 5 meters towards this charge? Well, the way you think about it is the field keeps changing, right? But we can assume over a very, very, very, very infinitely small distance, and let's call that infinitely small distance dr, change in radius, and as you can see, we're about to If you don't understand what any of this is, you might want to review or learn the calculus in the calculus playlist, but how much work does it require to move this particle a very, very small distance? Well, let's just assume over this very, very, very small distance, that the electric field is roughly constant, and so we can say that the very, very small amount of work to move over that very, very small distance is equal to Coulomb's constant q1 q2 over r squared times dr. Now before we move on, let's think about something for a second. Coulomb's Law tells us that this is the outward force that this charge is exerting on this particle or that the field is exerting on this particle. The force that we have to apply to move the particle from here to here has to be an inward force. to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters"
    },
    {
        "Q": "I have a question regarding the integral setup at 6:44. Mr. Khan said that the lower limit should be x=10 and upper limit x=5. However, would we still need the negative sign (attached to the force equation) since the integral takes care of direction? Or does the sign stuff actually matter (I'm not sure if work is considered a vector or scalar quantity)? Thank you and great video!\n",
        "A": "Yes,we would still require the negative sign as it specifies that we are moving in the opposite direction of the force..and integral has negative sign because we are going from 10 to 5.",
        "video_name": "CqsYCIjSm9A",
        "timestamps": [
            404
        ],
        "3min_transcript": "to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here."
    },
    {
        "Q": "At around 7:17 wasn't Sal meant to write 1/-k*q1*q2* the integral? This doubt arised due to Sal getting rid of the -k*q1*q2 by diving the integral by the term.\n",
        "A": "Here -k*q1*q2 is just a constant. Because of the linearity of the integration the integral of c*f is equal to c times the integral of f, where f is some function and c is a constant.",
        "video_name": "CqsYCIjSm9A",
        "timestamps": [
            437
        ],
        "3min_transcript": "to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here."
    },
    {
        "Q": "At the minute 2:34 I cannot understand why the middle C of the propyl group is a secondary carbon: ok, it's a secondary carbon if I don't consider its bond with the main chain (DECANE)... So I don't have to consider the bond with the main chain? Always?\n",
        "A": "Is it a secondary carbon because it is the second-to-last carbon in the group?",
        "video_name": "O9RPGJcAfJk",
        "timestamps": [
            154
        ],
        "3min_transcript": "So we have this group right over here. This has two carbons in it, one carbon, two carbons. And so, because it has two carbons, we would use the prefix eth. Remember, meth is one carbon, eth is two carbons. And since it's a group and we're not talking about the backbone, this is an ethyl group. And we have another ethyl group right over here. Two carbons attached right over here. This is also an ethyl group. And now, this group right over here is interesting. We can count the carbons in it. So it has 1, 2, 3 carbons. So you could think about, well, this has three carbons. Our prefix for three carbons is prop. So you could say, hey, maybe this is a propyl group. This right over here, you could say maybe this is propyl group. And you wouldn't be completely off base by saying that. But we have to be a little bit more careful when we name it. assume that you're attaching to one end of the propyl group. But we're not attaching to one end of the propyl group. We're attaching essentially to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached only one carbon, it would be a primary carbon. Since we're attached to the secondary carbon right over here, this is sometimes called a sec-propyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. And so the backbone right over here is ethyl. Let me write this right over here. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon, this is the two carbon. So this right over here, this is just one carbon group. This right over here is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methylethyl."
    },
    {
        "Q": "\nAt 4:15, Mr. Sal Khan mentioned the systematic name 1-methylethyl. Would it be wrong to name this compound in alphabetical order( i.e.. 1-ethylmethyl)?",
        "A": "Yes, it would be wrong. The longest chain in the group is 2 carbons long, so the base name is ethyl. There is a methyl group on C1, so the name of the group is 1-methylethyl.",
        "video_name": "O9RPGJcAfJk",
        "timestamps": [
            255
        ],
        "3min_transcript": "assume that you're attaching to one end of the propyl group. But we're not attaching to one end of the propyl group. We're attaching essentially to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached only one carbon, it would be a primary carbon. Since we're attached to the secondary carbon right over here, this is sometimes called a sec-propyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. And so the backbone right over here is ethyl. Let me write this right over here. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon, this is the two carbon. So this right over here, this is just one carbon group. This right over here is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methylethyl. Now, the systematic name, you might say, hey, why go through the pain of doing this for something so simple that we could just call isopropyl? This is useful if this was a much larger or a much more complex group that was attached to this main chain. But more typically, and this is why it's called the common name, you'll see this thing right over here just called isopropyl. And sometimes you would see it called sec-propyl even s-propyl. Now that we've named all of the groups, let's think about what carbons they are attached to and where we can start counting from. And the way that this is done is that you would start counting from the end of your carbon chain, this decane backbone, and you'd count from the end that bumps into the most groups faster. So, for example, if you count from this end, this would be the 1 carbon, 2 carbon, 3 carbon, 4 carbon,"
    },
    {
        "Q": "In the \"work example problems\" video, they say that in lifting an object at constant velocity, the net work is 0, since the change in kinetic energy is 0 (we give energy into the system to push it up, and the gravity does negative work...in total, no work is done). They don't even consider PEgravitational.\n\nBut here, they say work is the same as change in energy, so lifting an object changes its gravitational potential energy and thus wok is done (2:35).\n\nWas the first video wrong?\nThanks\n",
        "A": "The NET work includes the work you do to lift AND the work done by gravity, which is opposite of the work you do. So the KE doesn t change. But you did work against gravity so you added PE to the system.",
        "video_name": "sZG-zHkGR4U",
        "timestamps": [
            155
        ],
        "3min_transcript": "One way to find the amount of work done is by using the formula Fd cosine theta. But this number for the amount of work done represents the amount of energy transferred to an object. For instance, if you solve for the work done and you get positive 200 joules, it means that the force gave something 200 joules of energy. So if you have a way of determining the amount of energy that something gains or loses, then you have an alternate way of finding the work done, since the work done on an object is the amount of energy it gains or loses. For instance, imagine a 50-kilogram skateboarder that starts at rest. If a force starts the skateboarder moving at 10 meters per second, that force did work on the skateboarder since it gave the skateboarder energy. The amount of kinetic energy gained by the skateboarder is 2,500 joules. That means that the work done by the force on the skateboarder was positive 2,500 joules. It's positive because the force on the skateboarder If a force gives energy to an object, then the force is doing positive work on that object. And if a force takes away energy from an object, the force is doing negative work on that object. Now imagine that the skateboarder, who's moving with 10 meters per second, gets stopped because he crashes into a stack of bricks. The stack of bricks does negative work on the skateboarder because it takes away energy from the skateboarder. To find the work done by the stack of bricks, we just need to figure out how much energy it took away Since the skateboarder started with 2,500 joules of kinetic energy and ends with zero joules of kinetic energy, it means that the work done by the bricks on the skateboarder was negative 2,500 joules. It's negative because the bricks took away energy from the skateboarder. Let's say we instead lift the bricks, which are 500 kilograms, upwards a distance of four meters. To find the work that we've done on the bricks, we could use Fd cosine theta. We could just figure out the amount of energy that we've given to the bricks. The bricks gain energy here. And they're gaining gravitational potential energy, which is given by the formula mgh. If we solve, we get that the bricks gained 19,600 joules of gravitational potential energy. That means that the work we did on the bricks was positive 19,600 joules. It's positive because our force gave the bricks energy. This idea doesn't just work with gravitational potential energy and kinetic energy. It works for every kind of energy. You can always find the work done by a force on an object if you could determine the energy that that force gives or takes away from that object. [MUSIC PLAYING]"
    },
    {
        "Q": "where do you get the yellow thing that you used at 5:00?\n",
        "A": "those are alligator clips, they are wires with little alligator like clips on them.",
        "video_name": "Kq0Er6JBMmc",
        "timestamps": [
            300
        ],
        "3min_transcript": "for us to run on and allow us to run our high-current, high-voltage motors and control them with our low-power, low-current, low-voltage Arduino. So that's why we need the motor controller, and we go into more detail on that in the motor controller video. So let's get started taking our hair dryer motor apart, or taking it out of the hair dryer, I should say. And since that's going to be what we're using to move our craft around, we want to start to experiment with it and see how much power it's going to require and how much torque it's going to have and things like that. All right. so we're going to experiment with our motor and see what it's going to take to power it. We've got our alligator clips connected. And we're going to use a 1.5-volt AA battery to see if we can make the motor turn and to kind of get a sense for how much air it will pull through it at 1.5 volts. So we're running it, and we're holding the plastic up, and you can see it's not moving the plastic at all. So that's not going to work. We're going to need more volts than that. And we can increase the voltage by combining the cells in a battery holder, and that allows us to wire the cells in series so we go from 1.5 volts to over 12 volts because those cells are new. So let's see what impact that has on the motor. OK, so we're connecting our battery to our hair dryer motor. And whoa, you can see it's moving much more quickly now. Now it'll push the plastic completely out of the way. And we're getting a fair amount of air coming out of it. But I don't think we're going to use this method for moving our craft because, even though it's blowing a fair amount of air, it only works really efficiently in one direction. In the other direction, it doesn't work as well because it's only meant to blow air in one direction, out of the hair dryer. so we know exactly what we can run it on and how much voltage it needs. To do that, we're going to need to remove the propeller and the outer housing around the motor. So we're just going to trim that off with our hack saw there. And we're time lapsing this so you don't have to sit through all of it. But in any case, we're going to trim the propeller off. And then it's a really tough thing to get off because it's friction fitted on to this brass fitting on the end of the motor. And so it's really hard. They definitely did a good job of press fitting that onto the brass fitting so that it won't come off as the hair dryer moves around. So we're taking our nipper pliers here, and we're just going to trim the rest of the propeller off so we can get to the motor. And we'll move the end of it off there and then unscrew the last two screws and slide the ends off."
    },
    {
        "Q": "At 10:19 why does he say \"Then it goes to the heart, rubs up against some alveoli \"?\n",
        "A": "Well, rubbing up against some alveoli means that the oxygen diffuses from any alveolus (singular for alveoli) to enter in the blood stream as oxygenated blood.",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            619
        ],
        "3min_transcript": "And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs. And it's going away from the heart. It's the pulmonary artery and it is de-oxygenated. Then it goes to the heart, rubs up against some alveoli and then gets oxygenated and then it comes right back. Now this right here, we're going to the heart. So that's a vein. It's in the loop with the lungs so it's a pulmonary vein and it rubbed up against the alveoli and got the oxygen diffused into it so it is oxygenated. And then it flows into your left atrium. Now, the left atrium, once again, from our point of view, is on the right-hand side, but from the dude looking at it, it's his left-hand side."
    },
    {
        "Q": "At 3:56 he say blue one is an artery but at 7:15 Bro Sal says that blue one is a vein??\nEm confused! :/\n",
        "A": "Remember, blue simply represents deoxygenated blood. So, the pulmonary artery carries deoxygenated blood from the heart to the lungs. The vena cava veins carry deoxygenated blood from the rest of the body back to the heart.",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            236,
            435
        ],
        "3min_transcript": "So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose. we have the capillaries. So when we go away from the heart-- and we're going to delve a little bit into the heart in this video as well-- so when blood travels away from the heart, it's de-oxygenated. It's this blue color. So this right here is blood. This right here is blood traveling away from the heart. It's going behind these two tubes right there. So this is the blood going away from the heart. So this blue that I've been highlighting just now, these are the pulmonary arteries and then they keep splitting into arterials and all of that and eventually we're in capillaries-- super, super small tubes. They run right past the alveoli and then they become oxygenated and now we're going back to the heart. So we're talking about pulmonary veins. So we go back to the heart. Now we're going to go back to the heart. Hope you can see what I'm doing. And we're going to enter the heart on this side. You actually can't even see where we're entering the heart. We're going to enter the heart right over here-- and I'm going to go into more detail on that. Now we have oxygenated blood. And then that gets pumped out to the rest of the body. Now this is the interesting thing. When we're talking about pulmonary arteries and veins-- remember, the pulmonary artery was blue. As we go away from the heart, we have de-oxygenated blood, but it's still an artery. Then as we go towards the heart from the lungs, we have a vein, but it's oxygenated."
    },
    {
        "Q": "\nI thought the veins carried oxygenated blood, 9:08 says the vein carried deoxygenated blood ?? I'm confused",
        "A": "Arteries take blood away from the heart. Most arteries carry oxygenated blood away from the heart to other parts of the body. The veins take blood back to the heart. Veins carry oxygenated blood is carried in veins. The pulmonary artery takes blood away from the right ventricle to the lungs where it is oxygenated. The pulmonary vein takes oxygenated blood from the lungs to the left atrium - back to the heart.",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            548
        ],
        "3min_transcript": "So you don't see it. I'm going to do a detailed diagram in a second-- into the pulmonary artery. We're going away from the heart. This was a vein, right? This is a vein going to the heart. This is a vein, inferior vena cava vein. This is superior vena cava. They're de-oxygenated. Then I'm pumping this de-oxygenated blood away from the heart to the lungs. Now this de-oxygenated blood, this is in an artery, right? This is in the pulmonary artery. It gets oxygenated and now it's a pulmonary vein. And once it's oxygenated, it shows up here in the left-- let me do a better color than that-- it shows up right here in the left atrium. Atrium, you can imagine-- it's kind of a room with a skylight or that's open to the outside and in both of these cases, things are entering from above-- not sunlight, but blood is entering from above. And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs."
    },
    {
        "Q": "\nhow long dos it take to get your blood circulating 0:20 minutes",
        "A": "round about 1 sec actually time taken for one breath",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            20
        ],
        "3min_transcript": "Where I left off in the last video, we talked about how the hemoglobin in red blood cells is what sops up all of the oxygen so that it increases the diffusion gradient-- or it increases the incentive, we could say, for the oxygen to go across the membrane. We know that the oxygen molecules don't know that there's less oxygen here, but if you watch the video on diffusion you know how that process happens. If there's less concentration here than there, the oxygen will diffuse across the membrane and there's less inside the plasma because the hemoglobin is sucking it all up like a sponge. Now, one interesting question is, why does the hemoglobin even have to reside within the red blood cells? Why aren't hemoglobin proteins just freely floating in the blood plasma? That seems more efficient. You don't have to have things crossing through, in and out of, these red blood cell membranes. You wouldn't have to make red blood cells. What's the use of having these containers of hemoglobin? It's actually a very interesting idea. If you had all of the hemoglobin sitting in your the flow of the blood. The blood would become more viscous or more thick. I don't want to say like syrup, but it would become thicker than blood is right now-- and by packaging the hemoglobin inside these containers, inside the red blood cells, what it allows the blood to do is flow a lot better. Imagine if you wanted to put syrup in water. If you just put syrup straight into water, what's going to happen? The water's going to become a little syrupy, a little bit more viscous and not flow as well. So what's the solution if you wanted to transport syrup in water? Well, you could put the syrup inside little containers or inside little beads and then let the beads flow in the water and then the water wouldn't be all gooey-- and that's exactly what's happening inside of our blood. Instead of having the hemoglobin sit in the plasma and make it gooey, it sits inside these beads that we call red blood cells that allows the flow to still be So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose."
    },
    {
        "Q": "at \"0:42\" I could not hear it with my volume at full, what did he say?\n",
        "A": "He didn t really say anything important. He said I don t know and came up with (1/2)m^2 for the second area.",
        "video_name": "xlJYYM5TWoA",
        "timestamps": [
            42
        ],
        "3min_transcript": "Let's say I have a horizontal pipe that at the left end of the pipe, the cross-sectional area, area 1, which is equal to 2 meters squared. Let's say it tapers off so that the cross-sectional area at this end of the pipe, area 2, is equal to half a square meter. We have some velocity at this point in the pipe, which is v1, and the velocity exiting the pipe is v2. The external pressure at this point is essentially being applied rightwards into the pipe. The pressure at this end, the pressure 2-- that's the external pressure at that point in the pipe-- that is equal to 6,000 pascals. Given this information, let's say we have water in this pipe. We're assuming that it's laminar flow, so there's no friction within the pipe, and there's no turbulence. Using that, what I want to do is, I want to figure out what is the flow or the flux of the water in this pipe-- how much volume goes either into the pipe per second, or out of the We know that those are the going to be the same numbers, because of the equation of continuity. We know that the flow, which is R, which is volume per amount of time, is the same thing as the input velocity times the input area. The input area is 2, so it's 2v1, and that also equals the output area times output velocity, so it equals 1/2 v2. We could rewrite this, that v1 is equal to 1/2 R, and that v2 is equal to 2R. This immediately tells us that v2 is coming out at a faster rate, and this is based on the size of the openings. We know, because V2 is coming out at a faster rate, but we also know because we have much higher pressure at this end"
    },
    {
        "Q": "At around 5:15, why are the 3 branches drawn off of the monocyte cell branch? Could it have been drawn just coming from the myeloid (Is it just to save space)?\n",
        "A": "There is a reason to it :) The Common Myeloid Progenitor cell (the first red one) grows into 4 different cells; RBC, Mast cell, Megakaryocyte and MYELOBLAST. Myeloblast then later grows into; Neutrophils, Eosinphils, Basophils, and monocyte. He kinda skipped this step because it makes it a little bit more complicated.",
        "video_name": "ddifthdMNVc",
        "timestamps": [
            315
        ],
        "3min_transcript": "And those two lineages are the myeloid lineage and the lymphoid lineage. And each of these lineages gives rise to many different cells. The myeloid lineage gives rise to red blood cells, which are biconcave in shape. They are the most common of all blood cells. Now, the myeloid lineage also gives rise to a big cell called a megakaryocyte. Now, you might have never heard of this before, but the megakaryocytes themselves produce platelets, which I think that you've probably have heard of. They're little fragments of cells, They kind of squeeze out little pieces of cytoplasm that become platelets. And now I have a challenge for you. Do you think that a macrophage, which is an immune cell that likes to eat up invaders like bacteria, do you think that macrophages come from the myeloid lineage or the lymphoid lineage? So, I was surprised to find out that they actually come from the myeloid lineage. I was surprised because macrophages are immune cells, but they actually come from the same lineage as red blood cells and platelets. So here is a... This is actually not yet a macrophage, this is a monocyte. A lot of crazy words here, but this is a monocyte. Monocytes actually become macrophages once they settle down in the tissues. But, before that, while they're still circulating, they are monocytes. the myeloid lineage gives rise to three guys, one of whom you have heard of probably, two of whom you may not have heard of. I'll just draw them here. So, the one you might have heard of is the, I'm running out of space here, but it's the neutrophil. Neutrophils are the most common immune cell in the blood. The other two are called eosinophils, which are significantly more rare than neutrophils. And, even more rare than eosinophils, are something called the basophils. So, it's the three phils. So, now let's go over to the lymphoid lineage. There's three important cells that come from this one. Two of them you've probably heard of."
    },
    {
        "Q": "\nAt approximately 6:00, what was Sal implying by the term, \"superimposed\" when he was referring to enantiomers?",
        "A": "To superimpose means to put on top of eachother, so putting the image of one on top of the other and only seeing one since they re the same (or not since they re different.",
        "video_name": "z8M4EciPpYI",
        "timestamps": [
            360
        ],
        "3min_transcript": "in three dimensions. We don't just care about what's bonded to what or the constituents and actually this one is, as we'll see, is also a stereoisomer because this carbon is bonded to the same things in either case. So these are both, these are both situations, there are both stereoisomers, stereoisomers, and this particular variation of stereoisomer is called a cis trans isomer. Cis is when you have the two groups on the same side, cis, and trans is when you have the two groups on the opposite sides of the double bond. Cis trans isomers. Cis trans isomers. Isomers, and these are often called geometric isomers. Geometric, geometric isomers. So that's a subset, so when I'm talking about cis trans or geometric, I'm talking about these two characters over here. They are a subset of the stereoisomers. I have no double bond, I'm not talking about cis and trans. The carbon, as I've just said, is bonded to fluorine, chlorine, bromine, and a hydrogen, fluorine, chlorine, bromine, and a hydrogen. How are these two things different? And the way that they're different is if you were to actually try to superimpose them on each other. You will see that it is impossible. There are mirror images of each other and because there's four different constituents here, you can actually not superimpose this molecule onto this molecule over here and actually because of that, they actually have different chemical properties, and so this over here, these two characters, which is a subset of stereoisomers. Stereoisomers are concerned with how things are positioned in three dimensions, not just how their bonding is different, but this subset where you have these mirror images that cannot be superimposed, we call these enantiomers. So these two characters, these are enantiomers. Enantiomers, and enantio comes from Greek, So these are opposites of each other, they cannot be superimposed, they're mirror, they're mirror images. So all of these are different variations of isomers and once again, you might say, okay theses are clearly two different molecules that have different bonding, but even cis trans isomer will have different chemical properties. These two in particular, they aren't that different but they do have different chemical properties, but sometimes they're so different that one might be able to exist in a biological system while the other is not. One might be okay for your health, and the other might not be okay for your health. Same thing for enantiomers. One might be biologically active in a certain way and the other one might not be biologically active in that same way."
    },
    {
        "Q": "Perhaps I missed it in the video, but at 3:13 Sal is talking about the Helium absorbing light energy. Is this light coming from the Cepheid itself, or does he mean light from an outside source?\n",
        "A": "I think he meant from the inside of the cepheid.",
        "video_name": "X_3QAB3o4Vw",
        "timestamps": [
            193
        ],
        "3min_transcript": "is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw a neutral helium, neutral helium's got two protons, it's got two protons, two neutrons, two neutrons, and then two electrons and obviously this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium you knock off one of these electrons. And these type of things happen in stars when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color, this helium now has a net charge, we could write one plus here, but if you just write a plus you implicitly mean a positive charge of one. Now you can also double the ionized helium if the environment is hot enough. and doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus. It's really just a helium nucleus like this. This right here is doubly, doubly ionized helium. Now I just said in order to do this you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both these, this electron really doesn't want to leave, to take an electron off of something that's already positive is difficult. You have to have a lot of really pressure and temperature. This is cooler. And this is all relative, we're talking about the insides of stars. So, you know, it's hot, this is a hotter part of the star versus a cooler part of the star I guess is a way you think about this, it's still a very hot environment by our traditional, every day standards. Now the other thing about the doubly ionized helium is that it is more opaque. It is more opaque, it doesn't allow light to go through it, it actually absorbs light. It is more opaque, it absorbs light. It absorbs light. it absorbs that light energy that energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent. This is more transparent. More transparent, it allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it, or through it, or whatever. It allows them to go through it here the photons are going to actually heat up, heat up the ion. So let's think about how this might cause cepheid variable to pulsate. So assuming that cepheid variables have a large enough quantity, I should say, of these ions, we can imagine that when a cepheid variable is dim, so let me draw a dim cepheid variable, so I'll draw that like, I'll draw this in a dim color"
    },
    {
        "Q": "\nAt approximately 3:17 he states the peak at 1100cm^-1 corresponds to a C-O single bond. However, in prior videos he described the area below 1500 as the \"fingerprint\" area, and the area above 1500 as the diagnostics area. Can you explain the significance of this?",
        "A": "Some peaks are so strong and so characteristic that you can identify them even they are in the fingerprint region.",
        "video_name": "ALLSsIDhFdU",
        "timestamps": [
            197
        ],
        "3min_transcript": "are going to have different amounts of hydrogen bonding. Some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wavenumber decreases a little bit. But other molecules might have a lot of hydrogen bonding, and so we can decrease k even more, therefore we are going to decrease the wavenumber even more. You get a range of wavenumbers, and since you get a range of wavenumbers for the OH bond, when hydrogen bonding is present, you get a very broad signal on your IR spectrum. So, if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen. So we draw a line at 3000, and we know that just below 3000, we're talking about a carbon-hydrogen bond stretch, where the carbon is sp3-hybridized. But, this over here, this very broad signal right here, So let me go ahead and highlight that. This bond right here, this oxygen-hydrogen bond, gives us a very broad signal on our IR spectrum because of hydrogen bonding. So we get this very broad signal because of the different wavenumbers. And usually you're going to see this somewhere around 3500 to 2900. So if I find this is 31, 32, 33, 34, 35... so usually in this range, maybe even a little bit higher than that, you're going to find this very broad signal. In this case, the oxygen-hydrogen bond. And so you know immediately to think about the possibility of an alcohol functional group in your molecule. Also, we can draw a line at 1500 here, and this signal actually, so somewhere around 1100 wavenumbers, this is actually the carbon-oxygen single bond. Let me go ahead and highlight that. So we have a carbon-oxygen single bond, And that's where-- that's this stretch right here. Not always going to be super useful to you, but it's just thinking about what we talked about in the earlier video, I think we calculated the approximate wavenumber for a carbon-oxygen single bond. So that's what the typical spectrum for an alcohol is going to look like. Look for that broad signal there. Alright, let's compare this alcohol to another one here. So, this molecule is butylated hydroxytoluene, or BHT, and I drew two BHT molecules in there for a reason. Let's think about why. So, you might think at first, \"OK, I have another opportunity for hydrogen bonding.\" So, here's an opportunity for hydrogen bonding, so we're going to get a broad signal for this OH bond. So I'm going to highlight it here. I might expect, since I have hydrogen bonding,"
    },
    {
        "Q": "\n10:31 Why is keto form more stable than enol form ?",
        "A": "Good Question! Enols are less stable because the C=C (double bond) is weakened by the electronegativity of oxygen.",
        "video_name": "NdRl1C6Jr5o",
        "timestamps": [
            631
        ],
        "3min_transcript": "hybridized with tetrahedral geometry. Now, this carbon is SP two hybridized with trigonal planar geometry. Whatever stereochemical information we had over here on the left, whether it was the R or the S enantiomer, it's been lost now that we've formed the enol. The enol is achiral, it's flat, it's planar. When we reform the keto form, so one of the possibilities is to form the enantiomer that we started with but the other possibility is to form the other enantiomer. You can see that's what I've shown here. I've shown the hydrogen now going away from us and our R double prime group coming out at us. This is the enantiomer. Because we formed the enol we can get a mixture of enantiomers. Enolization can lead to racemization. We can get a mixture of enantiomers and if we wait long enough, we can get an equal mixture of these guys. with our enol form. That's something to think about if you have a chiral center at your alpha carbon. Let's look at two quick examples of keto and enol forms. Over here on the left we have cyclohexanone and on the right would be the enol version of it. You could think about one of these as being your alpha carbon, right, and you could move these electrons in here and push those electrons off. You could see that would give you this enol form. It turns out that the keto form is favored. The equilibrium is actually far to the left favoring formation of the keto form. Even under just normal conditions, so not acid or base-catalyzed. There's only a trace amount of the enol presence however, there are some cases where the enol is extra-stabilized and that's the case for this example down here. We have the keto form and we have the enol form. Once again, you could think about pushing those electrons off giving you your enol form. This is a specially-stabilized enol, right? This is phenol right here. We know that phenol has an aromatic ring. The formation of the enol form is extra-stabilize because of this aromatic ring. This time the equilibrium is actually to the right and much more of it is in the enol form than in the keto form. In this case, we have some special stabilization."
    },
    {
        "Q": "Would the reaction still be complete if we didn't \"open up the structure\"? (Meaning leave it at the epoxide Jay drew at 4:16)\n",
        "A": "Yes, if you simply want the epoxide, the reaction would be complete.",
        "video_name": "KfTosrMs5W0",
        "timestamps": [
            256
        ],
        "3min_transcript": "And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here, the electrons in this bond right here. These are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction for the formation of epoxide, and then we'll talk about how to form a diol from that. So if we start with cyclohexene-- let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? But it has one extra oxygen in there, so it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid, we're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example, but we'll go ahead and put in our epoxide using wedges here. And that must mean going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this up epoxide using acid. So just to refresh everyone's memory, go back up here. Now we're going to look at this second part where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide. And I'm going to redraw our epoxide"
    },
    {
        "Q": "1:30 How come the electrons between the O and H move twards the alkene's C? What triggers it? Thank you.\n",
        "A": "i think C is more electronegative than H so that due to inductive effect electrons move towards C rather than H",
        "video_name": "KfTosrMs5W0",
        "timestamps": [
            90
        ],
        "3min_transcript": "If you start with an alkene and add to that alkene a percarboxylic acid, you will get epoxide. So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed, and we're going to talk about an acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you could assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right. Let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid except it has an extra oxygen. And the bond between these two oxygen atoms is weak, so this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular confirmation that it's in. So this hydrogen ends up being very a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this confirmation, the mechanism will begin. This is a concerted eight electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight electron mechanism. So, of course, at the bottom here we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going"
    },
    {
        "Q": "\nAt 7:46, why does Jay say that there is only 1 pi bond? Are pi bonds made up of more than 2 valence electrons?",
        "A": "There s only one pi bond. It contains two electrons - one electron comes from the p-orbital on the left carbon and the other from the p-orbital on the right carbon.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            466
        ],
        "3min_transcript": "here's another head on overlap of orbitals. The carbon carbon bond, here's also a head on overlap of orbitals and then we have these two over here. We have a total of five sigma bonds in our molecules. Let me go ahead and write that over here. There are five sigma bonds. If I would try to find those on my dot structure this would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond and then these over here. A total of five sigma bonds and then we have a new type of bonding. These unhybridized P orbitals can overlap side by side. Up here and down here. We get side by side overlap of our P orbitals and this creates a pi bond. A pi bond, let me go ahead and write that here. A pi bond is side by side overlap. There is overlap above and below this sigma bond here When we're looking at the example of ethane, we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here, this pi bond is going to prevent rotations so we don't get different confirmations of the ethylene molecules. No free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds, I'm just gonna say it's this one right here. If you have a double bond, one of those bonds, the sigma bond and one of those bonds is a pi bond. We have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, let me go ahead and use a different color for that. The distance between this carbon and this carbon. It turns out to be approximately 1.34 angstroms, between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization."
    },
    {
        "Q": "@12:17 there's a bubble that pops up which says 'trigonal planar', as if in correction of what Jay says in the video. (He calls it planar). However, I think that what he says, planar, is correct, and not the box: the empty p orbital exists on BOTH sides of the molecule, and so the sp2 orbitals don't bend out of shape\n",
        "A": "No the word trigonal planar is correct as it lies in plane and it is not a 3D structure!",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            737
        ],
        "3min_transcript": "If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization."
    },
    {
        "Q": "at 2:00 aren't there 4 bonds on every carbon? so why don't we get sp3\n",
        "A": "To determine the hybridization, look at the # of sigma bonds + # of lone pairs, rather than the total number of bonds. Here, each carbon does have 4 bonds, but only 3 sigma bonds, so it is sp2.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            120
        ],
        "3min_transcript": "Voiceover: In an earlier video, we saw that when carbon is bonded to four atoms, we have an SP3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees. If you look at one of the carbons in ethenes, let's say this carbon right here, we don't see the same geometry. The geometry of the atoms around this carbon happens to be planar. Actually, this entire molecule is planar. You could think about all this in a plane here. And the bond angles are close to 120 degrees. Approximately, 120 degree bond angles and this carbon that I've underlined here is bonded to only three atoms. A hydrogen, a hydrogen and a carbon and so we must need a different hybridization for each of the carbon's presence in the ethylene molecule. We're gonna start with our electron configurations over here, the excited stage. We have carbons four, valence electron represented. One, two, three and four. In the video on SP3 hybridization, to make four SP3 hybrid orbitals. In this case, we only have a carbon bonded to three atoms. We only need three of our orbitals. We're going to promote the S orbital. We're gonna promote the S orbital up and this time, we only need two of the P orbitals. We're gonna take one of the P's and then another one of the P's here. That is gonna leave one of the our P orbitals unhybridized. Each one of these orbitals has one electron and it's like that. This is no longer an S orbital. This is an SP2 hybrid orbital. This is no longer a P orbital. This is an SP2 hybrid orbital and same with this one, an SP2 hybrid orbital. We call this SP2 hybridization. Let me go and write this up here. and use a different color here. This is SP2 hybridization because we're using one S Orbital and two P orbitals to form our new hybrid orbitals. and same with this carbon. Notice that we left a P orbital untouched. We have a P orbital unhybridized like that. In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here. We're taking one S orbital. We know S orbitals are shaped like spheres. We're taking two P orbitals. We know that a P orbital is shaped like a dumbbell. We're gonna take these orbitals and hybridized them to form three SP2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that. Once again, when we draw the pictures, we're going to ignore the smaller back lobe. This gives us our SP2 hybrid orbitals. In terms of what percentage character, we have three orbitals that we're taking here and one of them is an S orbital. One out of three, gives us 33% S character"
    },
    {
        "Q": "At 8:21 shouldn't it be A with a circle on the top? Not on the side? Or are both forms of the angstrom acceptable?\n",
        "A": "Yes. The symbol for angstrom is \u00c3\u0085. In cursive writing, we don t always get the little circle directly on top of the A. Of course, it would be preferable to use the SI units of picometres (pm) instead of angstroms.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            501
        ],
        "3min_transcript": "here's another head on overlap of orbitals. The carbon carbon bond, here's also a head on overlap of orbitals and then we have these two over here. We have a total of five sigma bonds in our molecules. Let me go ahead and write that over here. There are five sigma bonds. If I would try to find those on my dot structure this would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond and then these over here. A total of five sigma bonds and then we have a new type of bonding. These unhybridized P orbitals can overlap side by side. Up here and down here. We get side by side overlap of our P orbitals and this creates a pi bond. A pi bond, let me go ahead and write that here. A pi bond is side by side overlap. There is overlap above and below this sigma bond here When we're looking at the example of ethane, we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here, this pi bond is going to prevent rotations so we don't get different confirmations of the ethylene molecules. No free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds, I'm just gonna say it's this one right here. If you have a double bond, one of those bonds, the sigma bond and one of those bonds is a pi bond. We have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, let me go ahead and use a different color for that. The distance between this carbon and this carbon. It turns out to be approximately 1.34 angstroms, between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization."
    },
    {
        "Q": "At 11:45, jay tells us that boron acts a lewis acid because it has no electrons in its unhybridised porbital and hence can gain two electrons.\nbut the sp2 hybrid orbital have only 3 electrons and hence will they be able to gain 3 MORE electrons?\n",
        "A": "Each sp\u00c2\u00b2 orbital has one electron, and they each gain another electron when they form bonds to the fluorine atoms. That makes six electrons in the valence shell of boron. Boron needs two more electrons to complete its octet.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            705
        ],
        "3min_transcript": "If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization."
    },
    {
        "Q": "\nAt 2:37, why has he written 36?",
        "A": "This is a known error in the video. A box pops up and tells you it should be 81 and the final answer is sqrt(97)",
        "video_name": "gluN2wxqES0",
        "timestamps": [
            157
        ],
        "3min_transcript": ""
    },
    {
        "Q": "When you squared that negative number in 2:37, does it become non negative?\n",
        "A": "Yep. Squares of negative numbers are positive(non negative) since we are multiplying a negative with a negative which gives a positive result.",
        "video_name": "gluN2wxqES0",
        "timestamps": [
            157
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhen you squared that negative number in 2:37, does it become non negative?",
        "A": "yes...for example squaring is the same thing as multiplying by the same number ..so here -4 X -4=16 ..the negatives cancel out",
        "video_name": "gluN2wxqES0",
        "timestamps": [
            157
        ],
        "3min_transcript": ""
    },
    {
        "Q": "8:26 would even a metal chain stretch?\n",
        "A": "sure, it would stretch a bit. You might not be able to notice it.",
        "video_name": "QKXeZFwFPS0",
        "timestamps": [
            506
        ],
        "3min_transcript": "We divide by 3 kg, 'cause that's the mass. But I've still got a problem. I don't know this acceleration or this tension. So what do I do? You might notice, if you're clever you'll say wait, I've got my unknown on this side is acceleration and tension. My unknown on this side is acceleration and tension. It seems like I've got two equations, two unknowns, maybe we should combine them. And that's exactly how you do these. So I've got tension in both of these equations. Let me solve for tension over here, where it's kind of simple. And I'll just get the tension equals 5 kg, times the acceleration of the 5 kg mass in the X direction. So now I know what tension is. Tension is equal to this. And that tension over on this side is the same as the tension on this side. So I can take this and I can plug it in for this tension right here. And let's see what we get. We get that the acceleration of the 3 kg mass vertically, what am I gonna get? I'm gonna get, so T is the same as 5AX. So I'll plug in 5 kg times the acceleration of the 5 kg mass in the X direction. And then I get all of this stuff over here. So I'll get the rest of this right here. I'll just bring that down right there. Alright, now what do I have? I've got 3 kg on the bottom still, so I have to put that here. Are we any better off? Yeah, we're better, because now my only unknowns are acceleration. But these are not the same acceleration. Look, this acceleration here is the acceleration of the 3 kg mass, vertically. This acceleration here is the acceleration of the 5kg mass horizontally. Now here's where I'm gonna have to make an argument, and some people don't like this. But, it's crucial to figuring out this problem. And the key idea is this, if this 3 kg mass moves down, Well then this 5 kg mass had better move forward one meter. Because if it doesn't, then it didn't provide the one meter of rope that this 3 kg mass needed to go downward. Which means either the rope broke, or the rope stretched. And we're gonna assume that our rope does not break or stretch. All ropes are gonna stretch a little bit under tension. We're gonna assume that stretch is negligible. So the argument is that if this 3 kg mass moves downward a certain amount, this 5 kg mass has to move forward by that same amount in order to feed that amount of rope for this 3 kg mass to go downward by that amount. Otherwise, think about it. If this 5 kg mass just sat here and the 3 kg moved, or the 3 kg moved farther than the 5 kg mass, then this rope is stretching or breaking. So if you believe that, if you don't believe it, pause it and think about it. 'Cause you've gotta convince yourself of that. If you believe that then you can also convince yourself"
    },
    {
        "Q": "\nAcceleration of 3Kg box is downwards so it shotld be negative but you wrote it as positive in the equation above at 10:06",
        "A": "Typically the acceleration of the 3kg block would be considered to be negative, because downwards and to the left are typically considered to be the negative direction and upwards and to the right are typically considered to be positive directions. However in this case he just used different signs instead of following the normal convention. He chose downwards and leftwards to be positive and upwards and rightwards to be negative.",
        "video_name": "QKXeZFwFPS0",
        "timestamps": [
            606
        ],
        "3min_transcript": "Well then this 5 kg mass had better move forward one meter. Because if it doesn't, then it didn't provide the one meter of rope that this 3 kg mass needed to go downward. Which means either the rope broke, or the rope stretched. And we're gonna assume that our rope does not break or stretch. All ropes are gonna stretch a little bit under tension. We're gonna assume that stretch is negligible. So the argument is that if this 3 kg mass moves downward a certain amount, this 5 kg mass has to move forward by that same amount in order to feed that amount of rope for this 3 kg mass to go downward by that amount. Otherwise, think about it. If this 5 kg mass just sat here and the 3 kg moved, or the 3 kg moved farther than the 5 kg mass, then this rope is stretching or breaking. So if you believe that, if you don't believe it, pause it and think about it. 'Cause you've gotta convince yourself of that. If you believe that then you can also convince yourself at a certain speed, let's say two meters per second. Then the 5 kg mass had better also be moving forward two meters per second because otherwise it wouldn't be feeding rope at a rate that this 3 kg needs to move downward at that rate. And finally, if you believe all that, it's not too much harder to convince yourself that this 3 kg mass, no matter what its acceleration downward must be, this 5 kg mass had better have the same magnitude of acceleration forward so that it's again, feeding the rope so this rope doesn't break, or snap, or stretch. 'Cause we're gonna assume the rope doesn't do that. So what I'm saying is that the acceleration of the 3 kg mass in the Y direction had better equal the magnitude. So these magnitudes have to be the same. The sign doesn't have to be the same. So this 3 kg mass has a negative acceleration just 'cause it points down, and we're assuming up is positive, down is negative. This 5 kg mass has a positive acceleration 'cause is the positive horizontal direction. So, they can have different signs, but the magnitudes had better be the same so that your feeding this rope at a rate that the other one needs in order to move. And so we can say that the magnitudes are the same. In this case, since one is negative of the other, I can say that the acceleration of the 3 kg mass vertically downward is gonna be equal to, let's say negative of the acceleration of the 5 kg mass in the X direction. I could have written it the other way. I could have wrote that A of the 5 kg mass in the X direction is a negative A of the 3 kg mass in the Y direction. They're just different by a negative sign is all that's important here. Okay, so this is the link we need. This is it. So this allows us to put this final equation here in terms of only one variable. 'Cause I know I've got A3Y on this left hand side. I know A3Y should always be -A5X. If I take this and just plug it in for A3Y right here,"
    },
    {
        "Q": "why cant the outer shell electrons repel the inner shell electrons towards the nucleus? (5:57) the inner shell atoms could easily get sucked into the nucleus with the pushing( (repelling) force from the outer shell and the pulling force from positive charge of the nucleus, Right?\n",
        "A": "The thing about atoms is they have this property where they only can occupy certain states, so electrons can only jump between orbits if they are empty and energy levels. There is no orbit in the nucleus and thus they can t be there. I d suggest looking at the lesson in chemistry about the electronic structure of atoms.",
        "video_name": "rKoIcgBM4Vg",
        "timestamps": [
            357
        ],
        "3min_transcript": "charged inner shell electrons are going to repel it. So let me go ahead and highlight these guys right here. These are our inner shell electrons. Like charges repel. And so you could think about this electron right here wanting to push this outer electron that way, and this electron wanting to push this electron that way. And so the nucleus attracts a negative charge, and the inner shell electrons repel the outer electron. And then we call this shielding, because the inner shell electrons are shielding that magenta electron from the pole of the nucleus. So this is called electronic shielding or electron screening. Now, it's going to be important concepts. So now let's go ahead and draw the atom for beryllium, so atomic number 4. And so here's our nucleus for beryllium. With an atomic number of 4, that means there are four protons in the nucleus, so a charge of four plus in our nucleus. And we have four electrons to worry about this time, in my inner orbital in our first energy level. And then we have two electrons in our outer orbital, or our second energy level. And so again, this is just a rough approximation for an idea of what beryllium might look like. And so when we think about what's happening, we're moving from a charge of 3 plus with lithium to a charge of 4 plus with beryllium. And the more positive your charges, the more it's going to attract those outer electrons. And when you think about the idea of electron screening, so once again we have these electrons in green here shielding our outer shell electrons from the effect of that positively charged nucleus. Now, you might think that outer shell electrons could shield, too. So you might think that oh, this electron right here in magenta could shield the other electron in magenta. much the same distance from the nucleus, so outer shell electrons don't really shield each other. It's more of these inner shell electrons. And because you have the same number of inner shell electrons shielding as in the lithium example-- so let me go ahead and highlight those again. So we have two inner shell electrons shielding a beryllium. We also have two inner shell electrons shielding in lithium. Because you have the same number of shielding but you have a higher positive charge, those outer electrons are going to feel more of a pull from the nucleus. And they're going to be pulled in even tighter than you might imagine, or at least tighter than our previous example. So these electrons are pulled in even more. And because of that, you're going to get the beryllium atom as being smaller than the lithium atom, hence the trend. Hence as you go across the period, you're always going to increase in the number of protons and that increased whole is going to pull those outer electrons in closer, therefore decreasing the size of the atom."
    },
    {
        "Q": "at 9:07, wouldn't it be that the neutral atom is smaller than the cation? If not, how come?\n",
        "A": "Cations are positively charged because they lost an electron, so they have fewer electrons than protons and therefore the pull toward to nucleus is stronger per electron. The cation is therefore smaller than the neutral atom.",
        "video_name": "rKoIcgBM4Vg",
        "timestamps": [
            547
        ],
        "3min_transcript": "Let's look at ionic radius now. And ionic radius can be kind of complicated depending on what chemistry you are involved in. So this is going to be just a real simple version. If I took a neutral lithium atom again, so lithium-- so we've drawn this several times. Let me go ahead and draw it once more. So we have our lithium nucleus, which we have three electrons. So once again I'll go ahead and sketch in our three electrons real fast. Two electrons in the inner shell, and one electron in the outer shell like that. And let's say you were going to form a cation, so we are going to take away an electron from our neutral atom. So we have-- let me go ahead and draw this in here-- we had a three protons in the nucleus and three electrons those cancel each other out to be a neutral atom. And if we were to take away one of those electrons, so let's go ahead and show lithium losing an electron. So if lithium loses an electron, it's So the nucleus still has a plus 3 charge, because it has three protons in it. And we still have our two inner shell electrons like that, but we took away that outer shell electron. So we took away this electron in magenta, so let me go ahead and label this. So we lost an electron, so that's this electron right here, and so you could just show it over here like that. And by doing so, now we have three positive charges in our nucleus and only two electrons. And so therefore our lithium gets a plus 1 charge. So it's Li plus, it's a cation. And so we formed a cation, which is smaller than the neutral atom itself. And that just makes intuitive sense. If you take away this outer electron, now you have three positive charges in the nucleus and only two electrons here. So it's pulling those electrons in, you lost that outer electron, it's getting smaller. And so we've seen that neutral atoms will shrink when you convert them to cations, so it kind of makes sense that if you take a neutral atom and add an electron, it's going to get larger. And so that's our next concept here. So if we took something like chlorine, so a neutral chlorine atom, and we added an electron to chlorine, that would give it a negative charge. So we would get chlorine with a negative charge, or the chloride anion, I should say. And so in terms of sizes, let's go ahead and draw a representative atom here. So if this is our neutral chlorine atom and we add an electron to it, it actually gets a lot bigger. So the anion is bigger than the neutral atom. And let's see if we can think about why here. So if we were to draw an electron configuration, or to write a noble gas electron configuration for the neutral chlorine-- so you should already know how to do this-- you would just"
    },
    {
        "Q": "at 4:40 he says insulator is charged, how can an insulator get charged , i wont get an electric shock from it if i touch it\n",
        "A": "Yes you can get an electric shock from an non-conducting material. It is commonly known that if you walk on a carpet with socks on, and then tough a doorknob you feel a shock. Basically you are being shocked by the charge accumulated by your socks travelling through the point of contact. This is called electrostatic discharge or ESD. On a grander scale, lightning is also an example of ESD caused by charge build up.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            280
        ],
        "3min_transcript": "and this side of the atom would be more positive. Even though the electron doesn't move, and the electrons don't move, now because this is set up where the positive is shifted from the negative, this material, if you get all of them to do this or a lot of them, this can create an overall electrical effect where this insulator can interact with other charges nearby and exert forces on them. Even though the charges can't flow through an insulator, they can still interact electrically. Now, let's see what happens if we add extra charge to these insulators or conductors. I mean, the way they started off right here we had just as many positives in the nucleus as there are negatives surrounding them and that's true for the conductors and insulators. What happens if we add extra charge? Maybe we add extra negatives into here. Then what happens? Well, it'll get really messy if we try to draw it with all the atoms, so since these all cancel out their overall charge, I am not going to draw every atom and nucleus. I'm just going to pretend like those are there I'm just going to draw the actual extra charge. Let's say we added extra negative charges to this insulator. What would happen? Let's say I just add a negative charge here and a negative charge there, and here and there, I have added a bunch of negative charges to this insulator. What would happen? Well, we know these negatives can't move throughout and insulator. Charges can't flow through an insulator so they're stuck which means for an insulator, I could charge the whole thing uniformly if I wanted to where the charge is spread out throughout the whole thing or I could make them bunch up on one side if I wanted to and they'd be stuck there. The point is that they're stuck. For a conductor, what would happen if I tried to put a negative here and a negative there, some extra negative charge on a conductor? They don't have to stay here if they don't want to. If you put extra negatives in here, they are not going to want to because negatives repel each other So what are they going to do? Well, this negative is going to try to get as far away from this other negative as it can so go over here. This negative is going to try to get as far away as it can. It repels it. Now, it can't jump off the conductor. That takes a lot more energy, but it can go to the very edge. That's what charges do for conductors. You've got a solid conducting material, you put extra charge on it, it's all... All that charge is going to reside on the outside edge whether you've added extra negative or positive, always on the outside edge. You can only add charge to the outside edge for a conductor, because if it wasn't on the outside edge it will quickly find its way to the outside edge because all these negatives repel each other. I said this is true for positives or negative. You might wonder, \"How do we add a positive?\" Well, the way you add a positive is by taking away a negative. If you started off with a material that had just as many positives as negatives and you took away a negative,"
    },
    {
        "Q": "\nIn the example at 10:0, what would happen if the can wasnt connected through a wire to a metal but they were just touching. Would a transfer of electrons still occur?",
        "A": "yes but there would be virtually no resistance",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            600
        ],
        "3min_transcript": "Charge by induction says alright, first imagine I just take this and I bring it nearby but don't touch it. Just bring it near by this other piece of metal and I don't touch it. What would happen? There is negatives in here, I haven't drawn them. There's positives in here. The negatives can move if they wanted to. Do they want to? Yeah, they want to! These negatives are coming nearby, they want to get as far away from them as possible. Even though there are already some negatives here, a net amount of negatives are going to get moved over to this side. They were located with their atom on this side, but they want to get away from this big negative charge so they can move over here, which leaves a total amount of positive charge over here. I.E. There is a deficit of electrons over here, so this side ends up positively charged. You might think, \"Okay, well that's weird. \"Does anything else happen?\" Yeah because now these positives are closer to the negatives and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground? If you took a big metal pipe and stuck it in the ground that would count, or any other huge supply of electron, a place where you can gain, steal, basically take infinitely many electrons or deposit infinitely many electrons and this ground would not care. So the frame of your car, the actual metal, is a good ground because it can supply a ton of electrons or take them. Or a metal pipe in the earth. Some place you can deposit electrons or take them and that thing won't really notice or care. Now what would happen? If I bring this negative rod close to this rod that was originally had no net charge? Now instead of going to the other side of this, they say \"Hey, I can just leave. \"Let me get the heck out of here.\" These negatives can leave. A whole bunch of negatives can start leaving and what happens when that happens is that your rod is no longer uncharged. It has a net amount of charge now. They won't all leave. You're not going to get left with no electrons in here."
    },
    {
        "Q": "\nat 6:30 he says most plastics are insulator, can you give me an example of a plastic which is not an insulator?",
        "A": "exposure of few plastics in ionic beam shows conductance like metals, and some are made semi conductors.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            390
        ],
        "3min_transcript": "I'm just going to draw the actual extra charge. Let's say we added extra negative charges to this insulator. What would happen? Let's say I just add a negative charge here and a negative charge there, and here and there, I have added a bunch of negative charges to this insulator. What would happen? Well, we know these negatives can't move throughout and insulator. Charges can't flow through an insulator so they're stuck which means for an insulator, I could charge the whole thing uniformly if I wanted to where the charge is spread out throughout the whole thing or I could make them bunch up on one side if I wanted to and they'd be stuck there. The point is that they're stuck. For a conductor, what would happen if I tried to put a negative here and a negative there, some extra negative charge on a conductor? They don't have to stay here if they don't want to. If you put extra negatives in here, they are not going to want to because negatives repel each other So what are they going to do? Well, this negative is going to try to get as far away from this other negative as it can so go over here. This negative is going to try to get as far away as it can. It repels it. Now, it can't jump off the conductor. That takes a lot more energy, but it can go to the very edge. That's what charges do for conductors. You've got a solid conducting material, you put extra charge on it, it's all... All that charge is going to reside on the outside edge whether you've added extra negative or positive, always on the outside edge. You can only add charge to the outside edge for a conductor, because if it wasn't on the outside edge it will quickly find its way to the outside edge because all these negatives repel each other. I said this is true for positives or negative. You might wonder, \"How do we add a positive?\" Well, the way you add a positive is by taking away a negative. If you started off with a material that had just as many positives as negatives and you took away a negative, But again, the net positive charge, the net negative charge always resides on the outside edge of the conductor because charges try to get as far away from each other as possible. So what physical materials actually do this? What physical materials are insulators? These are things like glass is an insulator. Wood is an insulator. Most plastics are insulators. All of these display this kind of behavior where you can distribute charge and the charge can't flow through it. You can stick charge on it. In fact, you can stick charge on the outside edge and it will stay there. There's conductors. These are things like metals, like gold or copper is typically used because it's kind of cheap. Cheaper than gold, certainly. Or any other metal. Silver works very well. These are materials where charges can flow freely through them. Now that we see how conductors and insulators work,"
    },
    {
        "Q": "\nAt 8:26, why did you add the area of the triangle and the rectangle together? Wouldn't the displacement only be the area of the triangle because the base of the triangle is where the initial velocity is?",
        "A": "The initial velocity causes displacement, too, doesn t it? If the velocity didn t change, wouldn t there still be displacement?",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            506
        ],
        "3min_transcript": "and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity"
    },
    {
        "Q": "at 8:05, it says that the area of the triangle is 1/2 x 4 x 8. in the other videos it says that the area of the triangle is one half times base times height. why?\n",
        "A": "that s exactly the SAME 1/2x 4 x 8 = HALF= 1/2 BASE= 4 HIEGHT= 8",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            485
        ],
        "3min_transcript": "and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity"
    },
    {
        "Q": "@4:27 why did Sal cancel the \u00e2\u0080\u009cseconds\u00e2\u0080\u009d and the \u00e2\u0080\u009cseconds squared\u00e2\u0080\u009d to make it just \u00e2\u0080\u009cseconds\u00e2\u0080\u009d? I thought we were multiplying. Wouldn\u00e2\u0080\u0099t it be \u00e2\u0080\u009cseconds cubed\u00e2\u0080\u009d? PLEASE HELP!\n",
        "A": "If you have a number, let s say X^2 and you have that X^2 divided by X, you end up with just having X as the answer because X^2 is the same as X*X, So, if you divide X*X by X, you cancel out one X in the numerator, and the X in the denominator. Hope that helps!",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            267
        ],
        "3min_transcript": "So that is my time axis, time this is velocity This is my velocity right over there and I'm starting off with 5m/s, so this is 5m/s right over here So vi is equal to 5m/s And every second goes by it goes 2m/s faster that's 2m/s*s every second that goes by So after 1 second when it goes 2m/s faster it will be at 7 another way to think about it is the slope of this velocity line is my constant accleration, my constant slope here so it might look something like that So what has happend after 4s? So 1 2 3 4 this is my delta t So my final velocity is going to be right over there so this is v this is my final velocity what would it be? Well I'm starting at 5m/s So we are doing this both using the variable and concretes Some starting with some initial velocity I'm starting with some initial velocity Subscript i said i for initial and then each second that goes by I'm getting this much faster so if I gonna see how much faster have I gone I multiply the number of second, I will just multiply the number second it goes by times my acceleration, times my acceleration and once again, this right here, subscript c saying that is a constant acceleration, so that will tell my how fast I have gone If I started at this point and multiply the duration time with slope I will get this high, I will get to my final velocity I'm just taking this to make it concrete in your mind you have 5m/s plus 4s plus, I wanna do it in yellow plus 4s times our acceleration with 2m per second square and what is this going to be equal to? you have a second that is cancelling out one of the second down here You have 4 times, so you have 5m/s plus 4 times 2 is 8 this second gone, we just have 8m/s or this is the same thing as 13m/s which is going to be our final velocity and I wanna take a pause here, you can pause and think about it yourself this whole should be intuitive, we are starting by going with 5m/s every second goes by we are gonna going 2m/s faster"
    },
    {
        "Q": "\nat 8:15 with the answer equaling 77664439914.3 my calculator gets 7.77*10^-52 is there a fix around this?",
        "A": "With an answer of that magnitude different, you must have typed something into your calculator wrong. If you type the following into Google it gets pretty close to Jay s answer (not exact because he hasn t rounded the first number) 3.53E-20 * 2 / 9.11E-31",
        "video_name": "vuGpUFjLaYE",
        "timestamps": [
            495
        ],
        "3min_transcript": "to free the electron and so we've exceeded that minimum amount of energy, and so we will produce a photoelectron. So, this photon is high-energy enough to produce a photoelectron. So let's go ahead and find the kinetic energy of the photoelectron that's produced. So we're gonna use this equation right up here. So let me just go and get some more room, and I will rewrite that equation. So we have the kinetic energy of the photoelectron, kinetic energy of the photoelectron, is equal to the energy of the photon, energy of the photon, minus the work function. So let's plug in our numbers. The energy of the photon was 3.78 times 10 the negative 19 joules, and then the work function is right up here again, it's 3.43, so minus 3.43 times 10 to the negative 19 joules. So let's get out the calculator again. 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20. So let's go ahead and write that. This is equal to 3.5 times 10 to the negative 20 joules. This is equal to the kinetic energy of the photoelectron, and we know that kinetic energy is equal to one half mv squared. The problem asked us to solve for the velocity of the photoelectron. So all we have to do is plug in the mass of an electron, which is 9.11 times 10 to the negative 31st kilograms, times v squared. This is equal to 3.5 times 10 to the negative 20. So, let's do that math. So we take 3.5 times 10 to the negative 20, we multiply that by 2, 9.11 times 10 to the negative 31st, and this gives us that number, which we need to take the square root of. So, square root of our answer gives us the velocity of the electron, 2.8 times 10 to the 5th. So if you look at your decimal place here, this'll be one, two, three, four, five, so 2.8 times 10 to the 5th meters per second. So here's the velocity of the photoelectron produced, 2.8 times 10 to the 5th meters per second, and if you increased the intensity of this light, so you had more photons, they would produce more photoelectrons. So one photon knocks out one photoelectron if it has enough energy to do so. So let's think about this same problem,"
    },
    {
        "Q": "(about 0:30): If shining a light of the right frequency and wavelength on a substance can knock electrons loose, therefore creating ions, wouldn't that change the properties of the substance? Is the requirement for light to free electron a minimum amount of energy, or is it a range? Does it have anything to do with why things bleach in the sun, or why, in some museums, there is no flash-photography allowed?\n",
        "A": "When an electron is knocked loose, the remaining material is positively charged, which means you can t knock too many electrons loose because it will get harder and harder for them to move away. The positively charged object will attract electrons from the surroundings to neutralize itself. So the photoelectric effect is not going to make any significant change to the material. It has nothing to do with bleaching, which is a process of chemical breakdown due to the incoming energy from the sun.",
        "video_name": "vuGpUFjLaYE",
        "timestamps": [
            30
        ],
        "3min_transcript": "- Sometimes light seems to act as a wave, and sometimes light seems to act as a particle. And, an example of this, would be the Photoelectric effect, as described by Einstein. So let's say you had a piece of metal, and we know the metal has electrons. I'm gonna go ahead and draw one electron in here, and this electron is bound to the metal because it's attracted to the positive charges in the nucleus. If you shine a light on the metal, so the right kind of light with the right kind of frequency, you can actually knock some of those electrons loose, which causes a current of electrons to flow. So this is kind of like a collision between two particles, if we think about light as being a particle. So I'm gonna draw in a particle of light which we call a photon, so this is massless, and the photon is going to hit this electron, and if the photon has enough energy, it can free the electron, right? So we can knock it loose, and so let me go ahead and show that. So here, we're showing the electron being knocked loose let's just say, this direction, with some velocity, v, and if the electron has mass, m, we know that there's a kinetic energy. The kinetic energy of the electron would be equal to one half mv squared. This freed electron is usually referred to now as a photoelectron. So one photon creates one photoelectron. So one particle hits another particle. And, if you think about this in terms of classical physics, you could think about energy being conserved. So the energy of the photon, the energy that went in, so let me go ahead and write this here, so the energy of the photon, the energy that went in, what happened to that energy? Some of that energy was needed to free the electron. So the electron was bound, and some of the energy freed the electron. I'm gonna call that E naught, the energy that freed the electron, and then the rest of that energy must have gone into the kinetic energy of the electron, kinetic energy of the photoelectron that was produced. So, kinetic energy of the photoelectron. So let's say you wanted to solve for the kinetic energy of that photoelectron. So that would be very simple, it would just be kinetic energy would be equal to the energy of the photon, energy of the photon, minus the energy that was necessary to free the electron from the metallic surface. And this E naught, here I'm calling it E naught, you might see it written differently, a different symbol, but this is the work function. Let me go ahead and write work function here, and the work function is different for every kind of metal. So, it's the minimum amount of energy that's necessary to free the electron, and so obviously that's going to be different depending on what metal you're talking about. All right, let's do a problem. Now that we understand the general idea of the Photoelectric effect, let's look at what this problem asks us."
    },
    {
        "Q": "\nAt 9:30, Why do non bonding electrons take up more space than bonding electrons?",
        "A": "Bonding electrons have to spend most of their tine between two nuclei. Nonbonding electrons are attracted to a nucleus only from one side, so they are free to wander further away.",
        "video_name": "0na0xtIHkXA",
        "timestamps": [
            570
        ],
        "3min_transcript": "We put sulfur in the center here. We know sulfur is bonded to 4 fluorines. So we put our fluorines around like that. And let's see how many valence electrons we've shown so far- 2, 4, 6, and 8. So 34 minus 8 gives us 26 valence electrons we still need to account for on our dot structure."
    },
    {
        "Q": "\nAt 9:25 are these resonant structures? Is the second structure possible as later at 11:58 it is said that VSEPR theory predicts first one more?",
        "A": "They aren t resonance structures, they re two possible ways to arrange the electron pairs. The molecule will do whichever is more stable.",
        "video_name": "0na0xtIHkXA",
        "timestamps": [
            565,
            718
        ],
        "3min_transcript": "We put sulfur in the center here. We know sulfur is bonded to 4 fluorines. So we put our fluorines around like that. And let's see how many valence electrons we've shown so far- 2, 4, 6, and 8. So 34 minus 8 gives us 26 valence electrons we still need to account for on our dot structure."
    },
    {
        "Q": "\nAt 5:49, Sal explains just out of the blue how we reduced 2NAD+'s into 2NADH's when we went from the 2 pyruvates to Acetil-CoA, but nowhere in the previous minutes he explains how that came to happen. Any help? :(",
        "A": "Glucose has 12 hydrogens. Pyruvate has 3 hydrogens (X2=6 for two molecules). 4H from the glucose molecule go to produce the 2 NADH in the first step (2H are needed to produce one NADH). The remaining 2H are used with 2H from two coenzymeA molecules (H-S-CoA) to produce 2 NADH when pyruvate is converted to acetyl CoA.",
        "video_name": "9zoS5WGsmpc",
        "timestamps": [
            349
        ],
        "3min_transcript": "then the acetyl group, bonding with that sulfur, and by doing that, you form acetyl-CoA. And acetyl-CoA, just so you know, you only see three letters here, but this is actually a fairly involved molecule. This is actually a picture of acetyl-CoA, I know it's really small, but hopefully you'll appreciate that it's a more involved molecule. That, the acetyl group that we're talking about is just this part, right over here, and it's a coenzyme. It's really acting to transfer that acetyl group, and we'll see that in a second. But it's also fun to look at these molecules, because once again, we see these patterns over and over again in biology or biochemistry. Acetyl-CoA, you have an adenine right over here. It's hard to see, but you have a ribose, and you also have two phosphate groups. So this end of the acetyl-CoA is essentially, is essentially an ADP. But it's used as a coenzyme. Everything that I'm talking about, this is all going to be facilitated by enzymes, and the enzymes will have cofactors, coenzymes, if we're talking about organic And as we see, the acetyl group joins on to the coenzyme A, forming acetyl-CoA, but that's just a temporary attachment. The acetyl-CoA is, essentially, gonna transfer the acetyl group over to, and now we're going to enter into the citric acid cycle. It's gonna transfer these two carbons over to oxaloacetic acid, to form citric acid. So it's gonna transfer these two carbons to this one, two, three, four carbon molecule, to form a one, two, three, four, five, six carbon molecule. But before we go into the depths of the citric acid cycle, I wanna make sure that I don't lose track of my accounting, because, even that step right over here, where we decarboxylated the pyruvate, we went from pyruvate to acetyl-CoA, that also reduced some NAD to NADH. Now, this is gonna happen once for each pyruvate, but we're gonna- all the accounting we're gonna say, is for one glucose molecule. gonna happen for each of the pyruvates. So this is going to be times- This is going to be times two. So we're gonna produce two, two NADH's in this step, going from pyruvate to acetyl-CoA. Now, the bulk of, I guess you could say, the catabolism, of the carbons, or the things that are eventually going to produce our ATP's, are going to happen in what we call the citric acid, or the Kreb cycle. It's called the citric acid cycle because, when we transferred the acetyl group from the coenzyme A to the oxaloacetic acid, we formed citric acid. And citric acid, this is the thing that you have in lemons, or orange juice. It is this molecule right over here. And the citric acid cycle, it's also called the Kreb cycle, when you first learn it, seems very, very complex, and some could argue that it is quite complex. But I'm just gonna give you an overview of what's going on. The citric acid, once again, six-carboned, it keeps getting broken down, through multiple steps,"
    },
    {
        "Q": "At 3:05 Sal says that the most compact state of water is liquid. Why?\n",
        "A": "Because when water turns into a solid, it forms crystals, and those crystals take up a little more volume than when the water molecules are allowed to mix freely without forming crystals.",
        "video_name": "zjIVJh4JLNo",
        "timestamps": [
            185
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n~4:00 he says the CL is smaller than the Na. Why is it a smaller molecule? Doesn't it have more electrons?",
        "A": "as we go across the periodic table nuclear charge as well as no. of electrons increase BUT increase in nuclear charge dominates over increase in no. of electrons so overall atomic radius decreases and this is the case with Cl and Na.",
        "video_name": "zjIVJh4JLNo",
        "timestamps": [
            240
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:21, Sal said that some molecules will be hitting the wall and would have a change in momentum. Does the net energy of the molecules change with the change in momentum?",
        "A": "Of course energy changes if momentum changes.",
        "video_name": "tQcB9BLUoVI",
        "timestamps": [
            201
        ],
        "3min_transcript": "in that direction. They all have their own little velocity vectors, and they're always constantly bumping into each other, and bumping into the sides of the container, and ricocheting here and there and changing velocity. In general, especially at this level of physics, we assume that this is an ideal gas, that all of the bumps that occur, there's no loss of energy. Or essentially that they're all elastic bumps between the different molecules. There's no loss of momentum. Let's keep that in mind, and everything you're going to see in high school and on the AP test is going to deal with ideal gases. Let's think about what pressure means in this context. A lot of what we think about pressure is something pushing on an area. If we think about pressure here-- let's pick an arbitrary area. Let's take this side. Let's take this surface of its container. generated onto this surface? It's going to be generated by just the millions and billions and trillions of little bumps every time-- let me draw a side view. If this is the side view of the container, that same side, every second there's always these little molecules of gas moving around. If we pick an arbitrary period of time, they're always ricocheting off of the side. We're looking at time over a super-small fraction of time. And over that period of time, this one might end up here, this one maybe bumped into it right after it ricocheted and came here, this one changes momentum and goes like that. This one might have already been going in that direction, But what's happening is, at any given moment, since there's so many molecules, there's always going to be some molecules that are bumping into the side of the wall. When they bump, they have a change in momentum. All force is change in momentum over time. What I'm saying is that in any interval of time, over any period or any change in time, there's just going to be a bunch of particles that are changing their momentum on the side of this wall. That is going to generate force, and so if we think about how many on average-- because it's hard to keep track of each particle individually, and when we did kinematics and stuff, we'd keep track of the individual object at play. But when we're dealing with gases and things on a macro"
    },
    {
        "Q": "At 7:21 he says \"they have the same kinetic energy.\" But in order to \"squeeze\" the box, at least one wall must move inward, and while moving it will hit some of the particles and make them move faster. So wouldn't the kinetic energy increase (both the energy of some individual particles and the average energy)? Maybe this is negligible (especially if we squeeze the box very slowly)?\n",
        "A": "We don t squeeze the box it s more like transferring the particles from a larger box to a smaller box keeping the kinetic energy same.",
        "video_name": "tQcB9BLUoVI",
        "timestamps": [
            441
        ],
        "3min_transcript": "With that out of the way, let me give you a formula. I hope by the end of this video you have the intuition for why this formula works. In general, if I have an ideal gas in a container, the pressure exerted on the gas-- on the side of the container, or actually even at any point within the gas, because it will all become homogeneous at some point-- and we'll talk about entropy in future videos-- but the pressure in the container and on its surface, times the volume of the container, is equal to some constant. We'll see in future videos that that constant is actually proportional to the average kinetic energy of the molecules bouncing around. That should make sense to you. If the molecules were moving around a lot faster, then you would have more kinetic energy, and then they would be changing momentum on the sides of the surface a lot more, so you would have more pressure. why pressure times volume is a constant. Let's say I have a container now, and it's got a bunch of molecules of gas in it. Just like I showed you in that last bit right before I erased, these are bouncing off of the sides at a certain rate. Each of the molecules might have a different kinetic energy-- it's always changing, because they're always transferring momentum to each other. But on average, they all have a given kinetic energy, they keep bumping at a certain rate into the wall, and that determines the pressure. What happens if I were able to squeeze the box, and if I were able to decrease the volume of the box? in it, but I squeeze. I make the volume of the box smaller-- what's going to happen? I have the same number of molecules in there, with the same kinetic energy, and on average, they're moving with the same velocities. So now what's going to happen? They're going to be hitting the sides more often-- at the same time here that this particle went bam, bam, now it could go bam, bam, bam. They're going to be hitting the sides more often, so you're going to have more changes in momentum, and so you're actually going to have each particle exert more force on each surface. Because it's going to be hitting them more often in a given amount of time. The surfaces themselves are smaller. You have more force on a surface, and on a smaller surface, you're going to have higher pressure."
    },
    {
        "Q": "at 8:51 they have written cos60 equals 3^1/2/2but actually cos60 equals 1/2\n",
        "A": "Yup he took the value wrong.",
        "video_name": "KDHuWxy53uM",
        "timestamps": [
            531
        ],
        "3min_transcript": "product-- taking the dot product, to the force and the distance factor. And we know that the definition is the magnitude of the force vector, which is 100 newtons, times the magnitude of the distance vector, which is 10 meters, times the cosine of the angle between them. Cosine of the angle is 60 degrees. So that's equal to 1,000 newton meters times cosine of 60. Cosine of 60 is what? It's square root of 3 over 2. Square root of 3 over 2, if I remember correctly. So times the square root of 3 over 2. So the 2 becomes 500. So it becomes 500 square roots of 3 joules, whatever that is. I don't know 700 something, I'm guessing. Maybe it's 800 something. I'm not quite sure. But the important thing to realize is that the dot It applies to work. It actually calculates what component of what vector goes in the other direction. Now you could interpret it the other way. You could say this is the magnitude of a times b cosine of theta. And that's completely valid. And what's b cosine of theta? Well, if you took b cosine of theta, and you could work this out as an exercise for yourself, that's the amount of the magnitude of the b vector that's going in the a direction. So it doesn't matter what order you go. So when you take the cross product, it matters whether you do a cross b, or b cross a. But when you're doing the dot product, it doesn't matter what order. So b cosine theta would be the magnitude of vector b that goes in the direction of a. So if you were to draw a perpendicular line here, b cosine theta would be this vector. That would be b cosine theta. The magnitude of b cosine theta. So you could say how much of vector b goes in the same Or you could say how much of vector a goes in the same direction is vector b? And then multiply the two magnitudes. And now, this is, I think, a good time to just make sure you understand the difference between the dot product and the cross product. The dot product ends up with just a number. You multiply two vectors and all you have is a number. You end up with just a scalar quantity. And why is that interesting? Well, it tells you how much do these-- you could almost say-- these vectors reinforce each other. Because you're taking the parts of their magnitudes that go in the same direction and multiplying them. The cross product is actually almost the opposite. You're taking their orthogonal components, right? The difference was, this was a a sine of theta. I don't want to mess you up this picture too much. But you should review the cross product videos. And I'll do another video where I actually compare and But the cross product is, you're saying, let's multiply the magnitudes of the vectors that are perpendicular to each other, that aren't going in the same direction, that are"
    },
    {
        "Q": "At 6:38, why is that proton acidic?\n",
        "A": "The carbon atom is sp hybridized (50% s character). Since s electrons are held more tightly to the carbon nucleus, they are further from the hydrogen nucleus. The H atom is not tightly held, so it is more easily removed. The proton is acidic.",
        "video_name": "_-I3HdmyYfE",
        "timestamps": [
            398
        ],
        "3min_transcript": "is going to interact with that negatively charged carbanion like that. So that's so that's the first reaction, Formation of your alkynide anion. And then if you want to do an alkylation, it's a separate reaction. You take this, and let's react it with the ethyl bromide. So CH3CH2Br. If you think about what's going to happen, the lone pair of electrons on the carbon is going to attack this carbon, the one that's bonded to your halogen, like that. The halogen is going to leave, and you're going to put this alkyl group onto your alkyne. So you're going to end up with an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen and then carbon triple bonded to another carbon, and then we have to put our alkyl group on there. So a CH2CH3, like that. So we've alkylated our alkyne. This is a very useful reaction for organic synthesis. So let's take the molecule we just made, So if I took this-- let me go ahead and redraw it over here. So if I took this alkyne, so we just formed this. And let's react it. Let's first react it with our base again. So let's use sodium amide right here. And in our second step, we'll react it with a primary alkyl halide. So let's go ahead and draw a primary alkyl halide here, so that is our molecule. So we think to ourselves, what happens? I have a strong base. I still have an acidic proton left on my alkyne, right? So the proton over here on the left. So that's what the base is going to do. The base is going to take that proton forming a negatively charged carbanion, an alkynide anion. And then that anion is going to be our nucleophile for an SN2 reaction. So when you're thinking about it, these electrons in here that are going to be on that carbon giving a negative 1 formal charge are going to come all the way over to our carbon. So let's go ahead and draw the products of that. We're going to have our benzene ring. So let's go ahead and draw our benzene ring here, so let's put in our electrons going around my benzene ring. And then on that benzene ring is a CH2. So that CH2 is the red one that we marked right here, and this is the alkyl group that gets put onto your alkyne. So let's just go ahead and finish drawing our alkyne here. So we have now our triple bonds, right? Carbon triple bonded to another carbon. And then our ethyl group. So CH2CH3. So you'll see in later videos how we use the acidity of terminal alkynes to alkylate when we do a few different synthesis problems."
    },
    {
        "Q": "8:37 Howcome theres not actual recorded history we learn about, on humans from 200,000 years ago?\n",
        "A": "We have not found any records that go that far back. The earliest writing we have discovered is from around 3200 BCE (about 5,200 years ago).",
        "video_name": "MS7x2hDEhrw",
        "timestamps": [
            517
        ],
        "3min_transcript": "believe, a huge rock, a six-mile in diameter rock, colliding with what is now the Yucatan Peninsula in Mexico, or right off the coast of the Yucatan Peninsula. And it destroyed all of the large land life forms, especially the dinosaurs. And to put all of this in perspective-- and actually the thing that really was an aha moment for me-- it's, OK, plants are 450 million years ago. Grass, I kind of view as this fundamental thing in nature. But grass has only been around for about-- I've seen multiple estimates-- 40 to 70 million years. Grass is a relatively new thing on the planet. Flowers have only been around for 130 million years. So there was a time where you had dinosaurs, but you did not have flowers and you did not have grass. And so you fast forward all the way. And so when you look at this scale, it's kind of funny to look at this. This is the time period where the dinosaurs showed up. This whole brown line is where the mammals showed up. And then, of course, the dinosaurs died out here. Our ancestors, when the giant rock hit the Earth, must have been boroughed in holes and were able to stash some food away, or who knows what, and didn't get fully affected. I'm sure most of the large mammals were destroyed. But what's almost-- it's humbling, or almost humorous, or almost ridiculous, when you look at this chart is they put a little dot-- you can't even see it here, They say 2 million years ago, the first humans-- and even this is being pretty generous when they say first humans. These are really the first prehumans. The first humans that are the same as us, if you took one of those babies and your brought them up in the suburbs and gave them haircuts and stuff, they would be the same thing as we are, those didn't exist until 200,000 years ago, give or take. 200,000 to 400,000 years ago, I've seen estimates. So this is actually a very generous period of time to say first humans. It's actually 200,000 years ago. we are and how new evolution is, it was only 5 million years ago-- and I mentioned this in a previous video-- it was only 5 million years ago-- so this is just to get a sense. This is 0 years. Homo sapien sapien, only around for 200,000 years. The Neanderthals, they were cousin species. They weren't our ancestors. Many people think they were. They were a cousin species. We come from the same root. Although there are now theories that they might have remixed in with Homo sapiens. So maybe some of us have some Neanderthal DNA. And it shouldn't be viewed as an insult. They had big brains. Well, they didn't necessarily have big brains. They had big heads. But that seems to imply a big brain. But who knows? We always tend to portray them as somehow inferior. But I don't want to get into the political correctness of how to portray Neanderthals. But anyway, this is a very small period of time, 200,000. If you go 2 million years, then you"
    },
    {
        "Q": "\nAt 3:39, Sal says there is only one Snowball Earth. However, the chart says there are two.",
        "A": "He probably only mentioned one.",
        "video_name": "MS7x2hDEhrw",
        "timestamps": [
            219
        ],
        "3min_transcript": "layer to build up. Ozone is just three oxygen atoms. It is O3. And by the end of the Proterozoic Eon-- so we're talking, I don't know, maybe 550 million years ago, give or take tens of, or hundreds, or maybe 100 million years-- these are all moving targets-- the ozone layer was dense enough to protect the land from UV rays. We talked about that in the last video, that the Earth is being bombarded with UV rays. And the ozone layer is the only thing that really keeps us from being seriously irradiated by the Sun and allows land animals to actually live. And so coinciding with that time period, around 550 million years ago, you start to have life colonizing, especially significant life, colonizing land. So life colonizes land, colonizes the land. And this was kind of an interesting-- when I first You always assume that kind of trees and grasses are kind of part of the background. They come part and parcel with land. But it turns out that animals colonized land before plants did. Plants didn't come into the picture until about 450 million years ago, give or take a few tens of millions of years. And so we're now entering the end of the Proterozoic Eon. Life has started to colonize land. We now have an ozone layer. And what happens-- and actually there's another snowball glaciation or a snowball Earth near the end of the Proterozoic Era, Eon, I should say. And there's a bunch of theories about why it came about. And then why disappeared. Maybe there were volcanoes, greenhouse gases, who knows. But as we enter the end of that, we start seeing life began to flourish. And it starts to really flourish as we enter the Phaner-- I always have trouble And it's not even labeled here. The Phanerozoic Eon is this chunk of time right over here. And let me write it out. So this right over here is the Phanerozoic, the Phanerozoic Eon. And so this chart, these divisions right here are eons. And then they jump into, instead of doing eons here, they then break into eras. Eras are subsets of eons. They are hundreds of millions of years. So this is the Paleozoic Era, the Mesozoic Era, and the Cenozoic Era. And that's actually our current era. But perhaps the most interesting-- well, I don't want to pick favorites here. But it's one of the most interesting times in the geologic era-- is the first period in the Paleozoic Era, which is the first era in the Phanerozoic Eon."
    },
    {
        "Q": "\nAt 2:00, when Sal explains, why is there in the velocity formula, an \"s\" instead of a \"d\" can the letter \"s\" stand for space to be covered? in other words moving from one point to another?",
        "A": "the letters don t matter. You can define any variable to be anything you want. Focus on the concepts.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            120
        ],
        "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time."
    },
    {
        "Q": "Isn't there any other way to convert it? 7:26 Maths seems to be easier here in Brazil...\n",
        "A": "The conversion method is certified by the system international units so if we use it it would be better and can be applied on almost every sum used in maths and physics. by the way no math is difficult or easy in any country provided the way u think lol,",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            446
        ],
        "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always"
    },
    {
        "Q": "\nAt 11:16 in the video , sal calculates 5000/3600 which is somewhere 1.38888888888889. so my question is how did that 1.888888888889 become 1.39 ?",
        "A": "when we estimate 1.38888888 ,we round it off to 1.39 because it is better to write it shorter.if in case you had to multiply it with another number,it would be difficult .For example, 1.38888888888888.....*19 will be never ending ,so we do it.And by the way its 1.3888 not 1.88888",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            676
        ],
        "3min_transcript": "are in an hour. So that's your gut check. We should get a smaller number than this when we want to say meters per second. But let's actually do it with the dimensional analysis. So we want to cancel out the hours, and we want to be left with seconds in the denominator. So the best way to cancel this hours in the denominator is by having hours in the numerator. So you have hours per second. So how many hours are there per second? Or another way to think about it, 1 hour, think about the larger unit, 1 hour is how many seconds? Well, you have 60 seconds per minute times 60 minutes per hour. The minutes cancel out. 60 times 60 is 3,600 seconds per hour. or if you flip them, you would get 1/3,600 hour per second, or hours per second, depending on how you want to do it. So 1 hour is the same thing as 3,600 seconds. And so now this hour cancels out with that hour, and then you multiply, or appropriately divide, the numbers right here. And you get this is equal to 5,000 over 3,600 meters per-- all you have left in the denominator here is second. Meters per second. And if we divide both the numerator and the denominator-- I could do this by hand, but just because this video's already getting a little bit long, let me get my trusty calculator out. I get my trusty calculator out just for the sake of time. as 50 divided by 36, that is 1.3-- I'll just round it over here-- 1.39. So this is equal to 1.39 meters per second. So Shantanu was traveling quite slow in his car. Well, we knew that just by looking at this. 5 kilometers per hour, that's pretty much just letting the car roll pretty slowly."
    },
    {
        "Q": "at 2:20 he says you use d for calculus, but don't you put an arrow on top for displacement? Do you use an arrow for the dirrivative operation too?\n",
        "A": "Haha no, it just means that our alphabet ran out of letters, and it is generally a bad idea to use one letter for 2 purposes.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            140
        ],
        "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time."
    },
    {
        "Q": "\nAt 7:23 Sal says km is going to cancel out with the denominators but the numerator is meters.\nCould someone pls. help",
        "A": "What Sal meant was at the denominator, the unit was also km. in direct variation, you cancel out the units to change a value of a particular unit into a different value with a different unit. In other words, Sal was saying that in 5 km X 1000 m/ 1 km, the kilometers will cancel out, and you will be left with just the meters. He then multiplied it and came out with 5000 meters.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            443
        ],
        "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always"
    },
    {
        "Q": "Can you simplify velocity? At about 7:20, Sal stated 5 km per hour is the velocity. Usually, in physics, do you simplify final answers or leave it in their original measurements, the question asked with?\n",
        "A": "we should probably simplify them because we usually count them per 1 hour.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            440
        ],
        "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always"
    },
    {
        "Q": "Can someone please explAin the h2o part which sal talks about from 3:00. Thanks\n",
        "A": "What he s saying is that in each H-O bond two electrons are being shared. However, oxygen, being more electronegative than hydrogen, grabs more than its fair share of the electron density in the bond. (You can think of it as a tug-of-war between oxygen and hydrogen, with oxygen winning by dragging more electron density its way.) This means that the bonds are polarised, with oxygen having a partial negative charge and each of the hydrogens a partial positive charge.",
        "video_name": "Rr7LhdSKMxY",
        "timestamps": [
            180
        ],
        "3min_transcript": "So how badly wants to hog, and this is an informal definition clearly, hog electrons, keep the electrons, to spend more of their time closer to them then to the other party in the covalent bond. And this is how, how much they like electrons, or how much affinity they have towards electrons. So how much they want electrons. And you can see that these are very, these are very related notions. This is within the context of a covalent bond, how much electron affinity is there? Well this, you can think of it as a slightly broader notion, but these two trends go absolutely in line with each other. And to think about, to just think about electronegativity makes it a little bit more tangible. Let's think about one of the most famous sets of covalent bonds, and that's what you see in a water molecule. you have an oxygen atom, and you have two hydrogens. Each of the hydrogen's have one valence electron, and the oxygen has, we see here, at it's outermost shell, it has one, two, three, four, five, six valence electrons. One, two, three, four, five, six valence electrons. And so you can imagine, hydrogen would be happy if it was able to somehow pretend like it had another electron then it would have an electron configuration a stable, first shell that only requires two electrons, the rest of them require eight, hydrogen would feel, hey I'm stable like helium if it could get another electron. And oxygen would feel, hey I'm stable like neon if I could get two more electrons. And so what happens is they share each other's electrons. This, this electron can be shared in conjunction with this electron for this hydrogen. So that hydrogen can kind of feel like it's using it stabilizes the outer shell, or it stabilizes the hydrogen. And likewise, that electron could be, can be shared with the hydrogen, and the hydrogen can kind of feel more like helium. And then this oxygen can feel like it's a quid pro quo, it's getting something in exchange for something else. It's getting the electron, an electron, it's sharing an electron from each of these hydrogens, and so it can feel like it's, that it stabilizes it, similar to a, similar to a neon. But when you have these covalent bonds, only in the case where they are equally electronegative would you have a case where maybe they're sharing, and even there what happens in the rest of the molecule might matter, but when you have something like this, where you have oxygen and hydrogen, they don't have the same electronegativity. Oxygen likes to hog electrons more than hydrogen does. And so these electrons are not gonna spend an even amount of time. Here I did it kind of just drawing these, you know, these valence electrons as these dots. But as we know, the electrons are in this kind of blur around, around the,"
    },
    {
        "Q": "9:05 Is nuclear envelope and nuclear membrane the same thing?\n",
        "A": "Yes, they are just different terms. You could call my shirt clothing or a textile envelope, although most people say the first I prefer the latter.",
        "video_name": "mMCcBsSAlF4",
        "timestamps": [
            545
        ],
        "3min_transcript": "and now going down over here. Over here. And let me draw the microtubules that are really... well, I've said it multiple times, super involved in actual the movement going on. They're elongating, they're these motor proteins that are moving the chromosomes along, once again, they're connected at the kinetochores right over here. Connected at the kinetochores. Right over there. And now we're almost done, we're ready to move into telephase II. So we're now going to go into telephase II. Telephase II. Where my two cells are now becoming four cells, so telephase II, I'm gonna show the cytokinesis starting to happen. So telephase II. So turning into four cells, In this cell up here, I have this character, and has a little bit of magenta right over here. That's this right over there and then you have the shorter magenta one. And actually, they are starting to, they're starting to unravel into their chromatid form, so maybe I'll draw that a little bit, and then this one, right over here, is starting to unravel into its chromatid form. And so it this that one. Whoops, wanna do that in that magenta color. Starting to unravel into its chromatid form, I wanna do it over here, this one is starting to unravel. And so is this one. So is... (laughs) I'm having trouble changing colors. And so is that one, and then up here, this one's starting to unravel, this one over here, and... this longer, mostly magenta one Also starting to unravel. You start having your nuclear envelope formed again, so your nuclear envelope is forming again. Nuclear envelope is forming. Your microtubules are dissolving. Let me draw the centrosomes, they're outside of the nuclear envelope. Outside of the nuclear envelope. And of course, you're finally dividing the cells, your cytokinesis happens, so now you have your four, your four cells, each have a haploid number. They each have two chromosomes. Remember, you diploid number was four, the germ cell had four chromosomes, two pairs of homologous chromosomes. Now each of your resulting gametes, these are now gametes now, these are gametes, they have a haploid number. But we started with a haploid number at the beginning of meiosis II, so that's why meiosis II"
    },
    {
        "Q": "\nAt 7:04, how exactly do the glycolipids help the cell to be recognized or tagged? Do they form specific patterns across the cell membrane? Do they have specific components that signal different things?",
        "A": "One example is that immune cells have receptors that can attach to them. So a cell with a type of glycolipid that isn t normally present in your body (the glycolipid is then an antigen) will attach to the immune cell, and so be identified as an invader.",
        "video_name": "cP8iQu57dQo",
        "timestamps": [
            424
        ],
        "3min_transcript": "this is a protein, this is a protein, and I just drew some blobs to be indicative of the variety of proteins. But the important thing to realize is, if we think of cells, there's all of this diversity. There's all of this complexity that is on, or embedded, inside of its membrane. So instead of just thinking of it as just kind of as a uniform phospholipid bilayer, there's all sorts of stuff, maybe if we view this as a cross-section, there's all sorts of stuff embedded in it and we see it right over here in this diagram. You could say there's a mosaic of things embedded in it. A mosaic is a picture made up of a bunch of different components of all different colors, and you can see that you have all different components here, different types of proteins. You have proteins like this, that go across the membrane. We call these transmembrane proteins, they're a special class of integral protein. You have integral proteins like this, that might only interact with one part of the bilayer while these kind of go across it. So this right over here, this is a glycolipid, which is fascinating. It lodges itself in the membrane because it has this lipid end, so that's going to be hydrophobic. It's going to get along with all of the other hydrophobic things, but then it has an end that's really a chain of sugars and that part is going to be hydrophilic, it's going to sit outside of the cell. And these chains of sugars, these are actually key for cell-cell recognition. Your immune system uses these to differentiate between which cells are the ones that are actually from my body, the ones I don't want to mess with, the ones I want to protect and which cells are the ones that are foreign, the ones that I might want to attack. When people talk about blood type, they're talking about, well, what type of specific glycolipids do you have on cells. And there's all sorts of, that's not all we're talking about when we talk about glycolipids as a way or to be tagged in different ways. So it's a fascinating thing that these chains of sugars can lead to such complex behavior, and frankly, such useful behavior, from our point of view. But you don't just want to have sugar chains on lipids, you also have sugar chains on proteins. This, right over here, is an example of a glycoprotein. And as you can see, when you put all this stuff together, you get a mosaic, and I'm actually not even done. You have things like cholesterol embedded. Cholesterol is a lipid, so it's going to sit in the hydrophobic part of the membrane and that actually helps with the fluidity of the membrane, making sure it's not too fluid or not too stiff. So this is cholesterol, right over there. So you see this mosaic of stuff, but what about the fluid part? And I just talked about cholesterol's value in making sure that it's just the right amount of fluidity. What's neat about this, is this isn't a rigid structure."
    },
    {
        "Q": "\nAt 4:49, he states the concentration of the Hydronium ions is 5.0 x 10^-14\nWhere did he get that figure from? Thanks",
        "A": "Well, (1.0 * 10^-14) / (0.2) is the equal to 5.0 * 10^-14. All he did was divide both sides by 0.2 in order to isolate for the variable x . Hope this helps!",
        "video_name": "gsu4gjrFApA",
        "timestamps": [
            289
        ],
        "3min_transcript": "and magnesium hydroxide. Let's do this problem. Calculate the pH of a 0.20 molar solution of sodium hydroxide. Let's go ahead and write sodium hydroxide here is a strong base. You get 100% dissociation in water, so 100% dissociation. Sodium hydroxide consist of Na plus and OH minus so you get dissociation and you get the sodium cation and the hydroxide anion, so Na plus and OH minus in water. If we have .2 molar of NaOH, we're also going to get .20 molar of hydroxide ions in solutions and sodium hydroxide is a strong base and we have 100% dissociation. If the concentration of hydroxide ion is 0.2, and so one way to solve this problem is to use this equation. The concentration of hydronium times the concentration of hydroxide is equal to 1.0 times 10 to the negative 14 from an earlier video. We can plug in our concentration for hydroxide and we can say the concentration of hydronium is x. We have x times 0.20 is equal to 1.0 times 10 to the negative 14. If you do that math, all right, so if you do that math, x is equal to the concentration of hydronium ions. This would be five, right. This would be 5.0 times 10 to the negative 14. Now that we have the concentration of hydronium ions, we can now calculate the pH of our solution because pH is equal to negative log We can do this on our calculator here. Let's get out the calculator to find the pH. The pH is equal to the negative log of, this is 5.0 times 10 to the negative 14. We get a pH of 13.30. All right, so let's go ahead and write that down here. pH is equal to 13.30. There's another way to do this problem, all right, and that is when you find, let me use a different color for this. When you find the concentration of hydroxides, you can immediately find the pOH, right. The pOH is equal to the negative log of the concentration of hydroxide ions. All right, so that would be .2, so we could plug that into here to find the pOH. pOH is equal to the negative log of the concentration of our hydroxide and ion."
    },
    {
        "Q": "At 1:30, he says that the concentration of HNO3 andH3O are the same. Would the NO3- also have that same concentration of .03M? Why or why not?\n",
        "A": "It would, we just don t care about it",
        "video_name": "gsu4gjrFApA",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] Here we have some strong acids which ionize 100% in solution. HClO4 is perchloric acid. HCl is hydrochloric acid. HBr is hydrobromic acid. HI is hydriodic acid. H2SO4 is sulfuric acid and HNO3 is nitric acid. Let's do a calculation using nitric acid here. Calculate the pH of a 0.030 molar solution of nitric acid. Well nitric acid is a strong acid which means it ionizes 100% in solution. If we have HNO3 and H2O here, the nitric acid is gonna donate a proton to water and if water accepts a proton, water turns into H3O plus, the hydronium ion. If HNO3 loses a proton, we're left with NO3 minus or nitrate. Since nitric acid is strong, and everything turns into our products. If this is our concentration of HNO3, this would also be our concentration of hydronium ions. The concentration of hydronium ions is .030 molar since we're dealing with a strong acid. To calculate the pH, all we have to do is plug in to our definition of pH. The pH is equal to the negative log of the concentration of hydronium ions. All we have to do is plug this number into here so the pH is equal to the negative log of 0.030. Let's get out the calculator and do that. All right, so we have the negative log of .030. We get 1.52. so the pH is equal to 1.52. For our significant figures, we have two significant figures here so we have two to the right of our decimal point. Instead of writing it this way, all right, so instead of writing this, you could have written a shortened version. You could have just written HNO3, nitric acid ionizes 100% and so it turns into H plus and NO3 minus, so that's just a shortened version of the same thing we have above. If you're working with this, you could have said that the pH is equal to the negative log of the concentration of H+. That's the same thing as we did above. All right, next let's look at strong bases. We're only gonna talk about metal hydroxides as being strong bases in this video. Some common metal hydroxides, sodium hydroxide,"
    },
    {
        "Q": "At 11:40, why do we not square the concentration of OH? From previous lessons, I remember we have to raise the concentration to the power of the molar ratio. So shouldn't it be 1.0*10^-14/(0.012)^2 for OH? Thanks\n",
        "A": "That rule is only for finding the equilibrium constant. But here, when we are finding the pH, we are supposed to multiply it by 2 as we need to know the number of moles.",
        "video_name": "gsu4gjrFApA",
        "timestamps": [
            700
        ],
        "3min_transcript": "and let's do that calculation here. Let's get some room, we have .0030 divided by .250 and we get .012 molar. All right, so this is .012 molar for the concentration of hydroxide. Now that you have the concentration of hydroxide ions and solution, you can find the pH using one of the two ways that we talked about above. All right, so you could go for this equation, hydronium ion concentration times hydroxide ion concentration is equal to 1.0 times 10 to the negative 14. All right, and you could take this and plug it into here. You get x times 0.012 is equal to 1.0 times 10 to the negative 14 and let's get a little bit more room here We have 1.0 times 10 to the negative 14. We're gonna divide that by .0. Actually let me do that again. We're going to divide that by .012 and this is going to give us 8.3 times 10 to the negative 13. All right, so let's write this down, x is equal to 8.3 times 10 to the negative 13. Remember what x referred to, x referred to the concentration of hydronium ions so this is the concentration of hydronium ions which means we can now calculate the pH. All right, the pH be equal to the negative log of that concentration. The negative log of 8.3 times 10 to the negative 13. Let's do one more final calculation here. We're gonna do negative log 10 to the negative 13, and we're going to get 12.08. All right, so the pH is equal to 12.08 and we're finally done with our problem here. Notice the pH is in the base range, so calcium hydroxide is a base."
    },
    {
        "Q": "At 3:35 in the video, I attached my 2 hydrogens to the nitrogen at different places. I put one hydrogen on the top and one on the bottom. Is that incorrect? Would my way of drawing it be acceptable?\n",
        "A": "Your way is equally correct to Sal s way. These dot structures show connectivity, but not shape, so it doesn t matter where you drawn in your H s attached to the nitrogen.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            215
        ],
        "3min_transcript": "that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore, I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH. with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group VI right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in-- one, two, three, four, five, and six, like that. And then I'm going to put in the hydrogen, right? So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond."
    },
    {
        "Q": "How will I know which part will I put the hydrogen? For example @4:54. How did you know that the hydrogen is to be put at the right side of oxygen?\n",
        "A": "It doesn t matter exactly where as long as it s bonded to the oxygen. It would be just as valid to put it on the top or bottom of the oxygen instead of right.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            294
        ],
        "3min_transcript": "I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH. with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group VI right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in-- one, two, three, four, five, and six, like that. And then I'm going to put in the hydrogen, right? So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond. And the oxygen has bonded to this hydrogen as well. Again, we can check our octet rule. So the carbon has eight electrons around it. And so does the oxygen. So this would be two right here, and then four, and then six, and then eight. So oxygen is going to follow the octet rule. Now when you're drawing dot structures, you don't always have to do this step where you're drawing each individual atom and summing all of your valence electrons that way. You can just start drawing it. So for an example, if I gave you C2 H6, which is ethane, another way to do it would just be starting to draw some bonds here. And so I have two carbons. And it's a pretty good bet those two carbons are going to be connected to each other. And then I have six hydrogens. And if I look at what's possible around those carbons, I could put those six hydrogens around those two carbon atoms, like that. And if I do that, I'll have an octet around each carbon atom."
    },
    {
        "Q": "\nI don't understand the part after 1:34 . The part about carbon with 8 electrons and hydrogen with 2",
        "A": "Hydrogen can only have two electrons total but Carbon can have 8 valence electrons total.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            94
        ],
        "3min_transcript": "In this video, we're going to look at how to draw dot structures of simple organic molecules that have single bonds. So if I look at the molecular formula CH4, which is methane, and I want to draw a dot structure for the methane molecule, I would go over here to my organic periodic table and find carbon. And I can see carbon is in group IV. Therefore, carbon will have four valence electrons. So I can draw a carbon with its four valence electrons around it like that. Remember from general chemistry, valence electrons are the electrons in the outermost energy level. So carbon has four valence electrons in its outermost energy level. Next, I have to think about hydrogen. And hydrogen is in group I on the periodic table. Therefore, hydrogen will have one valence electron. And so I can go ahead and put a hydrogen in there with one valence electron. And I know I have to do that three more times. So I keep putting in hydrogens, each with one valence electron, so a total of four hydrogens. And now I can start connecting my dots. I know that two valence electrons equals one single covalent bond. There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now I can see that carbon is surrounded by eight electrons So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be two, four, six, and eight, like that. And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be one, two, three, and four. And to get to eight electrons, we would go five, six, seven, eight. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore,"
    },
    {
        "Q": "At 5:49, Sal forgot to right hemoglobin!\n",
        "A": "Everyone makes mistakes. :)",
        "video_name": "xKJ3txXIuQk",
        "timestamps": [
            349
        ],
        "3min_transcript": "to reproduce itself. But you immediately see on this picture how small the HIV virus is compared to the actual T-cell. Each of these small little things each of these small things, is an HIV virus, which we already saw is a lot bigger than something like a hemoglobin protein. And so a hemoglobin protein you wouldn't even be able to on this scale, maybe it would be a pixel, if that. And on a similar scale, is this T-cell, you have a, you have things like red blood cells. This is actually a comparison this side by side. This is using an electron, this is using an electron microscope you see a red blood cell right over here and they're roughly roughly on the same size, or at least the same order of magnitude size. And a red blood cell is going to be six to eight micrometers, micrometers wide. So this is six to eight millionths of a meter. So if we were to as the average. Seven millionths, seven seven millionths of a meter. Over here, we're talking about a millionth of a millimeter. Now we're talking about seven millionths of a meter. And just to get an appreciation for size, we already compared the virus, the HIV virus, to this cell. We're seeing it directly as a emerge from the cell. But each of these red blood cells are gonna contain roughly 280 million hemoglobin molecules. So there's gonna be 200 each of these there's gonna be 280 million of these. So 280 million, that's a million million hemoglobins in each one of these. So hopefully this starts to give you an appreciation for even though we categorize cells as these unimaginably small things, they're actually far larger they're ginormous compared to things And especially when you think of things on the molecular, or the atomic scale. And that's why cells are so interesting. They actually have a lot of complexity to them. But just to have an appreciation also for how small cells are even though we've just described these red blood cells and these T-cells there's these kind of worlds unto themselves. They are these incredibly complex things. If I were to draw the width of a human hair on this screen right now relative to the scale of these red blood cells, it would be about as wide as this video. So from if I were to draw a human hair it would go from there roughly to there and there's actually a lot of variance in the width of a human hair. But the width of a human hair would be just about like that. If you looked at the scale of if you looked at the scale of this picture right over here. If you looked at these scales it would be much much much bigger. And I encourage you the width of a human hair. Look at it. Put it on a piece of paper. It's hard to even discern the width."
    },
    {
        "Q": "\nAt 10:23, David talks about how voltage is the difference in Electric Potential. So is the analogy Volts:Electrical Potential::Impulse:Momentum correct in the sense that one is the measurement of the difference of the other?",
        "A": "eh, sort of, but that s not a very useful way to think of it. Impulse is a transfer of momentum. Volts are units that we use to measure electrical potential difference. Voltage is the electrical engineers terms for what physicists call electric potential difference . You can subtract any two quantitites you want from one another. That doesn t mean that any time you choose two pairs of quantities to subtract that there s some interesting comparison to make between the differences.",
        "video_name": "ks1B1_umFk8",
        "timestamps": [
            623
        ],
        "3min_transcript": "per Coulomb of charge that you put there. And it works for any point, if I picked a point twice as close, it's half as far away, let's say some point over here, let's say this r value here was only 4.5 centimeters, well I'm dividing this by r, so if the r is half as big this point over here will have a V value of 200 Joules per Coulomb and the closer I get, if I went even closer, if I went to a point that was three centimeters away, well this is a third as much as this other distance, so if I'm only dividing by a third as much distance as you get three times the result 'cause r is not squared, it's just r. So at this point, we'll have a V value of 300 Joules per Coulomb. This tells me, if I wanted to get a charge that have a whole bunch of Potential energy, I should stick it over here, this will give me a lot of Potential energy. Not quite as much, even less, the further I put my charge the less Potential energy it will have. There will be no Potential energy until there is a charge, there'll just be Electric Potential. But once you place another charge in that region to go with the first one, then you'll have Electric Potential energy and this will be a way to find it, Q times the V that you get out of this calculation. You gotta be careful though, sometimes people get sloppy, and V looks, you know, we use V for Electric Potential and we use V for Voltage, what's the difference? Are they the same? Hmmm, not quite. Sometimes you can treat them as the same but sometimes you do and messes you up. Voltage is a, technically a change in Electric Potential between two points, between two points in space, so it's got the same units 'cause the change in Electric Potential still gonna have units of Joule per Coulomb, it's just, when it's a change in we give this a new title, we call the Joule per Coulomb unit a Volt. So Joules per Coulomb are Volts, but the word Voltage specifically refers to a difference in Electric Potential, what am I talking about? Well, look at, this point is 300 Joules per Coulomb, this point over here 100 Joules per Coulomb, so the delta V, if I were to take delta V between these two points right here and I ask, what's the difference in V? Well the difference in V is 200, 200 Joules per Coulomb, that means the Voltage between those two points in space is 200 Volts, that's what it means. So, when you're talking about a difference"
    },
    {
        "Q": "\nWhat differentiates the blue star 3:55 lifecycle from the red giant or main sequence stars? What makes them burn hotter and faster?",
        "A": "Its all about mass. A massive star will be hotter and burn brighter. Every star was once main sequence. It does not matter how massive, as long as it is stably burning hydrogen, it is main sequence. Once a star runs out of hydrogen or leaves the main sequence, it becomes a red giant or supergiant.",
        "video_name": "w3IKEa_GOYs",
        "timestamps": [
            235
        ],
        "3min_transcript": "in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum probably not too different than own sun the ones that are in the yellowish white, closer to orange-yellowish-white part of the spectrum and the ones that look a little more bluish, or a little bit more greenish these are burning super fast let me see if i can find, this one looks a little big bluish to me, these are burning super super fast, and so the super massive stars, they burn kind of fast and furious and then just die out but the smaller stars, the ones with less mass they burn slower over a much much longer period of time so the ones that are burning fast are emitting a lot of energy at the smaller wavelength part of the light spectrum that's why they look bluer or greener and these are going to be more massive stars the ones that look white or bluer or greener"
    },
    {
        "Q": "At 2:40 Sal mentions a supernova that may have blown away a portion of the Eagle Nebula, and that we wont be able to see the effects of it for another 1000 years. Why only 1000 years and not 7,000? Do we know that said supernova occured 6,000 years ago?\n",
        "A": "Yes, from what we can see right now, which is the Eagle Nebula 7,000 years ago, a burst of energy from a nearby supernova is heading towards the Eagle Nebula. At such a rate, it has destroyed the Eagle Nebula 6,000 years ago.",
        "video_name": "w3IKEa_GOYs",
        "timestamps": [
            160
        ],
        "3min_transcript": "so this is an enormous amount of distance remember, the distance from earth to the nearest star was about 4 light years it would take voyager, if it were pointed in the right direction moving at 60 thousand kilometers per hour it would take Voyager 80 thousand years to go 4 light years just this pillar is 7 light years but i wanted to show you this because these type of nebulae, the plural of nebula are where stars can form. so this right here, you actually see, is actually a breeding ground for the birth of new stars this gas is condensing, just like we talked about a couple of videos ago. Until it gets to that critical temperature, the critical density, where you can actually get fusion of hydrogen so this is just a huge interstellar cloud of hydrogen gas and over here you can see its just this breeding ground for stars and we don't even, we think that this structure doesn't even exist anymore in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum"
    },
    {
        "Q": "\n2:40 \"There was another supernova that happened that could have blown away some of this dust\".\nHow do you know it happened if you can't see it yet?",
        "A": "Supernovae happen all the time, its very likely one has happened.",
        "video_name": "w3IKEa_GOYs",
        "timestamps": [
            160
        ],
        "3min_transcript": "so this is an enormous amount of distance remember, the distance from earth to the nearest star was about 4 light years it would take voyager, if it were pointed in the right direction moving at 60 thousand kilometers per hour it would take Voyager 80 thousand years to go 4 light years just this pillar is 7 light years but i wanted to show you this because these type of nebulae, the plural of nebula are where stars can form. so this right here, you actually see, is actually a breeding ground for the birth of new stars this gas is condensing, just like we talked about a couple of videos ago. Until it gets to that critical temperature, the critical density, where you can actually get fusion of hydrogen so this is just a huge interstellar cloud of hydrogen gas and over here you can see its just this breeding ground for stars and we don't even, we think that this structure doesn't even exist anymore in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum"
    },
    {
        "Q": "\nAt 5:46, Isn't (10^6) ^2 (ten to the sixth power squared) supposed to be 36?",
        "A": "Nope. (10^6)^2 is 10^12. Remeber, (a^m)^n = a^mn",
        "video_name": "391txUI76gM",
        "timestamps": [
            346
        ],
        "3min_transcript": "How many meters is that? It's 6 million meters, right? And then, you know, the extra meter to get to my center of mass, we can ignore for now, because it would be .001, so So it's 6-- and soon. I'll write it in scientific notation since everything else is in scientific notation-- 6.371 times 10 to the sixth 6,000 kilometers is 6 million meters. So let's write that down. So the distance is going to be 6.37 times 10 to the sixth meters. We have to square that. Remember, it's distance squared. So let's see if we can simplify this a little bit. Let's just multiply those top numbers first. Force is equal to-- let's bring the variable out. Mass of Sal times-- let's do this top part. So we have 6.67 times 5.97 is equal to 39.82. multiply the 10's. So 10 to the negative 11th times 10 to the negative 24th. We can just add the exponents. They have the same base. So what's 24 minus 11? It's 10 to the 13th, right? And then what does the denominator look like? It's going to be the 6.37 squared times 10 to the sixth squared. So it's going to be-- whatever this is is going to be like 37 or something-- times-- what's 10 to the sixth squared? It's 10 to the 12th, right? 10 to the 12th. So let's figure out what 6.37 squared is. This little calculator I have doesn't have squared, so I have to-- so it's 40.58. Sal times-- let's divide, 39.82 divided by 40.58 is equal to 9.81. That's just this divided by this. And then 10 to the 13th divided by 10 to the 12th. Actually no, this isn't 9.81. Sorry, it's 0.981. 0.981, and then 10 to the 13th divided by 10 to the 12th is just 10, right? 10 to the first, times 10, so what's 0.981 times 10? Well, the force is equal to 9.81 times the mass of Sal."
    },
    {
        "Q": "Dear Sal,\nat 3:09 of the video above, you said that you found out the Earth's mass on Wikipedia. My teachers always say not to trust wikipedia because it is based off of other people's information/opinions. After watching the whole video (which by the way was really helpful) I found out that the Earth's mass is 5.972 multiplied by 10 to the 24th with the unit as kilograms. On your video, did you just round up the decimal? I just want to make sure so I don't get it wrong on my test next week.\nSincerely,\nAva F.\n",
        "A": "Teachers should stop telling students not to trust wikipedia. It is quite reliable for scientific information.",
        "video_name": "391txUI76gM",
        "timestamps": [
            189
        ],
        "3min_transcript": "So that's simple enough. So let's play around with this, and see if we can get some results that look reasonably familiar to us. So let's use this formula to figure out what the acceleration, the gravitational acceleration, is at the surface of the Earth. So let's draw the Earth, just so we know what we're talking about. So that's my Earth. And let's say we want to figure out the gravitational acceleration on Sal. That's me. And so how do we apply this equation to figure out how much I'm accelerating down towards the center of Earth or the Earth's center of mass? The force is equal to-- so what's this big G thing? The G is the universal gravitational constant. Although, as far as I know, and I'm not an expert on this, It's not truly, truly a constant, or I guess when on different scales, it can be a little bit different. But for our purposes, it is a constant, and the constant in most physics classes, is this: 6.67 times 10 to the negative 11th meters cubed per kilogram seconds squared. I know these units are crazy, but all you have to realize is these are just the units needed, that when you multiply it times a mass and a mass divided by a distance squared, you get Newtons, or kilogram meters per second squared. So we won't worry so much about the units right now. Just realize that you're going to have to work with meters in kilograms seconds. So let's just write that number down. I'll change colors to keep it interesting. 6.67 times 10 to the negative 11th, and we want to know the acceleration on Sal, so m1 is the mass of Sal. And I don't feel like revealing my mass in this And then what's the mass 2? It's the mass of Earth. And I wrote that here. I looked it up on Wikipedia. This is the mass of Earth. So I multiply it times the mass of Earth, times 5.97 times 10 to the 24th kilograms-- weighs a little bit, not weighs, is a little bit more massive than Sal-- divided by the distance squared. Now, you might say, well, what's the distance between someone standing on the Earth and the Earth? Well, it's zero because they're touching the Earth. But it's important to realize that the distance between the two objects, especially when we're talking about the universal law of gravitation, is the distance between their center of masses. For all general purposes, my center of mass, maybe it's like three feet above the ground, because I'm not that tall. It's probably a little bit lower than that, actually. Anyway, my center of mass might be three feet above the ground, and where's Earth's center of mass? Well, it's at the center of Earth, so we have to know the radius of Earth, right? So the radius of Earth is-- I also looked it up on"
    },
    {
        "Q": "\nAt 0:52, Rishi draws a rectangular human cell. I thought that human cells are rounder and plant cells are the rectangular ones. Am I wrong?",
        "A": "Human cells have no specific shapes. It is just for diagramming.",
        "video_name": "MNKXq7c3eQU",
        "timestamps": [
            52
        ],
        "3min_transcript": "So let's talk about exactly how flu causes so much damage to ourselves and why it makes us feel so lousy whenever we get the flu. I'm going to start out by drawing the flu virus here. This is our influenza virus. And we have on influenza a couple of features we have to remember. So on the outside there's this little envelope, and what's on the inside of this envelope are eight bits of RNA. Eight pieces of RNA. And so this RNA is important to remember, because in the human cell, in our cells-- I'm going to draw one of our cells right here-- we have, instead of RNA, we have DNA. Remember. And so this is our nucleus, and on the inside of our nucleus is our DNA. So this is our DNA over here. So the virus has RNA, and we have DNA. And the outside of the human cell-- actually let me label this over here. This is human cell. something called sialic acid. They're these little strands over here that are coming off. I'm drawing them far larger than they are in real life. They're not nearly this big, but they're these little tiny little things called sialic acid. And this sialic acid becomes very important in understanding how the influenza virus gets into and out of our cells. So on the outside, remember, of the influenza virus, there were a couple of proteins. And I'm going to draw one of these proteins here, and I'm going to make it look like a hand. So this is a little hand, and this protein is called hemagglutinin. In fact, previously I had called it the H protein, and you can call it that if you want. But the full name is hemagglutinin. And what hemagglutinin does is that it actually holds onto sialic acid. In fact, that's an easy way to remember it, right? Because H and H go together. It holds sialic acid. allows it the first step towards getting into the cell. Now there's another protein on the outside here-- I'm going to make it look like a pair of scissors, because that will kind of remind us what this one does. And this is called neuraminidase. And I'm going to-- neuraminidase. And I'm going to pass on explaining what it does, just for the moment. I'll tell you in a little bit what it does. So then the first step to get into the cell is for hemagglutinin to hold on to sialic acid. And then there are a few other small molecular steps that happen, important ones. But I'm going to suffice to say it gets inside. And once the influenza virus gets inside, these RNA segments, they are let loose. So these segments are going to start making their way towards the nucleus. And so once they get into the nucleus, they're in that same kind of area that the DNA is,"
    },
    {
        "Q": "\nAt 1:28 why did Sal say that Force/Mass=Acceleration?",
        "A": "Because it does. F = m*a. Solve for a.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            88
        ],
        "3min_transcript": "What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is"
    },
    {
        "Q": "\nWhere did the formula at 2:32 come from?",
        "A": "its the formula to calculate gravity between 2 objects. m1 = mass object 1 (kg) m2 = mass object 2 (kg) r = distance between objects (m) G = 0.0000000000667 Your result is in m/s^2",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            152
        ],
        "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall"
    },
    {
        "Q": "\nAt 2:23, Sal said over the square of the distance between the two things but wrote : (r^2) but isn't the sign for distance d?",
        "A": "Either is acceptable. It s just a different letter that represents the same thing.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            143
        ],
        "3min_transcript": "What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is"
    },
    {
        "Q": "I'm just curious, in the equation at 3:17 is m2 referring to the mass of the projectile?\n",
        "A": "Indeed it is. Since m1 is the mass of the Earth, m2 must be the mass of the other body interacting with the Earth i.e the mass of the projectile.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            197
        ],
        "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall"
    },
    {
        "Q": "\n4:00 so the force of gravity is accelerating the object downwards (makes sense) then if you were to be nosediving the force of gravity is pushing against you, or in other words upwards, causing \"g\" forces. Why is this? Or if I'm wrong can someone explain g forces, like in the previous video when the pilot of the fighter jet was pushed back by gravity?",
        "A": "Gravity does not push against you, it pulls you down. When a pilot feels g forces, that s not really because of gravity, it s because of acceleration of his plane. It s called g force because it FEELS like he is getting heavier, but that s not because of gravity, it s because the seat of the plane is pushing on his backside.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            240
        ],
        "3min_transcript": "The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have"
    },
    {
        "Q": "Why did Sal use average velocity as opposed to just velocity in the equation: displacement = average velocity multiplied by change in time in 4:54?\n",
        "A": "Because the velocity is not constant through out the process. Acceleration is constant. If you plot a graph of Velocity vs. Time, you ll see more clearly why using avg. Velocity is pretty much the same as finding the area below the curve.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            294
        ],
        "3min_transcript": "If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have but not in terms of initial velocity and acceleration. We know that average velocity is the same thing as initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2) If we assume constant acceleration. We can only calculate Vavg this way assuming constant acceleration. Once again when were are dealing with objects not too far from the center of the earth we can make that assumption. Assuming that we have a constant acceleration Once again we don't have what our final velocity is. So, we need to think about this a little more. We can express our final velocity in terms of our initial velocity and time. Just dealing with this part, the average velocity. So we can rewrite this expression as the initial velocity plus something over 2. and what is final velocity?"
    },
    {
        "Q": "\nAt 3:00, Sal says, \"The little g, is really all of the business over here........\". Is it because m2's mass is really too small compared to the Earth's mass, we say: g = [ G X (Earth's mass) ]/r\u00c2\u00b2?\nI'm getting confused of to why we exclude the 'm2' in the equation?\n\nThanks,\nRamana",
        "A": "We don t exclude it. We have F = GMm/r^2. We don t want to write all of that every time. GM/r^2 doesn t change as long as you stay on the surface of the earth. So we give it a new name, g, and we re-write that equation as F = mg. The little m is still there.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            180
        ],
        "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall"
    },
    {
        "Q": "\n9:00 what is a delta t",
        "A": "delta means change in . Delta t means change in t.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            540
        ],
        "3min_transcript": "All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means \"change in\" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time."
    },
    {
        "Q": "\nAt 3:15, what distance should one no longer assume that the radius (the earth in this case) is constant (say, assuming 2 or 3 significant digits)?",
        "A": "g = GM/r^2 If you plug in a few numbers for r, you can decide for yourself when g varies enough from 9.8 so that you have to worry about changing r Note that even on the surface of earth, g varies in the second decimal place, due primarily to density variations but also due to changes in altitude",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            195
        ],
        "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall"
    },
    {
        "Q": "\nAt 9:15 shouldn't the systematic name be 1,3-bis(2-methylpropyl)cyclopentane ?",
        "A": "but in that case the longest chain will be of 2 carbon atoms. whereas it is of 3 carbon atoms.",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            555
        ],
        "3min_transcript": "with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core So if an ethyl is equal to two carbons, so this is two carbons right there. So let me draw a two carbon: one, two. That is two carbons right over there. I'm just drawing it at the three spot. I'll draw it also at the one spot, actually. So that is two carbons right there. That's the ethyl part. And then on 1,1, so if we number them, we number where it's connected, so it's one, two. This is saying 1,1-dimethyl. So on this ethyl chain, you have two methyls. Remember, methyl is equal to one, so this is one carbon. You have one carbon. That's what methyl is, but you have two of them. You have dimethyl. You have it twice at the one spot. So you have one methyl here and then you have another methyl there. You have 1-methyl on the one spot and then you have another 1-methyl on the one spot. And then you are connected at positions one and positions three, so you're connected there and you are connected And you're done, That's it. That is our structure. Now, if you did this with common naming, instead of this group being a 1,1-dimethylethyl, you might see that we're connected to a group that has one, two, three, four carbons in it. The carbon that we're connected to branches off to three other carbons. It is a tert-butyl. So you can also call this a 1,3-- let me just write it down. So another name for this would be 1,3-tert-- or sometimes people just write a t there-- t-butylcyclo-- no, actually I should say di-t-butyl, because we have two of them. 1,3-di-t-butylcyclopentane."
    },
    {
        "Q": "Wouldn't the structure at minute 5:20 be named 4,4,7,10-tetramethyldodecane? With this numbering we get a total of 25 vs. 27 with the 3,6,9,9-tetramethyldodecane.\n",
        "A": "There is no rule that says we sum the numbers together, the rule is we are looking for the lowest number at the first point of difference. How you determine this is by making a list of the numbers and compare them one at a time until we find a point of difference: yours: 4,4,7,10 the video s: 3,6,9,9 As 3 is lower than 4, the numbering in the video is correct.",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            320
        ],
        "3min_transcript": "It's three carbons, so it's going to be one, two, three, and the connection point to the main ring in this case is going to be in the middle carbon, so it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's, so it's going to be linked right over here. That's also going to happen at the ninth carbon, so at the ninth carbon we're going to have another isopropyl. We're going to have another isopropyl at the ninth carbon. All right, we've taken care of the 2,9-isopropyl. Then we have the 6-ethyl, which is just a two carbon. Remember, meth- is one, eth- is two, prop- is three. Let me write this down. So this is going to be prop- is equal to three. Isoprop- is equal to that type of shape right over there. So at six we have an ethyl group, so one, two, carbons, and it's connected at the six carbon on the main ring. And then finally we have a cyclopentyl. So if we look at-- let me find a color I haven't used yet-- cyclopentyl. so pent- is five, but it's five in a cycle, so this is a five-carbon ring that's branching off of the main ring. It's at the first spot. Let me draw a five-carbon rings, so pent- is equal to five, so it would look like this, one: two, three, four, five. It looks just like a pentagon. That's a cyclopentyl group and it's attached to the one carbon on my cyclohexadecane, so it is attached just like that. ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon,"
    },
    {
        "Q": "At 5:56 Sal says that you won't see meniscus in plastic because it doesn't have the same polarity as the glass. Does this mean there is no capillary action in a plastic straw?\n",
        "A": "There is capillary action in plastic straw. Ever seen a cold drink with a straw? When you insert the straw into the drink..the liquid rises over a height. But capillary action in glass is more than that observed in plastic straws. All plastics have different abilities for adhesion.",
        "video_name": "eQXGpturk3A",
        "timestamps": [
            356
        ],
        "3min_transcript": "Because its partially positive end, its partially positive end at the hydrogens. Let me do it in that green color. The partially positive end at the hydrogens would be attracted to the partially negative ends of the oxygens in the glass. And so it'll stick to it. This is actually a stronger partial charge than what you would actually see in the water because there's a bigger electronegativity difference between the silicon and the oxygen in the glass than the oxygen and the hydrogen in the water. So these things just keep bumping around. Maybe there's another water molecule that just get knocked in the right way. All of a sudden for, you know, a very brief moment it gets knocked up here. And then it's going to stick to the glass. And this phenomenon of something sticking to its container, we would call that adhesion. So what you see going on here, that is called adhesion, adhesion. And adhesion is the reason why you also see the water a little bit higher there. we call that cohesion. And that's what the hydrogen bonds are doing inside the water. So this right over here, that over there, that is co-, that is cohesion. So that's why we have things, why we observe a meniscus like this. But there's even more fascinating properties of adhesion. If I were to take, if I were to take a container of water. If I were to take a container of water. And just to be clear what's going on here with the mercury, the mercury is more attracted to itself than it is to the glass container, so it bulges right over there. But let's go back to water. So let's say that this is a big tub of water. I fill it. So, I fill the water right over here. And let's say I take a glass tube, and the material matters. It has to be a polar material. That's why you'll see the meniscus in glass, but you might not see it or you won't see it if you were dealing with a plastic tube because the plastic does not have that polarity. But let's say you were to take a glass tube, So much thinner than even a beaker. So you take a thin glass tube and you stick it in the water, you will observe something very cool. And I encourage you to do this if you can get your hands on a very thin glass tube. You will notice that the water is actually going to defy gravity and start climbing up this thin glass tube. And so that's interesting. Why is that happening? Well this phenomenon which we call capillary action. Capillary, capillary action. The word capillary, it'll refer to anything from you know, a very, very narrow tube and we also have capillaries in our circulation system. Capillaries are our thinnest blood vessels, those are very, very, very, very thin. And there's actually capillary action inside of our capillaries. But what we're seeing here, this is called capillary, capillary action. And it's really just this adhesion occurring more intensely because more of the water molecules are able to come in touch with the polar glass lattice. And so you can imagine we have glass here. If you also had glass over here."
    },
    {
        "Q": "\nAt 5:56 Sal says that you won't see meniscus in plastic because it doesn't have the same polarity as the glass. Is this true for every kind of plastic?",
        "A": "Not all plastics are exactly the same but they all have the same type of carbon-polymer-structure. As far as meniscus is concerned, none of them will cause a meniscus.",
        "video_name": "eQXGpturk3A",
        "timestamps": [
            356
        ],
        "3min_transcript": "Because its partially positive end, its partially positive end at the hydrogens. Let me do it in that green color. The partially positive end at the hydrogens would be attracted to the partially negative ends of the oxygens in the glass. And so it'll stick to it. This is actually a stronger partial charge than what you would actually see in the water because there's a bigger electronegativity difference between the silicon and the oxygen in the glass than the oxygen and the hydrogen in the water. So these things just keep bumping around. Maybe there's another water molecule that just get knocked in the right way. All of a sudden for, you know, a very brief moment it gets knocked up here. And then it's going to stick to the glass. And this phenomenon of something sticking to its container, we would call that adhesion. So what you see going on here, that is called adhesion, adhesion. And adhesion is the reason why you also see the water a little bit higher there. we call that cohesion. And that's what the hydrogen bonds are doing inside the water. So this right over here, that over there, that is co-, that is cohesion. So that's why we have things, why we observe a meniscus like this. But there's even more fascinating properties of adhesion. If I were to take, if I were to take a container of water. If I were to take a container of water. And just to be clear what's going on here with the mercury, the mercury is more attracted to itself than it is to the glass container, so it bulges right over there. But let's go back to water. So let's say that this is a big tub of water. I fill it. So, I fill the water right over here. And let's say I take a glass tube, and the material matters. It has to be a polar material. That's why you'll see the meniscus in glass, but you might not see it or you won't see it if you were dealing with a plastic tube because the plastic does not have that polarity. But let's say you were to take a glass tube, So much thinner than even a beaker. So you take a thin glass tube and you stick it in the water, you will observe something very cool. And I encourage you to do this if you can get your hands on a very thin glass tube. You will notice that the water is actually going to defy gravity and start climbing up this thin glass tube. And so that's interesting. Why is that happening? Well this phenomenon which we call capillary action. Capillary, capillary action. The word capillary, it'll refer to anything from you know, a very, very narrow tube and we also have capillaries in our circulation system. Capillaries are our thinnest blood vessels, those are very, very, very, very thin. And there's actually capillary action inside of our capillaries. But what we're seeing here, this is called capillary, capillary action. And it's really just this adhesion occurring more intensely because more of the water molecules are able to come in touch with the polar glass lattice. And so you can imagine we have glass here. If you also had glass over here."
    },
    {
        "Q": "Am I missing something? At 2:27 Sal says that each Silicon is paired with two Oxygen but the video shows four Oxygens for every Silicon (though two of the Oxygen are shared in a covalent bond).\n",
        "A": "No, well Sal did not draw the complete picture here, Silicon dioxide is the complete molecule here and what you see above is a silicon dioxide lattice which is just many silicon dioxide forming bonds with each other. You saw silicon bonded to 4 oxygen bonds but actually that silicon is bonded to 2 of them and the other 2 oxygen is bonded to some other silicon atom.",
        "video_name": "eQXGpturk3A",
        "timestamps": [
            147
        ],
        "3min_transcript": "- If you were to take a glass beaker, so let me draw it right over here. If you were to take a glass beaker and you were to fill it up with water, you might expect that the surface of the water would be flat. But it's actually not the case and I encourage you to try it. You might have even observed this before. The surface of the water will not be flat. The surface of the water will actually be higher near the glass than it is when it's away from the glass. It forms a shape that looks something like that. And so the first thing we might ask is what'll we call this thing. And this right over here is called a meniscus. Meniscus. And in particular this meniscus, because the fluid is higher near the container than it is when you're away from the container, we would call this a concave, concave meniscus. And you might say, \"Well if this is a concave meniscus, \"are there any situations where might have \"a convex meniscus?\" Well sure, you can have a convex meniscus. If you were take that same glass beaker, instead of filling it with water If you filled it with mercury, you would get a meniscus that looks like this where there's a bulge near the center when you're further away from the container than when you're at the container. And so let me just label this. This is a convex, convex meniscus. But it's one thing to just observe this and to name them. To say, \"Hey this is a meniscus.\" So this is a concave meniscus. But a more interesting question is why does it actually happen. And so you might imagine this concave meniscus is because the fluid is more attracted to the container than it is to itself. And you might be saying, \"Wait, wait. \"Hold on, hold on a second here. \"We've been talking about how water \"has this polarity, it has partial negative end. \"Each water molecule has a partially negative \"and has partially positive ends at the hydrogens.\" So let me write this down. Partial positive charges at the hydrogens. And that causes this hydrogen bonding to form all of these special properties. \"You're telling me that it's more attracted to the glass than it is to itself?\" And I would say, \"Yes, I am telling you that.\" And you could imagine why it is going to be more attracted to the glass than itself, because glass actually has, the molecules in glass actually are quite polar. Glass, typically made up of silicon oxide lattice. For every one silicon atom, you have two oxygen atoms. You see that right over here. For every one silicon, you have two oxygen atoms. And it turns out that the electronegativity difference between oxygen and silicon is even higher than the electronegativity difference between oxygen and hydrogen. Silicon is even less electronegative than hydrogen. So the oxygens are really able to hog silicon's electrons. Especially the ones that are involved in the bonding. So you have partial charges, partial positive charges form at the silicon and then you still have"
    },
    {
        "Q": "At 9:30 and onward, it is stated that the accelerations for each of the block would have a magnitude of 1.84 m/s^2. But doesn't that contradict from the previous video, which stated that the accelerations would have a magnitude of 3.68 m/s^2? Could someone please explain why this is? Thanks! :)\n",
        "A": "No contradiction. In the previous examples, friction was ignored. In the last example, where the 1.84 m/s^2 was calculated, friction was taken into consideration. And this makes sense - if we have friction, then there s an opposing force to a moving object.",
        "video_name": "UrfLAlk2b_8",
        "timestamps": [
            570
        ],
        "3min_transcript": "that's just causing this mass to sit on the table and for two, it's cancelled by that normal force. So those cancel anyway, even though they're external forces. This is it, this is the only one that drives the system. So I put that in here and I divide by my total mass 'cause that tells me how much my system resists through inertia, changes in velocity, and this is what I get. I get the same thing I got before, I get back my three point six eight meters per second squared, and I get in one line. I mean, this trick is amazing and it works, and it works in every example where two masses or more masses are forced to move with the same acceleration. So this is great. This'll save you a ton of time. This is supposed to be a three here. And to show you how useful it is, let say there was friction, let's say there was a coefficient of friction of zero point three. Well, now I'd have a frictional force so there'd be an external frictional force here. It'd be applied this five kilogram mass. So if I get rid of this--- So it's not gonna be three point six eight anymore. I'm gonna have a force of friction that I have to subtract. So minus mu K so the force of friction--- I'll just put force of friction. And so to solve for the force of friction, the force of friction is gonna be equal to--- Well, I know three times nine point eight is--- Let me just write this in here, 29.4 Newtons minus the force of friction it's given by. So there's a formula for force of friction. The force of friction is always mu K FN. So the force of friction on this five kilogram mass is gonna be mu K which is point three. So it's gonna be zero point three times the normal force, not the normal force on our entire system. I don't include this three kilogram mass. It's only the normal force on this five kilogram mass that's contributing to this force of friction here. So even though we're treating this system as a whole, we still have to find individual forces So it won't be the entire mass that goes here. The normal force on the five kilogram mass is just gonna be five kilograms times nine point eight meters per second squared. I divide by my total mass down here because the entire mass is resisting motion through inertia. And if I solve this from my acceleration of the system, I get one point eight four meters per seconds squared. So this is less, less than our three point six eight and that makes sense. Now, there's a resistive force, a resistive external force, tryna prevent the system from moving. But you have to be careful. What I'm really finding here, I'm really finding the magnitude of the acceleration. This is just giving me the magnitude. If I'm playing this game where positive forces are ones that make it go and negative forces are ones that resist motion, external forces that is, I'm just getting the magnitude of the acceleration. Individual boxes will have that magnitude"
    },
    {
        "Q": "At around 6:30, when he starts solving the problem, is he assuming the penny immediately just falls down or something? Because the penny will move in some parabolic motion and have its own change in distance, which would then be added to the height of the hill.\n",
        "A": "First of all, Sal does not assume that the penny falls down immediately, because he said that it has an initial velocity of positive 30 m/s, which means that it has an upward direction. Second, Sal assumed that the object was thrown straight upward and not at an angle, and therefore, it would not have a parabolic path.",
        "video_name": "emdHj6WodLw",
        "timestamps": [
            390
        ],
        "3min_transcript": "here, and that might simplify things. If we multiply both sides by 2a, we get-- and I'm just going to switch this to distance, if we assume that we always start at distances equal to 0. di, or initial distance, is always at point 0. We could right 2ad-- I'm just multiplying both sides by 2a-- is equal to vf squared minus vi squared, or you could write it as vf squared is equal to vi squared plus 2ad. I don't know what your physics teacher might show you or written in your physics book, but of these variations will show up in your physics book. The reason why I wanted to show you that previous problem first is that I wanted to show you that you could actually figure out these problems without having to always memorize formulas and resort to the formula. With that said, it's probably not bad idea to memorize some form of this formula, although you should understand how it Now that you have memorized it, or I showed you that maybe you don't have to memorize it, let's use this. Let's say I have the same cliff, and it has now turned purple. It was 500 meters high-- it's a 500 meter high cliff. This time, with the penny, instead of just dropping it straight down, I'm going to throw it straight up at positive 30 meters per second. The positive matters, because remember, we said negative is down, positive is up-- that's just the convention we use. Let's use this formula, or any version of this formula, to figure out what our final velocity was when we hit the bottom of the ground. This is probably the easiest formula to use, because it We can say the final velocity vf squared is equal to the initial velocity squared-- so what's our initial velocity? It's plus 30 meters per second, so it's 30 meters per second squared plus 2ad. So, 2a is the acceleration of gravity, which is minus 10, because it's going down, so it's 2a times minus 10-- I'm going to give up the units for a second, just so I don't run out of space-- 2 times minus 10, and what's the height? What's the change in distance? Actually, I should be correct about using change in distance, because it matters for this problem. In this case, the final distance is equal to minus 500, and the initial distance is equal to 0. The change in distance is minus 500."
    },
    {
        "Q": "\nAt 0:28 Sal said that the sun orbits around the Milky Way Galaxy. So does the sun also move and orbit just like the planets do?",
        "A": "Yep! Our whole solar system orbits around the center of the Milky Way Galaxy, just like how our planets orbit the Sun. Though at a much higher velocity of 828,000 km/hr",
        "video_name": "FEF6PxWOvsk",
        "timestamps": [
            28
        ],
        "3min_transcript": "What I want to do in this video is give a very high-level overview of the four fundamental forces of the universe. And I'm going to start with gravity. And it might surprise some of you that gravity is actually the weakest of the four fundamental forces. And that's surprising because you say, wow, that's what keeps us glued-- not glued-- but it keeps us from jumping off the planet. It's what keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, the Sun in orbit around the center of the Milky Way galaxy. So if it's a little bit surprising that it's actually the weakest of the forces. And that starts to make sense when you actually think about things on maybe more of a human scale, or a molecular scale, or even atomic scale. Even on a human scale, your computer monitor and you, have some type of gravitational attraction. But you don't notice it. Or your cell phone and your wallet, there's gravitational attraction. But you don't see them being drawn to each other the way you might see two magnets drawn to each other And if you go to even a smaller scale, you'll see the it matters even less. We never even talk about gravity in chemistry, although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravity is our weakest. So if we move up a little bit from that, we get-- and this is maybe the hardest force for us to visualize. Or it's, at least, the least intuitive force for me-- is actually the weak force, sometimes called the weak interaction. And it's what's responsible for radioactive decay, in particular beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium-137-- 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137. Now, the weak interaction is what's responsible for one of the neutrons-- essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that. And the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton. But we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not the kind of this traditional things pulling"
    },
    {
        "Q": "At 5:05, why is scandium's electron configuration the same as argon's?\n",
        "A": "It isn t, but each electron configuration builds on from the last. We use the noble gas from the previous row in square brackets to represent all of its electron configuration which saves us time and space. Argon has the electron configuration: 1s2 2s2 2p6 3s2 3p6 Scandium has the electron configuration: 1s2 2s2 2p6 3s2 3p6 3d1 4s2 When we use the shorthand scandium s electron configuration can be written as: [Ar] 3d1 4s2 [Ar] just means all of argon s electron configuration from earlier",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            305
        ],
        "3min_transcript": "Well, sodium is going to have the same electron configuration as neon. Then it's going to go 3s1. Once again, it has one valence electron, one electron in its outermost shell. All of these elements in orange right over here, they have one valence electron and they're trying to get to the octet rule, this kind of stable nirvana for atoms. You could imagine is that they're very reactive and when they react they tend to lose this electron in their outermost shell. That is the case. These alkali metals are very, very reactive. Actually they have very similar properties. They're shiny and soft. Because they're so reactive it's hard to find them where they haven't reacted with other things. Let's keep looking at the other groups. If we move one over to the right this group two right over here, these are called the alkaline earth metals. alkaline earth metals. Once again, they have very similar ... They have very similar properties and that's because they have two valence electrons, two electrons in their outermost shell. Also for them, not as quite as reactive as the alkaline metals. Let me write this out, alkaline earth metals. But for them it's easier to lose two electrons than to try to gain six to get to eight. And so these tend to also be reasonably reactive and they react by losing those two outer electrons. Now something interesting happens as you go to the D block. We studied this when we looked at electron configurations, but if you look at the electron configuration for say scandium right over here, the electron, let me do it in magenta, the electron configuration for scandium, so scandium, scandium's electron configuration It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1."
    },
    {
        "Q": "\n3:05 what is a valence electron?",
        "A": "A valence electron is electrons in the outermost shell of the atom. They determine properties of the atom and bonding behavior. The Octet Rule states that the outermost energy shell has 8 electrons.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            185
        ],
        "3min_transcript": "but for the most part the elements in the column have very, very, very similar properties. That's because the elements in a column, or the elements in a group tend to have the same number of electrons in their outermost shell. They tend to have the same number of valence electrons. And valence electrons are electrons in the outermost shell they tend to coincide, although there's a slightly different variation. The valence electrons, these are the electrons that are going to react, which tend to be the outermost shell electrons, but there are exceptions to that. There's actually a lot of interesting exceptions that happen in the transition metals in the D block. But we're not gonna go into those details. Let's just think a little about some of the groups that you will hear about and why they react in very similar ways. If we go with group one, group one ... And hydrogen is a little bit of a strange character because hydrogen isn't trying to get to eight valence electrons. Hydrogen in that first shell just wants to get to two Hydrogen is kind of ... It doesn't share as much in common with everything else in group one as you might expect for, say, all of the things in group two. Group one, if you put hydrogen aside, these are referred to as the alkali metals. And hydrogen is not considered an alkali metal. These right over here are the alkali. Alkali metals. Now why do all of these have very similar reactions? Why do they have very similar properties? Well, to think about that you just have to think about their electron configurations. For example, the electron configuration for lithium is going to be the same as the electron configuration of helium, of helium. Then you're going to go to your second shell, 2s1. It has one valence electron. It has one electron in its outermost shell. Well, sodium is going to have the same electron configuration as neon. Then it's going to go 3s1. Once again, it has one valence electron, one electron in its outermost shell. All of these elements in orange right over here, they have one valence electron and they're trying to get to the octet rule, this kind of stable nirvana for atoms. You could imagine is that they're very reactive and when they react they tend to lose this electron in their outermost shell. That is the case. These alkali metals are very, very reactive. Actually they have very similar properties. They're shiny and soft. Because they're so reactive it's hard to find them where they haven't reacted with other things. Let's keep looking at the other groups. If we move one over to the right this group two right over here, these are called the alkaline earth metals."
    },
    {
        "Q": "\nat 6:42, carbon is described to have same configuration to helium, but then he says that carbon has 4 valence electrons or valency 4. how can it be similar to helium? is there a difference between configuration and valence electrons? I just couldn't understand..",
        "A": "He didn t mean carbon has EXACTLY the same electron configuration as helium, but what he is showing is a shorthand way of writing out electron configurations. You take the noble gas from the previous row and put its symbol inside square brackets, this represents the electron configuration for that noble gas, eg [He] means the same thing as 1s^2, [Ne] means the same thing as 1s^2 2s^2 2p^6. This is for our convenience later on when full electron configurations get VERY long.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            402
        ],
        "3min_transcript": "It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1. We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group. bonding behavior to silicone, to the other things in its group. We could keep going on, for example, oxygen and sulfur. These would both want to take two electrons from someone else because they have six valence electrons and they want to get to eight. They have similar bonding behavior. You go to this yellow group right over here. These are the halogens. There's special name for them. These are the halogens. These are highly reactive because they have seven valence electrons. They would love nothing more than to get one more valence electron. They love to react. In fact, they especially love to react with the alkali metals over here. Then finally you get to kind of your atomic nirvana in the noble gases here. The noble gases, that's the other name for the group, 18 elements, noble gases."
    },
    {
        "Q": "\nAt 08:30, are there lots of different types of bacteria or just one?",
        "A": "There are billions of species, some are harmful to us, some are beneficial.",
        "video_name": "TDoGrbpJJ14",
        "timestamps": [
            510
        ],
        "3min_transcript": "In bacteria,which are what people originally just classify it on whether or not you have a nucleus, in bacteria,there is no membrane surrounding the DNA. So what they have is just a big bundle of DNA. They just have this big bundle of DNA. It's sometimes in a loop all in one circle called a nucleoid. Now,whenever we look at something,and we say,oh, we have this thing;it doesn't;there's this assumption that somehow we're superior or we're more advanced beings. But the reality is that bacteria have infiltrated far more ecosystems in every part of the planet than Eukarya have, and there's far more diversity in bacteria than there is in Eukarya. these are the more successful organisms. If a comet were to hit the Earth--God forbid-- the organisms more likely to survive are going to be the bacteria than the Eukarya,than the ones with the larger--not always larger, but the organisms that do have this nucleus and have membrane-bound organelles like mitochondria and all that. We'll talk more about it in the future. Bacteria,for the most part,are just big bags of cytoplasm. They have their DNA there. They do have ribosomes because they have to code for proteins just like the rest of us do. Some of those proteins,they'll make some from-- bacteria,they'll make these flagella, which are tails that allow them to move around. They also have these things called pili. Pili is plural for pilus or pee-lus,so these pili. And we'll see in a second that the pili are kind of introducing genetic variation into their populations. Actually,I'll take a little side note here. I'm pointing out bacteria as not having a cell wall. There's actually another class that used to be categorized as type of a bacteria,and they're called Archaea. I should give them a little bit of justice. They're always kind of the stepchild. They used to be called Archaea bacteria, but now people realize,they've actually looked at the DNA, because when they originally looked at these,they said,OK, these guys also have no nucleus and a bunch of DNA running around. These must be a form of bacteria. But now that we've actually been able to look into the DNA of the things,we've seen that they're actually quite different. But all of these,both bacteria and Archaea,are considered prokaryotes. And this just means no nucleus."
    },
    {
        "Q": "At 1:45 sal says that if we put bacteria in milk it becomes yogurt. But then if you eat the yogurt wont the bacteria get in your body and make you sick.How is that a good thing.\n",
        "A": "There are predominately two bacteria used in making yogurt. One is non-probotic so it does not survive the stomach. The other, which is probotic, in non pathogenic, but aids in fermentation (common in the intestines). It is important to remember that not all bacteria make you sick, and to make it more complicated different sub-specie of a bacteria CAN make you sick, while a different subspecie of the same bacteria is ESSENTIAL to stay alive, such as E. Coli.",
        "video_name": "TDoGrbpJJ14",
        "timestamps": [
            105
        ],
        "3min_transcript": "I think we've all heard of the word bacteria. And we normally associate it with negative things. You say bacteria,those are germs. So we normally associate those with germs,and they indeed are germs, and they cause a whole set of negative things. Or at least from the standard point of view, people believe that they cause a whole bunch of negative things. So let's just list them all just to make sure we know about them, we're all on the same page.So the bad things they do, they cause a lot of diseases:tuberculosis,Lyme disease. I mean,I could go on and on. You know,pretty much any time--well,I'll be careful here. Whenever people talk about an infection, it's often caused by a bacteria.It can also be caused by a virus. An infection is,in general,anything entering you and taking advantage of your body to kind of replicate itself, and in the process,making you sick. And this whole perception of bacteria being a bad thing is probably a good reason why almost any soap you see now will say antibacterial on it. Because the makers of the soap know that in conventional thinking, bacteria are viewed as a negative thing. And you're like,OK,Sal,I know where you're going with this. Bacteria isn't all bad.There are some good traits of bacteria. For example,I could stick some yogurt in some-- or I could stick some bacteria in some milk and it'll help produce some yogurt,sometimes spelled yoghurt. And that's obviously a good thing.It's a delicious thing to eat. And these are all true,but you're like,look,you know, on balance,I still think bacteria is a bad thing. I'm not going to take sides on that debate, as I tend to avoid taking sides on debates in these science videos. Maybe I'll do a whole playlist where I do nothing but take sides on debates,but here I won't take any sides on that. But I'll just point out that you are to a large degree made up of bacteria.It's not just your gut. It's not just the gut or the yogurt you might eat or the plaque on your teeth,which is caused by bacteria. It's this kind of film that's created by bacteria Bacteria actually represents a majority of the cells on your body. And it's not just the pimples on your face. Bacteria actually represents a majority of the cells on your body. So for every--and this is kind of an astounding fact."
    },
    {
        "Q": "\nAt the start off the video, and at 2:06, why does Sal use the word \"rate\"? Is't distance divided by time just \"speed\" ?",
        "A": "Rate is used here as shorthand for rate of change. An object s speed is the rate of change of its position over time, so he could have used speed here. Using rate helps make it clear that the same mathematical tools can be used for non-speed rates of change (e.g. water filling a tank, population growth, and so on).",
        "video_name": "6FTiHeius1c",
        "timestamps": [
            126
        ],
        "3min_transcript": "- [Voiceover] Let's say that something is traveling at a constant rate of five meters per second. That's its velocity in one dimension. If it was negative, we'd be moving to the left. If it's positive, it's moving to the right. Let's say that we care about what is our change in distance over, the delta symbol represents change, over a change in time of four seconds. Over four seconds. I could say from t equal zero to t is equal to four. That's our change in time. That's our four-second interval that we care about. Well, one way to think about it is what a rate by definition is nothing but a change in some quantity. In this case, it's distance over a change in some other quantity. In this case, we're thinking about time. Or another way to think about it, if we multiply both sides times change in time, you get your change in distance is equal to your rate times change in time. from pre-algebra, distance is equal to rate times time. Time, that just comes from the definition of what a rate is. It's a change in one quantity with respect to another quantity. If you just apply this, you can say, \"Okay, my rate is a constant five meters \"per second and my delta t is four seconds.\" So times four seconds. Well, that's just going to give you 20. That's just going to get you 20. Let me do that in same color that I had for the change in distance. That's going to be 20. Then the seconds, cancel the seconds, 20 meters. So my total change in distance over those four seconds is going to be 20 meters. Nothing new here. Nothing too fancy. But I want to do now is connect this to the area under the rate function over this time period. So let's graph that. That's my rate axis. This is my time axis. This is going to be in seconds. This is going to be in meters per second. Let's see, one, two, three, four, five. Let's see if it's about enough, and then I go one, two, three, four, five. Our rate, at least in this example, is a constant, is a constant five meters per second. It's a constant five meters per second. That is my r of t in this example. What did we just do here? We just multiplied our change in time times our constant rate. We just multiplied our change in time. So from time equal zero seconds to four seconds. It's this length here, if we think on that axis. Now we multiply it times our constant rate. We multiply it times this right over here. If I multiply this base times this height, what am I going to get? I'm going to get this area under the rate function."
    },
    {
        "Q": "At 1:55, what is a mole of a molecule?\n",
        "A": "The first one is easy, and I hope this explains it for you. A mole consists of an Avogadro s Number of atoms, molecules, or whatever. That number is 6.02 x 10\u00c2\u00b2\u00c2\u00b3 So if you have 6.02 x 10\u00c2\u00b2\u00c2\u00b3 molecules of water, then you have EXACTLY one mole of it.",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            115
        ],
        "3min_transcript": "Let's say I have some weak acid. I'll call it HA. A is a place holder for really a whole set of elements that I could put there. It could be fluorine, it could be an ammonia molecule. If you add H it becomes ammonium. So this isn't any particular element I'm talking about. This is just kind of a general way of writing an acid. And let's say it's in equilibrium with, of course, and you've seen this multiple times, a proton. And all of this is in an aqueous solution. Between this proton jumping off of this and its conjugate base. And we also could have written a base equilibrium, where we say the conjugate base could disassociate, or it could essentially grab a hydrogen from the water and create OH. And we've done that multiple times. But that's not the point of this video. So let's just think a little bit about if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principal, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress. So let's say that the stress that I apply to the system --Let me do a different color. I'm going to add some strong base. That's too dark. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen."
    },
    {
        "Q": "\nat 4:29 Sal says when you increase the OH, you decrease the pOH and it increases the pH? I'm a bit confused by this, can someone please explain? Thanks",
        "A": "The same way that you you increase the concentration of H+ the pH goes down, the pOH would also go down. Don t forget that pOH=-log10([OH-]). Looking at [OH-]=.01 and .001 respectively, the pOH is 2 and 3.",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            269
        ],
        "3min_transcript": "Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen. So this hydrogen goes down initially, and then it starts getting to equilibrium very fast. But this is going to go down. This is going to go up. And then this is going to go down less. Because sure, when you put the sodium hydroxide there, it just ate up all of the hydrogens. But then you have this -- you can kind of view as the spare hydrogen capacity here to produce hydrogens. And when these disappear, this weak base will disassociate more. The equilibrium we'll move more in this direction. So immediately, this will eat all of that. But then when the equilibrium moves in that direction, a lot of the hydrogen will be replaced. So if you think about what's happening, if I just threw this sodium hydroxide in water. So if I just did NaOH in an aqueous solution so that's just throwing it in water -- that disassociates completely into the sodium cation and hydroxide anion. increase the quantity of OHs by essentially the number of moles of sodium hydroxide you're adding, and you'd immediately increase the pH, right? Remember. When you increase the amount of OH, you would decrease the pOH, right? And that's just because it's the negative log. So if you increase OH, you're decreasing pOH, and you're increasing pH. And just think OH-- you're making it more basic. And a high pH is also very basic. If you have a mole of this, you end up with a pH of 14. And if you had a strong acid, not a strong base, you would end up with a pH of 0. Hopefully you're getting a little bit familiar with that concept right now, but if it confuses you, just play around with the logs a little bit"
    },
    {
        "Q": "\nat 0:54 Sal says that this oxygen could grab a hydrogen from the water essentially creating OH. What is OH?",
        "A": "OH is a hydroxyl radical. But when A\u00e2\u0081\u00bb grabs a proton from the water it forms OH\u00e2\u0081\u00bb, which is hydroxide ion.",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            54
        ],
        "3min_transcript": "Let's say I have some weak acid. I'll call it HA. A is a place holder for really a whole set of elements that I could put there. It could be fluorine, it could be an ammonia molecule. If you add H it becomes ammonium. So this isn't any particular element I'm talking about. This is just kind of a general way of writing an acid. And let's say it's in equilibrium with, of course, and you've seen this multiple times, a proton. And all of this is in an aqueous solution. Between this proton jumping off of this and its conjugate base. And we also could have written a base equilibrium, where we say the conjugate base could disassociate, or it could essentially grab a hydrogen from the water and create OH. And we've done that multiple times. But that's not the point of this video. So let's just think a little bit about if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principal, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress. So let's say that the stress that I apply to the system --Let me do a different color. I'm going to add some strong base. That's too dark. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen."
    },
    {
        "Q": "@13:15 how does -log ([HA]/[A-]) turn into log ([HA/A-])^-1 ?\n",
        "A": "That is an application of a basic identity of logarithms. log (a/b) = log(a) - log (b) = -log (b) + log(a) = - [log(b) - log (a)] = - log (b/a) Thus, log (a/b) = - log(b/a)",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            795
        ],
        "3min_transcript": "we can multiply both sides by the reciprocal of this right here. And you get hydrogen concentration. Ka times --I'm multiplying both sides times a reciprocal of that. So times the concentration of our weak acid divided by the concentration of our weak base is equal to our concentration of our hydrogen. Fair enough. Now. Let's take the negative log of both sides. So the negative log of all of that stuff, of your acidic equilibrium constant, times HA, our weak acid divided by our weak base, our hydrogen concentration. Which is just our pH, right? Negative log of hydrogen concentration is --that's the definition of pH. I'll write the p and the H in different colors. You know a p just means negative log. Minus log. That's all. Base 10. Let's see if we can simplify this any more. So our logarithmic properties. We know that when you take the log of something and you multiply it, that's the same thing as taking the log of this plus the log of that. So this can to be simplified to minus log of our Ka minus the log of our weak acid concentration divided by its conjugate base concentration. Is equal to the pH. which is just the negative log of its equilibrium constant. So this is just the pKa. And the minus log of HA over A. What we can do is we could make this a plus, and just take this to the minus 1 power. Right? That's just another logarithm property, and you can review the logarithm videos if that confused you. And this to the minus 1 power just means invert this. So we could say, plus the logarithm of our conjugate base concentration divided by the weak acid concentration is equal to the pH. And this right here, this is called the Hendersen-Hasselbalch Equation. And I really encourage you not to memorize it. Because if you do attempt to memorize it, within a few hours, you're going to forget whether this was a plus over here. You're going to forget this, and you're going to forget whether you"
    },
    {
        "Q": "\nAt 3:53 Sal says that the electric field has a 3 n/c electric field. However, doesn't the electric field depend on the distance between the two particles? Isn't the equation K*Q/D^2?",
        "A": "It does except in this example Sal gave where it is a constant electric field. If it were not constant, you would use the equation you have defined provided the charge on the plate is defined (Q). Then you would solve in same manner by multiplying E times the test charge to get Force then multiply force times distance",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            233
        ],
        "3min_transcript": "in magnitude it's pushing out, because we assume when we draw field lines that we're using a test charge with a positive charge so it's pushing outward. Let's say I have a 1-coulomb charge. Actually, let me make it 2 coulombs just to hit a point home. Say I have a 2-coulomb charge right here, and it's positive. A positive 2-coulomb charge, and it starts off at 3 meters away, and I want to bring it in 2 meters. I want to bring it in 2 meters, so it's 1 meter away. So what is the electric-- or electrical-- potential energy difference between the particle at this point and at this point? amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy--"
    },
    {
        "Q": "\nAt 8:18 Sal said voltage is regardless of how small or big the charge is\nbut to find the potential we divide the work done by the charge\nso isn't the potential dependent upon the size of charge?",
        "A": "The potential is not dependent on the size of charge, but the potential energy is. The potential is a property of the field. It s similar to gravity. g does not depend on the mass of an object on the surface of the earth. But the gravitational potential energy does depend on the mass (mgh)",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            498
        ],
        "3min_transcript": "does it take to move any charge per unit charge from here to here? Well, in our example we just did, the total work to move it from here to here was 12 joules. But how much work did it take to move it from there to there per charge? Well, work per charge is equal to 12 joules for what? What was the charge that we moved? Well, it was 2 coulombs. It equals 6 joules per coulomb. That is the electric potential difference between this point and this point. So what is the distinction? Electric potential energy was associated with a particle. How much more energy did the particle have here than here? When we say electric potential, because we essentially divide by the size of the particle, it It actually just depends on our position. So electric potential, we're just saying how much more potential, irrespective of the charge we're using, does this position have relative to this position? And this electric potential, that's just another way of saying voltage, and the unit for voltage is volts. So 6 joules per coulomb, that's the same thing as 6 volts. And so if we think of the analogy to gravitation, we said gravitational potential energy was mgh, right? This was distance, right? Electric potential is essentially the amount of gravitational-- if we extend the analogy, the amount of gravitational potential energy per mass, right? So if we wanted a quick way of knowing what the gravitational the mass, we divide by the mass, and it would be the acceleration of gravity times height. Ignore that if it confused you. So what is useful about voltage? It tells us regardless of how small or big or actually positive or negative a charge is, what the difference in potential energy would be if we're at two different points. So electric potential, we're comparing points in space. Electric potential energy, we're comparing charges at points in space. Hopefully, I didn't confuse you. In the next video, we'll actually do a couple of problems where we figure out the electric potential difference or the voltage difference between two points in space as opposed to a charge at two different points in space. I will see you in the next video."
    },
    {
        "Q": "\n5:29\nIs that to say the electric potential energy of electrons is dependent on its position? In an electric circuit, how do we change the electric energy per charge? When electrons pass through a load in a circuit, such as a lamp, how is electrical energy lost (converted to light energy)? How does it affect current and amperes?",
        "A": "Potential energy ALWAYS relates to position. That s what PE is. The battery provides chemical energy to separate charges and give them PE. The PE is converted to KE when the circuit is complete and the charges flow as current. When they pass through the load, they bump into the atoms of the load and transfer some of their KE to the load.",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            329
        ],
        "3min_transcript": "amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy-- The electrical potential energy difference between this point and this point is 12 joules. Or another way to say it is-- and which one has a higher potential? Well, this one does, right? Because at this point, we're closer to the thing that's trying to repel it, so if we were to just let go, it would start accelerating in this direction, and a lot of that energy would be converted to kinetic energy by the time we get to this point, right? So we could also say that the electric potential energy at this point right here is 12 joules higher than the electric potential energy at this point. Now that's potential energy. What is electric potential? Well, electric potential tells us essentially how much work is necessary per unit of charge, right? Electric potential energy was just how much total work is needed to move it from here to here."
    },
    {
        "Q": "\nAt 5:12, Sal mentioned other species in the Homo genus, Neanderthals. What are Neanderthals?",
        "A": "Neanderthals were an extinct species of human, or possibly a subspecies. They are very closely related to modern humans, differing in DNA by only about 0.1%. Neanderthals were larger and heavier than modern humans. They went extinct sometime between 30,000 to 45,000 years ago, possibly due to inability to adapt to the changing climate of the era, or they simply interbred with humans.",
        "video_name": "oHvLlS_Sc54",
        "timestamps": [
            312
        ],
        "3min_transcript": "of deciding how close two animals are. But to a large degree, a lot of these categories-- deciding where to divide along kingdom, phylum, class, order, family, tribe-- these are somewhat arbitrary. These are just picked based on early taxonomists, including Carl Linnaeus, and saying, well, this looks like a grouping right over here. But they could have grouped at a broader level or a deeper So these things right over here are somewhat arbitrary. A more analytical way is just to see how much DNA you have in common and then use that as a measure of how far apart two animals are. Or really, I should say, two species are, because this taxonomy doesn't only apply just to animals. It applies to plants and bacteria and Archaea and all sorts of things, so it's actually a broader thing than just animals. Now, with that out of the way, what I thought would be fun-- just so that we could really get a sense of where modern taxonomy is, where the field that was essentially fathered by Carl Linnaeus, where it is now, how we-- and use that to figure out where we And obviously, I'm drawing just a small fraction of the universe of the organisms that we even know about right now. But at least it frames the picture in terms of something we understand-- in particular, us. In particular, humans. Now, our species, we call ourselves humans. But we're really Homo sapiens. And the sapiens is the species part, and then Homo is the genus. And what I'm doing right over here is I'm saying, well, if Homo is the genus, what other species were inside of Homo? And the reality is-- or at least as far as we know-- there are no other living species inside of Homo. We probably killed them all off. Or maybe we interbreeded with them somehow, which might have argued that maybe they weren't different species. But more likely, they were competing in the same ecosystems, and they became endangered species very quickly when they competed with our ancestors. But the most recent other species within the genus that we know about are the neanderthals, Now, if we go further up the tree of life, further up the taxonomy-- and you'll sometimes see tribe mentioned. Sometimes you won't. And we tend to get a little bit more granular the closer we get to humans. When we go further away in the tree of life, we get a little bit less granular sometimes. But that's not always the case as well. You go a little bit further up, then you get Hominini. And I'm sure I'm mispronouncing some of this as well. But another species that's in Homonini that is not in Homo-- and I'm definitely not listing all of them, and that's why I'm showing all of these other branches over here-- is what we call the common chimpanzee. And their species name is-- their genus is Pan, and their species is troglodytes. So you would refer to them as Pan troglodytes. And that's also another convention that Carl Linnaeus came up with, is that you refer to a particular species by its genus and then its species. And you capitalize the genus, and you lowercase a species."
    },
    {
        "Q": "Why are sharks fish and dolphins and whales mammals? (brought up at 10:02)\n",
        "A": "This is due to very notable differences between the two. Sharks have cartilaginous bones, gills, and a swim bladder, all of which are associated with fish. Whales and Dolphins have dense bones, lungs, and are warm blooded, things associated with mammals. Ocean mammals also have vestigial hip bones even though they lack legs and they are incapable of drinking salt water. Like many desert mammals, ocean mammals get their water from the food that they eat.",
        "video_name": "oHvLlS_Sc54",
        "timestamps": [
            602
        ],
        "3min_transcript": "from things that first lived in trees, and that's why their hands and their feet look the way they do. Now you get to even a broader level of classification. You get to the mammals. And once again, probably something you're used to thinking about. Mammals are air-breathing animals, and they tend to have fur or hair. They tend to provide some form of milk for their young. They have active mammary glands. There's other things we can talk about, what makes a mammal. I'm not going to go into the rigorous definition. But just to give you an example of a mammal that is not a primate, I could show you this polar bear right over here. This is a mammal that is not a primate. And I could do other things. I could show you a tiger, or I could show you a giraffe or a horse. And so by no stretch of the imagination am I being comprehensive. But let's keep getting broader. Now let's go to the class-- we're already at the class of Mammalia. Now let's go to the phylum. we are in the phylum chordates. And chordates, we're actually in the subphylum, which I didn't write here, vertebrates, which means we have a vertebra. We have a spinal column with a spinal cord in it. Chordates are a little bit more general. Chordates is a phylum where-- kind of the arrangement of where the mouth is, where are the digestive organs, where the anus is, where the spinal column is, where are the brains, where are the eyes, where are the mouth. They're kind of all in the same place. And if you think about it, everything I've listed here kind of has the same general structure. You have a spinal column. You have a brain. You have a mouth. Then the mouth leads to some type digestive column. And at the end of it, you have an anus over there. And you have eyes in front of the brain. And so this is a general way-- and I'm not being very rigorous here, is how you describe a chordate. And to show a chordate that is not a mammal, you would just have to think of a fish or sharks. So this right over here is a non-mammal chordate Now, let's go even broader. As you'll see, now we're going to things that are very, very not human-like. So you go one step broader. Now we're in Animalia, the kingdom of animals. And this is the broadest category that Carl Linnaeus thought about. Well actually, he did go into trees as well. But when you think of kingdom animals and you think of things that aren't chordates, you start going into things like insects. And you start going into things like jellyfish. If you go even broader, now we're talking about the domain. You go to Eukarya. So these are all organisms that have cells. And inside those cells, they have complex structures. So if you're a Eukarya, you have cells with complex structures. If you're a Prokarya, you don't have complex structures inside your cell. But other Eukarya that are not animals include things like plants. And obviously, I'm giving no justice"
    },
    {
        "Q": "At 1:16 , The force with which we pull will be stored as potential energy and it will be converted to kinetic energy once we leave the object.... But Sal said that the kinetic energy is converted into potential energy.. If I'm wrong, please someone correct me..\n",
        "A": "The energy bounces back and forth between KE and PE.",
        "video_name": "Nk2q-_jkJVs",
        "timestamps": [
            76
        ],
        "3min_transcript": "Let's see if we can use what we know about springs now to get a little intuition about how the spring moves over time. And hopefully we'll learn a little bit about harmonic motion. We'll actually even step into the world of differential And don't get daunted when we get there. Or just close your eyes when it happens. Anyway, so I've drawn a spring, like I've done in the last couple of videos. And 0, this point in the x-axis, that's where the spring's natural resting state is. And in this example I have a mass, mass m, attached to the spring. And I've stretched the string. I've essentially pulled it. So the mass is now sitting at point A. So what's going to happen to this? Well, as we know, the force, the restorative force of the spring, is equal to minus some constant, times the x position. The x position starting at A. So initially the spring is going to pull back this way, right? The spring is going to pull back this way. It's going to get faster and faster and faster and faster. And we learned that at this point, it has a lot of potential energy. At this point, when it kind of gets back to its resting energy, but very little potential energy. But then its momentum is going to keep it going, and it's going to compress the spring all the way, until all of that kinetic energy is turned back into potential energy. Then the process will start over again. So let's see if we can just get an intuition for what x will look like as a function of time. So our goal is to figure out x of t, x as a function of time. That's going to be our goal on this video and probably the next few. So let's just get an intuition for what's happening here. So let me try to graph x as a function of time. So time is the independent variable. And I'll start at time is equal to 0. So this is the time axis. Let me draw the x-axis. This might be a little unusual for you, for me to draw the x-axis in the vertical, but that's because x is the dependent variable in this situation. Or we could say x of t, just so you know x is a function of time, x of t. And this state, that I've drawn here, this is at time equals 0, right? So this is at 0. Let me switch colors. So at time equals 0, what is the x position of the mass? Well the x position is A, right? So if I draw this, this is A. Actually, let me draw a line there. That might come in useful. This is A. And then this is going to be-- let me try to make it relatively-- that is negative A. That's minus A. So at time t equals 0, where is it? So this is where the graph is, right? Actually, let's do something interesting. Let's define the period."
    },
    {
        "Q": "\nAt 1:09, shouldn't the carbonyl O be protonated first because H2O is a relatively weak nucleophile?",
        "A": "Not necessarily, assuming you don t have other stronger electrophiles in your reaction vessel. H2O is actually a pretty decent nucleophile (which is why for most SN2 reactions where it is NOT the intended substituent, we take great care to remove it from our system).",
        "video_name": "632MAqIB14E",
        "timestamps": [
            69
        ],
        "3min_transcript": "Voiceover: Here's an example of a nucleophilic addition reaction to an aldehyde or a ketone. So over here on the left, it could be an aldehyde, or we could change that to form a ketone. And if you add water to an aldehyde or ketone, you form this product over here on the right, which is called a hydrate, or also called a gem-diol, or geminal diol because these two OHs here are on the same carbon, so like they're twins. And this reaction is at equilibrium. So let's think about the aldehyde, or the ketone. We know the carbonyl on the aldehyde or ketone is polarized, so we know that the oxygen has more electronegatives than carbons, so it withdraws some electron densities. So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that. And therefore the carbonyl carbon, since it's partially positive, is electrophillic, so it wants electrons. And it can get electrons from water. So let's go ahead and draw the water molecule right here. Water can function as a nucleophile. It has two lone pairs of electrons, this oxygen here is partially negative, and so we're going to get a nucleophile So a lone pair of electrons on the oxygen is going to attack our carbonyl carbon like that, So the nucleophile attacks the electrophillic portion of the molecule, and these pi electrons here kick off onto the oxygen. So let's go ahead and draw the results of our nucleophilic attack here, and so we now have our oxygen bonded to this carbon, and this oxygen still has two hydrogens bonded to it, so I'm gonna go ahead and draw in those two hydrogens. There's still a lone pair of electrons on that oxygen, which gives that oxygen a +1 formal charge. And then this carbon here is bonded to another oxygen, which had two lone pairs of electrons around it, and now it picked up another one, so a -1 formal charge on this oxygen, and there's still an R group bonded to it, and a hydrogen over here, like that. And so let's try to follow some electrons here. So one of the lone pairs of electrons on the oxygen formed a bond with our carbon, so I'm And then we can think about our pi electrons. So, our pi electrons in here, as kicking off onto our oxygen, so it doesn't really matter which one of these three it is, right, let's say it's that one, and we get this as our intermediate. And so next, we can think about an acid-base reaction. So, another water molecule comes along right here, and so we know water can function as an acid or a base, and so this lone pair of electrons could take, let's say it takes this proton right here, and leaves these electrons behind on our oxygen. So let's go ahead and draw the result of that acid-base reaction, and so we would have our oxygen here, would now be bonded to only one hydrogen, and, now let's see, we still have our negatively charged oxygen over here on the right, and then we have our R group, and our hydrogen like that, and we still have this lone pair of electrons,"
    },
    {
        "Q": "at 6:40, you talk about how adding Cl groups causes it to be more reactive by withdrawing electronegativity, why is it then that Ketones are more stable than Aldehydes as stated in the reactivity video? Wouldn't the extra R group also withdraw electronegativity and cause it to be more reactive?\n",
        "A": "R groups mean they re carbon substituents and assuming they have C-H bonds they re electron donating through hyperconjugation. Obviously if you replaced all the C-H bonds with something highly electron withdrawing like C-Cl bonds you will make a very reactive ketone.",
        "video_name": "632MAqIB14E",
        "timestamps": [
            400
        ],
        "3min_transcript": "this as our product for our hydrate. Except, we know that ketones are not as reactive as aldehydes, and so this time the equilibrium is to the left. It favors the formation of the ketone. Let's look at another one. So this is acetaldehyde. And so if we add water to acetaldehyde over here, we form this as our hydrate product. And once again, we know that these reactions occur because of our carbonyl carbon right here being partially positive, so the oxygen withdraws some electron density like that. So, we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon, and one thing you could do, is add an electronegative atom like halogen, so let's go ahead and do that. Let's add three halogens here. Let's add three chlorines to this carbon, the one adjacent to our carbonyl carbon. And those electron withdrawing groups, withdraw some electron density, so they're gonna withdraw electron density this way, once again, away from our carbonyl carbon, and so this carbonyl carbon gets even more partially positive by the addition of these electronegative atoms. And the more positive you make that carbonyl carbon, the more electrophilic you make it, and therefore, the more the nucleophile, which is water, is going to attack, and so you make it even more reactive by adding these, and so you can push the equilibrium even more to the right. You can form more of your product. And so let's go ahead and draw the product of that reaction, so we would put three chlorines on here like that, and so you could also do this with ketones, and you could make ketones much more reactive by doing that. And so this particular reaction is a little bit famous. Over here on the left is trichloroacetaldehyde, and then, once you've formed the hydrate and this is famous for being knock-out drops. And so some of the old references to it are, \"slip someone a Mickey Finn.\" So you could slip chloral hydrate into someone's drink, and it was a knock-out drop situation. And so just a little bit of an interesting reaction here for formation of hydrates."
    },
    {
        "Q": "\nAt 0:26, Sal is naming the alkene. He notes the methyl group at Carbon-2, but not Carbon-1 or Carbon-5-- so I'm a bit confused. We don't have to name those?\nAs in: wouldn't it be 1,2,5-methylpent-2-ene?\n\nThanks for any help!",
        "A": "They re not methyl groups, they are part of the main chain. They ve already been accounted for by calling it pentene.",
        "video_name": "O_yeKo6-qIg",
        "timestamps": [
            26
        ],
        "3min_transcript": "- [Voiceover] Anytime you're trying to come up with a mechanism for a reaction, it's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different. So what we're starting with, we could call this one, two, three, four, five, so this is, let's see, we have methyl group on the number two carbon, it is a pentene, and that is double-bond between the number two and number three carbons, so this is two-methyl-pent-2-ene. So that's what we start with, we're in the presence, we're in an acidic environment, we've got what's gonna be catalyzed by our hydronium here, and we end up with this, and how is our product different from what we started with? Well the double bond is now gone, the number three carbon gains this hydrogen, and now the number two carbon gains a hydroxyl group. in the presence of an acid, it's acid-catalyzed, we have gained two hydrogens and an oxygen, which is what we've gained, what could be used to make a water. And this is actually called an acid-catalyzed addition of water. The water isn't sitting on one part of the molecule, but if you take the hydrogen we added, and the hydroxyl we added, if you combine them, that's what you need to make a water. So let's think about how we can, how this actually happens in the presence of our hydronium. So let me redraw this molecule right over here. So we copy and paste it, so that's not exactly it yet, that is just with the single bond. So let me draw, woops, wrong tool. Let me draw the double bond there. And now let me put it in the presence of some hydronium. Alright, so we have an oxygen bonded, two. but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton, thus making the entire molecule positive, because the hydrogen proton is positive. So there you go, this now has a positive charge. And this can be pretty reactive, 'cause we know that oxygen is quite electronegative, it lives to keep its electrons. So what is there was a way, what if there's a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond. Well what if one of these carbons, especially the ones that have the double bonds, what if some of the electrons from this double bond could be used to snab, to take that hydrogen proton, and then oxygen can hog its electrons again. And you might say, \"oh that's reasonable, but which of these carbons would actually do it?\" And to think about which of those carbons would do it,"
    },
    {
        "Q": "\nAt 7:40, if two oxygen electrons are shared, then why isn't a double bond formed??",
        "A": "2 electrons is a single bond 4 electrons is a double bond",
        "video_name": "O_yeKo6-qIg",
        "timestamps": [
            460
        ],
        "3min_transcript": "And so this is now just neutral water, and we see that we have a conservation of charge here, this was positive in charge, now our original molecule is positively charged. And what feels good about this is we're getting, we're getting close to our end product, at least on our number three carbon, we now have, we now have this hydrogen. Now we need to think about, well how do we get a hydroxyl group added right over here? Well we have all this water, we have all this water floating around, let me, I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules, we're in an aqueous solution, so let me draw another water molecule here. So the water molecules are all equivalent, but let me draw another water molecule here. And you can imagine, if they just pump into each other in just the right way, this is, water is a polar molecule, it has a because the oxygen likes to hog the electrons, and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged, so you can imagine the oxygen end might be attracted to this tertiary carbocation, and so just bumping it in just the right way, it might form a bond. So let me say these two electrons right over here, let's say they form a bond with this, with that number two carbon, and then what is going to result? So let me draw, so what is, what is going to result, let me scroll down a little bit, and let me paste, whoops, let me copy and paste our original molecule again. So, here we go. This is the one we constructed actually, so we have the hydrogen there. We have the hydrogen, now this character, so we have the water molecule, so oxygen bonded to two hydrogens, you have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Woops, let me do it in that orange color. It now, it now forms an actual bond. And we're really close to our final product, we have our hydrogen on the number three carbon, we have more than we want on our number two carbon, we just want a hydroxyl group, now we have a whole water bonded to the carbon. So somehow we have to get one of these other hydrogens swiped off of it, well that could happen with just another water molecule. So let's draw that. So another water molecule someplace, I'll do the different color just to differentiate, although as we know water, well it's hard to see what color is water if you're looking at the molecular scale."
    },
    {
        "Q": "\nAt 3:22, shouldn't it be -ke^2/r instead of ke^2/r?",
        "A": "the thing is, the final result of this calculation doesn t result in a vector, so we really don t need the energy, we only need the magnitude. In the second case however, it is a vector, so sign is required for vector algebra",
        "video_name": "7Zin8hG9Nhw",
        "timestamps": [
            202
        ],
        "3min_transcript": "with where that electron is. So we know the electron is also attracted to the nucleus. There's an electric force, alright, so this electron is pulled to the nucleus, this is an attractive force. This is the electric force, this is a centripetal force, the force that's holding that electron in a circular orbit around the nucleus here. And, once again, we talked about the magnitude of this electric force in an earlier video, and we need it for this video, too. We're gonna use it to come up with the kinetic energy for that electron. So the electric force is given by Coulomb's Law, the magnitude of the electric force is equal to K, which is a constant, \"q1\", which is, let's say it's the charge on the proton, times \"q2\", charge on the electron, divided by \"r squared\", where \"r\" is the distance between our two charges. We know that Newton's Second Law: force is equal to the mass times the acceleration. so the mass of the electron times the acceleration of the electron. The electric force is a centripetal force, keeping it in circular motion, so we can say this is the \"centripetal acceleration\". Alright, let's go ahead and write down what we know. \"K\" is a constant, we'll write that in here, \"q1\", \"q1\" is the charge on a proton, which we know is elemental charge, so it would be positive \"e\"... \"q2\" is the charge on the electron. The charge on the electron is the same magnitude as the charge on the proton, but it's a negative value. So we have negative \"e\", is the charge on the electron, divided by \"r squared\", is equal to the mass of the electron times the centripetal acceleration. So, centripetal acceleration is equal to \"v squared\" over \"r\". So, we did this in a previous video. We're gonna do the exact same thing we did before. because we already know the direction is always going to be towards the center, and therefore, we only care... we don't care about this negative sign here. We can also cancel one of the \"r\"s. So if we don't care about... if we only care about the magnitude, on the left side, we get: Ke squared over r is equal to mv squared, on the right side. And you can see, we're almost to what we want. Our goal was to try to find the expression for the kinetic energy, that's 1/2 mv squared. Here, we have mv squared, so if we multiply both sides by 1/2, right, multiply both sides by 1/2, now we have an expression for the kinetic energy of the electron. So: 1/2 mv squared is equal to the kinetic energy. So we know the kinetic energy is equal to: 1/2 Ke squared over r Alright, so we will come back to the kinetic energy. Next, we're gonna find the potential energy."
    },
    {
        "Q": "Whoa, hold on\u00e2\u0080\u0094that value of 0.15 at 10:40 is not a probability! Sal picked the box size arbitrarily. If he had picked a box that were ten times greater in volume, then the \"probability\" would be 1.5; that is, there would be a 150% \"chance\" of there being a hydrogen molecule in that box. That's obviously not right. Instead, it represents the expected, or average, number of hydrogen molecules per box of that volume. To go from an average value to a probability is quite hand-wavy indeed.\n",
        "A": "The volume of that box is 1 Liter; because SI units.",
        "video_name": "psLX080RQR8",
        "timestamps": [
            640
        ],
        "3min_transcript": "So let's see, if I take the twenty-third -- so let me write it out here. So my hydrogen per box -- So my concentration of hydrogen per dV, is equal to 12 times 10 -- whoops! That's not helping when my pen malfunctions. Let me get that right. 12 times 10 to the twenty-third power times 0.125 times 10 to the minus twenty-seventh power. All of that divided by 10 to the minus 3, right? That's 1 times 10 to the minus 3. So let's cancel out some exponents. If we get rid of the minus 3 here, you divide by minus 3, then this becomes minus 24. And then the minus 24 and the minus what's 12 times 1.25? So times 12 is equal to 1.5. So the 12 times the 1.25 is equal to 1.5 times -- and then 10 to the twenty-third times 10 to the minus twenty-fourth is equal to 10 to the minus 1, right? So it's just divided by 10. So on average, your concentration of hydrogen in a little cube that's half a nanometer in each direction is equal to 0.15 molecules -- not Moles anymore -- of hydrogen molecule per my little dV, my little box. And so this is a probability, right? This is a probability, because obviously I can't have 0.15 molecules in every box. there's a 0.15 chance that I have a hydrogen molecule in my box. So if I want to go back here to this, this is 0.15, this is 0.15, this is 0.15. But how did we get this 0.15? We multiplied the concentration of hydrogen, which was this right here. That's the concentration of the hydrogen -- I should've written it in a more vibrant color -- times just a bunch of scaling factors, right? We could just say that, well, this was just equal to the concentration of hydrogen times, based on how I picked my dV, I had to do all of this scaling. But it was times some constant of scaling, scaling to my appropriate factor. So if we want to figure out each of these, this is just the concentration of hydrogen times"
    },
    {
        "Q": "at 7:15, what did sal mean by scaling factor?\n",
        "A": "He scales the units so they are all the same. It is basically converting a unit to another. For example 0.01 l is equal to 1 ml, so the scaling here would be 1 ml * 10^-3 = 0.01 l",
        "video_name": "psLX080RQR8",
        "timestamps": [
            435
        ],
        "3min_transcript": "Let's pick my dV to be -- I looked up the diameter of an ammonia molecule. It was about 1/10 of a nanometer. If this was a nanometer box, you could put 10 in each direction, so you can almost fit 1,000 if you packed them really tightly. So let's make this half a nanometer in each direction. So if I pick my dV -- and remember, I don't know if this is the right distance. I'm just trying to give you the intuition behind the equilibrium formula. But if I pick this as being 0.5 nanometers by 0.5 nanometers by 0.5 nanometers, what is my volume? So my little volume is going to be 0.5 times 10 to the minus 1/9 meters-- that's a nanometer because we're dealing with cubic meters. So this is equal to 0.5 to the 1/3 power. 0.5 times 0.5 is 0.25 times 0.5 is 0.125. I want to do the math right, so let me just make sure I got that right. 0.5 to the 1/3 power. Right, 0.125 times -- negative 9 to the 1/3 power is minus 27 -- 10 to the minus 27 meters cubed. So that's my volume. Now, we know the concentration. Let's figure out what's the probability. So this is the probability in the box, right? That's what we're concerned with, the probability in the box. Well, the probability in the box, that's the probability that I have one hydrogen in the box, times the probability that I have another hydrogen in the box, times the probability that I have another hydrogen in the box --these are all in-the-box probabilities -- times the probability that I'll do the nitrogen in a different color just to ease -- oh, I should've done these in the orange because those are the color of the molecules up there. And I'll do this one in purple. What's the probability of having hydrogen in the box? Well, we know hydrogen's concentration at equilibrium is 2 Molar. So concentration of hydrogen, we know hydrogen's concentration is equal to 2 Molar, which is 2 Moles per liter, which is equal to -- 2 Moles is just 2 times 6 times 10 to the 23rd power -- Moles is just a number -- divided by liters. So 1 liter is-- we could write it in meters cubed, or we could just make the conversion. Actually, let me just do this for you. 1 liter is equal to 1 times 10 to the minus 3 meters cubed."
    },
    {
        "Q": "\nAt 8:13 Jay says that the Bohr model is incorrect. Is this because the Uncertainty Principle showed that the position of the electron would be at a distance greater than 2 times the radius. Whereas the Bohr model had the position of the electron at r1?",
        "A": "The Bohr model could not be correct because it only worked for H and even for H it did not properly explain all the observations about the emission spectrum of H. That s why physicists continued to search for a better model, and through the work of DeBroglie, Schrodinger, Heisenberg, Born and others, the modern model was developed.",
        "video_name": "PZIoFD_Z73M",
        "timestamps": [
            493
        ],
        "3min_transcript": "we had the uncertainty in the position of the electron, times the uncertainty in the momentum of the electron must be greater than or equal to Planck's Constant divided by four pi. So we can take that uncertainty in the momentum and we can plug it in here. So now we have the uncertainty in the position of the electron in the ground state of the hydrogen atom times 2.0 times 10 to the negative 25. This product must be greater than or equal to, Planck's Constant is 6.626 times 10 to the negative 34. Alright, divide that by four pi. So we could solve for the uncertainty in the position. So, Delta X must be greater than or equal to, let's go ahead and do that math. So we have Planck's Constant, 6.626 times 10 to the negative 34, and then we need to divide by the uncertainty in the momentum. So we also need to divide by the uncertainty in momentum, that's 2.0 times 10 to the negative 25, and that gives us 2.6 times 10 to the negative 10. So the uncertainty in the position must be greater than or equal to 2.6 times 10 to the negative 10 and if you worked our your units, you would get meters for this. So the uncertainty in the position must be 2.6 times 10 to the negative 10 meters. Let's go back up here to the picture of the hydrogen atom. 2.6 times 10 to the negative 10 meters, that's greater than the diameter of our hydrogen atom, so the uncertainty would be greater than this diameter. the diameter of the hydrogen atom, using the Bohr model. So the Bohr model is wrong. It's telling us we know the electron is orbiting the nucleus at a certain radius, and it's moving at a certain velocity. The uncertainty principle says this isn't true. If we know the velocity fairly accurately, we don't know the position of the electron, the position of the electron is greater than the diameter, according to the Bohr model. So this just one reason why the Bohr model is wrong. But again, we keep the Bohr model around because it is useful as a simple model when you're just starting to get into chemistry. But this concept of the uncertainty principle goes against our natural intuitions. So our every day life doesn't really give us any experience with the uncertainty principle. For example, if we had a particle, let's make it a much bigger particle here, so a much bigger particle than an electron, so something that we can actually see in our real life, and so this has a much bigger mass,"
    },
    {
        "Q": "\nAt 3:46, what does he mean by the electron being in the ground state?",
        "A": "Ground state means the lowest energy state",
        "video_name": "PZIoFD_Z73M",
        "timestamps": [
            226
        ],
        "3min_transcript": "so the uncertainty in the momentum must increase to four, because one times four is equal to four. If I decrease the uncertainty in the position even more, so if I lower that to point five, I increase the uncertainty in the momentum, that must go up to eight. So point five times eight gives us four. And so, what I'm trying to show you here, is as you decrease the uncertainty in the position, you increase the uncertainty in the momentum. So another way of saying that is, the more accurately you know the position of a particle, the less accurately you know the momentum of that particle. And that's the idea of the uncertainty principle. And so let's apply this uncertainty principle to the Bohr model of the hydrogen atom. So let's look at a picture of the Bohr model of the hydrogen atom. Alright, we know our negatively charged electron orbits the nucleus, like a planet around the sun. And, let's say the electron is going this direction, so there is velocity going in that direction. Alright, the reason why the Bohr model is useful, is because it allows us to understand things like quantized energy levels. And we talked about the radius for the electron, so if there's a circle here, there's a radius for an electron in the ground state, this would be the radius of the first energy level, is equal to 5.3 times 10 to the negative 11 meters. So if we wanted to know the diameter of that circle, we could just multiply the radius by two. So two times that number would be equal to 1.06 times 10 to the negative 10 meters. And this is just a rough estimate of the size of the hydrogen atom using the Bohr model, with an electron in the ground state. Alright, we also did some calculations to figure out the velocity. So the velocity of an electron in the ground state we calculated that to be 2.2 times 10 to the six meters per second. And since we know the mass of an electron, we can actually calculate the linear momentum. So the linear momentum P is equal to the mass times the velocity. Let's say we knew the velocity with a 10% uncertainty associated with that number. So a 10% uncertainty. If we convert that to a decimal, we just divide 10 by 100, so we get 10% is equal to point one. So we have point one here. If I want to know the uncertainty of the momentum of that electron, so the uncertainty in the momentum of that particle, momentum is equal to mass times velocity. If there's a 10% uncertainty associated with the velocity, we need to multiply this by point one. So let's go ahead and do that. So we would have the mass of the electron is 9.11 times 10 to the negative 31st."
    },
    {
        "Q": "at 7:02 jay took away 2 pi electrons, one from each corresponding carbon atom. how is it possible to take 2 pi electrons from different carbon atoms. i do know about breaking a pi bond where one carbon attains a positive charge where other attains a negative charge for showing resonance. i am finding it hard to believe that two corresponding carbon atoms attains a + charge. pls do explain why.\n",
        "A": "You need a reagent that really wants those electrons, like SbF\u00e2\u0082\u0085 in SO\u00e2\u0082\u0082Cl\u00e2\u0082\u0082",
        "video_name": "I6wzan4hNc4",
        "timestamps": [
            422
        ],
        "3min_transcript": "have three molecular orbitals. But these are my antibonding molecular orbitals. Those are higher energy. And my two points of intersection that are right on the center line here, represent two non-bonding molecular orbitals like that. So when I fill my molecular orbitals, again it's analogous to electron configurations. I have a total of eight pi electrons that I need to worry about for a planar cyclooctatetraene molecule. And so I can go ahead and start to fill in my pi electrons So that takes care of six of them and I have two more. And since this is analogous to electron configurations, I'm going to follow Hund's rule and not pair up my electrons in an orbital here. So that represents my eight pi electrons. And since I have unpaired electrons, I have two unpaired electrons, that predicts a very unstable molecule if it were to adopt a planar confirmation. And if I think about in terms of Huckel's rule, I know it doesn't follow Huckel's rule. Huckel's rule is 4n plus 2, where n is an integer. have this orbital down here. And I do have 4n right here. But I don't have 4n, where n is an integer for these two electrons up here. And so this is where it breaks down. And so I have a total of eight pi electrons, which does not follow Huckel's rule. And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar confirmation. And so cyclooctatetraene has a tub confirmation, and not planar. It is not aromatic. It is considered to be non-aromatic because of the violation of the first criteria. But it is possible to react cyclooctatetraene. It's possible to oxidize it. And so let's see what happens when we do that. So if we take cyclooctatetraene and we oxidize it. So it's going to lose some electrons. So I'm going to say that these pi electrons are going to stay. And we're going to lose the pi electrons on the left. So if I take away a bond from these two carbons that they're going to be positively charged. And so I could draw a resonance structure for this. I could move these electrons over here. And so if I go ahead and show that resonance structure, then this carbon still has a plus 1 charge. And the other positive charge moves over here to this carbon, And you could continue drawing resonance structures for this molecule. I am not going to do that. I just want to show you that the positive charges are spread out throughout the entire ion here. And so one way to represent that would be to just show the electrons are spread out throughout this entire ion. And the whole thing has a 2 plus charge like that. And so when I analyze this dication that I got from cyclooctatetraene, I realize that all the carbons are sp2 hybridized. So if I look at these, my carbocations those are sp2 Everything with a double bond on it is sp2 hybridized. And so I have eight sp2 hybridized carbons."
    },
    {
        "Q": "''5:38\"\n\nno ones dug that far?\ncool!\nhow deep is the earths core?\ndoes anyone at khan academy know?\n",
        "A": "did u guy ever see ice age continental drift. if so do u remember when scrat the little weird animal that loves acorns. well he went to the earths core and he didnt burn up and die.",
        "video_name": "KL0i1RSnpfI",
        "timestamps": [
            338
        ],
        "3min_transcript": "We don't know yet whether the inner part is liquid or solid. Now, the next point of evidence is how do we know that there's an inner core? And we can use P-waves for that. A P-wave can travel through anything, but remember, in general for the same type of material if you get denser material it's going to move faster, so it's going to refract outwards like we've seen over here. But if it goes into a liquid, in general, sound waves, or I should say P-waves, seismic waves move slower in liquids. And so the refraction patterns we get when we do measure from seismograph stations around the world is that it looks like the P-waves are kind of doing what you would expect in the mantle, but then they're getting refracted as if they're going to a slower medium as they go through the outer core. And we see that right over here. And then they get refracted again Now, that is just what you would expect if it was all liquid, but if you go to stations that are even further out it looks like, if you just look at the refraction patterns, and you can now model this with fancy computers and get all the data points, but you could say, well, the only way that reality can fit the data that we get based on when things reach here is if the P-waves are being first refracted through the outer core, but then they're refracted in a way that they're going through denser material, significantly denser material than the inner core. And then they're just continuing to refract the way you would expect. So it's really the refraction pattern of the P-waves. And frankly, the fact that there's this what you call a P-wave shadow. The P-wave shadow by itself, all that tells you is that kind of roughly crazy things are happening someplace in the core. But the real way to know that we have an inner core that's solid, as opposed to the whole thing being liquid, is that the P-waves is the pattern of when and how the P-waves reach And then you can kind of, based on modeling how waves would travel through different densities and different types of mediums, you could say, well, there's got to be an inner core right over here. And obviously, it's a lot more math than I'm going into. But if you do the math based on the shadow, and you know the speed of the material, and all of that type of thing, then you can figure out the depth at which these transitions occur. We know that we have a transition from mantle to outer core here. And then a transition from outer core to core there. So hopefully that satiates your questions about how do we know what the composition of the earth is without ever having to dig down there, because we've never even gotten below our crust."
    },
    {
        "Q": "At 5:36, why did he move the hydrogen to the other side of the periodic table?\n",
        "A": "He showed that the hydrogen could belong to the alkali metals (because it could lose one electron like them and have a complete outer shell) OR it could gain one electron and still have a full outer shell (like the halogens which are group 7 elements). Remember the first electron shell only holds 2 electrons, then each shell after that holds 8.",
        "video_name": "CCsNJFsYSGs",
        "timestamps": [
            336
        ],
        "3min_transcript": "it just needs one electron. So in theory, hydrogen could have been put there. So hydrogen actually could typically could have a positive or a negative 1 oxidation state. And just to see an example of that, let's think about a situation where hydrogen is the oxidizing agent. And an example of that would be lithium hydride right Now, in lithium hydride, you have a situation where hydrogen is more electronegative. A lithium is not too electronegative. It would happily give away an electron. And so in this situation, hydrogen is the one that's oxidizing the lithium. Lithium is reducing the hydrogen. Hydrogen is the one that is hogging the electron. So the oxidation state on the lithium here is a positive 1. And the oxidation state on the hydrogen here is a negative. to make sure we get the notation. Lithium has been oxidized by the hydrogen. Hydrogen has been reduced by the lithium. Now, let's give an example where hydrogen plays the other role. Let's imagine hydroxide. So the hydroxide anion-- so you have a hydrogen and an oxygen. And so essentially, you could think of a water molecule that loses a hydrogen proton but keeps that hydrogen's electron. And this has a negative charge. This has a negative 1 charge. But what's going on right over here? And actually, let me just draw that, because it's fun to think about it. So this is a situation where oxygen typically has-- 1, 2, 3, 4, 5, 6 electrons. And when it's water, you have 2 hydrogens like that. And then you share. over there sharing that pair, covalent bond sharing that right over there. To get to hydroxide, the oxygen essentially nabs both of these electrons to become-- so you get-- that pair, that pair. Now you have-- let me do this in a new color. Now, you have this pair as well. And then you have that other covalent bond to the other hydrogen. And now this hydrogen is now just a hydrogen proton. This one now has a negative charge. So this is hydroxide. And so the whole thing has a negative charge. And oxygen, as we have already talked about, is more electronegative than the hydrogen. So it's hogging the electrons. So when you look at it right over here, you would say, well, look, hydrogen, if we had to, if we were forced to-- remember, oxidation states is just an intellectual tool which we'll find useful. If you had to pretend this wasn't a covalent bond, but an ionic bond, you'd say, OK, then maybe this hydrogen would fully lose an electron,"
    },
    {
        "Q": "\nAt 14:03, how did the answer become -6.37 m/s^2? I substituted the same values in the exact same kinematics equation and I got -62.83 m/s^2. Did I possibly enter something wrong with my calculator (it was in radians mode).",
        "A": "I know the answer to your question! Ok, so when you put them into the calculator, ALWAYS PUT PARENTHESIS FOR PI! It is so important! If you just divide -(40rad/s)^2 by 80pi (2*40pi), then you will get -62.8319... If you divide -(40rad/s)^2 by (80pi), you will get the right answer -6.3662...",
        "video_name": "TBlDBaUGqNc",
        "timestamps": [
            843
        ],
        "3min_transcript": "we got how fast in meters per second. It was going 160 meters per second. And the next part asks, what was the angular acceleration of the bar? All right, this one we're gonna have to actually use a kinematic formula for. We'll bring these back, put 'em over here. Again the way you use these, you identify what you know. We know the initial angular velocity was 40. So this time we know omega initial 40 radians per second. Set it revolved 20 revolutions. That's delta theta, but again, we can't just write 20. We've gotta right this in terms of radians if we're gonna use these radians per second. They have to all be in the same unit. So it's gonna be 20 revolution times two pi radians per revolution. So that's 40 pi radians. What's our third known? You always need a third known to use a kinematic formula. It's this. It says it decelerates to a stop, which means it stops. That means omega final, the final angular velocity is zero. And we want the angular acceleration, that's alpha. We wanna know alpha. We know the rest of these variables. Again to figure out which equation to use, I figure out which one got left out. And that's the time. I was neither give the time nor was I asked to find the time. Since this was left out, I'm gonna look for the formula that doesn't use time at all. And that's not the first one. That's not the second or the third, it's actually the fourth. So I'm gonna use this fourth equation. So what do we know? We know omega final was zero. So I'm gonna put a zero squared. But zero squared is still zero, equals omega initial squared. That's 40 radians per second squared. And then it's gonna be plus two times alpha. We don't know alpha, but that's what we wanna find, so I'm gonna leave that as a variable. And then delta theta we know. Delta theta was 40 pi radians since it was 20 revolutions. And if you solve this algebraically for alpha, you move the 40 over to the other side. So you'll subtract it. You get a negative 40 radians per second squared. And then you gotta divide by this two as well as the 40 pi radians, which gives me negative 6.37 radians per second squared. Well this thing slowed down to a stop. So this angular acceleration has gotta have the opposite sign to the initial angular velocity. We called this positive 40, that means our alpha's gonna be negative. So recapping, these are the rotational kinematic formulas that relate the rotational kinematic variables. They're only true if the angular acceleration is constant. But when it's constant, you can identify the three known variables and the one unknown that you're trying to find and then use the variable that got left out of the mix to identify which kinematic formula to use, since you would use the formula that does not involve that variable that was neither given nor asked for."
    },
    {
        "Q": "\nAfter the third resonant structure at 6:58,what happens??",
        "A": "The third resonance structure can become the second resonance structure and then the first resonance structure, so on and so forth. The blue pi bond is more than one carbon away from the positive charge so it can t move to the carbocation to form a fourth resonance structure.",
        "video_name": "fpq0eICjuSI",
        "timestamps": [
            418
        ],
        "3min_transcript": "If this electron goes there, then it would look like this. Let me redraw it. I'll draw the resonance structures quickly. You have your hydrogen. You have your electrophile. That's not an electrophile anymore, but you have that E that's now been added. You have that hydrogen. You have a double bond here. Let me draw a little bit neater. You have this hydrogen. You have this hydrogen, this hydrogen and this hydrogen. What I said is, this is stabilized. So an electron here can actually jump over here. So if this electron jumps over here, the double bond is now over there If that goes over there like that, the double bond is now over here. Now this guy lost his electron and it would have a positive charge. And then that is resonance stabilized. It can either go back to this guy, or this electron over here can jump over there. Let me redraw the whole thing over again. This right here, you have the E and the hydrogen. You have a hydrogen here, hydrogen here, hydrogen here, hydrogen here. And normally you don't worry about the hydrogens, but one of the hydrogens is going to be nabbed later on in this mechanism, so I want to draw all the hydrogens just so you know that they are there. But as I said, this is resonance stabilized. If this electron right here jumps over there, then this double bond is now this double bond. And now this guy over here lost an electron, so it would have a positive charge. And again, once you had this double bond up here, this double bond up there is that double bond. So we can go back and forth between these. The electrons are just swishing around the ring. So it's not going to be maybe as great as the situation that we had when we had a nice benzene ring that was completely aromatic. and around the ring, stabilize the structure, but this is still a relatively stable carbocation, because the electrons can move around. You can kind of view it as a positive charge that gets dispersed between this carbon, this carbon, and that carbon over there. As I said, it's still not a great situation. The molecule wants to go back to being aromatic, wants to go to that really stable state. And the way it can go back to that really stable state is somehow an electron can be added to this thing. And the way that an electron can be added to this thing is, if we have some base flying around, and that base nabs this proton, this proton right here that's on the same carbon as where the electrophile is attached. So if this base nabs a proton, so it just nabs the hydrogen nucleus, then that electron that the hydrogen had, that"
    },
    {
        "Q": "\n2:28 Why are Ischemic Strokes are more common then Hemorrhagic strokes?",
        "A": "It is simply more common for blood clots to form and plaques to break off and cause ischemic strokes. The rupture of a vessel is just less likely.",
        "video_name": "xbyfeEW56Nc",
        "timestamps": [
            148
        ],
        "3min_transcript": "You've probably heard of people having a stroke, and you're probably familiar with the notion that it has something to do with the brain, and you'd be right. In particular, it's a rapid loss of brain function because of something strange happening with the blood flow to the brain. And let me show you that in a little bit more detail. And to do that, the 2 major types of strokes. There's the ischemic strokes, and the other type of stroke is hemorrhagic, and these can kind of be sub-categorized, but I won't go into all of the details there. And if I really just define ischemia and hemorrhaging to you, I think you'll have an idea of how these strokes are different and how they interrupt the blood flow to different parts of the brain. You know from the videos on stenosis and ischemia and the videos on heart attacks that ischemia is to certain body tissues. So an ischemic stroke is actually very, very similar to what we saw in a heart attack, except it's not occurring in a coronary blood vessel, it's occurring in a blood vessel in the brain. So let me draw that right over here. Let's say that this is a blood vessel in the brain. And let's say that blood is flowing in that direction (this is an artery). And so you could imagine that maybe there is a big blood clot that forms in some part of the brain. Let me do the blood clot in magenta. This blood clot might form because -- no, that's not magenta -- the blood clot might form because maybe there's a plaque there, maybe the plaque got ruptured, either way, this clot is restricting the flow of blood. And we know that this blood clot -- we can call this a thrombus, or we could say that thrombosis has occurred over here-- either way, the blood flow is restricted, is not going to get its oxygen, and it might die; it might experience infarction. And that's why ischemic strokes are also sometimes called cerebral infarctions. These are all very fancy words, but I think, hopefully, they're becoming a little bit more common in our vocabulary, they keep showing up over and over again. And I also want to be clear: most strokes are actually ischemic strokes. The numbers I looked up, they say, 87% of strokes are ischemic. Now, the other type of way that you could have ischemia in one of these blood vessels, and this is completely analogous to what we saw in the heart, when we had heart attacks, is: you could have thrombosis, or you could also have an embolism. Whenever someone says thrombosis, or a thrombus, or thrombi, they're talking about blood clots. Whenever someone talks about an embolus, or emboli, or embolism,"
    },
    {
        "Q": "7:17- Are the valves in the veins tethered to the walls; or do they not need to be because of the low pressure?\n",
        "A": "The valves are attached to the walls, as in the are not free moving. But there are not any other attachments like there are in the heart i.e. the chordae tendineae stoping the valves from flipping back.",
        "video_name": "iqRTd1NY-pU",
        "timestamps": [
            437
        ],
        "3min_transcript": "because the blood that's coming into our arteries is under, let's not forget, high pressure. So the arterial system we know is a high-pressure system. So this makes perfect sense that the first few arteries, those large arteries and even those medium-sized arteries, are going to be able to deal with the pressure really well. Now, let me draw a little line here just to keep it straight. The small artery and the arteriole, these two are actually sometimes called the muscular arteries. And the reason, again, if you just want to look at the wall of the artery, you'll get the answer. The wall of the artery is actually very muscular. In fact, specifically, it's smooth muscle. So not the kind of muscle you have in your heart or in your biceps, but this is smooth muscle that's in the wall of the artery. And there's lots of it. So again, if you have a little blood vessel like this, if you imagine tons and tons of smooth muscle on the outside-- If those bands decide that they want to contract down, that they want to squeeze down, you're going to get something that looks like a little straw, because those muscles are now tight. They're tightly wound, so you're going to create like a little straw. And this process is called vasoconstriction. Vaso just means blood vessel. And constriction is kind of tightening down. So vasoconstriction, tightening down of the blood vessel. And what that does is it increases resistance. Just like if you're trying to blow through a tiny, tiny little straw, there's a lot of resistance. Well, it's the same idea here. And actually, a lot of that resistance and change in the vasoconstriction is happening at the arteriole level. So that's why they're very special and I want you to remember them. From there, blood is going to go through the capillaries. I didn't actually label them the first time, but let me just write that here. Some, as they call them, capillary beds. And then it's going to go and get collected in the venules and eventually into the veins. And the important thing about the veins-- I'm going to stop right here and just talk about it very briefly-- is that they have these little valves. And these valves make sure that the blood continues to flow in one direction. So one important thing here is the valves. And remember, the other important thing is that they are able to deal with large volumes. So unlike the arterial side where it was all about large pressure, down here with the vein side, we have to think about large volumes. Remember about 2/3 of your blood at any point in time is sitting in some vein or venule somewhere."
    },
    {
        "Q": "At 1:32 it says that the very small arteries are called arterioles. When does an artery turn into an arteriole, and likewise for veins and venules? Does it depend on the size or the structure of the blood vessel or something else?\n",
        "A": "Size does define the classification of artery versus arterioles. When there is two or less medial layers of smooth muscle. This is typically less than 0.1 mm diameters. Structurally, arteries contain more elastic tissue and the arterioles contain more smooth muscle. Functionally, arterioles contribute more to restriction of blood flow and consequently control total peripheral resistance. There is also some gas exchange in arterioles.",
        "video_name": "iqRTd1NY-pU",
        "timestamps": [
            92
        ],
        "3min_transcript": "I want to figure out how blood gets from my heart, which I'm going to draw here, all the way to my toe. And I'm going to draw my foot over here and show you which toe I'm talking about. Let's say this toe right here. Now, to start the journey, it's going to have to go out of the left ventricle and into the largest artery of the body. This is going to be the aorta. And the aorta is very, very wide across. And that's why I say it's a large artery. And from the aorta-- I'm actually not drawing all the branches of the aorta. But from the aorta, it's going to go down into my belly. And it's going to branch towards my left leg and my right leg. So let's say we follow just the left leg. So this artery over here on the top, it's going to get a little bit smaller. And maybe I'd call this a medium-sized artery by this point. This is actually now getting down towards my ankle. Let's say we've gone quite a distance down in my ankle. And let's just follow the branch that goes towards my foot, which is this top one. Let's say this one goes towards my foot, and this is going to be now an even smaller artery. Let's call it small artery. From there, we're actually going to get into what we call arterioles, so it's going to get even tinier. It's going to branch. Now, these are very, very tiny branches coming off my small artery. And let's follow this one right here, and this one is my arteriole. So these are all the different branches I have to go through. And finally, I'm going to get into tiny little branches. I'm going to have to draw them very, very skinny just to convince you that we're getting smaller and smaller. Let me draw three of them. Let's draw four just for fun. And this is actually going to now get towards my little toe cells. to convince you that I actually have gotten there. Let's say one, two over here, and maybe one over here. These are my toes cells. And after the toe cells have kind of taken out whatever they need-- maybe they need glucose or maybe they need some oxygen. Whatever they've taken out, they're also going to put in their waste. So they have, of course, some carbon dioxide waste that we need to drag back. This is now going to dump into what we call a venule. And this venule is going to basically then feed into many, many other venules. Maybe there's a venule down here coming in, and maybe a venule up here coming in maybe from the second toe. And it's going to basically all kind of gather together, and again, to a giant, giant set of veins. Maybe veins are dumping in here now, maybe another vein dumping in here. And these veins are all going to dump into an enormous vein that we call the inferior vena cava."
    },
    {
        "Q": "At 9:04, where does the joules sign go?\n",
        "A": "Sal forgot to write it in, but it should still be there!",
        "video_name": "lsXcKgjg8Hs",
        "timestamps": [
            544
        ],
        "3min_transcript": "Once again, that's x grams. They cancel out. So the ice will absorb 333.55 joules as it goes from zero degree ice to zero degree water. Or 333.55x joules. Let me put the x there, that's key. So the total amount of heat that the ice can absorb without going above zero degrees... Because once it's at zero degree water, as you put more heat into it, it's going to start getting warmer again. If the ice gets above zero degrees, there's no way it's going to bring the water down to zero degrees. The water can't get above zero degrees. So how much total heat can our ice absorb? So heat absorbed is equal to the heat it can absorb when it goes from minus 10 to zero degrees ice. And that's 20.5x. Plus the amount of heat we can absorb as we go from zero degree ice to zero degree water. And that's 333.55x. And of course, all of this is joules. So this is the total amount of heat that the ice can absorb without going above zero degrees. Now, how much real energy does it have to absorb? Well it has to absorb all of this 125,340 joules of energy out of the water. Because that's the amount of energy we have to extract from the water to bring it down to zero degrees. So the amount of energy the ice absorbs has to be this 125,340. So that has to be equal to 125,340 joules. We can do a little bit of algebra here. Add these two things. 20.5x plus 333.55x is 354.05x. Yeah, 330 plus 30 is 350. Then you have a 3 with a 0.5 there. 354.05x and that is equal to the amount of energy we take out of the water. You divide both sides. So x is equal to 125,340 divided by 354.05. I'll take out the calculator for this. The calculator tells me 125,340, the amount of energy that has to be absorbed by the ice, divided by 354.05 is equal to 354 grams. Roughly, there's a little bit extra. So actually, just to be careful maybe I'll take 355 grams of ice."
    },
    {
        "Q": "\nAt 1:32, what is spin (in terms of chemistry) ?",
        "A": "If you want to think of it like a ball rotating clockwise or anticlockwise that is more than likely going to be fine for this level of understanding. The major thing it means for chemistry is that each orbital can only have at most 2 electrons, each with opposite spins (ie one spin up and one spin down per orbital)",
        "video_name": "u1eGSL6J6Fo",
        "timestamps": [
            92
        ],
        "3min_transcript": "Let's remind ourselves a little bit of what we already know about orbitals and I've gone over this early on in the regular chemistry playlist. Let's say that this is the nucleus of our atom, super small, and around that we have our first orbital, the 1s orbital. The 1s orbital, you can kind of just view it as a cloud around the nucleus. So you have your 1s orbital and it can fit two electrons, so the first electron will go into the 1s orbital and then the second electron will also go into the 1s orbital. For example, hydrogen has only one electron, so it would go into 1s. Helium has one more, so that will also go into the 1s orbital. After that is filled, then you move onto the 2s orbital. The 2s orbital, you can view it as a shell around the 1s orbital, and all of these, you can't really view it in our conventional way of thinking. You can kind of view it as a probability cloud of where you might find the electrons. kind of a shell cloud around the 1s orbital. So imagine that it's kind of a fuzzy shell around the 1s orbital, so it's around the 1s orbital, and your next electron will go there. Then the fourth electron will also go there, and I drew these arrows upward and downward because the first electron that goes into the 1s orbital has one spin and then the next electron to go into 1s orbital will have the opposite spin, and so they keep pairing up in that way. They have opposite spins. Now, if we keep adding electrons, now we move to the 2p orbitals. Actually, you can view it as there are three 2p orbitals and each of them can hold two electrons, so it can hold a total of six electrons in the 2p orbitals. Let me draw them for you just so you can visualize it. So if we were to label our axis here, so think in three So imagine that that right there is the x-axis. Let me do this in different colors. Let's say that this right here is our y-axis and then we have a z-axis. I'll do that in blue. Let's say we have a z-axis just like that. You actually have a p orbital that goes along each of those axes. So you could have your two-- let me do it in the same color. So you have your 2p sub x orbital, and so what that'll look like is a dumbbell shape that's going in the x-direction. So let me try my best attempt at drawing this. It's a dumbbell shape that goes in the x-direction, in kind of both directions, and it's actually symmetric. I'm drawing this end bigger than that end so it looks like it's coming out at you a little bit, but let me draw it a little bit better than that. I can do a better job."
    },
    {
        "Q": "At 12:08 he says that the Carbon has 1s and 3sp3 orbital, so my question is that ; that each of those 3 sp3 orbitals would contains a single electron ? Am i right?\n",
        "A": "Instead of having one s and three p orbitals, the carbon atom has four sp\u00c2\u00b3 orbitals. Each of those four sp\u00c2\u00b3 orbitals contains a single electron.",
        "video_name": "u1eGSL6J6Fo",
        "timestamps": [
            728
        ],
        "3min_transcript": "just-- the first one doesn't look like it's just in the s orbital and then the p and y and z for the other three. They all look like they're a little bit in the s and a little bit in the p orbitals. Let me make that clear. So instead of this being a 2s, what it really looks like for carbon is that this looks like a 2sp3 orbital. This looks like a 2sp3 orbital, that looks like a 2sp3 orbital, that looks like a 2sp3 orbital. They all look like they're kind of in the same orbital. This special type of-- it sounds very fancy. This sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital. It has a 25% s nature and a 75% p nature. You can imagine it as being a mixture of these four things. That's the behavior that carbon has. So when you mix them all, instead of having an s cross-section, an s orbital looks like that and the p orbital looks something like that in cross-section. So this is a an s and that is a p. When they get mixed up, the orbital looks like this. An sp3 orbital looks something like this. This is a hybridized sp3 orbital. Hybrid just means a combination of two things. A hybrid car is a combination of gas and electric. A hybridized orbital is a combination of s and p. Hybridized sp3 orbitals are the orbitals when carbon bonds with things like hydrogen or really when it bonds with anything. So if you looked at a molecule of methane, and people talk about sp3 hybridized orbitals, all they're saying is that you have a carbon in the center. And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. Let's say this is the big lobe that is kind of pointing near us, and then it has a small lobe in the back. Then you have another one that has a big lobe like that and a small lobe in the back. Then you have one that's going back behind the page, so let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up, and it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them is behind like that and it's pointing straight up, So a three-legged stool with something-- it's kind of like a tripod, I guess is the best way to think about it."
    },
    {
        "Q": "At 5:57 when you is not dissolved, do you mean it is not dissolved in water or in gas.\n",
        "A": "In this particular reaction, all the other components are gases, so it means that the C is not a gas. In other words, the carbon is not dissolved in the gaseous phase.",
        "video_name": "TsXlTWgyItw",
        "timestamps": [
            357
        ],
        "3min_transcript": "rate forward, is going to be dependent on some forward constant times just the concentration of the boron trifluoride. The water's everywhere, so you don't have to multiply it times the concentration of water, whatever that means, because the water's everywhere. So the denominator here, you do not put the solvent. So the correct answer for this one is you only put whatever is actually dissolved in the solution. Because frankly, the concentration doesn't actually makes sense for everything else, and if you think about it from the probability point of view, that also makes sense, because there's always water around. If you said, OK, what's the probability of finding water at any small volume of fluid so you could just multiply it by a 1 there, but that doesn't make a difference. Now, what about the following reaction? Any equilibrium where you have different states of matter is called a heterogeneous equilibrium. So let's say I have H2O in the gaseous state and that's essentially steam -- so it's not going to be the solvent this time-- plus carbon in the solid state. And let's say that that's an equilibrium with hydrogen in the gas state plus carbon dioxide in the gaseous state. This is a heterogeneous equilibrium because you have things in the gaseous and the solid state. And solid state, by definition, it can't be dissolved either into the gas or into the -- when we talk about solutions, we talked about colloids and suspensions and mixtures before, but we're talking about solutions. By definition, if this is in the solid state, it's not dissolved. If this was dissolved, we would write an aq here. So if you talk about the forward reaction, what's the forward reaction going to be dependent on? So the rate forward, well, the solid, there's a big block of carbon sitting there. There's a big cube of carbon there, and there's steam, there's water gas all around it. So if you pick any volume, especially if you pick some volume near the boundary of the carbon, you're always going to have carbon around. It's just what matters is the concentration of the water gas. That's what's going to drive the forward rate, so the forward rate is going to be dependent on some constant times the concentration of the water gas. And, of course, the backwards rate, so you need to get some h2, some molecules of -- let me draw it like that, because it has 2 hydrogen molecules plus a carbon dioxide,"
    },
    {
        "Q": "At 7:12 Sal mentioned about buoyancy effect. What do you mean by that ?\n",
        "A": "buoyancy is upthrust acting on a body",
        "video_name": "R5CRZONOHCU",
        "timestamps": [
            432
        ],
        "3min_transcript": "And then I want to multiply that times the mass of Earth, which is right over here. That is 5.9722 times 10 to the 24th. So times 10 to the 24th power. And we want to divide that by the radius of Earth squared. So divided by the radius of Earth is-- so this is in kilometers. And I just want to make sure that everything is the same units. So 6,371 kilometers-- actually, let me scroll over. Well, you can't see the kilometers right now. But this is kilometers. It is the same thing as 6,371,000 meters. If you just multiply this by 1,000. Or you could even write this as 6.371. 6.371 times 10 to the sixth meters. That's the radius of the Earth. The distance between the center of mass of Earth and the center of mass of this object, which is sitting at the surface of the Earth. And so let's get our drum roll. And we get 9.8. And if we round, we actually get something a little bit higher than what the textbooks give us. We get 9.82. Let's just round. So we get 9.82-- 9.82 meters per second squared. And so you might say, well, what's going on here? Why do we have this discrepancy between what the universal law of gravitation gives us and what the average measured acceleration due to the force of gravity at the surface of the Earth. And the discrepancy here, the discrepancy between these two numbers, is really because Earth is not a uniform sphere of uniform density. And that's what we have to assume over here when we use the universal law of gravitation. It's actually a little bit flatter than a perfect sphere. The different layers of the Earth have different densities. You have all sorts of different interactions. And then you also, if you measure effective gravity, there's also a little bit of a buoyancy effect from the air. Very, very, very, very negligible, I don't know if it would have been enough to change this. But there's other minor, minor effects, irregularities. Earth is not a perfect sphere. It is not of uniform density. And that's what accounts for the bulk of this. Now, with that out of the way, what I'm curious about is what is the acceleration due to gravity if we go up 400 kilometers? So now, the main difference here, g will stay the same. The mass of Earth will stay the same, but the radius is now going to be different. Because now we're placing the center of mass of our object-- whether it's a space station or someone sitting in the space station, they're going to be 400 kilometers higher. And I'm going to exaggerate what 400 kilometers looks like. This is not drawn to scale. But now the radius is going to be the radius of the Earth plus 400 kilometers."
    },
    {
        "Q": "\n8:50 I did not understand the equation. It sounds logical that if we have parallel flow the resistance is lower, but how do we get to the equation?",
        "A": "We have less resistance when there is less flow. When we double the amount of identical paths, we halve the amount of flow through each individual path. By doubling the amount of paths, we halve the resistance. Opening more doors makes it easier to exit the theater. Even if only one person gets out through the new door, everyone else experiences less competition for all the other doors.",
        "video_name": "E-q9JpkGc-8",
        "timestamps": [
            530
        ],
        "3min_transcript": "Here it stays at 2. But here in the middle, it goes from 2 to 32 because it's 16 times greater. So you end up increasing the resistance in the middle section by a lot. So let me just write that out for you. So 2 times 16 gets us to 32. So here the resistance is 32. And so if I wanted to calculate the total resistance, I'd get something like this-- 32 plus 2 plus 2 is 36. So I actually went from 6 to 36 when this blood clot came and clogged up part of that vessel. So just keep that in mind. We'll talk about that a little bit more, but I just wanted to use this example and also kind of cement the idea of what you do with resistance in a series. Let's contrast that to a different situation. And this is when you have resistance in parallel. through all of my vessels, I could also do something like this-- I could say, well, let's say, I have three vessels again. And this time, I'm going to change the length and the radius. And let's say this one's really big. And the resistance here, let's say, is 5, here is 10, and here is 6. So you've got three different resistances. And the blood now can choose to go through any one of these paths. It doesn't have to go through all three. So how do I figure out now what the total resistance is? So what is the total resistance? Well, the total resistance this time is going to be 1 over 1 R1, plus 1 over R2, plus 1 over R3. And you can go on and on just as before. So let's just put that there, that there, and that there. And I can figure this out pretty easily. So I can say 1 over 1 over 6 plus 1 over 10 plus 1/5. And the common denominator there is 30. So I could say 5/30. This is 3/30, and this would be 6/30. And adding that up together, I get 1 over 14/30 or 30 over 14, which is 2 and let's say 0.1. So 2.1. So the total resistance here is 2.1. Putting all three of these together is pretty interesting. And I want you to realize that the resistance in total is actually less than any component part."
    },
    {
        "Q": "\nAt 2:53 why we only use centripetal acceleration on the left side of the equation? Shouldn't it be centripetal acceleration plus acceleration due to gravity? When the ball is in the air isn't it always accelerating downwards due to Earth's gravity?",
        "A": "I believe, it is because the direction chosen in the free body/force diagram is towards the centre of the circular path. The effect of gravity is included in the weight or gravity force( m x g ).",
        "video_name": "2lcaBPLLoLo",
        "timestamps": [
            173
        ],
        "3min_transcript": "going four meters per second? And if it's a force you want to find, the first step always is to draw a quality force diagram. So let's do that here. Let's ask what forces are on this yo-yo. Well, if we're near Earth and we're assuming we're going to be near the surface of the Earth playing with our yo-yo, there's gonna be a force of gravity and that force of gravity is gonna point straight downward. So the magnitude of that force of gravity is gonna be m times g, where g is positive 9.8. g represents the magnitude of the acceleration due to gravity, and this expression here represents the magnitude of the force of gravity. But there's another force. The string is tied to the mass, so this string can pull on the mass. Strings pull, they exert a force of tension. Which way does that tension go? A lot of people want to draw that tension going upward, and that's not good. Ropes can't push. If you don't believe me, go get a rope, try to push on something. You'll realize, oh yeah, it can't push, but it can pull. So that's what this rope's gonna do. This rope's gonna pull. How much? I don't know. This is gonna be the force of tension right here, and we'll label it with a capital T. We could have used F with the sub T. There's different ways to label the tension, but no matter how you label it, that tension points in towards the center of the circle 'cause this rope is pulling on the mass. So, after you draw a force diagram, if you want to find a force, typically, you're just gonna use Newton's second law. And we're gonna use this formula as always in one dimension at a time so vertically, horizontally, centripetally, one dimension at a time to make the calculations as simple as possible. And since we have a centripetal motion problem, we have an object going in a circle, and we want to find one of those forces that are directed into the circle, we're gonna use Newton's second law for the centripetal direction. So we'll use centripetal acceleration here and net force centripetally here. So in other words, we're gonna write down that the centripetal acceleration is gonna be equal to the net centripetal force exerted on the mass that's going around in that circle. So because we chose the centripetal direction, the centripetal acceleration with the formula for centripetal acceleration. The centripetal acceleration's always equivalent to v squared over r, the speed of the object squared divided by the radius of the circle that the object is traveling in. So we set that equal to the net centripetal force over the mass, and the trickiest part here, the part where the failure's probably gonna happen is trying to figure out what do you plug in for the centripetal force. And now we gotta decide what is acting as our centripetal force and plug those in here with the correct signs. So, let's just see what forces we have on our object. There's a force of tension and a force of gravity. So, when you go try to figure out what to plug in here, people start thinking, they start looking all over. No, you drew your force diagram. Look right there. Our force diagram holds all the information about all the forces that we've got as long as we drew it well. And we did draw it well. We included all the forces, so we'll just go one by one. Should we include, should we even include the force of gravity in this centripetal force calculation?"
    },
    {
        "Q": "\nin 1:42 the example, does that actually work?",
        "A": "Try it! Put a pot of water on a stove and only heat one side. See what happens!",
        "video_name": "f8GK2oEN-uI",
        "timestamps": [
            102
        ],
        "3min_transcript": "What I want to do in this video is talk a little bit about maybe why the plates are actually moving in the first place. And nothing I'm talking about in this video has been definitively proved. This is just kind of the current thinking, the leading thinking, on why plates are actually moving. Although we haven't seen the definitive evidence yet, and it's probably a combination of a bunch of things. Now, before we even talk about plates, let's just talk about convection. And you might already be familiar with the term, but just in case you're not, let's do a little bit of review of convection. So let's say I have a pot over here. So that is my pot. And it contains some water. So I have water in my pot. And let's say I only heat one end of the pot. So I put a flame right over at that end of the pot. So what's going to happen? Well, the water that's right over the flame is going to be warmed up more than any of the other water. So this water is going to get warm. But when it gets warm, it also becomes less dense. the molecules are vibrating more. They have more kinetic energy. They're going to a bounce further distances away from each other. They will become less dense. And if you have something that's less dense, and it's surrounded by things that are more dense, and we're dealing in kind of a fluid state right here, that warm, less dense water is going to move upwards. It's going to move upwards. Well, when it moves upwards something has to replace it. So you're going to have cooler water from this side of the container kind of replacing where that water was. Now, this water, as it rises, what's going to happen to it? Well, it's going to cool down. It's going to get further from the flame. It's going to mix with maybe some of the other water, or transfer some of its kinetic energy to the neighboring water. So it'll cool down. But once it cools down, what's it going to want to do? Remember, in general, the closer you are to the flame. is going to be warmer. So all of this stuff is going to be warmer, and all of this stuff up here, like the coldest water, is always going to be furthest from the flame. And so the coldest water is going to be over here. But remember, the coldest water is also the densest water. So this water over here is dense. And so it will sink. It's denser than the water around it. And it also helps replace the water that's going here to get warmed up again. And so what you do is you have this cycle here. Warm water rises, moves over to the right down here, and then goes back down as it cools down, and it's dense, and then it gets warmed up again. And so this process, essentially what it's doing is it's transferring the heat. It's allowing the heat to be transferred from this one spot throughout the fluid. And so we call this process, this is convection. Now, the reason why we think the plates are moving is because we think that there are similar types of convection"
    },
    {
        "Q": "\nYou know, that is something I wondered. At 4:54, Sal says \"degrees Kelvin\", even though it's supposed to be just plain \"Kelvin\". I know that. I'm cool with that. But why is that? Why are the other temperature readings in degrees but Kelvin is not?",
        "A": "Because the other scales don t actually measure a physical quantity, just relative warmth. 0 degrees F is not zero somethings . Same for Celsius. But zero kelvin represents zero thermal energy, and 100 K is twice as hot as 50 K.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            294
        ],
        "3min_transcript": "So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here."
    },
    {
        "Q": "At 3:24 Sal says the lowest possible temperature in C is -273.15 Degrees C, and I was wondering, what would happen if we could quickly freeze a person to -273.15 Degrees C? Then if we could return their temperature back to a normal temperature, would they be ok?\nI know it sounds kind of odd, but I was considering forensic science, and just wondering....\nThanks!- MP\n",
        "A": "Technically speaking, you cannot actually reach absolute zero, because the would be a state of no thermal energy at all. But we can get very close to it. No, the person would die.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            204
        ],
        "3min_transcript": "So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is, So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale"
    },
    {
        "Q": "At 6:12 I understand why you would pick a certain R value, but how are they derived? On my chemistry worksheet they aren't given or anything.\n",
        "A": "R is nothing more than the Boltzmann constant multiplied by Avogadro s constant with a few unit conversion factors thrown in to adjust for whatever units you are using.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            372
        ],
        "3min_transcript": "and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here. Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out."
    },
    {
        "Q": "At 6:11, where did Sal come up with the numbers for R?\n",
        "A": "R is the universal or ideal gas constant in PV = nRT. These problems are called ideal gas equations because it is assumed that the gas behaves in an ideal manner, which would allow us to use the constant that Sal shows. It is a constant that makes the equation work... you ll either have a table or have to memorize this number, unfortunately.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            371
        ],
        "3min_transcript": "and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here. Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out."
    },
    {
        "Q": "At 1:37, how big is one atm?\n",
        "A": "One atm is the amount of air pressure that is normally on you if you are near sea level.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            97
        ],
        "3min_transcript": "In the last video we hopefully learned the intuition behind the ideal gas equation, that pressure times volume is equal to the number of molecules we have times some constant times the temperature. And that's all nice and it hopefully it makes sense to you how all of these fit together. That pressure should be inverse to volume and that's why you're multiplying both sides by each other. You could take volume and put it on this side of the equation. Or that pressure should be proportional to the number of particles and the temperature. But now let's apply it and actually do some problems. Because just knowing this isn't good enough. So let's say that I have a two liter container, or let's say a two liter balloon, containing hydrogen gas. So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is,"
    },
    {
        "Q": "\nThis might be a silly question but at 7:42, how come the (.082 L . atm/mole . K) becomes 1/mole??\n\nCheers",
        "A": "Just look at the units. If you ve used consistent units then RT/PV has the units of 1/mol because the pressure and volume units have cancelled out. Or, a more useful way of looking at it is : PV = nRT divide by RT PV/RT = n Thus, PV/RT has units of mol, with all the other units cancelling out.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            462
        ],
        "3min_transcript": "Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out. you can treat units like numbers. So if you divide both sides of this equation by atmospheres, the atmospheres cancel out. Divide both sides of this equation by liters, liters cancel out. You have a Kelvin in the numerator, Kelvin in the denominator, that cancels out. And so we have 2 times 2 is equal to n times 0.082 times 303. And then we have just a per mole and a 1 over the mole. So to solve for n, or the number of moles, what we do is we divide both sides of this equation by all of this stuff. So we get 2 times 2 is 4. 4 divided by 0.082 divided by 303. dividing both sides by it. And when you divide by a per mole, if you put a 1 over a mole here, that's the same thing as multiplying by a mole. So it's good, the units all worked out. We're getting n in terms of moles. And so we just have to get the calculator out and figure out how many moles we're dealing with. So we have 4 divided by 0.082 divided by 303 is equal to 0.16. If we wanted to go more digits, 0.161, but we'll just round. So this is equal to 0.16 moles of H2. I am telling you actually here, the exact number of hydrogen molecules. But if you wanted a number, you'd just multiply this times 6.02 times 10 to the 23 and then you would have a number in kind of the traditional sense. And of course, if you wanted to know"
    },
    {
        "Q": "at 4:04 why does it say peed in orange?\n",
        "A": "I think it reads PE = 0. That is, the potential energy is equal to zero.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            244
        ],
        "3min_transcript": "It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it."
    },
    {
        "Q": "\nAt 4:04, Sal said you could solve this problem using kinematic formulas but how could you? Don't you only know distance and acceleration?",
        "A": "s = 1/2*a*t^2",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            244
        ],
        "3min_transcript": "It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it."
    },
    {
        "Q": "At 2:15 how does the object have no energy?I thought that everything always has energy\n",
        "A": "everything has some combination of potential energy, kinetic energy, and internal energy...the internal energy of the block (related to the material at the atomic level) doesn t change during this example so it got left out of the explanation...but, yes, everything does always have some energy in the real world",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            135
        ],
        "3min_transcript": "Welcome back. At the end of the last video, I left you with a bit of a question. We had a situation where we had a 1 kilogram object. This is the 1 kilogram object, which I've drawn neater in this video. That is 1 kilogram. And we're on earth, and I need to mention that because gravity is different from planet to planet. But as I mentioned, I'm holding it. Let's say I'm holding it 10 meters above the ground. So this distance or this height is 10 meters. And we're assuming the acceleration of gravity, which we also write as just g, let's assume it's just 10 meters per second squared just for the simplicity of the math instead of the 9.8. So what we learned in the last video is that the potential energy in this situation, the potential energy, which equals m times g times h is equal to the mass is 1 kilogram times the acceleration of gravity, which is 10 I'm not going to write the units down just to save space, although you should do this when you do it on your test. And then the height is 10 meters. And the units, if you work them all out, it's in newton meters or joules and so it's equal to 100 joules. That's the potential energy when I'm holding it up there. And I asked you, well when I let go, what happens? Well the block obviously will start falling. And not only falling, it will start accelerating to the ground at 10 meters per second squared roughly. And right before it hits the ground-- let me draw that in brown for ground-- right before the object hits the ground or actually right when it hits the ground, what will be the potential energy of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of gravity stay the same, but the height is 0. So they're all multiplied by each other. So down here, the potential energy is going to be equal to 0. And I told you in the last video that we have the law of conservation of energy. It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot."
    },
    {
        "Q": "\nAt 9:00- 9:30 Sal talks about how the velocity can be determined, but wouldn't that be affected by the shape of the slope? Which we haven't taken into account? An upslope at that particular height would cause a different velocity than a downslope surely?\nPlease clarify",
        "A": "Your argument is totally correct ,but according to the question we have to only find the velocity of the body on the surface and not on the iceberg.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            540,
            570
        ],
        "3min_transcript": "But what else can this do for me? And this is where it's really cool. Not only can I figure out the velocity when all of the potential energy has disappeared, but I can figure out the velocity of any point-- and this is fascinating-- along this slide. So let's say when the box is sliding down here, so let's say the box is at this point. It changes colors too as it falls. So this is the 1 kilogram box, right? It falls and it slides down here. And let's say at this point it's height above the ground is 5 meters. So what's its potential energy here? So let's just write something. All of the energy is conserved, right? So the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. I'm just saying energy is conserved here. Well the potential energy is 100 and the kinetic energy is 0 because it's stationary. I haven't dropped it. I haven't let go of it yet. It's just stationary. So the initial energy is going to be equal to 100 joules. That's cause this is 0 and this is 100. So the initial energy is 100 joules. At this point right here, what's the potential energy? Well we're 5 meters up, so mass times Mass is 1, times gravity, 10 meters per second squared. Times height, times 5. So it's 50 joules. That's our potential energy at this point. And then we must have some kinetic energy with the velocity going roughly in that direction. Plus our kinetic energy at this point. And we know that no energy was destroyed. It's just converted. So we know the total energy still has to be 100 joules. So essentially what happened, and if we solve for this-- the kinetic energy is now also going to be equal to 50 joules. Halfway down, essentially half of the potential energy got converted to kinetic energy. And we can use this information that the kinetic energy is 50 joules to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared is equal to 100. The velocity is 10 meters per second along this crazy, icy slide. And that is something that I would have challenged you to solve using traditional kinematics formulas, especially considering that we don't know really much about the surface of this slide. And even if we did, that would have been a million times harder than just using the law of conservation of energy and realizing that at this point, half the potential energy is now kinetic energy and it's going along the direction of the slide."
    },
    {
        "Q": "At 6:15 , all the potential energy gets converted to K.E.\nLet us say it fell straight down. At the last instant all the P.E. would be converted into K.E.\nBut then it hits the ground and will come to a rest. Then its height = 0 and so is its velocity. So both P.E. and K.E. are zero, then where did that energy go?\n",
        "A": "That energy went into impacting the ground - making noise, deforming it, heating it.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            375
        ],
        "3min_transcript": "solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down."
    },
    {
        "Q": "\nAt 6:10 when the object is sliding , some of the energy will be converted to heat energy due to fiction right?? So, all the energy will not be converted to kinetic energy right?",
        "A": "That s correct",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            370
        ],
        "3min_transcript": "solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down."
    },
    {
        "Q": "\nIn example at 9:00 we evaluate velocity knowing PE, height and mass. But what about a slope? If you would drown the slope differently, let's say an infinite straight line at height 5 m. , then an object would have stopped after some amount of time due to friction or air resistance. Does Sal assumes that there are no other forces but gravity?",
        "A": "yes, he s ignoring air resistance and friction so that you can focus on the concepts of PE and KE.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            540
        ],
        "3min_transcript": "But what else can this do for me? And this is where it's really cool. Not only can I figure out the velocity when all of the potential energy has disappeared, but I can figure out the velocity of any point-- and this is fascinating-- along this slide. So let's say when the box is sliding down here, so let's say the box is at this point. It changes colors too as it falls. So this is the 1 kilogram box, right? It falls and it slides down here. And let's say at this point it's height above the ground is 5 meters. So what's its potential energy here? So let's just write something. All of the energy is conserved, right? So the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. I'm just saying energy is conserved here. Well the potential energy is 100 and the kinetic energy is 0 because it's stationary. I haven't dropped it. I haven't let go of it yet. It's just stationary. So the initial energy is going to be equal to 100 joules. That's cause this is 0 and this is 100. So the initial energy is 100 joules. At this point right here, what's the potential energy? Well we're 5 meters up, so mass times Mass is 1, times gravity, 10 meters per second squared. Times height, times 5. So it's 50 joules. That's our potential energy at this point. And then we must have some kinetic energy with the velocity going roughly in that direction. Plus our kinetic energy at this point. And we know that no energy was destroyed. It's just converted. So we know the total energy still has to be 100 joules. So essentially what happened, and if we solve for this-- the kinetic energy is now also going to be equal to 50 joules. Halfway down, essentially half of the potential energy got converted to kinetic energy. And we can use this information that the kinetic energy is 50 joules to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared is equal to 100. The velocity is 10 meters per second along this crazy, icy slide. And that is something that I would have challenged you to solve using traditional kinematics formulas, especially considering that we don't know really much about the surface of this slide. And even if we did, that would have been a million times harder than just using the law of conservation of energy and realizing that at this point, half the potential energy is now kinetic energy and it's going along the direction of the slide."
    },
    {
        "Q": "\nAround 9:05 Sal says that glycolysis is aerobic but , my understanding is that a cell without enough oxygen will go through fermentation instead of glycolysis. So, how is glycolysis still anaerobic?",
        "A": "fermentation happens instead of the TCA cycle and oxidative phosphorylation and thus after glycolysis. Glycolysis does not require oxygen, and as such it is considered anaerobic; however, unlike fermentation it still occurs in the presence of oxygen.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            545
        ],
        "3min_transcript": "that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we You might be familiar with the idea of aerobic exercise. The whole idea of aerobic exercise is to make you breathe hard because you need a lot of oxygen to do aerobic exercise. So anaerobic means you don't need oxygen. Aerobic means it needs oxygen. Anaerobic means the opposite. You don't need oxygen. So, glycolysis anaerobic. And it produces two ATPs net. And then you go to the Krebs cycle, there's a little bit of setup involved here. And we'll do the detail of that in the future. But then you move over to the Krebs cycle, which is aerobic. It is aerobic. It requires oxygen to be around. And then this produces two ATPs. And then this is the part that, frankly, when I first learned it, confused me a lot. But I'll just write it in order the way it's Then you have something called-- we're using the same colors too much-- you have something called the electron"
    },
    {
        "Q": "At 13:05, it is said that each NADH produces 3 ATPs in the electron transport chain. So, glycolysis and Kreb's cycle give us 10 NADHs, which amounts to 10 \u00c3\u0097 3 = 30 ATPs in the electron transport chain. However, the electron transport chain produces 34 ATPs as said. May anyone please tell me from where the other 4 ATPs come from (All mechanisms proceed ideally, of course)? Thank you for any help.\n",
        "A": "The Krebs cycle also produces 2 FADH2, which produce 1.5-2 ATP each during the electron transport chain = ~4 more ATP",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            785
        ],
        "3min_transcript": "That's called alcohol fermentation. And we, as human beings, I guess fortunately or unfortunately, our muscles do not directly produce alcohol. They produce lactic acid. So we do lactic acid fermentation. Let me write that down. Lactic acid. That's humans and probably other mammals. But other things like yeast will do alcohol fermentation. So this is when you don't have oxygen. It's actually this lactic acid that if I were to sprint really hard and not be able to get enough oxygen, that my muscles start to ache because this lactic acid starts to build up. But that's just a side thing. If we have oxygen we can move to the Krebs cycle, get our two ATPs, and then go on to the electron transport chain and produce 34 ATPs, which is really the bulk of what happens in respiration. Now I said this as an aside, that to some degree this isn't fair. producing these other molecules. They're not producing them entirely, but what they're doing is, they're taking-- and I know this gets complicated here, but I think over the course of the next few videos we'll get an intuition for it-- in these two parts of the reaction, glycolysis and the Krebs cycle, we're constantly taking NAD-- I'll write it as NAD plus-- and we're adding hydrogens to it to form NADH. And this actually happens for one molecule of glucose, this happens to 10 NADs. Or 10 NAD plusses to become NADHs. And those are actually what drive the electron transport chain. And I'll talk a lot more about it and kind of how that happens and why is energy being derived and how is this an oxidative reaction and all of that. And what's getting oxidized and what's being reduced. But I just wanted to give due credit. These guys aren't just producing two ATPs in each of They're also producing, actually combined, 10 NADHs, which each produce three ATPs in an ideal situation, the And they're also doing it to this other molecule, FAD, which is very similar. But they're producing FADH. Now I know all of this is very complicated. I'll make videos on this in the future. But the important thing to remember is cellular respiration, all it is is taking glucose and kind of repackaging the energy in glucose, and repackaging it in the form of, your textbooks will tell you, 38 ATPs. If you're doing an exam, that's a good number to write. It tends to, in reality be a smaller number. It's also going to produce heat. Actually most of it is going to be heat. But 38 ATPs, and it does it through three stages. The first stage is glycolysis, where you're just literally splitting the glucose into two. You're generating some ATPs. But the more important thing is, you're generating some NADHs that are going to be used later in the electron transport chain. Then those byproducts are split even more in the Krebs"
    },
    {
        "Q": "\nAt 13:11, Sal says that FAD is being reduced to FADH during these processes. Isn't the reduced form of FAD actually FADH2?",
        "A": "You are correct. I think if you go to that spot in the video, a small box should come up with the correction on the bottom right corner.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            791
        ],
        "3min_transcript": "That's called alcohol fermentation. And we, as human beings, I guess fortunately or unfortunately, our muscles do not directly produce alcohol. They produce lactic acid. So we do lactic acid fermentation. Let me write that down. Lactic acid. That's humans and probably other mammals. But other things like yeast will do alcohol fermentation. So this is when you don't have oxygen. It's actually this lactic acid that if I were to sprint really hard and not be able to get enough oxygen, that my muscles start to ache because this lactic acid starts to build up. But that's just a side thing. If we have oxygen we can move to the Krebs cycle, get our two ATPs, and then go on to the electron transport chain and produce 34 ATPs, which is really the bulk of what happens in respiration. Now I said this as an aside, that to some degree this isn't fair. producing these other molecules. They're not producing them entirely, but what they're doing is, they're taking-- and I know this gets complicated here, but I think over the course of the next few videos we'll get an intuition for it-- in these two parts of the reaction, glycolysis and the Krebs cycle, we're constantly taking NAD-- I'll write it as NAD plus-- and we're adding hydrogens to it to form NADH. And this actually happens for one molecule of glucose, this happens to 10 NADs. Or 10 NAD plusses to become NADHs. And those are actually what drive the electron transport chain. And I'll talk a lot more about it and kind of how that happens and why is energy being derived and how is this an oxidative reaction and all of that. And what's getting oxidized and what's being reduced. But I just wanted to give due credit. These guys aren't just producing two ATPs in each of They're also producing, actually combined, 10 NADHs, which each produce three ATPs in an ideal situation, the And they're also doing it to this other molecule, FAD, which is very similar. But they're producing FADH. Now I know all of this is very complicated. I'll make videos on this in the future. But the important thing to remember is cellular respiration, all it is is taking glucose and kind of repackaging the energy in glucose, and repackaging it in the form of, your textbooks will tell you, 38 ATPs. If you're doing an exam, that's a good number to write. It tends to, in reality be a smaller number. It's also going to produce heat. Actually most of it is going to be heat. But 38 ATPs, and it does it through three stages. The first stage is glycolysis, where you're just literally splitting the glucose into two. You're generating some ATPs. But the more important thing is, you're generating some NADHs that are going to be used later in the electron transport chain. Then those byproducts are split even more in the Krebs"
    },
    {
        "Q": "at 7:35 it needs 2 ATPs for cellular respiration?\n",
        "A": "2ATP is needed to start cellular repiration (in glycollosis). In short, the 2ATP breaks into 2ADP and the remaining phosphate bonds to the gluclose molecule (and the other remianing phophate later in the cycle). This gives the energy required for glycollosis to occur and produce energy which results in a gain in at (38 in the full cycle for one glucose)",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            455
        ],
        "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we"
    },
    {
        "Q": "I don't understand what Sal means by \"on a net basis\" at 7:37. Someone please explain this concept to me...\n",
        "A": "He summarized the glycolyse reaction - in some stage of this reaction you need to use 2ATPs [-2] and in some you produce 4ATPs [+4]. And, summing up, you generate ( produce ) 2 ATPs [-2 +4 = +2].",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            457
        ],
        "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we"
    },
    {
        "Q": "\nAt 7:34, what does it mean by net ATP?",
        "A": "Some ATP are used in the production of ATP. Net is the produced - the used",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            454
        ],
        "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we"
    },
    {
        "Q": "\nIn my textbook, it is said that in cellular respiration, energy is produced in the form of ATP. But, in this video, around 3:19, Sal said that energy is used to produce ATP. Which one is correct ? I am confused. :(",
        "A": "ATP is the energy currency in the cell and can be thought of as chemical form of energy. One way to think about is to think about it like a rechargeable battery. ATP can be used, giving up its stored energy in the phosphate bond and then recharged with the addition of energy to reform the phosphate bond.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            199
        ],
        "3min_transcript": "mole of glucose, if you had six moles of molecular oxygen running around the cell, then-- and this is kind of a gross simplification for cellular respiration. I think you're going to appreciate over the course of the next few videos, that one can get as involved into this mechanism as possible. But I think it's nice to get the big picture. But if you give me some glucose, if you have one mole of glucose and six moles of oxygen, through the process of cellular respiration-- and so I'm just writing it as kind of a big black box right now, let me pick a nice color. So this is cellular respiration. Which we'll see is quite involved. But I guess anything can be, if you want to be particular enough about it. Through cellular respiration we're going to produce six moles of carbon dioxide. Six moles of water. going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly. So if I were to break down this energy portion of cellular respiration right there, some of it would just be heat. You know, it just warms up the cell. And then some of it is used-- and this is what the textbooks will tell you. The textbooks will say it produces 38 ATPs. It can be more readily used by cells to contract muscles or to generate nerve impulses or do whatever else-- grow, or divide, or whatever else the cell might need. So really, cellular respiration, to say it produces energy, a little disingenuous. It's really the process of taking glucose and producing ATPs, with maybe heat as a byproduct. But it's probably nice to have that heat around. We need to be reasonably warm in order for our cells to operate correctly. So the whole point is really to go from glucose, from one mole of glucose-- and the textbooks will tell you-- to 38 ATPs."
    },
    {
        "Q": "At 8:54, he said that the structure on right is not going to contribute to the hybrid as much, what does he mean by that? If the structure is supposed to be a hybrid of both, does that mean the resonance structure is closer to the structure on the left?\n",
        "A": "It means the true structure of the molecule is closer to the left one than the right. The bond lengths in acetone are consistent with a C-O double bond which gives a bit of weight to what he s saying.",
        "video_name": "UHZHkZ6_H5o",
        "timestamps": [
            534
        ],
        "3min_transcript": "so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become."
    },
    {
        "Q": "at 9:40 , why is the resonance structure on the right a minor contributor compared to the one on the left when the right one has both negative and positive charges assigned corresponding to the electro negativity of the atoms in the compound?\n",
        "A": "Because in the right structure those full formal charges have been introduced where the left structure has none Less formal charges are generally better structures",
        "video_name": "UHZHkZ6_H5o",
        "timestamps": [
            580
        ],
        "3min_transcript": "so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become."
    },
    {
        "Q": "\nAt 4:00 min when sal names the 7,7, dibromo oct - 5 - yn - 4 -ol isn't there a chiral center on the number 4 carbon?",
        "A": "You re right, there is a chiral center on C4. Since it s drawn flat (no wedges or dashes) there is no way to tell whether it s R or S and we d just assume that there is a racemic mixture (which you could indicate explicitely with +/- or R/S in a more complete name). If, for example, the OH was shown on a wedge or a dash, then we would indicate whether it was R or S in the name.",
        "video_name": "nQ7QSV4JRSs",
        "timestamps": [
            240
        ],
        "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex."
    },
    {
        "Q": "at 5:20, is the informal name phenol?\n",
        "A": "its not a benzene either way because he doesnt identify the conjugated double bonds. so it is only cyclic but not aromatic",
        "video_name": "nQ7QSV4JRSs",
        "timestamps": [
            320
        ],
        "3min_transcript": "multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex. It's a cyclohexane. But then of course, the hydroxide or the hydroxy group I should call it, takes dominance. It's a hexanol. So this is a cyclohexanol. And once again, that comes from the OH right there. And you don't have to number it. Because no matter what carbon it's on, it's on the same one. If you had more than one of these OH groups, then we would have to worry about numbering them. Let's just do this one right over here. So once again, what is our carbon chain? We have 1, 2, 3 carbons. And we have the hydroxy group attached to the 1 and the 3 carbon. Prop is our prefix. It is an alkane. So we would call this-- and there's a couple of ways to do this. We could call this 1 comma 3 propanediol."
    },
    {
        "Q": "at 4:22 why is it not OCT-3-yn-5-ol since the double bond ought to take precedence over the -OH group? thanks!\n",
        "A": "The double bond does not take precedence over the alcohol. Carboxylic acids > esters > amides > ketones > aldehydes > alcohols > amines > alkenes > alkynes > R = OR = X = Ph (those of equal precedence go in alphabetical order.",
        "video_name": "nQ7QSV4JRSs",
        "timestamps": [
            262
        ],
        "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex."
    },
    {
        "Q": "At 8:55, I don't understand clearly why i1 equals (V1-V2)/R1.\n",
        "A": "V1 and V2 are node voltages . That means they are measured with respect to the reference node (ground). This is suggested by the two orange arrows that start at the ground node and curve up to each node. Resistor R1 is connected between node 1 and node 2. The element voltage that appears across R1 is the difference of the two node voltages, V1-V2. Using Ohm s Law you compute the current through R1 as i = Voltage across resistor / R i = (V1 - V2) / R",
        "video_name": "2lY757QaaKs",
        "timestamps": [
            535
        ],
        "3min_transcript": "let me move up a little bit. Step Four is write the Kirchoff's Current Law equations directly from the circuit. We're going to do this in a special way, We're going to perform at this node here, at node two. We're going to write the current law for this. That means we got to identify the currents. There's a current, we'll call that a current, and that's a current. Let me give some names to these currents just to be clear. We'll call this one I1 because it goes through R1. We'll call this one here, I2 because it goes through R2. This one is already Is. Now let's write Kirchoff's Current Law just in terms of I, and we'll say all the currents flowing into the node so they're going to get negative signs when we write Kirchoff's Current Law. Let's do that right here. And we write I1 minus I2 minus Is equals to zero. So right now we're working on Step Four. This is the essence of the Node Voltage Method. This is where we do something new that we haven't done before. We're going to write these currents in terms of the node voltages. So we can write I1, I1 is current flowing this way through this current. I1 equals V1 minus V2 over R1. in terms of node voltages. The current flowing down through I2, now we have to subtract I2, so we just apply Ohm's Law directly, which means that the current in I2 is equal to V2 divided by R2. The last current is Is, minus Is. We'll write that in terms of Is, like that, and that equals zero. This means we have now completed Step Four. That is KCL written using the terminology of node voltages. We could check off that we've done Step Four."
    },
    {
        "Q": "at 5:09, can plasma be considered to be the fourth state of matter?\n",
        "A": "Yes, it is the fourth state of matter.",
        "video_name": "WenwtcuqOj8",
        "timestamps": [
            309
        ],
        "3min_transcript": "where does it exist? Well, probably closest to home, it exists in lightning. And that's worthy of an entire video. But the idea is that you start having a huge potential difference between the clouds and the ground. And then because you have this huge voltage difference between the two, you have electrons that are essentially wanting to go into the ground. You have a build-up of electrons up here that want to go into the ground. They can't because air is normally a fairly bad conductor. It's an insulator. But what happens is because there's so much electropotential here, the electrons that are close in the molecules up here, at least how I visualize it, So their electrons start to want to move away in the air molecules. Whether you're talking about the air is a mixture of oxygen, and nitrogen, and carbon dioxide. They start wanting to get away from the clouds. So they start disassociating and start forming this ionized air. And eventually, at some point, this happens to such a degree that you can actually get conduction from the cloud to the ground. And that conduction is when the air is in a plasma state. The conduction allows extremely high temperatures and the electrons to flow all the way to the ground. The other common example, you might see something like this, well actually not like this, but at least a plasma state, is in stars. And that's because you have extremely strong electromagnetic fields, extremely high pressure, and in that type of environment, once again, I'm old super over simplifying it, you can get to a state where the electrons can get disassociated from things that otherwise wouldn't I thought I would touch on that because it's an interesting subject. And it exists in the universe. On the universal level, because stars are pretty much all plasma, it is actually the most common state of matter in the universe. Although in our everyday life, we probably encounter solids, liquids and gases a lot more. And one other thing I want to maybe clarify from the last video is, I talk about the bonding between water molecules. And let's say we're talking in the solid state. So I have an oxygen, a hydrogen, a hydrogen. And I have some electrons here, some electrons here. Let's say there's another hydrogen here, an oxygen, and a hydrogen. Maybe there's an oxygen here. That has hydrogen. And then this has a hydrogen. And it has two electrons, two electron pairs."
    },
    {
        "Q": "7:46: Sal mentions NH3 then at 8:09 Sal mentions HF and water is H2O. Is there a certain way to label a compound? Because the H appears in different spots in the examples provided.\n",
        "A": "In general it is convention to have carbon first then hydrogen then followed by other compounds (C6H12O6). However, there is also a convention that in binary acidic compounds the hydrogen is at the beginning (HCl, HF, HBr etc.), and at the end of organic acids (R-COOH). Long story short there are several different formula conventions and naming conventions and they use separate rules, so you wind up with some variations in writing classic formulas.",
        "video_name": "WenwtcuqOj8",
        "timestamps": [
            466,
            489
        ],
        "3min_transcript": "That oxygen is so much more electronegative that it hogs the electrons. And so the oxygen side starts to have a partial negative charge. While the hydrogen side starts to have a partial positive side. Because with the hydrogen, essentially all of its electrons are hanging out close to the oxygen, hydrogen ends up just becoming like this proton that's floating out there. Because we said it doesn't even have neutrons in most cases. So this has a slightly positive charge. This one has a positive charge. And the positive polar end of the water molecule is attracted to the negative polar end. And I called it polar bonds, and it shows you my memory from high school chemistry is not ideal, I really should have called it hydrogen bonds. So this is a hydrogen bond, and this is a hydrogen bond. It's just a matter of the name I used. I just want to clarify that because that is what's typically used in your chemistry class. And that is just the bond that exists from a partially positive hydrogen atom. Because its electrons are hanging out near the oxygen. And a partially negative oxygen atom in the water molecule. Because it has stolen all of these electrons from the hydrogen. You draw it like that, it's called a hydrogen bond. And hydrogen bonds tend to form between hydrogen or really only a handful of super electronegative atoms. And that's nitrogen, fluorine, and oxygen. And these are actually the three most electronegative atoms. So the nitrogen, NH3, when it bonds with hydrogen, is essentially so electronegative that you have the same situation. All the electrons hang out here, so you have a partial negative charge, partial positive on the hydrogen ends. You get the same HF. You get the same type of hydrogen bonds. And so in this case, these guys would be attracted to the nitrogen part of other molecules and would form I just want to get that out of the way. And with that done, I think we can return to some of the ideas of the last video and actually do some problems. Let's take the case with water. Actually, let me just state the problem first. So let's say that we have a"
    },
    {
        "Q": "at 0:58, I don't understand what a plasma is\n",
        "A": "Plasma is when an atom is vibrating so fast that the electrons are shaken off. This can be accomplished with high heat, low pressure, or an electromagnetic field.",
        "video_name": "WenwtcuqOj8",
        "timestamps": [
            58
        ],
        "3min_transcript": "In the last video we touched on the three states of matter that are really most familiar to our everyday experience. The solid, the liquid, and the gas. And I kind of hinted that there is a fourth state, which I don't cover, because it's usually not the domain of an introductory chemistry course. But a little bit of a discussion ensued on the message board for that video. So I thought I would at least touch on that fourth stage. And that's plasma. I'll do it in a suitably bright color. Plasma. And people consider it a fourth state because it has some properties of gases. In some ways it's almost a subset of gases. But it also has properties of conductivity that you normally wouldn't associate with a gas. And just so you know, when you first hear it you think, oh that's a fairly exotic thing, plasma. And in the first video, I said it's only something that occurs at high temperatures, which isn't exactly 100% right. It doesn't have to be at high temperatures. I really should have said that under extenuating electromagnetic field. Or something has to happen to essentially bump the the electrons, or move the electrons off of gases that would've otherwise have kept their electrons. So it's kind of analogous to what happens in metal. When we talk about metal bonds, we talk about this notion of a sea of electrons. Let's say if we talked about iron. What happens with most metals is that they have so many electrons, and they are so willing to give them, that the electrons just kind of float outside of the atoms themselves and create this kind of big sea of electrons. And then the atoms themselves become positively charged ions. Because they essentially donated some So they're attracted to the sea and that's what makes them malleable and even more importantly what allows them to conduct electricity. But they're all really packed closely together and it's a very dense structure. Plasma is a situation where if you take gases, and remember, So you take a bunch of gases and they have high kinetic energy. Although, they don't have to be, that could be under very But they're moving around and bumping into each other. But they're not close to each other. They don't have a fixed structure with each other. Or they're not rubbing against each other like in the case of a liquid. But what happens in a plasma, or one situation is, that you apply such a strong electromagnetic field that the electrons want to disassociate. So let's say these electrons start bumping off of the plasma. And so a solid has its own shape. A plasma will take the shape of its container like a gas. And sometimes it is described as an ionized gas. And it's described as ionized because electrons are bumped off. And when the electrons are bumped off, the otherwise neutral atoms now have positive charges. And what this allows is, essentially a conduction of electricity. Because now these electrons are free to move."
    },
    {
        "Q": "at 4:45 hank mentions that the hagfish doesn't have a vertebrae yet it is a vertebrate. how?\n",
        "A": "He actually doesn t classify it as a vertebrate, just a chordate. Chordates are known to have spinal chords be their main attraction. But to be classified you also need a skull, a post-anus tail at one point in one s life cycle, the ability to make mucus, and other minor details. The hagfish, while not containing most of these has at least a skull and a post-anus tail so it can be grouped into this section because there nowhere better to put it.",
        "video_name": "c7Yy9v8dH8s",
        "timestamps": [
            285
        ],
        "3min_transcript": "These four traits all began to appear during the Cambrian explosion, more than 500 million years ago. Today, they're shared by members of all three chordate sub-phyla, even if the animals in those sub-phyla look pretty much nothing like each other. For instance, our new friends here in cephalochordata are the oldest living sub-phyla, but you can't forget the other invertebrate group of chordates, the urochordata - literally, tail cords. There are over 2,000 species here, including sea squirts. If you're confused about why this ended up in a phylum with us, it's because they have tadpole-like larva with all four chordate characteristics. The adults, which actually have a highly-developed internal structure, with a heart and other organs, retain the pharyngeal slits, but all the other chordate features disappear or reform into other structures. The third and last, and most complex sub-phylum, is the vertebrata, and has the most species in it, because its members have a hard backbone, which has allowed for an explosion in diversity, You can see how fantastic this diversity really is when you break down vertebrata into its many, many classes, from slimy sea snake-y things to us warm and fuzzy mammals. As these classes become more complex, you can identify the traits they each develop that gave them an evolutionary edge over the ones that came before. For example, how's this for an awesome trait - a brain. Vertebrates with a head that contains sensory organs and a brain are called craniates. They also always have a heart with at least two chambers. Since this is science, you're gonna have to know that there's an exception for every rule that you're gonna have to remember. The exception in this case is the myxini, or hagfish, the only vertebrate class that has no vertebra, but is classified with us because it has a skull. This snake-like creature swims by using segmented muscles to exert force against its notochord. Whatever, hagfish. Closely related to it is the class petromyzontida, otherwise known as lampreys, the oldest-living lineage of vertebrates. Now, these have a backbone made of cartilage, With the advent of a backbone, we see vertebrates getting larger, developing more complex skeletons, and becoming more effective at catching food and avoiding predators. But, did you notice anything missing? Lampreys, and other early vertebrates, are agnafins, literally no jaws, and if you want to be able to chew food, it really helps to have a jaw and teeth. Most scientists think that the jaw evolved from structures that supported the first two pharyngeal slits near the mouth. And teeth? Well, the current theory is that they evolved from sharp scales on the face. Gnathostomes, or jaw mouths, arrived on the scene 470 million years ago, and one of the oldest, most successful groups of gnathostomes that have survived to present day are the class chondrichthyes - the cartilage fish. You know them as the sharks, and skates and rays, and as their name says, their skeleton is made up mostly of cartilage, but they show the beginnings of a calcified skeleton. Chondrichthyes haven't changed much in the past 300 million years or so, and their success stems from the paired fins that allow for fish and swimming,"
    },
    {
        "Q": "\nhow can I find the plane of symmetry in the meso compound in 9:30....it is impossible to find it?!",
        "A": "Carbons on each end, and a OH group attached on both. Try to imagine drawing a line of symmetry DIAGONALLY.",
        "video_name": "zNAL1R-hZr0",
        "timestamps": [
            570
        ],
        "3min_transcript": "our chiral centers here, so this one has an OH coming out at us, and then that one has it going away from us, this one has an OH coming out at us and this one has it going away from us, so we have opposite configurations at both chirality centers. Let's look at three and four next. So what is the relationship between these two? Well, first, we might think these could be enantiomers because at this carbon, we have OH on a wedge, and then here we have OH on a dash. And then here we have OH on a dash, and here we have it on a wedge. So that might be your first guess. But let's look at the video and let's look at the model sets to help us out. Remember, I'm leaving the hydrogens off the methyl groups and the hydrogens off the oxygens in the video just to help us see the molecule more clearly. On the left, we have a model of drawing three. So here's our carbon chain with an OH going away from us in space, and an OH coming out at us in space. Here's our carbon chain with an OH coming out at us in space and an OH going away from us in space. So I'll hold the two models and we'll compare them. First let's see if they are mirror images of each other. So I'll take the one on the right and I'll rotate it and I'll hold it up next to the one on the left. And now we can see that these two are mirror images of each other. So next, let's see if one is superimposable on the other. So I'll go back to the starting point and I'll rotate it around like that, and let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left. And notice that we can. All of the atoms line up. So all of the hydrogens, carbons and oxygens are in the same place. So this is a compound that has chirality centers, but it is achiral, the mirror image is superimposable on itself. So we should be able to find a plane of symmetry. So I'll just pick one of these models, doesn't matter which one because they represent so we can see a plane of symmetry. So right there is our plane of symmetry. This is a meso compound. So three and four actually represent the same compound. So this is one meso compound. So these two are the same, and we have one meso compound. It's not really obvious looking at these bond line structures that these represent the same molecule. So definitely get a model set and try this out for yourself. So we thought there might be four stereoisomers, but actually there are only three. We have a pair of enantiomers, and we have one meso compound."
    },
    {
        "Q": "In the last section of the video with the astronaut glove won't the reaction force of the heavy box push you even further away from the space arm contraption? slightly puzzled , Thankyou in advance\n(Sorry the timing is 1:09 and onwards)\n",
        "A": "no. newton s third law says that every action has an equal and opposite reaction. So, pushing the box away from the space arm results in a force applied in the opposite direction, which is towards the space arm.",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            69
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:16, when the astronaut is pushing an object away from their body, will the force increase if the astronaut increases the amount of time it takes to throw the object? (i.e. will the object have a greater equal and opposite force if the astronaut takes a long time to throw the object than if it was thrown quickly)\n",
        "A": "No, the force is determined by how hard the astronaut pushes, not how long. The total momentum for the astronaut is a combination of how hard he pushes and how long",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            436
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At, 7:40, will the astronaut have to let go of the object to accelerate in the opposite direction or not?\n",
        "A": "Of course she has to let go.",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            460
        ],
        "3min_transcript": ""
    },
    {
        "Q": "3:15 - 4:30. Would it not mean if you took your foot away (presuming that your foot is not part of you), the sand would accelerate upwards. In the same way as: two persons pushing against each other and one of them moves out of the way. It's obvious that the person that didn't move out the way would fly forwards, unbalanced. Is this not the same thing as what Newton is saying. I'm not disregarding Newton's Law, I just don't understand it.\n",
        "A": "The sand pushes up on you foot and your foot pushes down on the sand. If you foot does not push on the sand, the sand does not push on your foot. For the sand to accelerate upwards, something would have to be pushing upward on the sand. The sand cannot push on itself.",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            195,
            270
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 4:09 , Sal said that the molecules of the rock would align themselves to the poles , but how do people track molecular alignment?",
        "A": "In this instance, the alignment is evident because the rock has become magnetized.",
        "video_name": "6EdsBabSZ4g",
        "timestamps": [
            249
        ],
        "3min_transcript": "but it is kind of a curious thing to look at. And not only is there this ridge. There's lot of underwater volcanic activity. You have magma flowing out and lava flowing out into the water, and it's kind of forming this ridge that really goes across the whole Atlantic Ocean. There are other ridges in the world like that, underwater ridges. You have one over here in the Pacific Ocean. You have these here in the Indian Ocean. That's just a little clue, but that by itself doesn't tell you that these plates are actually moving apart at the ridge. The more conclusive-- this is just the beginning of the clue-- but what made this conclusive is one, the separate discovery. And this is what's interesting is that you have these separate discoveries in different domains that eventually let you come to a pretty neat conclusion. So you've had a separate discovery that if you look at different eras of magnetic rock, or maybe I should say magnetic rock from different periods And you can tell where they are in geologic time by how they're layered. So this would be newer rock. And then this would be a little bit older. And then this would be even, even older. Geologists noticed something interesting. If I were to take magnetic rock, and if it was molten lava, and if it were to harden, remember it's magnetic rock so it would want to align with the poles the same way a compass would. So if I had a bunch of magnetic-- so let's say this is some lava right here. And so the molecules can align themselves. Since it's a liquid and they can align themselves, they are going to naturally want to align with the poles. So they'll naturally all want to align in one direction because of Earth's magnetic field. that alignment will kind of be frozen. Now, if Earth's magnetic field was constant over time, then when you look at magnetic rocks from any period, you would expect them all to be aligned in the same direction. So since we're taking a cross section of rock here let's say an alignment towards the North pole looks like this. And I draw it like that. That's kind of an arrow pointing into our screen. And let's say an alignment pointing to the South pole would look like this. This would be an arrow pointing out of our screen. So what you would expect is the newer rock that kind of the alignment, the field, the alignment of the rock, would go into the screen, and then the older rock, it would still go older into the screen. So if I were to draw a top view-- Let me draw it like this just so I make sure that everyone is on the same page. So let me just draw a cross section like this so that we know what we're talking about."
    },
    {
        "Q": "\ni have studied in my text books that index finger should point towards the magnetic field.so i am a bit confused cz in this video at approx 6:40 and 10:00 ,sal has said that index finger should be pointing in the direction of the current....",
        "A": "There are many different versions of these rules. Some use the right hand, some use the left, some define fingers one way, some do it another. They all lead to the same answer. Use whatever one works for you, and stick with that.",
        "video_name": "l3hw0twZSCc",
        "timestamps": [
            400,
            600
        ],
        "3min_transcript": "It's just a notation. I've seen professors do it either way, I've seen it written either way, as well. Cross the magnetic field that it's in. What's the magnetic field that it's in? The magnetic field-- I'll do it in magenta, because it's the magnetic field created by current 1. So it's magnetic field 1, which is this magnetic field. So before going into the math, let's just figure out what direction is this net force going to be in? So here we say, well, the current is a scalar, so that's not going to affect the direction. What's the direction of L2? This is L2. I didn't label it L2 on the diagram. What's the direction of L2? Well, it's up. And then the direction of B1, the magnetic field created by current 1, is going into the page here. So here we just do the standard cross product. Let me see if I can pull this off. This is actually an easy one to draw. And then I put my middle finger in the direction of the field. So my middle finger's going to point straight down into this page. My other fingers just do what they would naturally do. And then my thumb would go in the direction of the net force. This is just the cross product. You'll see teachers teach the cross product other ways, where they tell you to put your thumb in the direction of the field, and this and that, your palm-- those are all valid. They're just different variations of the same thing. I find this one easier to remember. Because when I take the cross product, index finger is the first term of the cross product. Middle finger is the second term of the cross product. Thumb is the direction of the cross product. So anyway, this is the direction of L2. The magnetic field, we already know, goes into the page. So my middle finger is going into the page. And my thumb is in the direction of the force on the magnetic field. So that's the direction of the force. So there you have it. is-- we know from this wrap around rule that pops out here and it goes in here-- the effect that it has on this other wire is that where the current is going in the same direction, is that it will be attracted. So the net force you is going in that direction. We could say the force from 1 on 2. That's just my convention. Maybe other people would have written it the force given to 2 by 1. That's the force given by 1 to 2. That's how I'm writing it. Now what's going to be the force on current 1 from I2? Well, it's going to be the current-- well, it's going to be the force there. Well it's going to be the same thing. Let me draw I2's magnetic field. You do the wrap around rule, it's going to look the same. So I2, sure, on this side its field is going to be going into the page. But what's I2's field going to be doing here?"
    },
    {
        "Q": "\nAt 8:40, you talked about voltage drop. So, does the voltage differ from resistor to resistor? what does voltage drop mean?",
        "A": "The voltage drop refers to the difference in electric potential at opposite ends of the resistor. If the resistor is carrying current there must be a potential difference, otherwise why are the charges moving. The PD is given by V = IR, where I is the current through the resistor.",
        "video_name": "7vHh1sfZ5KE",
        "timestamps": [
            520
        ],
        "3min_transcript": "That's a 1, not an I. I1 times R1, right? And similarly, if I measured the voltage from here to here, that voltage is going to be equal to I2 times R2. Let's say this is where I3 is. So the voltage, if I were to measure it from here to here-- But anyway, if we look at the voltage from here to here, it's going to be I3 times R3. So what we see is that the voltage across the entire circuit, which I can write as V-total, is going to be equal to the potential drops, the total potential drop across each of these devices. So the way to think about it is that-- well, let's think about the electrons. The electrons here, they really want to get here. get here, they've experienced some potential drop. So the electrons here actually are a little bit less eager to get here. And then once they've gone through here, maybe they're just tired of bumping around so much. And once they're here, they're a little bit less eager to get here. So there's a voltage drop across each device, right? So the total voltage is equal to the voltage drop across each of the devices. And now let's go back to the convention, and we'll say that the current is going in that direction. The total voltage drop is equal to V1 plus V2 plus V3, so the total voltage drop is equal to I1 R1 plus I2 R2 plus I3 R3. And what's the total voltage drop? Well, that's equal to the total current through the whole system. I-total, or we just call it I, times the total resistance is Well, we know that all the I's are the same. Hopefully, you can take it as, just conceptually it makes sense to you that the current through the entire circuit will be the same. So all these I's are the same, so we can just cancel them out. Divide both sides by that I. We assume it's non-zero, so I, I, I, I, and then we have that the total resistance of the circuit is equal to R1 plus R2 plus R3. So when you have resistors in series like this, the total resistance, their combined resistance, is just equal to their sum. And that was just a very long-winded way of explaining something very simple, and I'll do an example. Let's say that this voltage is-- I don't know. Let's say it's 20 volts. Let's say resistor 1 is 2 ohms. Let's say resistor 2 is"
    },
    {
        "Q": "\nAt around 10:50 in the video sal goes from 20= current * 10 ohms to current = 2 AMPERES. How did he get to amperes from ohms?",
        "A": "Voltage = Current x Resistance in terms of units: (volts)=(amperes) x (ohms) (ohms)=(volts) / (amperes) with a little rearranging, you get: (amperes)=(volts) / (ohms)",
        "video_name": "7vHh1sfZ5KE",
        "timestamps": [
            650
        ],
        "3min_transcript": "Well, we know that all the I's are the same. Hopefully, you can take it as, just conceptually it makes sense to you that the current through the entire circuit will be the same. So all these I's are the same, so we can just cancel them out. Divide both sides by that I. We assume it's non-zero, so I, I, I, I, and then we have that the total resistance of the circuit is equal to R1 plus R2 plus R3. So when you have resistors in series like this, the total resistance, their combined resistance, is just equal to their sum. And that was just a very long-winded way of explaining something very simple, and I'll do an example. Let's say that this voltage is-- I don't know. Let's say it's 20 volts. Let's say resistor 1 is 2 ohms. Let's say resistor 2 is total resistance through this circuit? Well, the total resistance is 2 ohms plus 3 ohms plus 5 ohms, so it's equal to 10 ohms. So total resistance is equal to 10 ohms. So if I were to ask you what is the current going through this circuit? Well, the total resistance is 10 ohms. We know Ohm's law: voltage is equal to current times resistance. The voltage is just equal to 20. 20 is equal to the current times 10 ohms, right? We just added the resistances. Divide both sides by 10. You get the current is equal to 2 amps or 2 coulombs per second. So what seemed like a very long-winded explanation actually results in something that's very, very, very easy to apply. I will see you in the next video."
    },
    {
        "Q": "At 13:00, why did you choose this as the longest chain? they're all the same.. can we choose any of the three lines as our main group? And the rest is methyl ? or is there a specific rule to follow?\nThank you so much.\n",
        "A": "Right. They are all the same length. You can choose any one of them. Whichever one you choose, it will be an ethyl group with two methyl groups on C-1 (1,1-dimethylethyl).",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            780
        ],
        "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming."
    },
    {
        "Q": "\nAt 12:42, why is it necessary to say \"dimethyl\" instead of simply \"methyl\" if it's already indicated by the \"1,1\" that there are two methyls branching out?",
        "A": "It works the other way. You count the methyl groups first, then you tell where each of them is. The name says, You have two methyl groups, and they are each on carbon-1 .",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            762
        ],
        "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming."
    },
    {
        "Q": "I am not sure but at 11:50 wont the compound name be (2,2-dimethylpropyl)cyclopentane ?\n",
        "A": "In the older systematic name you use the point of connection to the parent chain as carbon #1. So from that carbon the longest chain is 2 carbons long which is why it is ethyl. There s also two other carbons coming off that first carbon so that is why it is 1,1-dimethyl. All together that is (1,1-dimethylethyl)",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            710
        ],
        "3min_transcript": "Now, that describes just the group. 1-methylpropyl describes just this part right here. That describes just that right over there. And then to have the whole compound, to describe the whole compound, you put this in parentheses, so this is the systematic naming. So 1-methyl-- I put an L there. Let me do it in the same color. 1-methyl, because you're starting where you're attaching. So 1-methyl, you have a methyl group right there on that first carbon. It's a propyl chain. One, two, three, propyl, and then you would say cyclopentane. That's the systematic name for that. Now if you look at this one right here, and the common name is iso-butyl, what you do is you Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached"
    },
    {
        "Q": "\nIs there any reason that Isobutylcyclopentane can't be written as i-butylcyclopentane?\nOr is it just one of those, \"we don't do this because they tell us not to,\" kind of science situations? (8:48 in video)",
        "A": "No. There s no reason. isobutylcyclopentane, i-butylcyclopentane, and 3-methylpropylcyclopentane are all IUPAC names. You can write any of these.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            528
        ],
        "3min_transcript": "the difference here. The common naming, it's easier to say and easier to spell, but it's sometimes a little confusing. This is just straight up butyl so you would call this butylcyclopentane. This is sec-butyl, because you have this guy connected to two carbons. That's where the sec- comes from. Sometimes it'll be s-butyl. So this could be called sec-butylcyclopentane or s-butylcyclopentane. This, because we're attached to the end away from this branching off, is still a butyl group, since we have four carbons. But since we're attached here, this is iso-butyl, so this is iso-butylcyclopentane, And then finally, since the carbon we're attaching to is attached to one, two, three other carbons, it is a tert-butyl or a t-butyl group. So this is t-butylcyclopentane. That's the common naming. So maybe I should clear out systematic here just so it's clear to you that everything we've done So let me write it down. It won't hurt to write them down again because the more familiar you are with these, the better. So this is just butylcyclopentane. This is s- or sec- butylcyclopentane. And this is iso-butylcyclopentane. I'm going off the screen here. And then finally, this is tert-butyl, or t-butylcyclopentane. Now, I said these are the common naming. What are the systematic naming? Well, in the systematic, this is still butylcyclopentane. So let me write this down. Systematic, this is still This is very clearly a cyclopentane. This is very clearly a butyl group. But in the systematic naming, what we try to do is we try to name this group right here just as we would name a traditional chain, but we ended it with an -yl. So if you look at this right here, what we do is we just consider the chain where we attach. We attached over here, so the longest chain from that point is there and there. So if you look at it like that, it looks like you have one, two, three carbons, and you have one carbon attached on the beginning. So this little group right here in the systematic naming, this looks like a one, two, three. Three carbons, that's the prop- prefix, so we're dealing with a prop-, and it's all going to be one group, so it's a propyl group. This is a propyl group, but it has a methyl-- remember, meth- is one carbon. It has a methyl group attached on the first carbon."
    },
    {
        "Q": "\nIN 4:30 WHAT DOES 'sec' actually means?",
        "A": "It means that the bond coming from the rest is attached to 2 carbons.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            270
        ],
        "3min_transcript": "this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here. And you do have a butyl group on it, so we do have a butyl group, but because we are bonded-- we aren't bonded to the first carbon. We're bonded to a carbon that is bonded to two other carbons. We call this sec-butylcyclopentane, so this is sec-. And everything I'm doing is obviously free-hand. If you were to see this in a book, the sec- would be italicized, or sometimes it would be written as s-butylcyclopentane. And this sec- means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group, and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same"
    },
    {
        "Q": "at 4:22, why do you call it secbutyl cyclopentane instead of 2-butyl-cyclopentane ?\n",
        "A": "It is called sec-butyl cyclopentane because the sec shows how many carbons are attached to the carbon in the butyl chain that attaches to the cyclopentane, in this case, 2 carbons (on either side of the attached carbon). If it were 2-butyl cyclopentane, it would mean that you had a butane substituent in the second position of the cyclopentane..",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            262
        ],
        "3min_transcript": "this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here. And you do have a butyl group on it, so we do have a butyl group, but because we are bonded-- we aren't bonded to the first carbon. We're bonded to a carbon that is bonded to two other carbons. We call this sec-butylcyclopentane, so this is sec-. And everything I'm doing is obviously free-hand. If you were to see this in a book, the sec- would be italicized, or sometimes it would be written as s-butylcyclopentane. And this sec- means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group, and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same"
    },
    {
        "Q": "At 13:05 near the end of the video, when describing the molecule. Is it not important to label the molecule in alphabetical order? For example. the end result is (1,1 dimethylethyl)cyclopentane. Would it not be more correct to have it as (ethyl - 1,1 -dimethyl)cyclopentane? as E is before M in the alphabet.\n",
        "A": "(1,1-dimethylethyl) does not need to go alphabetically. Ethyl is the parent chain of this group, the 2 methyl groups are coming off of that ethyl, so the name needs to show that. It would be similar to naming 4-ethyldecane as decane-4-ethyl because d comes before e.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            785
        ],
        "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming."
    },
    {
        "Q": "I had learned that while naming, the functional groups should be added in alphabetic order, . In 13:01, you named t-butyl cyclopentane as (1,1-dimethyl ethyl) cyclopentane. shouldn't it be (ethyl 1,1, dimethyl)cyclopentane because 'e' comes before 'm'?\nThanks!\n",
        "A": "Hu hu. they re right",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            781
        ],
        "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming."
    },
    {
        "Q": "\nAt 9:55 where did he get the square root of 3/2 from?",
        "A": "The cosine of 30 degrees is equal to the square root of three over two.",
        "video_name": "_UrfHFEBIpU",
        "timestamps": [
            595
        ],
        "3min_transcript": "a little confusing to you. We just said, this point is stationery. It's not moving up or down. It's not accelerating up or down. And so we know that there's a downward force of 100 Newtons, so there must be an upward force that's being provided by these two wires. This wire is providing no upward force. So all of the upward force must be the y component or the upward component of this force vector on the first wire. So given that, we can now solve for the tension in this first wire because we have T1-- what's sine of 30? Sine of 30 degrees, in case you haven't memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal to 100 Newtons. Divide both sides by 1/2 and you get T1 is equal to 200 Newtons. second wire is. And we also, there's another clue here. This point isn't moving left or right, it's stationary. So we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction. And that force in the opposite direction is the x component of the first wire's tension. So it's this. So T2 is equal to the x component of the first wire's tension. And what's the x component? Well, it's going to be the tension in the first wire, 200 Newtons times the cosine of 30 degrees. It's adjacent over hypotenuse. And that's square root of 3 over 2. So it's 200 times the square root of 3 over 2, which equals 100 square root of 3. completely offsets to the left and the x component of this wire is 100 square root of 3 Newtons to the right. Hopefully I didn't confuse you. See you in the next video."
    },
    {
        "Q": "\nAt 16:00, Sal talks about all the different variations, 2 to the 46th. how does this work in twins? My cousins are identical but one has slightly curlier hair than the other, how can this work if they have exactly the same DNA, there both very different from each other in personality too. I know there monozigotic, but I don't know what that means.",
        "A": "Even identical twins are not completely identical. There were slight differences in the womb, there continue to be differences when they are babies. Genes can be expressed more or expressed less depending on what other genes do and depending on environmental factors.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            960
        ],
        "3min_transcript": "to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so But there's 2 to the twenty-third combinations here, because there are 23 pairs that I'm collecting from. Now, my wife's egg is going to have the same situation. There are 2 to the 23 different combinations of DNA that she can contribute just based on which of the homologous pairs she will contribute. So the possible combinations that just one couple can produce, and I'm using my life as an example, but this applies to everything. This applies to every species that experiences sexual reproduction. So if I can give 2 to twenty-third combinations of DNA and my wife can give 2 to the 23 combinations of DNA, then we can produce 2 to the forty-sixth combinations. Now, just to give an idea of how large of a number this is, this is roughly 12,000 times the number of human beings on So there's a huge amount of variation that even one couple can produce. And if you thought that even that isn't enough, it turns out that amongst these homologous pairs, and we'll talk about when this happens in meiosis, you can actually have DNA recombination. And all that means is when these homologous pairs during meiosis line up near each other, you can have this thing called crossover, where all of this DNA here crosses over and touches over here, and all this DNA crosses over and touches over there. So all of this goes there and all of this goes there. What you end up with after the crossover is that one DNA, the one that came from my mom, or that I thought came from my mom, now has a chunk that came from my dad, and the chunk that came from my dad, now has a chunk that came from my mom. Let me do that in the right color. It came from my mom like that. And so that even increases the amount of variety even more."
    },
    {
        "Q": "\nAt 18:26 why did Professor Sal think that Sexual Reproduction is variation of population?",
        "A": "Sexual reproduction is a huge contributor to genetic variation in a population. Because each parent only donates half of their genome, each indivudual offspring is guaranteed to have a unique genome of its own (Except in the case of identical twins). Mutation also contributes to variation in an important but less immediately noticeable way.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            1106
        ],
        "3min_transcript": "So there's a huge amount of variation that even one couple can produce. And if you thought that even that isn't enough, it turns out that amongst these homologous pairs, and we'll talk about when this happens in meiosis, you can actually have DNA recombination. And all that means is when these homologous pairs during meiosis line up near each other, you can have this thing called crossover, where all of this DNA here crosses over and touches over here, and all this DNA crosses over and touches over there. So all of this goes there and all of this goes there. What you end up with after the crossover is that one DNA, the one that came from my mom, or that I thought came from my mom, now has a chunk that came from my dad, and the chunk that came from my dad, now has a chunk that came from my mom. Let me do that in the right color. It came from my mom like that. And so that even increases the amount of variety even more. chromosomes that you're contributing where the chromosomes are each of these collections of DNA, you're now talking about-- you can almost go to the different combinations at the gene level, and now you can think about it in almost infinite form of variation. You can think about all of the variation that might emerge when you start mixing and matching different versions of the same gene in a population. And you don't just look at one gene. I mean, the reality is that genes by themselves very seldom code for a specific-- you can very seldom look for one gene and say, oh, that is brown hair, or look for one gene and say, oh, that's intelligence, or that is how likable someone is. It's usually a whole set of genes interacting in an incredibly complicated way. You know, hair might be coded for by this whole set of genes on multiple chromosomes and this might be coded for a whole set of genes on multiple chromosomes. And so then you can start thinking about all of the different combinations. And then all of a sudden, maybe some combination that never existed before all of a sudden emerges, and that's But I'll leave you to think about it because maybe that combination might be passed on, or it may not be passed on because of this recombination. But we'll talk more about that in the future. But I wanted to introduce this idea of sexual reproduction to you, because this really is the main source of variation within a population. To me, it's kind of a philosophical idea, because we almost take the idea of having males and females for granted because it's this universal idea. But I did a little reading on it, and it turns out that this actually only emerged about 1.4 billion years ago, that this is almost a useful trait, because once you introduce this level of variation, the natural selection can start-- you can kind of say that when you have this more powerful form of variation than just pure mutations, and maybe you might have some primitive form of crossover before, but now that you have this sexual reproduction and you have this variation, natural selection can occur in a"
    },
    {
        "Q": "\n14:01. Why is it 2 to the power 23 of possible combinations?",
        "A": "Statistics: you have 23 [blanks] to be filled in a new chromosome. In each [blank] you can have genetic material from one or the other type (2 options). Therefore, the total combination is 2^23.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            841
        ],
        "3min_transcript": "another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so"
    },
    {
        "Q": "At 12:22, it mentions different versions of allele genes. How many are there?\n",
        "A": "We don t know. A gene can have any number of different alleles. Some alleles may have no phenotype at all, others may have a very significant phenotype.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            742
        ],
        "3min_transcript": "reproduction have this complete set of chromosomes in it, which I find amazing. But only certain chromosomes-- for example, these genes will be completely useless in my fingernails, because all of a sudden, the straight and the curly don't matter that much. And I'm simplifying. Maybe they will on some other dimension. But let's say for simplicity, they won't matter in certain places. So certain genes are expressed in certain parts of the body, but every one of your body cells, and we call those somatic cells, and we'll separate those from the sex sells or the germs that we'll talk about later. So this is my body cells. So this is the great majority of your cells, and this is opposed to your germ cells. And the germ cells-- I'll just write it here, just so you get a clear-- for a male, that's the sperm cells, and for female that's the egg cells, or the ova. and what I want to give you the idea is that for every trait, I essentially have two versions: one from my mother and one from my father. Now, these right here are called homologous chromosomes. What that means is every time you see this prefix homologous or if you see like Homo sapiens or even the word homosexual or homogeneous, it means same, right? You see that all the time. So homologous means that they're almost the same. They're coding for the most part the same set of genes, but they're not identical. They actually might code for slightly different versions of the same gene. So depending on what versions I get, what is actually another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to"
    },
    {
        "Q": "How can an organism be both a male and a female? @4:29\n",
        "A": "Many organisms in nature are hermaphordites, possessing both male and female reproductive organs. The worm C. elegans is a common example of a hermaphrodite.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            269
        ],
        "3min_transcript": "these code for proteins or they code for things that control other proteins, but maybe you have a change in one of them. Maybe this cytosine for whatever reason becomes a guanine randomly, or maybe these got deleted, and that would change the DNA. But you could imagine, if I went to someone's computer code and just randomly started changing letters and randomly started inserting letters without really knowing what I'm doing, most of the time, I'm going to break the computer program. Most of the time, the great majority of the time, this is going to go nowhere. It'll either do nothing, for example, if I go into someone's computer program and if I just add a couple of spaces or something, that might not change the computer program, but if I start getting rid of semicolons and start changing numbers and all that, it'll probably make the computer program break. So it'll either do nothing or it'll actually kill the organisms most of the time. Mutations: sometimes, they might make the actual cell kind of run amok, and we'll do a whole maybe series of videos organism as well as a whole, although if it occurs after the organism has reproduced, it might not be something that selects against the organism and it also wouldn't be passed on. But anyway, I won't go detailed into that. But the whole point is that mutations don't seem to be a satisfying source of variation. They could be a source or kind of contribute on the margin, but there must be something more profound than mutations that's creating the diversity even within, or maybe I should call it the variation, even within a population. And the answer here is really it's kind of right in front of us. It really addresses kind of one of the most fundamental things about biology, and it's so fundamental that a lot of people never even question why it is the way it is. And that is sexual reproduction. And when I mean sexual reproduction, it's this notion that have nucleuses-- and we call those eukaroytes. Maybe I'll do a whole video on eukaryotes versus prokaryotes, but it's the notion that if you look universally all the way from plants-- not universally, but if you look at cells that have nucleuses, they almost universally have this phenomenon that you have males and you have females. In some organisms, an organism can be both a male and a female, but the common idea here is that all organisms kind of produce versions of their genetic material that mix with other organisms' version of their genetic material. If mutations were the only source of variation, then I could just bud off other Sals. Maybe just other Sals would just bud off from me, and then randomly one Sal might be a little bit different and"
    },
    {
        "Q": "\nsal u just told(15:07)that we get the chromosome from our parents and that means our parent too follow this then this will mean that the chromosome that the first human ever had are shill carried.am i right?",
        "A": "Unless the first human didn t have children.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            907
        ],
        "3min_transcript": "So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so But there's 2 to the twenty-third combinations here, because there are 23 pairs that I'm collecting from. Now, my wife's egg is going to have the same situation. There are 2 to the 23 different combinations of DNA that she can contribute just based on which of the homologous pairs she will contribute. So the possible combinations that just one couple can produce, and I'm using my life as an example, but this applies to everything. This applies to every species that experiences sexual reproduction. So if I can give 2 to twenty-third combinations of DNA and my wife can give 2 to the 23 combinations of DNA, then we can produce 2 to the forty-sixth combinations. Now, just to give an idea of how large of a number this is, this is roughly 12,000 times the number of human beings on"
    },
    {
        "Q": "2:35\ncan we modify DNA accroding to our wish.\ncan we design the codons on our own.\n",
        "A": "Yes, technically in I think it was late 2007 a group a microbiologists created a new species of bacteria that was blue. The blue bit was just so that they could see them easier if it worked, but it s true. However the technology, like CarlBiologist said, is in its infancy and when they created the bacteria it was impossible for them to create anything much more complex then that. Medium-complexity bacteria was their limit, just because of how much trouble it is to assemble long strands of DNA.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            155
        ],
        "3min_transcript": "And if they are drastic enough, maybe these guys start becoming dominant and start not liking these guys, because they're so different or whatever else. We could see a lot of different reasons. This could eventually turn into a different species. Now, the obvious question is what leads to this variation? In a population what leads to this-- in fact, even in our population, what leads to one person having dirty blonde hair, one person having brown hair, one person having black hair, and we have the spectrum of skin complexions and heights is pretty much infinite. What causes that? And then one thing that I kind of point to, we talked about this a little bit in the DNA video, is this notion of mutations. DNA, we learned, is just a sequence of these bases. So adenine, guanine, let's say I've got some thymine going. I have some more adenine, some cytosine. And that these code, if you have enough of these in a row, these code for proteins or they code for things that control other proteins, but maybe you have a change in one of them. Maybe this cytosine for whatever reason becomes a guanine randomly, or maybe these got deleted, and that would change the DNA. But you could imagine, if I went to someone's computer code and just randomly started changing letters and randomly started inserting letters without really knowing what I'm doing, most of the time, I'm going to break the computer program. Most of the time, the great majority of the time, this is going to go nowhere. It'll either do nothing, for example, if I go into someone's computer program and if I just add a couple of spaces or something, that might not change the computer program, but if I start getting rid of semicolons and start changing numbers and all that, it'll probably make the computer program break. So it'll either do nothing or it'll actually kill the organisms most of the time. Mutations: sometimes, they might make the actual cell kind of run amok, and we'll do a whole maybe series of videos organism as well as a whole, although if it occurs after the organism has reproduced, it might not be something that selects against the organism and it also wouldn't be passed on. But anyway, I won't go detailed into that. But the whole point is that mutations don't seem to be a satisfying source of variation. They could be a source or kind of contribute on the margin, but there must be something more profound than mutations that's creating the diversity even within, or maybe I should call it the variation, even within a population. And the answer here is really it's kind of right in front of us. It really addresses kind of one of the most fundamental things about biology, and it's so fundamental that a lot of people never even question why it is the way it is. And that is sexual reproduction. And when I mean sexual reproduction, it's this notion"
    },
    {
        "Q": "\nAround 14:10, Sal says he has 2^23 forms of contributing to his \"son or daughter\" . Wouldn't it be 2^22 since one chromosome indicates the sex?",
        "A": "No, he will contribute a chromosome that will help decide what gender the child could be.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            850
        ],
        "3min_transcript": "another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so"
    },
    {
        "Q": "At 2:45 what does delta T mean?\n",
        "A": "Say we heat something from 20 Celsius to 50 Celsius, the delta T would be 50 - 20 = 30, i.e. 30 is the change in temperature. Now say we cool something from 60 Celsius to 10 Celsius, the change in temp is 60 - 10 = 50 celsius. It doesnt matter if we heat or cool something, we still have to use energy to change the state of the substance; the only difference is when we heat something, it takes in energy, when it cools it releases energy, but the calculation is the same. I hope u understand",
        "video_name": "H7nrVDV8ahc",
        "timestamps": [
            165
        ],
        "3min_transcript": "And it's going to be a little mathy, but I think the math is pretty straightforward, especially if you've taken a first-year course in calculus. And this is actually a pretty neat application of it. So let's just think a little bit about the rate of change, or the probability, or the number particles that are changing at any given time. So if we say, the difference or change in our number of particles, or the amount of particles, in any very small period of time, what's this going to be dependent on? This is the number particles we have in a given period time. This is our rate of change. So one thing, we know that our rate of change is going down. We know it's a negative number. We know that, in the case of radioactive decay, I could do the same exercise with compounding growth, where I would say, oh no, it's not a negative number, that our growth is dependent on how much we have. In this case the amount we're decaying is proportional, but it's going already have. Let me explain that. So what I'm saying is, look, our amount of decay is proportional to the amount of the substance that we already are dealing with. And just to maybe make that a little bit more intuitive, imagine a situation here where you have 1 times 10 to the 9th. You have a billion carbon atoms. And let's say over here you have 1 times 10 to the 6th carbon atoms. And if you look at it at over some small period of time, let's say, if you look at it over one second, let's say our dt. dt as an infinitesimally small time, but let's say it's a change in time. It's a delta t. And let's say over one second, you observe that this sample had, I don't know, let's say you saw 1000 carbon particles. You really wouldn't see that with carbon-14, but this is just for the sake of our intuition. Let's say over one second you saw 1000 carbon particles per Well here you have 1000th of the number particles in this sample as this one. So, for every thousand particles you saw decaying here, you'd really expect to see one carbon particle per second here. Just because you have a smaller amount. Now I don't know what the actual constant is. But we know that no matter what substance we're talking about, this constant is dependent on the substance. Carbon's going to be different from uranium, is going to be different from, you know, we looked at radon. They're all going to have different quantities right here. And we can see that. We'll actually do it in the next video, you can actually calculate this from the half-life. But the rate of change is always going to be dependent on the number of particles you have, right? I mean, we saw that here with half-life. When you have 1/2 the number of particles, you lose 1/2 as much. Here, if we start with 100 particles here, we went to 50 particles, then we went to 25. When you start with 50, in a period of time you lose 25."
    },
    {
        "Q": "At 1:00 where do the 2 hydrogen cations come from as showed on the products side of glycolysis?\n",
        "A": "The H+ comes from NADH",
        "video_name": "ArmlWtDnuys",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Voiceover] So let's give ourselves an overview of glycolysis. and glycolysis is an incredibly important biochemical pathway. It occures in practically all life as we know it and it's all about taking glucose as a fuel and, in the process of breaking it up, lycing the glucose, glycolysis, breaking it up into two pyruvate molecules. Glucose is a six carbon molecule. Each of the pyruvates are three carbon molecules. In the process of doing that, you produce two ATPs net. It actually turns out that you need to use two ATPs and then you produce four. So you use two ATPs. That's often called the investment phase and we'll talk about that in a second. And then you produce four ATPs for a net of plus two ATPs and that's what we see right over here. You see a net of two ATPs being produced directly the reduction of NAD to NADH. Remember, reduction is all about gaining electrons, and over here, NAD, that's nicotinamide adinine dinucleotide, we have other videos on that, it's an interesting molecule, it's actually a fairly decent-sized molecule, we see this positive charge, but then we see that not only does it gain a hydrogen, but it loses its positive charge. It gains a hydrogen and an electron. You can think on a net basis it's gaining a hydride. Now a hydride anion's not going to typically be all by itself, but on a net basis, you can think about that's what's happening. And so it's gaining a hydrogen and an extra electron and so this, the NAD+, this is going to get reduced. That is going to get reduced to NADH. So this is getting reduced to NADH. be oxidized in the electron transport chain. We'll study that later on when we think about oxidative phosphorylation, to produce even more ATPs. But on a very high-level, simple basis. Glucose being broken down in pyruvate, six carbons, three carbons each of these pyruvates, now there's other things attached to the carbons, and we'll see that in a little bit. Two ATPs net generated, and you have the reduction of two NADs to two NADHs, and those can be used later on to produce more ATPs. Now, glycolysis is typically just the beginning of cellular respiration. If oxygen is around, then you have these products, some of these moving into the mitochondria where you can have the citric acid cycle, Krebs cycle, and the oxidative phosphorylation occur. If you don't have oxygen around, then you're going to do anaerobic respiration, or you're"
    },
    {
        "Q": "Around 9:00 Hank says \"Haeckel influenced from Darwin and Darwin disagreed with him\".\n\nWhat does that mean? Is it a mistake?\n",
        "A": "It is not a mistake, it just means that Haeckel was inspired by Darwin, but Darwin didnt agree with where Haeckel was taking his (Darwins) ideas.",
        "video_name": "cstic6WHr2E",
        "timestamps": [
            540
        ],
        "3min_transcript": "the esophagus and stomach and colon and stuff. And in addition, some of the cells start breaking off between the endoderm and the ectoderm and form another layer called the mesoderm. These cells will eventually end up as the muscles and the circulatory system and the reproductive systems and in the case of vertebrates, most of the bone. So, what's our embryo looking like now? Awesome! From here, this little guy is gonna go on to fulfill his destiny as a ladybug or a walrus or whatever. And now this seems to me like a great time to take a look at a completely dis-proven theory that biologists hold in the highest contempt, but which is actually a kind of useful way to think about the way that an animal embryo develops into a fully formed animal. Plus, it makes for a great biolo-graphy! (music) Back in the mid-1800's a German zoologist named Ernst Haeckel tried to prove what we now refer to as recapitulation theory. Basically, and this is not basic at all, \"Ontogeny recapitulates phylogeny.\" Uhh! In other words, ontogeny, or the growth and development of an embryo recapitulates or sums up phylogeny, which is the evolutionary history of a species. So this means, for instance, that a human embryo over the course of it's development will go through all of the hundreds of millions of years worth of evolutionary steps that it took for a single celled organism to evolve into a fully tricked out person. Haeckel was a contemporary of Darwin and on the origin of species, made a giant impression on him, especially a section of it that notes how cool it is that all vertebrate embryos look pretty similar to one another, regardless of whether they're a mammal or bird or reptile. Darwin, however, cautioned that this probably wasn't a very good way of reconstructing the history of evolution. He just thought it meant that the embryological similarities were evidence of common ancestry between species. Well, Haeckel was kind of a spaz and he definitely heard the first part of Darwin's idea, and very quickly wrote a couple of books about how the development of an embryo mirrors the evolutionary development of adults of a species, which is exactly what Darwin said was not happening. Anyway, Haeckel did spend a lot of time looking at embryos and observed that the slits in the neck of a human embryo resembled the gill slits of fish, which he took to mean that we must have, at one point, had a fish-like ancestor. He drew tons of figures of different animal embryos in different stages of development to prove his theory and his illustrations of embryos started to make their way into textbooks all over the world. Haeckel is exactly the sort of person who really ticks other scientists off because real science loving scientists like to sit and think about stuff and find out all the problems with an idea before they start publishing books about it. And here, Haeckel was firing off volume after volume and before long all the \"data\" he had \"collected\" convinced a bunch of other people,"
    },
    {
        "Q": "\n09:09 PE=m*g*h right? why he did not multiply it to the gravity?",
        "A": "The weight of 10 N is already the result of Fg = m * g.",
        "video_name": "vSsK7Rfa3yA",
        "timestamps": [
            549
        ],
        "3min_transcript": "of this machine is equal to 10 Newtons. Mechanical advantage is the output over the input, so the mechanical advantage is equal to the force output by the force input, which equals 10/5, which equals 2. And that makes sense, because I have to pull twice as much for this thing to move up half of that distance. Let's see if we can do another mechanical advantage problem. Actually, let's do a really simple one that we've really been working with a long time. Let's say that I have a wedge. A wedge is actually considered a machine, which it took me a little while to get my mind around that, but a wedge is a machine. And why is a wedge a machine? Because it gives you mechanical advantage. So if I have this wedge here. And this is a 30-degree angle, if this distance up here, distance going to be? Well, it's going to be D sine of 30. And we know that the sine of 30 degrees, hopefully by this point, is 1/2, so this is going to be 1/2D. You might want to review the trigonometry a little bit if that doesn't completely ring a bell for you. So if I take an object, if I take a box-- and let's assume it has no friction. We're not going to go into the whole normal force and all that. If I take a box, and I push it with some force all the way up here, what is the mechanical advantage of this system? Well, when the box is up here, we know what its potential energy is. Its potential energy is going to be the weight of the box. So let's say this is a 10-Newton box. The potential energy at this point is going to be 10 Newtons times its height. So potential energy at this point has to equal 10 Newtons And that's also the amount of work one has to put into the system in order to get it into this state, in order to get it this high in the air. So we know that we would have to put 5 joules of work in order to get the box up to this point. So what is the force that we had to apply? Well, it's that force, that input force, times this distance has to equal 5 joules. So this input force-- oh, sorry, this is going to be-- sorry, this isn't 5 joules. It's 10 times 1/2 times the distance. It's 5D joules. This isn't some kind of units. It's 10 Newtons times the distance that we're up, and that's 1/2D, so it's 5D joules. Sorry for confusing you. And so the force I'm pushing here times this distance has to also equal to 5D joules."
    },
    {
        "Q": "\nAround 9:22, Sal mentioned that PE is 10kg times height. What about gravity? I thought PE equals Mxgxheight?? Please help. Thanks.",
        "A": "I think you misheard. He didn t say 10kg he said 10N. N would already include any relevant gravitational force.",
        "video_name": "vSsK7Rfa3yA",
        "timestamps": [
            562
        ],
        "3min_transcript": "of this machine is equal to 10 Newtons. Mechanical advantage is the output over the input, so the mechanical advantage is equal to the force output by the force input, which equals 10/5, which equals 2. And that makes sense, because I have to pull twice as much for this thing to move up half of that distance. Let's see if we can do another mechanical advantage problem. Actually, let's do a really simple one that we've really been working with a long time. Let's say that I have a wedge. A wedge is actually considered a machine, which it took me a little while to get my mind around that, but a wedge is a machine. And why is a wedge a machine? Because it gives you mechanical advantage. So if I have this wedge here. And this is a 30-degree angle, if this distance up here, distance going to be? Well, it's going to be D sine of 30. And we know that the sine of 30 degrees, hopefully by this point, is 1/2, so this is going to be 1/2D. You might want to review the trigonometry a little bit if that doesn't completely ring a bell for you. So if I take an object, if I take a box-- and let's assume it has no friction. We're not going to go into the whole normal force and all that. If I take a box, and I push it with some force all the way up here, what is the mechanical advantage of this system? Well, when the box is up here, we know what its potential energy is. Its potential energy is going to be the weight of the box. So let's say this is a 10-Newton box. The potential energy at this point is going to be 10 Newtons times its height. So potential energy at this point has to equal 10 Newtons And that's also the amount of work one has to put into the system in order to get it into this state, in order to get it this high in the air. So we know that we would have to put 5 joules of work in order to get the box up to this point. So what is the force that we had to apply? Well, it's that force, that input force, times this distance has to equal 5 joules. So this input force-- oh, sorry, this is going to be-- sorry, this isn't 5 joules. It's 10 times 1/2 times the distance. It's 5D joules. This isn't some kind of units. It's 10 Newtons times the distance that we're up, and that's 1/2D, so it's 5D joules. Sorry for confusing you. And so the force I'm pushing here times this distance has to also equal to 5D joules."
    },
    {
        "Q": "At min 9:00 you say that the heater is switched off by the bimetallic strip if it gets too hot. That means that the fan switches off too, right? Because as you said, the heater works as a resistor to the fan motor....\nThis implies that the dryer stops from working if it gets hot, which obviously cant be true!\nI am loving this video btw :D\n",
        "A": "If the circuit is broken then both the heater and the motor would stop functioning, however metal takes time to cool down. I do not know the schematic of the circuit",
        "video_name": "Vq7EOmvU1eQ",
        "timestamps": [
            540
        ],
        "3min_transcript": "So a lot of times when you heat up other metals, they'll oxidize, they'll rust. And then you get problems with that. This doesn't do that. It just heats up, and then it cools back down and there's no oxidization. So it works really well as a heater. And on the front, here, you can see there's this 90 degree opposed bracket that holds everything 90 degrees apart, which keeps everything working as far as air flow and stuff like that. It also holds this mica sheet separated. And then you can see on the inside, we have these brass contacts. And they're little brass rivets. And they distribute the power around to different parts of the heater. And right here, this part is a bimetallic strip. So that bimetallic strip is made out of two pieces of metal. And what it does is when the metal is heated to a certain point, it causes the bimetallic strip to bend. than the other piece of metal. And it causes the strip to bend. So maybe, this one on the outside-- if this was the bimetallic strip. This one expands faster than this one, and it causes it to bend. So you could have two different types of metal like, say, an alloy called invar and another piece of metal called copper. And the copper is going to respond to heat at a certain rate, and it's going to expand. And the invar is going to not expand nearly as fast. And it's going to cause that switch to open up. And that will shut down the power to the heater. So if it gets too hot, the bimetallic strip will expand, and it will pull itself away from the contact and shut down the electricity to the heater. You can see, we also have a diode, right here, in line. That just controls the flow of electricity. . It's like a little electrical valve. And then we have a thermal resistor, here. This is called a thermal fuse. And the thermal fuse, basically, is another safety precaution. If temperatures get too high, the fuse will blow. And it will shut down the hair dryer and prevent the housing from melting, or the hair dryer from getting too hot and potentially blowing air out that could burn you. And so, remember, the heater has functioned as a resistor. And it has dropped the voltage down to 12 volts. But it's still AC power, which means alternating current. Alternating current, basically, functions as a sine wave, It flows back and forth. And the motor that we have here is designed to run on direct current, so just flowing in a continuous loop. And the way that this unit, this hair dryer, deals with that is it has these four diodes. You can see them-- one, two, three, four. And so what those four diodes do is they function like a bridge rectifier."
    },
    {
        "Q": "\nAt 5:20, why does the lone pair repel the other bonds more strongly? Didn't it act just like a regular bond in the previous videos?",
        "A": "When you are drawing dot structures, you are not dealing directly with geometries. When you start talking about bond angles, the difference between the repulsion from two electrons belonging to a single atom (lone pair) vs the repulsion from two electrons being shared by two atoms (sigma bond) becomes important.",
        "video_name": "BM-My1AheLw",
        "timestamps": [
            320
        ],
        "3min_transcript": "and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately 109 point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four. So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen, and let's go ahead and draw in all four of those. those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding housed in an SP three hybridized orbital. So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of electrons up here, at the top. So, this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example, and because it's going to repel those electrons a little bit more strongly, you're not gonna get a bond angle of 109 point five; it's going to decrease the bond angle. So let me go ahead, and use the same color we used before, so this bond angle is not 109 point five; it goes down a bit, because of the extra repulsion, so it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say \"tetrahedral\"; we say \"trigonal-pyramidal.\" So let me go ahead, and write that here, so the geometry of the ammonia molecule is trigonal-pyramidal, and let's analyze that a little bit."
    },
    {
        "Q": "at around 4:50 minutes sal said the volume of 1 pound lead is more than volume of 1 pound lead .shouldn't it be the reverse of it?\n",
        "A": "he says that the volume of one pound of feathers will be a lot more than the volume of 1 pound of lead because feathers have a smaller density than lead",
        "video_name": "5EWjlpc0S00",
        "timestamps": [
            290
        ],
        "3min_transcript": "Actually, I shouldn't say water-- let me change this, because I said that this is going to be some random liquid, and the mass is a liquid. The force down is going to be equal to the mass of the liquid times gravity. What is that mass of the liquid? Well, now I'll introduce you to a concept called density, and I think you understand what density is-- it's how much there is of something in a given amount of volume, or how much mass per volume. That's the definition of density. The letter people use for density is rho-- let me do that in a different color down here. rho, which looks like a p to me, equals mass per volume, The units are kilograms per meter cubed-- that is density. I think you might have an intuition that if I have a cubic meter of lead-- lead is more dense than marshmallows. Because of that, if I have a cubic meter of lead, it will have a lot more mass, and in a gravitational field, weigh a lot more than a cubic meter of marshmallows. Of course, there's always that trick people say, what weighs more-- a pound of feathers, or a pound of lead? Those, obviously, weigh the same-- the key is the volume. A cubic meter of lead is going to weigh a lot more than a cubic meter of feathers. Making sure that we now know what the density is, let's go back to what we were doing before. We said that the downward force is equal to the mass of the liquid times the gravitational force, and so We could use this formula right here-- density is equal to mass times volume, so we could also say that mass is equal to density times volume. I just multiply both sides of this equation times volume. In this situation, force down is equal to-- let's substitute this with this. The mass of the liquid is equal to the density of the liquid times the volume of the liquid-- I could get rid of these l's-- times gravity. What's the volume of the liquid? The volume of the liquid is going to be the cross-sectional area of the cylinder times the height. So let's call this cross-sectional area A. A for area-- that's the area of the cylinder or the foil that's floating within the water."
    },
    {
        "Q": "at 7:13 howcome ligers arnt species\n",
        "A": "Organisms are considered part of a species if they can breed amongst themselves. Since ligers are sterile and unable to produce offspring, then that s your explanation.",
        "video_name": "Tmt4zrDK3dA",
        "timestamps": [
            433
        ],
        "3min_transcript": "And that is a different hybrid that has slightly different properties than a liger. I encourage you to look up what a tiglon is. Similarly, you give me a male donkey. And donkeys are clearly a species by themselves, because if you give me a male donkey and a female donkey they can reproduce, produce another donkey, and then that donkey can mate with other donkeys to produce more and more donkeys. So not only can a donkey interbreed with another donkey, but that product, that child donkey, can keep interbreeding with other donkeys. Similarly, horses, they can interbreed and produce fertile offspring. But if you give me a male donkey and a female horse they can mate and they can produce a mule. But once again, like the ligers, mules are not, at least as far as I know, mules they're not fertile. They cannot interbreed with each other. And because, even though donkeys and horses can breed and produce mules, their offspring aren't fertile. We don't consider donkeys and horses part of the same species. And we would consider mules, like a liger or a tiglon, we would consider them a hybrid. So these are all hybrids, or we would call cross. In general, the word hybrid is used when you have two things, two different types that are somehow coming together, somehow having a combination. And once again, like the case with the tiglon, you might say, well, what if I had a female donkey and a male horse? And then you would actually produce something called a hinny, which isn't as common as a mule. And people like to use mules, they're actually very good work animals because they have some of the good properties of both donkeys and horses. Hinnies are less common, but once again, it is possible. And they have different properties than mules. And I do want to emphasize this idea. to think about, well, how can we classify things? And we said, hey, maybe things that look and act similar, we can call a species. And maybe things that look and act different, we shouldn't call them species. But I want to show you a very typical case, one that's really all around us all the time, where this definition-- animals that can interbreed and the offspring are fertile-- really does seem to become much, much more important than just some notion of animals that look alike or animals that act the same. And the best example of that is with dogs. As I said, this is a very typical species here, because dogs-- and I just took a sample of some of the different types of breeds of dogs-- they can look very, very different. It's obvious, look at the difference between these dogs. For example, this little chihuahua here and this dog right over here. Obviously, they're size-wise, their look, and even how they act are much, much more different than maybe how this donkey would act relative to this horse,"
    },
    {
        "Q": "At,08:23,why is the upper limit zero ,as we are putting r^infinity which will be zero so what happened to the h^2?\n",
        "A": "As r gets larger, theta gets larger. As theta gets larger, the vertical force component gets shorter. At point infinity, theta becomes a right angle and the force is effectively flat (0 vertical).",
        "video_name": "TxwE4_dXo8s",
        "timestamps": [
            503
        ],
        "3min_transcript": "H squared plus r squared is u. We do that by definition. So u to the 3/2, which is equal to the antiderivative of-- we could write this as u to the minus 3/2 du. This is just kind of reverse the exponent rule. So that equals minus 2u to the minus 1/2, and we can confirm, right? If we take the derivative of this, minus 1/2 times minus 2 is 1, and then subtract 1 from here, we get minus 3/2. And then we could add plus c, but since we're eventually going to do a definite integral, the c's all cancel out. Or we could say that this is equal to-- since we made that substitution-- minus 2 over-- minus 1/2, that's the same thing as over the square root of h squared So all of the stuff I did in magenta was just to figure out the antiderivative of this, and we figured it out to be this: minus 2 over the square root of h squared plus r squared. So with that out of the way, let's continue evaluating our definite integral. So this expression simplifies to-- this is a marathon problem, but satisfying-- K-- let's get all the constants-- Kh pi sigma-- we can even take this minus 2 out-- times minus 2, and all of that, and we're going to evaluate the definite integral at the two boundaries-- 1 over the square root of h squared plus r squared evaluated at infinity minus it evaluated at 0, right? What is 1 over the square root of h squared plus infinity, right? What happens when we evaluate r at infinity? Well, the square root of infinity is still infinity, and 1 over infinity is 0, so this expression right here just becomes 0. When you evaluate it at infinity, this becomes 0 minus this expression evaluated at 0. So what happens when it's at 0? When r squared is 0, we get 1 over the square root of h squared, right? So let's write it all out. This becomes minus 2Kh pi sigma times 0 minus 1 over the square root of h squared. Well this equals minus 2Kh pi sigma times-- well, 1 over the"
    },
    {
        "Q": "6:14\nHe integrates u^(-3/2) to -2u^(-1/2). Shouldn't it instead be (-2^(-1/2))/3 ? (since you have to divide by -3/2)\n",
        "A": "Try taking the derivative of your answer and see if you get the original equation",
        "video_name": "TxwE4_dXo8s",
        "timestamps": [
            374
        ],
        "3min_transcript": "You'll see why in a second, but I'm going to take all the other constants out that we're not integrating across. So it's equal to Kh pi sigma times the integral from zero to infinity of what is this? So what did I leave in there? I left a 2r, so we could rewrite this as-- well, actually, I'm running out of space. 2r dr over h squared plus r squared to the 3/2, or we could think of it as the negative 3/2, right? So what is the antiderivative of here? Well, this is essentially the reverse chain rule, right? I could make a substitution here, if you're more comfortable using the substitution rule, but you We could make the substitution that u is equal to-- if we just want to figure out the antiderivative of this-- if u is equal to h squared plus r squared-- h is just a constant, right-- then du is just equal to-- I mean, the du dr-- this is a constant, so it equals 2r, or we could say du is equal to 2r dr. And so if we're trying to take the antiderivative of 2r dr over h squared plus r squared to the 3/2, this is the exact same thing as taking the antiderivative with this substitution. 2r dr, we just showed right here, that's the same thing as du, right? H squared plus r squared is u. We do that by definition. So u to the 3/2, which is equal to the antiderivative of-- we could write this as u to the minus 3/2 du. This is just kind of reverse the exponent rule. So that equals minus 2u to the minus 1/2, and we can confirm, right? If we take the derivative of this, minus 1/2 times minus 2 is 1, and then subtract 1 from here, we get minus 3/2. And then we could add plus c, but since we're eventually going to do a definite integral, the c's all cancel out. Or we could say that this is equal to-- since we made that substitution-- minus 2 over-- minus 1/2, that's the same thing as over the square root of h squared"
    },
    {
        "Q": "At 5:12, David says 'from 3 to 5'. Wouldn't that be -5?\n",
        "A": "Yes, that s correct. Since the turtle never makes it to +5m, you can assume that whenever he says five in this video, he means -5m.",
        "video_name": "GtoamALPOP0",
        "timestamps": [
            312
        ],
        "3min_transcript": "Sorry. Now what happens? Turtle at some later time, four seconds, is at negative five meters. That's all the way back here. So between two seconds and four seconds, this turtle rocketed back this way. That's also awkward. Turn down the reverse booster. What a noob, ah turtle. Here we go. Made it all the way back to here. Then what does the turtle do? After that point, turtle rockets forward. Makes it back to zero at this point. And then all the way back to three meters, so this turtle rockets forward back to three meters. That's what the turtle did. That's what this graph is representing, and that's how you read it. But there's more than that in here. I told you there was a lot of information and there is. So one piece of information you can get is the displacement of the turtle. And the displacement I'm gonna represent this with a delta x. And remember the displacement is the final position. Minus the initial position. You can find the displacement between any two times here, for the total time shown on the graph. But I could've found it between zero and like four seconds. Let's just do zero to 10, the whole thing. So what's the final position? The final position would be the position the turtle has. At 10 seconds she was at three meters. At 10 seconds 'cause I read the graph right there. Minus initially, 'cause we're considering the total time, at zero seconds, the turtle was also at three, that means the total displacement was zero. And that makes sense 'cause this turtle started at three. Rocketed back to five, well actually started at three, stood there for a second or two, rocketed back to five, rocketed back to three, ended at the same place she started, no total displacement. What else can we find? We can figure out the total distance. For the total distance traveled remember distance is the sum of all the path links traveled. So for this first path link, there was no distance traveled. That was the awkward part. We're not gonna talk about that. Then, so this is zero meters, plus between two seconds and four seconds, the turtle went from three to five. That's a distance traveled of eight meters. And should we make that negative? Nope. Distance is always positive. We make all these path links positive, we round them all up. So eight meters. Because the turtle went from three all the way back to five. That's the total distance of eight meters traveled. Plus between four seconds and 10 seconds, the turtle made it from negative five meters all the way back to three meters. That means she traveled another eight meters. That means the total distance traveled was 16 meters for the whole trip. Again you could have found this for two points any two points on here. Alright what else can you figure out? You can figure out the say average velocity, sometimes people represent that with a bar. Sometimes they just say the AVG."
    },
    {
        "Q": "\nWhy at 4:04 does the .0100 turn into acetate?",
        "A": "As you use up the reactants, you are making products. So, as CH3COOH and OH are being used up( hence the (-) sign), CH3COO is being made ( hence the (+) sign).",
        "video_name": "WbDL7xN-Pn0",
        "timestamps": [
            244
        ],
        "3min_transcript": "So notice we have the same number of moles of acid as we do of base. And the base is going to neutralize the acid. Let's go ahead and write the reaction. Let's write the neutralization reaction. If we have, we start with some acetic acid. And to our acidic solution we add our sodium hydroxide. So, we're adding some sodium hydroxide here. The hydroxide ions are going to take the acidic proton. So, a hydroxide ion takes this acidic proton right here. H plus and OH minus give us H2O. If you take away the acidic proton from acetic acid, you're left with acetate, CH 3 COO minus. And we're starting with .01 moles of acetic acid. So, let's color coordinate here. So, we're starting with 0.0100 moles. So, 0.0100 moles of acetic acid. And that's the same number of moles of base. 0.0100, so we have 0.0100 moles of base. Notice our mol ratio is one to one, so we have one to one here. So, all of the base is going to react. It's going to completely neutralize the acid that we originally had present. So, all of our base reacts, and we end up with zero here, and all of our acid has been completely neutralized. We lose all of this. So, we lose all of that, and so we've neutralized all of our acid too. So, this is our equivalence point for this titration. And if we're losing acetic acid, we're converting acetic acid into acetate. So, if we think about starting with zero moles of acetate, and we lose that turns into acetate. So, we have to write plus 0.0100 over here, so we're making that. So, we end up with 0.0100 moles of our acetate anion. All right, next. If we have moles of our acetate, and we have a concentration. So, if we find the total volume of our solution, we can find the concentration of acetate anions. So, let's do that next. We'll go back up here. What is the total volume now? We started with 50 mL, and we added 200 more. So, 50.0 and 200.0 give us 250.0 mL. So, that's our total volume now. And 250.0 mL would be, move our decimal place three, 0.25 liters. So, next we find our concentration of acetate. So, what is our concentration of acetate now?"
    },
    {
        "Q": "at 00:27, whats velosity\n",
        "A": "first of all it is velocity not velosity velocity is the rate of change of the position of an object ,equivalent to a specification of its speed and direction of motion or Or it is speed in a specific direction",
        "video_name": "zAx61CO5mDw",
        "timestamps": [
            27
        ],
        "3min_transcript": "Let's say we have some object that's moving in a circular path Let's say this is the center of the object path, the center of the circle So the object is moving in a circular path that looks something like that counterclockwise circular path--you could do that with clockwise as well I want to think about how fast it is spinning or orbiting around this center how that relates to its velocity? So let's say that this thing right over here is making five revolutions every second So in 1 second, 1 2 3 4 5. Every second it's making 5 revolutions So how could we relate that to how many radians it is doing per second? Remember radians is just one way to measure angles You could do with how degrees per second If we do it with radians, we know that each revolution is 2 pi radians which is really just you say you've gone 2 pi radii, whatever the radius of the circle is and that's where actually the definition of the radian comes from So if you go 5 revolutions per second and they're 2 pi per revolution then you can do a little bit of dimensional analysis. These cancel out and you get 5 times 2 pi which gets us to 5 times 2 pi gets us 10 pi radians per second And it works out the dimensional analysis and hopefully it also makes sense to you intuitively If you're doing five revolution a second, each of those revolutions is 2 pi radians so you're doing 10 pi radians per second. You're going 1 2 3 4 5, so that gives us 10, or 2 pi 2 pi 2 pi 2 pi 2 pi radians So this right here, either five revs per second or 10 pi radians per second they're both essentially measuring the same thing how fast are you orbiting around this central point? And this measure of how fast you're orbiting around a central point is called angular velocity It's called angular velocity because if you think about it this is telling us how fast is our angle changing, or speed of angle changing When you're dealing with it in two dimensions and this is typically when in a recent early physics course how we do deal with it Even though it's called the angular velocity it tends to be treated as angular speed It actually is a vector quantity and it's a little unintuitive that the vector's actually popping out of the page for this. It's actually a pseudo-vector and we'll talk more about that in the future So it is a vector quantity and the direction of the vector"
    },
    {
        "Q": "\nIs angular velocity a pseudo scalar quantity or a pseudo vector quantity? He mentions it as a (pseudo) scalar at 3:47 but as a pseudo vector at 2:57. Which one is correct?",
        "A": "Angular velocity is a vector. Sal mentions at 2:57 that angular velocity is a vector quantity and at the 3:47 he says it is often treated as a scaler not that it is a scaler.",
        "video_name": "zAx61CO5mDw",
        "timestamps": [
            227,
            177
        ],
        "3min_transcript": "So this right here, either five revs per second or 10 pi radians per second they're both essentially measuring the same thing how fast are you orbiting around this central point? And this measure of how fast you're orbiting around a central point is called angular velocity It's called angular velocity because if you think about it this is telling us how fast is our angle changing, or speed of angle changing When you're dealing with it in two dimensions and this is typically when in a recent early physics course how we do deal with it Even though it's called the angular velocity it tends to be treated as angular speed It actually is a vector quantity and it's a little unintuitive that the vector's actually popping out of the page for this. It's actually a pseudo-vector and we'll talk more about that in the future So it is a vector quantity and the direction of the vector when it's spinning in a counterclockwise direction there is a vector, the real angular vector does pop out of the page We start thinking about operating in three dimensions And if it's going clockwise, the angular velocity vector would pop into the page The way you think about that, right-hand rule Curl your fingers of your right hand in the direction that it's spinning and then your thumb is essentially pointing in the direction that the actual vector or the pseudo-vector's gonna going We'll not think too much about that For our purposes, when we're just thinking about two-dimensional plane right over here we can really think of an angular velocity as a--the official term is a pseudo-scaler but we can include that as a scaler quantity, as long as we do specify which way it is rotating So this right over here, this 10 pi radians per second we could call this its angular velocity a lower case omega right there Upper case omega looks like this So there's a couple of ways you could think about it You could say angular velocity is equal to change in angle over a change in time So for example, this is telling us 10 pi radians per second Or if you want to do in the calculus sense and take instantaneous angular velocity it would be the derivative of your angle with respect to time How the angle is changing with respect to time With that out of the way, I want to see if we can see how this relates to speed How does this relate to the actual speed of the object? So to get the speed of the object, we just have to think about how far is this object traveling"
    },
    {
        "Q": "\nAt 2:04 you were talking about net force. I do not know what the net force is. So please tell me.",
        "A": "Net force is a resultant force or the vector sum of all the forces onto the body.",
        "video_name": "1E3Z_R5AHdg",
        "timestamps": [
            124
        ],
        "3min_transcript": "What I want to do in this video is think about the two different ways of interpreting lowercase g. Which as we've talked about before, many textbooks will give you as either 9.81 meters per second squared downward or towards the Earth's center. Or sometimes it's given with a negative quantity that signifies the direction, which is essentially downwards, negative 9.81 meters per second squared. And probably the most typical way to interpret this value, as the acceleration due to gravity near Earth's surface for an object in free fall. And this is what we're going to focus on this video. And the reason why I'm stressing this last part is because we know of many objects that are not in free fall. For example, I am near the surface of the Earth right now, and I am not in free fall. What's happening to me right now is I'm sitting in a chair. And so this is my chair-- draw a little stick drawing on my chair, and this is me. And let's just say that the chair is supporting all my weight. So I have-- my legs are flying in the air. So this is me. And so what's happening right now? If I were in free fall, I would be accelerating towards the center of the Earth at 9.81 meters per second squared. But what's happening is, all of the force due to gravity is being completely offset by the normal force from the surface of the chair onto my pants, and so this is normal force. And now I'll make them both as vectors. is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll"
    },
    {
        "Q": "At 1:56,what is the normal force?\n",
        "A": "The normal force is the force from the surface perpendicular to the object. In this case, it is the force of the chair pushing up on the person.",
        "video_name": "1E3Z_R5AHdg",
        "timestamps": [
            116
        ],
        "3min_transcript": "What I want to do in this video is think about the two different ways of interpreting lowercase g. Which as we've talked about before, many textbooks will give you as either 9.81 meters per second squared downward or towards the Earth's center. Or sometimes it's given with a negative quantity that signifies the direction, which is essentially downwards, negative 9.81 meters per second squared. And probably the most typical way to interpret this value, as the acceleration due to gravity near Earth's surface for an object in free fall. And this is what we're going to focus on this video. And the reason why I'm stressing this last part is because we know of many objects that are not in free fall. For example, I am near the surface of the Earth right now, and I am not in free fall. What's happening to me right now is I'm sitting in a chair. And so this is my chair-- draw a little stick drawing on my chair, and this is me. And let's just say that the chair is supporting all my weight. So I have-- my legs are flying in the air. So this is me. And so what's happening right now? If I were in free fall, I would be accelerating towards the center of the Earth at 9.81 meters per second squared. But what's happening is, all of the force due to gravity is being completely offset by the normal force from the surface of the chair onto my pants, and so this is normal force. And now I'll make them both as vectors. is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll"
    },
    {
        "Q": "At 1:05 , if the air resistance isn't assumed negligible,is vertical velocity going to slow down,too?\n\n(sorry if grammar mistakes... not very good at English...)\n",
        "A": "yes. Both vertical and horizontal velocities will be affected by air resistance",
        "video_name": "sTp4cI9VyCU",
        "timestamps": [
            65
        ],
        "3min_transcript": "In the last video, I told you that we would figure out the final velocity of when this thing lands. So let's do that. I forgot to do it in the last video. So let's figure out the final velocity-- the vertical and the horizontal components of that final velocity. And then we can reconstruct the total final velocity. So the horizontal component is easy, because we already know that the horizontal component of its velocity is this value right over here, which we-- this 30 cosine of 80 degrees. And that's not going to change at any point in time. So this is going to be the horizontal component of the projectile's velocity when it lands. But what we need to do is figure out the vertical component of its velocity. Well, one thing we did figure out in the last video, we figured out what the time in the air is going to be. And we know a way of figuring out our final velocity from an initial velocity given our time in the air. We know that a change in velocity -- and we're only dealing in the vertical now-- we're only dealing with the vertical, We've assumed that air resistance is negligible. So we're only dealing with the vertical component right over here. We know that the change in velocity-- or, we could say the horizontal-- the vertical component of the change in velocity, is equal to the vertical component of the acceleration times time. Now, we know what the change in time is, we know it is-- I'll just write down times our time. And what is our change in velocity? Well, our change in velocity is our final vertical velocity minus our initial vertical velocity. And we know what our initial vertical velocity is, we solved for it. Our initial vertical velocity, we figured out, was 29.54 meters per second. That's 30 sine of 80 degrees, 29.54 meters per second. is equal to-- our acceleration in the vertical direction is negative, because it's accelerating us downwards, negative 9.8 meters per second squared. And our time in the air is 5.67 seconds. Times 5.67 seconds. And so we can solve for the vertical component of our final velocity. So once again, this is the vertical component. This isn't the total one. So, the vertical component. Let me-- well I wrote vertical up here. So there's the vertical component. So let's solve for this. So if you add 29.54 to both sides, you get the vertical component of your final velocity. Well, this is a vertical component, I didn't mark it up here properly-- is equal to 29.54 meters per second plus 9.8 plus -- or I should say minus-- meters per second. Minus 9.8 meters per second squared, times 5.67 seconds."
    },
    {
        "Q": "@3:40 What don't we use velocity as a vector quantity?\n",
        "A": "Sal does. It is showing the direction ( displacement ). At least this is my understanding. You can try to ask Sal.",
        "video_name": "vZOk8NnjILg",
        "timestamps": [
            220
        ],
        "3min_transcript": "I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time and the change in velocity actually continues changing but hopefully you're going to see the pattern here So between those two points in time, this is our change in velocity And let me translate that vector right over there It would look something like that change in velocity So what do you see, if I were to keep drawing more of these change in velocity vectors you would see at this point, the change in velocity would have to be going generally in that direction At this point, the change in velocity would have to be going generally in that direction So what do you see? What's the pattern for any point along this circular curve? Well, the change in velocity first of all, is perpendicular to the direction of the velocity itself And we haven't proved it, but it at least looks like it Looks like this is perpendicular And even more interesting, it looks like it's seeking the center"
    },
    {
        "Q": "at 3:00 sal starts drawing change in velocity vectors. they are essentially the difference of the 2 velocity vectors, right?!\n",
        "A": "Essentially, yes. The 2nd velocity vector minus the 1st velocity vector.",
        "video_name": "vZOk8NnjILg",
        "timestamps": [
            180
        ],
        "3min_transcript": "in particular the direction of the force would have to act on this object in order for the velocity vector to change like that? This remind ourselves if there was no force acting on this body this comes straight from Newton's 1st Law of motion then the velocity would not change neither the magnitude nor the direction of the velocity will change If there were no force acting on the subject it would just continue going on in the direction it was going it wouldn't curve; it wouldn't turn; the direction of its velocity wasn't changing Let's think about what the direction of that force would have to be and to do that, I'm gonna copy and paste these velocity vectors and keep track of what the direction of the change in velocity has to be Copy and paste that So that is our first velocity vector Copy all of these. This is our second one right over here I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time"
    },
    {
        "Q": "\n@3:04 when the tail of the two vectors are joined together, shouldn't the resultant vector be the diagnol (parallelogram law of addition) ?",
        "A": "Sal is calculating change in velocity, not overall velocity. If it was overall velocity, then yes, you would draw the head of one of the arrows adjoining the tail of the other, and calculate the diagonal.",
        "video_name": "vZOk8NnjILg",
        "timestamps": [
            184
        ],
        "3min_transcript": "in particular the direction of the force would have to act on this object in order for the velocity vector to change like that? This remind ourselves if there was no force acting on this body this comes straight from Newton's 1st Law of motion then the velocity would not change neither the magnitude nor the direction of the velocity will change If there were no force acting on the subject it would just continue going on in the direction it was going it wouldn't curve; it wouldn't turn; the direction of its velocity wasn't changing Let's think about what the direction of that force would have to be and to do that, I'm gonna copy and paste these velocity vectors and keep track of what the direction of the change in velocity has to be Copy and paste that So that is our first velocity vector Copy all of these. This is our second one right over here I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time"
    },
    {
        "Q": "\nat 5:00 why did you preform SN1 even though the nucleophile carries a negative charge and it is supposed to be a strong nucleophile so it must go SN2 ??",
        "A": "The answer is: steric hindrance. There isn\u00e2\u0080\u0099t enough room for the nucleophile to attack the back side of the carbon bearing the leaving group.",
        "video_name": "sDZDgctzbkI",
        "timestamps": [
            300
        ],
        "3min_transcript": "when the nucleophile attacked from the left, and on the right I'm holding when the nucleophile attacks from the right. So what's the relationship between these two? Well they're mirror images of each other, but if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm going to take the product on the left, and I'm going to turn it so that the S-H is coming out at me in space. So here we can see the S-H coming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product, and this time if we're going to keep the same carbon chain, the methyl group's coming out at us in space, the S-H is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two products that we got from the video, let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer by putting the S-H on a wedge, so let me just draw that in here. So here's our S-H on a wedge, which means the methyl group must be going away from us in space. I would have to show the S-H going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm going to say here approximately 50% is this enantiomer, and approximately 50% of our product is this enantiomer. Finally let's go through the hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral, right? It's sp3 hybridized so it has tetrahedral geometry. When we formed our carbocation, the carbon in red is now sp2 hybridized, so it has planar geometry. But for our products we're back to an sp3 hybridized carbon"
    },
    {
        "Q": "\nAt 1:40, where is the blood pressure taken?",
        "A": "on the upper arm where the brachial artery is. this is because the artery is close to the skin here and it is more convenient than taking it from the neck or the leg",
        "video_name": "J97G6BeYW0I",
        "timestamps": [
            100
        ],
        "3min_transcript": "So recently I went into my doctor's office, and I was told that my blood pressure was 115/75. So I thought we would talk about exactly what this means and try to figure out how to think about blood pressure in general, using these numbers and this experience as kind of a launching point. So the way I think about blood pressure is I always imagine kind of a tube and I imagine blood going through that tube. And this tube is like a blood vessel. So there's blood trying to get its way from one side to the other. And on its way, the neat thing that it's doing is as it flows, it's pushing out. So it's forcing against these walls, and specifically what I mean by that is there are cells and there's plasma, and all that stuff is pushing out against the walls of the blood vessels. So you've got a force, and that force So it's force over a surface area. And any time you see force over an area, you know that equals a pressure. And in this case, it's a blood pressure because it's the blood that's doing that work. So this is how I think of blood pressure, specifically as those little blue arrows. And the two questions that kind of pop into my mind anytime I'm thinking about blood pressure are where is the blood pressure being taken, and when are you taking it? So let's start with the first question, where? And by that I mean where in the circulatory system. So you've got the heart-- and this is my Valentine's Day heart-- and you've got the aorta coming off of the heart. And it's got lots of branches, but I'm going to just draw one branch, which is the artery leading off to my arms. This is called the brachial artery going off to my arm. maybe even more, the blood pressure that you're getting recorded, or the number that's being told to you, is being checked at this point. I marked it with a little x because that's kind of the upper arm. So that's usually where they're checking the blood pressure. And again, they're checking the force that the blood is putting on the vessel walls. So these little blue arrows. So that answers the where question. And certainly, you can imagine if I checked blood pressure let's say at some other spot, let's say over here or over here, you might get a different blood pressure reading than if I checked it at the yellow x. So really were just talking about that one spot. Now the other question is when are you checking it? So for this, let me show you a little table or a figure, So imagine that over time, time is this way, you have different recordings for blood pressure. So this will be blood pressure."
    },
    {
        "Q": "\nAt 02:06, why blood pressure is measured on brachial artery",
        "A": "It s easy to get to, in most people it s a very reliable measurement, and most people don t mind having the cuff around their arm during the measurement process.",
        "video_name": "J97G6BeYW0I",
        "timestamps": [
            126
        ],
        "3min_transcript": "So recently I went into my doctor's office, and I was told that my blood pressure was 115/75. So I thought we would talk about exactly what this means and try to figure out how to think about blood pressure in general, using these numbers and this experience as kind of a launching point. So the way I think about blood pressure is I always imagine kind of a tube and I imagine blood going through that tube. And this tube is like a blood vessel. So there's blood trying to get its way from one side to the other. And on its way, the neat thing that it's doing is as it flows, it's pushing out. So it's forcing against these walls, and specifically what I mean by that is there are cells and there's plasma, and all that stuff is pushing out against the walls of the blood vessels. So you've got a force, and that force So it's force over a surface area. And any time you see force over an area, you know that equals a pressure. And in this case, it's a blood pressure because it's the blood that's doing that work. So this is how I think of blood pressure, specifically as those little blue arrows. And the two questions that kind of pop into my mind anytime I'm thinking about blood pressure are where is the blood pressure being taken, and when are you taking it? So let's start with the first question, where? And by that I mean where in the circulatory system. So you've got the heart-- and this is my Valentine's Day heart-- and you've got the aorta coming off of the heart. And it's got lots of branches, but I'm going to just draw one branch, which is the artery leading off to my arms. This is called the brachial artery going off to my arm. maybe even more, the blood pressure that you're getting recorded, or the number that's being told to you, is being checked at this point. I marked it with a little x because that's kind of the upper arm. So that's usually where they're checking the blood pressure. And again, they're checking the force that the blood is putting on the vessel walls. So these little blue arrows. So that answers the where question. And certainly, you can imagine if I checked blood pressure let's say at some other spot, let's say over here or over here, you might get a different blood pressure reading than if I checked it at the yellow x. So really were just talking about that one spot. Now the other question is when are you checking it? So for this, let me show you a little table or a figure, So imagine that over time, time is this way, you have different recordings for blood pressure. So this will be blood pressure."
    },
    {
        "Q": "At 8:24, why is there a layer of non-fusing helium? isn't all helium supposed to fuse and form carbon and oxygen? Help!\n",
        "A": "That layer isn t at a temperature/pressure where it can fuse.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            504
        ],
        "3min_transcript": "But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening, is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember,"
    },
    {
        "Q": "\nAt 3:15 is are sun a \"main sequence\" star?",
        "A": "Yes, for now. In 5 billion years, it will run out of hydrogen and move away from the main sequence line and into the red giant phase.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            195
        ],
        "3min_transcript": "So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. Red giants are much, much larger than main sequence stars. But the whole time that this is getting more dense, the rest of the star is, you could kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler-- let me write that a little neater-- is actually cooler than the surface of a main sequence star. This right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today."
    },
    {
        "Q": "At 1:41, it is mentioned that the sun today is 'brighter and hotter' as it is fusing faster than it was when it was born 4.5 to 4.6 billion years ago. Is this an antithesis for global warming?\n",
        "A": "No, the Sun warming process is much slower than global warming appears to be. The Sun will get 1% brighter in the next 100 million years.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            101
        ],
        "3min_transcript": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you have hydrogen fusion going on. So that is hydrogen fusion, and then outside of the core, you just had hydrogen. You just hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like this soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense, because helium is a more massive atom. It is able to pack more mass in a smaller volume. So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger."
    },
    {
        "Q": "At 11:58 Sal says it would take a long time for a white dwarf to turn into a black dwarf, but do we have any estimation for the time it would take? And how long does it take for a planetary nebula to disappear?\n",
        "A": "If we define a black dwarf as a white dwarf cooled to 5 K, then it would take 10\u00c2\u00b9\u00e2\u0081\u00b5 years to form (which is considerable larger than the current age of the universe). Planetary nebulae, however, are very short-lived. They only last for tens of thousands of years on average.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            718
        ],
        "3min_transcript": "All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely it'll just be this superdense ball of carbon and oxygen, at which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive can imagine the more massive. There would be so much pressure on these things, because you have so much mass around it, that these would begin to fuse into heavier and heavier elements until we get to iron."
    },
    {
        "Q": "\nAt 8:38 why did Sal say 'you have helium that is not fusing'- whatever helium is there outside the core is continuously fusing due to the pressure right?",
        "A": "No, that s what he just explained to you. The helium outside the core may not be fusing because the pressure may not be high enough.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            518
        ],
        "3min_transcript": "is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember, and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out."
    },
    {
        "Q": "What part(s) of a star are luminous? Does the outside gas layer of a star produce light? (Such as at 11:11).\n",
        "A": "The core of the star produces light from fusion. The rest the gas just puts pressure on the core and scatter the light emitted by fusion.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            671
        ],
        "3min_transcript": "and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out. All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely"
    },
    {
        "Q": "at 2:30 why does the pulse turns upside down after reflection?\n",
        "A": "on reflection, there is a change in phase. (If the medium is fixed... if it is free to move, then there is no phase change)",
        "video_name": "gT0IqL1dyyk",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Instructor] If you've got a medium and you disturb it, you can create a wave. And if you create a wave in a medium that has no boundaries, in other words, a medium that's so big, this wave basically never meets the boundary, then there's nothing really stopping you from making a wave of any wave length or frequency whatsoever. In order words, there's not really any naturally preferred wave lengths, they're all pretty much as good as any other wave length. However, if you confine this wave into a medium that has boundaries, this wave is gonna reflect when it meets the boundary and that means it's gonna overlap with itself. And when this happens, you can create something that are called standing waves. And we'll talk about what these mean in a minute but the reason we care about them, is because when standing waves happen, they select preferred wave lengths and frequencies. Only particular wave lengths and frequencies are gonna set up these standing waves and what ends up happening is that these often become dominant and that's why these standing waves are important to study. So, let's study some standing waves. Let's take a particular example. Let's say you've got a string, whoa, not that many strings. this string down at both ends. So, you're gonna prevent any motion from happening at the end of this string. This string can wiggle in the middle but it can't wiggle at the end points. And this isn't that crazy. A guitar string is basically a string fixed at both ends. Piano strings are strings fixed at both ends. So, the physics behind standing waves determines the types of notes you're gonna get on all of these instruments. And by the way, this point over here, we're basically making sure that it has no motion. So, by nailing it down, what I really mean is that there's gong to be no motion at this end point and no motion at this end point. And instead of calling those no motion points, physicists came up with a name for that. They call these nodes. So node is really just a fancy word for not moving at that point. So, for this string, there's gonna be nodes at each end. And we'll see, when you set up a standing wave, it's possible that there's nodes in the middle as well but there don't have to be. For this string though, we're making sure that there have to be nodes at each end. Well, let's say, you give the end of the string here a little pluck and you cause a disturbance, that disturbance is gonna move down the line because that's what waves do. It's gonna come over here. Once it meets a boundary, it's gonna reflect back to the left. Now, it turns out, when a string hits a boundary where it's fixed, when it hits a node, in other words, it gets flipped over. So, you might have tried this before with the hose. If you send a pulse down the line and you try to see how it reflects, it gets reflected upside down. It doesn't matter too much for our purposes, but every time it's gonna reflect, it flips its direction and it keeps bouncing. Now, let's say instead of sending in a single pulse, we send in a whole bunch of pulses, right. We send in like a simple harmonic wave. Now when this thing reflects, it's gonna reflect back on top of itself because this leading edge will get reflected upside down this way and it's gonna meet all the rest of the wave behind it and overlap with it, creating some total wave"
    },
    {
        "Q": "\nat 1:47 does anyone know what galaxies the circled ones are?",
        "A": "I don t think anyone would bother to name all the billions of galaxys that Hubble can see. If they had names, they would be numbers.",
        "video_name": "Wl4re38deh0",
        "timestamps": [
            107
        ],
        "3min_transcript": "The whole point of this video is really just to look at, what in my mind is one of the coolest pictures ever taken by anything, and this was actually taken by the Hubble telescope and what they did is, they pointed the telescope at this area of our night sky and obviously the Hubble telescope, it's out in orbit, so it doesn't have to worry about all the interference from our actual atmosphere, so it gets a nice good look at things, but it's right over here, relative to the moon, and obviously the moon is moving around, but on that day, it was here, relative to the moon and they picked this location. They picked this location right over here because there weren't that many nearby stars there so it really allowed the telescope, because if there were nearby stars that light would've kind of outshone things that are behind it further away maybe perhaps galaxies. So just keep in mind everything you're going to see is in this little patch of the night sky, and I think the main point for showing the moon here. Obviously the moon is moving around so I'm not going to tell you which exact patch of sky this is. But to really give you an idea of how small of a patch of sky that really is. You really could've blocked things, but the galaxies are there, beyond that. The clusters of galaxies, and the super clusters of galaxies. So with that said out of the way, just remember everything we're talking about in this video is in this little patch of sky right over here. The whole point of this, once again like all of these videos is really to kind of just blow your mind. So this right here is what the Hubble telescope saw in that patch. Everything, everything that I'm showing you right here. I just want to be clear, some of these things are nearby stars, but most of these things are galaxies. That's a galaxy. That is a galaxy. That is a galaxy. That's a galaxy, and that's a galaxy. The reason I want to draw you this, obviously in the last video I showed you our local group, I showed you the Virgo super cluster, I showed you the kind of clusters of clusters and I even showed you the depiction of the observable universe. But what is amazing galaxy. Oh look there's another galaxy up here. That you can keep doing that forever and this is just in that little patch of sky, and this is obviously not all of the galaxies of the universe. These are just the ones that we could see. There ones that might be even further, or there definitely are ones that are further back and their light is just even more. We could probably even focus on a patch of sky like that and see that many galaxies again. You can kind of keep, keep zooming in, but I just this thing, and I encourage you I mean there's so many unbelievable images you can look up. They're all on NASA's website, a lot of these are on Wikipedia. But these images are just unbelievable. You can see a galaxy, another galaxy, another galaxy, another galaxy. I suspect some of this stuff might actually be clusters of galaxies. Galaxy, another galaxy. And remember I'm just circling these galaxies, just left and right, and you kind of lose sight of what each galaxy actually is. These galaxies have hundred of billions"
    },
    {
        "Q": "\nAt 0:36, she says that there is a dynamic equilibrium. I thought we were discussing chemical equilibrium. How do chemical and dynamic equilibriums relate?",
        "A": "They are the same thing - the chemical equilibrium is dynamic, as per Le Chatelier s Principle.",
        "video_name": "5gujU2QcGcY",
        "timestamps": [
            36
        ],
        "3min_transcript": "- [Voiceover] In this video we're going to go through an example reaction that uses Le Chatelier's principle. So what we're gonna do is, we're gonna apply Le Chatelier's principle to look at various changes to this reaction when we perturb our reaction from equilibrium. Just as a reminder, what do I mean when I say, This reaction is at equilibrium.? So what that means is we have a reversible reaction. We have the forward reaction which has some rate K forward. The reverse reaction has some rate K backward and a dynamic equilibrium. These rate are equal to each other, so all of the concentrations are going to stay constant, and then what we do is we decide to see what happens when we add some carbon dioxide gas. If we add carbon dioxide gas, the concentration of carbon dioxide gas will go up, or you can think about it as a partial presh going up. that if we had a reaction at equilibrium and then we perturbed it by adding more CO2, it will shift to try to reduce the effect of that change. It will favor the reverse reaction, so if we add CO2, what happens is, we favor our reactants. Another way we can see this is by looking at the equilibrium constant for this reaction. We can write our equilibrium constant K, where this is a capital K. It's kind of confusing, but I will try to make this look like a capital K. (laughs) We can write it in two ways, we can write it in terms of the molar concentration, and if we write Kc, the expression will be the product concentration, our CO2 gas concentration. And that's it, because when we write out Kc, we write out concentrations of gases but we don't include solids. Kc is just the concentration of CO2 at equilibrium. I'm going to write an eq there just to show that's the equilibrium concentration. I said that you can also write it in terms of partial pressures, so there's our fancy capital K with a p subscript, which means that instead of concentrations, we're writing everything for gases in terms of partial pressures. We have the partial pressure of CO2 and again, that's it, because everything else is a solid, so we don't include those in our equilibrium expression. Writing these expressions out will be really helpful for our second condition. We're going to think about what happens, when you increase the volume of our container. We can rewrite the partial pressure, actually, in terms of the volume, so if you use the ideal gas law, the partial pressure of CO2 is equal to"
    },
    {
        "Q": "At 5:36 there is a mention of when you add calcium carbonate (solid), you don't interrupt equilibrium. However thinking about that makes me wonder... For example, you add a scoop of calcium carbonate, which has a certain volume into this 'fixed' volume, isn't it so that it will change the pressure and/or volume slightly? And by doing so change the equilibrium a bit? Hope I'm clear. Thanks\n",
        "A": "That s a great question Sander Sloof! The assumption made in the video, which I unfortunately failed to state, is that changing the amount of solid did not change the total volume of the container significantly. I apologize for any confusion! Your thinking is entirely correct-if enough solid was added to change the volume of the container, that would also change gas pressures and perturb the reaction from equilibrium. Thanks for commenting!",
        "video_name": "5gujU2QcGcY",
        "timestamps": [
            336
        ],
        "3min_transcript": "Argon gas is an inert gas. We don't expect it to react to anything. One thing that will happen when you add the argon gas is it will increase the overall pressure of your container, so it will increase the total pressure. But that actually doesn't tell us what it does to our equilibrium. Let's look back at our equilibrium expressions Kc and Kp. We can see that the partial pressure for Kp only depends on the moles of our CO2 and our volume. Since we didn't change the moles of CO2 and we also didn't change the volume. Even though we increased P-total, the partial pressure of CO2 stayed the same. That means we didn't perturb our reaction from equilibrium and since we didn't perturb it from equilibrium, We are still at equilibrium. The concentrations will still stay the same. What happens when we add more calcium carbonate? That's our starting material and it is a solid. Our equilibrium expressions are determined by our CO2 concentrations, so adding more calcium carbonate, which is a solid, isn't actually going to perturb our reaction from equilibrium. Our reaction is still going to be at equilibrium and we will get no shift in concentrations. We're actually gonna look at one more thing. We're going to think about what happens when you add a catalyst. Let's say we want to speed up this reaction. We can envision what's going on here when we add the catalyst by using an energy diagram. If we have an energy diagram. We have energy on the Y axis between our reactants, or starting materials, with our products and I just sort of made up these relative energies. The way that I have this drawn here, we can see that our product is lower energy than our starting material and our forward rate, Kf, which is is up here, is determined by the size of this activation barrier between our starting material and our transition state. Our backward rate, Kb, is determined by the size of this energy barrier, so the difference in energy between the product and our transition state. If we add a catalyst to our reaction, we can think about it as lowering the activation energy for our reaction and that means we have a lower energy barrier for our forward reaction, so our forward reaction is gonna get faster,"
    },
    {
        "Q": "\nat 0:22 my beloved Mr.Khan said if there is g the situation will be little bit different but from equation T = 2pi sqrt(m/k) , must not g have any effect on this case?",
        "A": "Good observation :-). He said that because the figure that he made wouldn t be as simple as it is if we were dealing with gravity. There would be the weight of the object acting downwards (which would affect its position) and a whole lot other things. But eventually the formula would turn out to be the same.(T=2pi sqrt(m/k)). So I think he said that so that we don t get confused by the diagram. That s it. :-)",
        "video_name": "oqBHBO8cqLI",
        "timestamps": [
            22
        ],
        "3min_transcript": "And if you were covering your eyes because you didn't want to see calculus, I think you can open your eyes again. There shouldn't be any significant displays of calculus in this video. But just to review what we went over, we just said, OK if we have a spring-- and I drew it vertically this time-- but pretend like there's no gravity, or maybe pretend like we're viewing-- we're looking at the top of a table, because we don't want to look at the effect of We just want to look at a spring by itself. So this could be in deep space, or something else. But we're not thinking about gravity. But I drew it vertically just so that we can get more intuition for this curve. Well, we started off saying is if I have a spring and 0-- x equals 0 is kind of the natural resting point of the spring, if I just let this mass-- if I didn't pull on the spring at all. But I have a mass attached to the spring, and if I were to stretch the spring to point A, we said, well what happens? Well, it starts with very little velocity, but there's a restorative force, that's going to be pulling it back towards this position. So that force will accelerate the mass, accelerate the mass, accelerate the mass, until it gets right here. And then it'll have a lot of velocity here, but then it'll And then it'll decelerate, decelerate, decelerate. Its velocity will stop, and it'll come back up. And if we drew this as a function of time, this is what happens. It starts moving very slowly, accelerates. At this point, at x equals 0, it has its maximum speed. So the rate of change of velocity-- or the rate of change of position is fastest. And we can see the slope is very fast right here. And then, we start slowing down again, slowing down, until we get back to the spot of A. And then we keep going up and down, up and down, like that. And we showed that actually, the equation for the mass's position as a function of time is x of t-- and we used a little bit of differential equations to prove it. But this equation-- not that I recommend that you memorize anything-- but this is a pretty useful equation to memorize. Because you can use it to pretty much figure out anything-- about the position, or of the mass at any given anything else. Even the velocity, if you know a little bit of calculus, you can figure out the velocity at anytime, of the object. And that's pretty neat. So what can we do now? Well, let's try to figure out the period of this oscillating system. And just so you know-- I know I put the label harmonic motion on all of these-- this is simple harmonic motion. Simple harmonic motion is something that can be described by a trigonometric function like this. And it just oscillates back and forth, back and forth. And so, what we're doing is harmonic motion. And now, let's figure out what this period is. Remember we said that after T seconds, it gets back to its original position, and then after another T seconds, it gets back to its original position. Let's figure out with this T is. And that's essentially its period, right? What's the period of a function? It's how long it takes to get back to your starting point. Or how long it takes for the whole cycle to happen once."
    },
    {
        "Q": "\nat 9:28, Sal used the formular Qr=2*pi*r*dr*sigma (i'm sorry i can't write the sigma symbol on my laptop) to calculate the area of the ring but i don't quite get it. Can anyone show me how to understand it? Thank you very much.",
        "A": "2*pi*r shows the circumference of the ring which the width can be ignored. cut the ring into infinite tiny rectangles, each rectangle has a width of dr. so the total area can be expressed as 2*pi*r*dr sigma. Hope this would help.",
        "video_name": "prLfVucoxpw",
        "timestamps": [
            568
        ],
        "3min_transcript": "So now let's see if we can figure out what the magnitude of the electric field is, and then we can put it back into this and we'll figure out the y-component from this point. And actually, we're not just going to figure out the electric field just from that point, we're going to figure out the electric field from a ring that's surrounding this. So let me give you a little bit of perspective or draw it with a little bit of perspective. So this is my infinite plate again. I'll draw it in yellow again since I originally drew it in yellow. This is my infinite plate. It goes in every direction. And then I have my charge floating above this plate someplace at height of h. And this point here, this could have been right here maybe, but what I'm going to do is I'm going to draw a ring that's of an equal radius around this point right here. Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on"
    },
    {
        "Q": "at 9:52 sal mentioned test charge what does that mean?\n",
        "A": "A test charge is a small charge that we imagine placing somewhere in a field to help us think about the direction and magnitude of the field at that point.",
        "video_name": "prLfVucoxpw",
        "timestamps": [
            592
        ],
        "3min_transcript": "Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on Well, this could be one of the points on the ring and this could be another one, right? And this is like a cross-section. So the distance at any point, this distance right here, is once again by the Pythagorean theorem because this is also r. This distance is the square root of h squared plus r squared. It's the same thing as that. So it's the distance squared and that's equal to k times the charge in the ring times our test charge divided by Well, distance is the square root of h squared plus r squared, so if we square that, it just becomes h squared plus r squared. And if we want to know the electric field created by that ring, the electric field is just the force per test charge, so if we divide both sides by Q, we learned that the electric field of the ring is equal to Coulomb's constant"
    },
    {
        "Q": "\nAt 8:50, \"Acyl\" groups are introduced. But what is the difference between an acyl group and a carbonyl group? Don't they have the same structures?",
        "A": "Great question, very tricky to answer too! A carbonyl has the structure of C=O only. An acyl has the structure of R-C=O, where R is any hydrocarbon. A carbonyl is present in a acyl group. Example: (CH3)2-C=O is has an acyl group, which is CH3-C=O. It also has a carbonyl group, C=O.",
        "video_name": "OpyTJbzA7Fk",
        "timestamps": [
            530
        ],
        "3min_transcript": "is this green bond. And this second green bond is this green bond. And this third green bond is that green bond. The way I've drawn it right now, each of these oxygens haven't let go of its hydrogens. And that could actually happen before or after or all at the same time. Chemistry actually is not a clean thing. But I could, if I want, I could draw the hydrogens here. I could draw the hydrogens. I could draw the hydrogens over here and then these would have a positive charge. These would have a positive charge. But then you could imagine another water molecule coming by and kind of taking one of these hydrogen protons, taking the hydrogen protons away from each of those oxygens. this molecule right over here. And remember we produced three water molecules. So that's one, two and three water molecules. And now this molecule, if you ignore the water molecules out there, this is a triglyceride. Let me write it again. Actually, let me write the slightly more technical term. Sometimes referred to as triacylglycerol. Well we know where the glycerol comes from. It has this glycerine or this glycerol backbone right over here. Now what is \"acyl\" mean? Well acyl is a functional group where you have a carbon that's part of a carbonyl group. It can be bound to a kind of an organic chain And then it's bound to something else. And so we have three acyl groups so \"triacyl\". So this right over here, this right here is an acyl group. This right over here is an acyl group. This right over here is an acyl group. Each of them, they're bound to an oxygen right over here. And actually that gives us more practice with another functional group. When you have a situation-- Let me get another color out. When you have a carbonyl group and let's say you have just an organic chain right over here and then you have an oxygen and then you have another organic chain right over there. This thing is called an ester. And you actually see three esters right over here. So you see this ester. This ester right over here."
    },
    {
        "Q": "\nat 7:48 why would Oxygen have a positive charge?",
        "A": "2 electrons in the S orbital (-2), and 1 lone pair of electrons (-2), and 3 covalent bonds (3 x-1), give a total electron charge of -7 and oxygen has 8 protons so it ends up with a net positive charge.",
        "video_name": "OpyTJbzA7Fk",
        "timestamps": [
            468
        ],
        "3min_transcript": "and then just as he's leaving, this guy comes back. So there's different ways that all of these could happen but this is the general idea. And then you have it happening a third time down here. One of these lone pairs come and form a bond with this carbon. This carbon in the carbonyl group, part of this carboxyl group. And so once again, this guy can take those two electrons away and maybe share that pair with a hydrogen proton. Once again, this is forming a water molecule, this is forming a water molecule. So three water molecules are going to be produced. This is why we call it dehydration synthesis. We're losing three water molecules in order to form these bonds. So what's it going to look like after this has happened? Well, let me scroll down here. And actually let me just zoom. Actually just let me scroll down here. So this green bond over here is going to now-- is this green bond. And this second green bond is this green bond. And this third green bond is that green bond. The way I've drawn it right now, each of these oxygens haven't let go of its hydrogens. And that could actually happen before or after or all at the same time. Chemistry actually is not a clean thing. But I could, if I want, I could draw the hydrogens here. I could draw the hydrogens. I could draw the hydrogens over here and then these would have a positive charge. These would have a positive charge. But then you could imagine another water molecule coming by and kind of taking one of these hydrogen protons, taking the hydrogen protons away from each of those oxygens. this molecule right over here. And remember we produced three water molecules. So that's one, two and three water molecules. And now this molecule, if you ignore the water molecules out there, this is a triglyceride. Let me write it again. Actually, let me write the slightly more technical term. Sometimes referred to as triacylglycerol. Well we know where the glycerol comes from. It has this glycerine or this glycerol backbone right over here. Now what is \"acyl\" mean? Well acyl is a functional group where you have a carbon that's part of a carbonyl group. It can be bound to a kind of an organic chain"
    },
    {
        "Q": "I noticed how you said, at 3:10, \"The period does not change this way.\" This statement was referring to the horizontal period I presume. Is there a vertical period I am unaware of?\n",
        "A": "The period is the time for one oscillation. There s no horizontal or vertical to it.",
        "video_name": "6M_bjRzyUn0",
        "timestamps": [
            190
        ],
        "3min_transcript": "Some of you might say, yes, it should increase the period because look, now it has farther to travel, right? Instead of just traveling through this amount, whoa that looked horrible, instead of just traveling through this amount back and forth, it's gotta travel through this amount back and forth. Since it has farther to travel, the period should increase. But some of you might also say, wait a minute. If we pull this mass farther, we know Hooke's law says that the force is proportional, the force from the spring, proportional to the amount that the spring is stretched. So, if I pulled this mass back farther, there's gonna be a larger force that's gonna cause this mass to have a larger velocity when it gets to you, a larger speed when it gets to the equilibrium position, so it's gonna be moving faster than it would have. So, since it moves faster, maybe it takes less time for this to go through a cycle. But it turns out those two effects offset exactly. In other words, the fact that this mass and the fact that it will now be traveling faster offset perfectly and it doesn't affect the period at all. This is kinda crazy but something you need to remember. The amplitude, changes in the amplitude do not affect the period at all. So pull this mass back a little bit, just a little bit of an amplitude, it'll oscillate with a certain period, let's say, three seconds, just to make it not abstract. And let's say we pull it back much farther. It should oscillate still with three seconds. So it has farther to travel, but it's gonna be traveling faster and the amplitude does not affect the period for a mass oscillating on a spring. This is kinda crazy, but it's true and it's important to remember. This amplitude does not affect the period. In other words, if you were to look at this on a graph, let's say you graphed this, put this thing on a graph, if we increase the amplitude, what would happen to this graph? Well, it would just stretch this way, right? We'd have a bigger amplitude, but you can do that and there would not necessarily If you leave everything else the same and all you do is change the amplitude, the period would remain the same. The period this way would not change. So, changes in amplitude do not affect the period. So, what does affect the period? I'd be like, alright, so the amplitude doesn't affect it, what does affect the period? Well, let me just give you the formula for it. So the formula for the period of a mass on a spring is the period here is gonna be equal to, this is for the period of a mass on a spring, turns out it's equal to two pi times the square root of the mass that's connected to the spring divided by the spring constant. That is the same spring constant that you have in Hooke's law, so it's that spring constant there. It's also the one you see in the energy formula for a spring, same spring constant all the way. This is the formula for the period of a mass on a spring. Now, I'm not gonna derive this because the derivations typically involve calculus. If you know some calculus and you want to see how this is derived,"
    },
    {
        "Q": "At 1:58 why is it that although there are two sets of genetic code, the sister chromatids are still considered as only 1 chromosome?\n",
        "A": "The sister chromatids are still attached to each other by a centromere. Only when separated are they considered 2 chromosomes. You usually visualize a chromosome as X-shaped, do you not? That is a chromosome that has undergone DNA replication therefore it has two sets of genetic code.",
        "video_name": "TKGcfbyFXsw",
        "timestamps": [
            118
        ],
        "3min_transcript": "- In the previous video, we talked about interphase which is the bulk of a cell's life cycle as it grows and its DNA replicates, and it grows some more. And now, we're gonna talk about the actual cell division. We're gonna talk about mitosis. And if you wanna be precise, mitosis is the process by which this one nucleus will turn into two nuclei that each have the original genetic information. Now, as we exit mitosis, we get into cytokineses which will then split each of the nuclei into a separate cell when we split the cytoplasm right over here. We split or the cell as it turns into to cells. Well, let's see how all of this happens. So the first phase, and I'll leave the end of interphase right over here. We have this big cell, our DNA has been replicated. We have two centrosomes right over here. The first phase of mitosis involves the cell enough space here. So, involve. So, this is the cell right over here. So we're gonna go to this phase right over here. And a few things start happening. One, the DNA, the chromosomes go from being in their chromatin form where they're all spread out to kind of a more condensed form that you can actually see from a light microscope. So for example, that magenta chromosome which is now made up of two sister chromatids after replication, we talked about that in the interphase video. It might look something like this in a ... It might look something like this if you were to look in a microscope. It's unlikely to be magenta but it's gonna have kind of that classic chromosome shape that you're used to seeing in textbooks. And it has the centromere that connects these two sister chromatids. Right now, both of these two sister chromatids combined are considered to be one chromosome. the magenta stuff was still considered to be one chromosome. And we can draw the blue chromosome. Once again, it's now in the condensed form. That's one sister chromatid right over there. That's another sister chromatid. They are connected at the centromere. So they're condensing now as we enter into mitosis. And the nuclear membrane starts to go away. So the nuclear membrane is starting to go away. And these two centrosomes are starting to migrate to opposite sides of the cell. So one of them is going over here and one of them is maybe going to go over here. So they're migrating to opposite sides of the cell. And it's pretty incredible. It's easy to say, \"Oh, this happens and then that happens.\" Remember, this cell doesn't have a brain. This is all happening through chemical and thermodynamic"
    },
    {
        "Q": "\nat 8:36 what is happening here how the two events are experienced at different times\nin A and B frame of reference",
        "A": "Time is relative.",
        "video_name": "2BVGig1LXLs",
        "timestamps": [
            516
        ],
        "3min_transcript": "if you think about the time axis to the left and the right. And since they're equally skewed, the time dilation relative to this rest frame is the scaling is going to be the same. So we could put both of these on, both of these on the same scale. And you could see that this neutral frame of reference, this ct prime prime frame of reference, I drew them over here, that if you look at either of A or B's frame of reference, they're going to be in between A and B's frame of reference. But here it's the neutral one. It's the one where we're drawing the time or we're drawing the two axes being perpendicular to each other. And now if you look at these two events, so if you look at this first event, where you have our delta ct, so you look at this first event where you have a delta in ct between right when the spaceships pass and right over there. You see that if you were to look at that, if you were to look at the ct prime coordinate for that event, when you go parallel to the x prime axis, So this is, that is your change in ct prime. And likewise, if you have that other event that I drew in that blue-green color, right over here. And this is your change in ct prime. This is your change in ct prime, well then if you think of it from A's frame of reference, well we just follow the x-axis not the x prime. We go parallel to it. We end up right over there. Now what's really interesting about this, what's really interesting about this is from A's frame of reference, the yellow event happens before the blue event. But in B's frame of reference, the blue event happens before the yellow event. So really, really, really fascinating things going on but what I really like about this diagram is that A and B's frame of reference are going to have the same scale since they're both equally skewed to the left and the right if we're thinking about the time axis. And this type of diagram is called a Loedel diagram. a variation of a Minkowski diagram but it lets us really appreciate the symmetry between these frames of reference."
    },
    {
        "Q": "\nOK, at 2:20, why did you divide acceleration by 2?",
        "A": "mm. alight thnx",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            140
        ],
        "3min_transcript": "Just want to follow up on the last video, where we threw balls in the air, and saw how long they stayed up And we used that to figure out how fast we initially threw the ball and how high they went in the air. And in the last video, we did it with specific numbers. In this video, I just want to see if we can derive some interesting formulas so that we can do the computations really fast in our brains, while we're playing this game out on some type of a field, and we don't necessarily have any paper around. So let's say that the ball is in the air for delta t. Delta t is equal to time in the air. Then we know that the time up is going to be half that, which is the same thing as the time down. The time up is going to be equal to delta t-- I'm going to do that in the same color-- is going to be equal to the time in the air divided by 2. So what was our initial velocity? Well, all we have to do is remind ourselves that the change in velocity, which is the same thing as the final velocity minus the initial velocity. we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation."
    },
    {
        "Q": "I don't understand Sal's \"Oil Rig\" reference at 8:57, it is something to do with NADPH's function in cellular respiration?\n",
        "A": "It refers to Redox reactions! Its a little acronym to remember the difference between oxidation and reduction. O-oxidation i-is l-lose and R-reductuion i-is g-gain. I was taught Leo the lion says Ger. Which is the same principle! (Loss electrons= oxidation (LEO) and Gain electrons=reduction (GER))",
        "video_name": "-rsYk4eCKnA",
        "timestamps": [
            537
        ],
        "3min_transcript": "although they do occur when the sun is out. They don't need those photons, but they need the byproducts from the light reaction to occur, so that's why it's called the light-independent reaction. They occur while the sun is out, but they don't need the sun. This needs the sun, so let me make it very clear. So this requires sunlight. This requires photons. And let me just make a very brief overview of this. This'll maybe let us start building a scaffold from which we can dig deeper. So the light reactions need photons, and then it needs water. So water goes into the light reactions and out of the other side of the light reactions. We end up with some molecular oxygen. So that's what happens in the light reactions, and I'm going to go much deeper into what actually occurs. And what the light the actions produce is ATP, which we know It produces ATP and it produces NADPH. Now, when we studied cellular respiration, we saw the molecule NADH. NADPH is very similar. You just have this P there. You just have this phosphate group there, but they really perform similar mechanisms. That this agent right here, this molecule right here, is able to give away-- now let's think about what this means-- it's able to give away this hydrogen and the electron associated with this hydrogen. So if you give away an electron to someone else or someone else gains an electron, that something else is being reduced. Let me write that down. This is a good reminder. OIL RIG. Oxidation is losing an electron. Your charge is reduced when you gain an electron. It has a negative charge. So this is a reducing agent. It gets oxidized by losing the hydrogen and the electron with it. I have a whole discussion on the biological versus chemistry view of oxidation, but it's the same idea. When I lose a hydrogen, I also lose the ability to hog that hydrogen's electron. So this right here, when it reacts with other things, it's a reducing agent. It gives away this hydrogen and the electron associated with it, and so the other thing gets reduced. So this thing is a reducing agent. And what's useful about it is when this hydrogen, and especially the electron associated with that hydrogen, goes from the NADPH to, say, another molecule and goes to a lower energy state, that energy can also be used in the"
    },
    {
        "Q": "At 10:35, Sal says that the dark reactions are called the Calvin Cycle, aren't light reactions called the Calvin cycle and dark reactions, the Hill reactions?\n",
        "A": "The Light Reactions are called the Light Reactions and the Calvin Cycle is sometimes called the dark reactions, but it is more accurate to describe the Calvin Cycle as the light independent reactions because the Calvin Cycle occurs whether the sun is shining or not.",
        "video_name": "-rsYk4eCKnA",
        "timestamps": [
            635
        ],
        "3min_transcript": "Your charge is reduced when you gain an electron. It has a negative charge. So this is a reducing agent. It gets oxidized by losing the hydrogen and the electron with it. I have a whole discussion on the biological versus chemistry view of oxidation, but it's the same idea. When I lose a hydrogen, I also lose the ability to hog that hydrogen's electron. So this right here, when it reacts with other things, it's a reducing agent. It gives away this hydrogen and the electron associated with it, and so the other thing gets reduced. So this thing is a reducing agent. And what's useful about it is when this hydrogen, and especially the electron associated with that hydrogen, goes from the NADPH to, say, another molecule and goes to a lower energy state, that energy can also be used in the And we saw in cellular respiration the very similar molecule, NADH, that through the Kreb Cycle, or actually more importantly, that through the electron transport chain, was able to help produce ATP as it gave away its electrons and they went to lower energy states. But I don't want to confuse you too much. So the light reactions, you take in photons, you take in water, it spits out oxygen, and it spits out ATP and NADPH that can then be used in the dark reactions. And the dark reactions, for most plants we talk about, it's called the Calvin Cycle. And I'll go into a lot more detail of what actually occurs in the Calvin Cycle, but it takes in the ATP, the NADPH, and it produces-- it doesn't directly produce glucose. It produces-- oh, you probably saw this. You could call it PGAL. These all stand for-- let me write these down-- this is phosphoglyceraldehyde. My handwriting broke down. Or you could call it glyceraldehyde 3-phosphate. Same exact molecule. You can almost imagine it as-- this is a very gross oversimplification-- as three carbons with a phosphate group attached to it. But this can then be used to produce other carbohydrates, including glucose. If you have two of these, you can use those two to produce glucose. So let's just take a quick overview again because this is super important. I'm going to make videos on the light reactions and the"
    },
    {
        "Q": "at 9:20, the molecule has 3 carbons instead of 4, right? because the double bond is between two carbon molecules\n",
        "A": "Yes that molecule only has three carbons",
        "video_name": "8x8tA4YPhJw",
        "timestamps": [
            560
        ],
        "3min_transcript": "So let me draw in some electrons here and I'll use red, so this bond is two electrons, and then we have a bunch of electrons in here, so how many electrons around oxygen do we have so far? Well, we would have three, just these three, and we need five, which means we need two more, which means we need a lone pair of electrons on the oxygen. So that's a little bit too complicated, I think, for figuring it out, but you could use that method, or you could just learn this pattern, right, and eventually, you'll have to have this pattern down pretty well. Let's look at another example for assigning formal charge to oxygen. So our goal is to find the formal charge on oxygen in this example, and we put in our electrons in this bond. Each bond represents two electrons, so the formal charge on oxygen, is equal to the number of valence electrons that oxygen is supposed to have which is six, minus the number of valence electrons and in this example, right, we would take one of these electrons from this bond, and how many electrons is that total around oxygen? This would be one, two, three, four, five, six, seven, so six minus seven gives us a formal charge of negative one. So I could re-draw that over here, I could say oxygen has three lone pairs of electrons, and a negative one formal charge, so our pattern, this time, our pattern is one bond. Here's the one bond, and then three lone pairs of electrons. So let me write that down. So the pattern is one bond, when oxygen has one bond, and three lone pairs of electrons the formal charge is negative one, just like we saw up here with the calculation. So we could leave those electrons off if you wanted to save some time, we could just say, oh, this is oxygen with a negative one formal charge, three lone pairs of electrons on that oxygen. Let's look at one more example, where formal charge is negative one. So right here, this oxygen has a negative one formal charge, and we can see it already has one bond to it. And so the pattern, of course, is one bond plus three lone pairs of electrons. So we already have the one bond. In order for that oxygen to have a negative one formal charge, we need three lone pairs of electrons. So we could re-draw this, so that is one way to represent that ion, and we could also represent it like this, with putting three lone pairs of electrons on that oxygen with a negative one formal charge. So again, become familiar with these patterns."
    },
    {
        "Q": "\nat 8:22, why doesnt the n\n-14 also have a half life",
        "A": "Some atoms are stable, others aren t. C-14 is not, so it decays, but N is stable, so it doesn t decay.",
        "video_name": "9REPnibO4IQ",
        "timestamps": [
            502
        ],
        "3min_transcript": "chance that any of the guys that are carbon will turn to nitrogen. So if you go back after a half-life, half of the atoms will now be nitrogen. So now you have, after one half-life-- So let's ignore this. So we started with this. All 10 grams were carbon. 10 grams of c-14. This is after one half-life. And now we have five grams of c-14. And we have five grams of nitrogen-14. Fair enough. Let's think about what happens after another half-life. Well we said that during a half-life, 5,740 years in the case of carbon-14-- all different elements have a different half-life, if they're radioactive-- over atom-- there's a 50% chance it'll decay. So if we go to another half-life, if we go another half-life from there, I had five grams of carbon-14. So let me actually copy and paste this one. This is what I started with. Now after another half-life-- you can ignore all my little, actually let me erase some of this up here. Let me clean it up a little bit. After one one half-life, what happens? Well I now am left with five grams of carbon-14. Those five grams of carbon-14, every one of those atoms still has, over the next-- whatever that number was, 5,740 years-- after 5,740 years, all of those once again have a 50% chance. And by the law of large numbers, half of them will So we'll have even more conversion into nitrogen-14. So now half of that five grams. So now we're only left with 2.5 grams of c-14. And how much nitrogen-14? Well we have another two and a half went to nitrogen. So now we have seven and a half grams of nitrogen-14. And we could keep going further into the future, and after every half-life, 5,740 years, we will have half of the carbon that we started. But we'll always have an infinitesimal amount of carbon. But let me ask you a question. Let's say I'm just staring at one carbon atom. Let's say I just have this one carbon atom. You know, I've got its nucleus, with its c-14. So it's got its six protons. 1, 2, 3, 4, 5, 6. It's got its eight neutrons. It's got its six electrons. 1, 2, 3, 4, 5, 6, whatever."
    },
    {
        "Q": "At about 7:25, why didn't he just add T2 to the zero in the blue equation so he would get sqrt(3)T1 = T2 and plug that into any T2s as opposed to him multiplying the second equation from the bottom by the sqrt(3) and making it more complicated? I got the same answer for both. Is there a mathematical reason he didn't do that?\n",
        "A": "That s just the strategy that he used to solve this calculus , your strategy is correct too.I think he did this way because it makes more easy to solve the system of equations by elimination.",
        "video_name": "zwDJ1wVr7Is",
        "timestamps": [
            445
        ],
        "3min_transcript": "to be 10 Newtons. Because it's offsetting this force of gravity. So what's this y component? Well, this was T1 of cosine of 30. This should start to become a little second nature to you that this is T1 sine of 30, this y component right here. So T1-- Let me write it here. T1 sine of 30 degrees plus this vector, which is T2 sine of 60 degrees. You could review your trigonometry and your SOH-CAH-TOA. Frankly, I think, just seeing what people get confused on is the trigonometry. But you can review the trig modules and maybe some of the earlier force vector modules that we did. And hopefully, these will make sense. I'm skipping a few steps. And these will equal 10 Newtons. And let's rewrite this up here where I substitute the values. Actually, let me do it right here. What's the sine of 30 degrees? The sine of 30 degrees is 1/2 so we get 1/2 T1 plus the sine of 60 degrees, which is square root of 3 over 2. Square root of 3 over 2 T2 is equal to 10. And then I don't like this, all these 2's and this 1/2 here. So let's multiply this whole equation by 2. So 2 times 1/2, that's 1. So you get T1 plus the square root of 3 T2 is equal to, 2 times 10 , is 20. Similarly, let's take this equation up here and let's multiply this equation by 2 and bring it down here. So this is the original one that we got. So if we multiply this whole thing by 2-- I'll do it in this color so that you know that it's a different equation. So if you multiply square root of 3 over 2 times 2-- I'm just doing this to get rid of the 2's in the denominator. So you get square root of 3 T1 minus T2 is equal to 0 because And let's see what we could do. What if we take this top equation because we want to start canceling out some terms. Let's take this top equation and let's multiply it by-- oh, I don't know. Let's multiply it by the square root of 3. So you get the square root of 3 T1. I'm taking this top equation multiplied by the square root of 3. This is just a system of equations that I'm solving for. And the square root of 3 times this right here. Square root of 3 times square root of 3 is 3. So plus 3 T2 is equal to 20 square root of 3. And now what I want to do is let's-- I know I'm doing a lot of equation manipulation here."
    },
    {
        "Q": "\n7:25 \u00c2\u00bfand according with that graph how many electrons will lose the element with tahat energy 1 or all the shell?",
        "A": "That graph is for the first ionisation energy only, so the energy required for an atom to lose one electron only.",
        "video_name": "5CBs36jtZxY",
        "timestamps": [
            445
        ],
        "3min_transcript": "they don't want their electron configurations messed with. So, it would be very hard... Neon on down has their eight electrons that (mumbling) Octet Rule. Helium has two which is full for the first shell, and so it's very hard to remove an electron from here, and so it has a very high ionization energy. Low energy, easy to remove electrons. Or especially the first electron, and then here you have a high ionization energy. I know you have trouble seeing that H. So, this is high, high ionization energy, and that's the general trend across the periodic table. As you go from left to right, you go from low ionization energy to high ionization energy. Now, what about trends up and down the periodic table? Well, within any group, if we, even if we look at the Alkali, if we look at the Alkali Metals right over here, if we're down at the bottom, if we're looking at, if we're looking at, say, Cesium right over here, that electron in the, one, two, three, four, five, six, further from that one electron that Lithium has and its second shell. So, it's going to be, it's going to be further away. It's not going to be as closely bound to the nucleus, I guess you could say. So, this is going to be even, that one electron's gonna even easier to remove than the one electron in the outermost shell of Lithium. So, this one has even lower, even lower, even lower... And that's even going to be true of the Noble Gases out here that Xenon, that it's electrons in its outermost shell, even though it has eight valence electrons, they're further away from the nucleus, and so they're a little, the energy required to remove them is still going to be high but it's going to be lower than the energy from, from say Neon or Helium. So, this is low. So, once again, ionization energy low to high as we go from left to right, and low to high as we go from bottom to top. that if we go from the bottom left to the top right we go from low ionization energy, very easy to remove an electron from these characters right over here to high ionization energy, very hard to move, remove an electron from these characters over here. And you can see it if, you could see in a trend of actual measured ionization energies and I like to see charts like this because it kind of show you where the periodic table came from when people noticed these kind of periodic trends. It's like, hey, it looks like there's some common patterns here. But on this one in particular we see on this axis we have ionization energy and electron volts, that's actually, it's literally a, this is units of energy. You could convert it to Joules if you like. Then over here, we're increasing the atomic numbers. So, we're (mumbling), we're starting with Hydrogen then we go to Helium, and we keep, and then we go, go from Hydrogen to Helium to Lithium and let me show you what's happening right over here."
    },
    {
        "Q": "At 2:26, how do you know what to dram for the ion size?\n",
        "A": "These are just schematics (pictures not intended to realistically depict something or indicate actual values)!",
        "video_name": "HBi8xjMchZc",
        "timestamps": [
            146
        ],
        "3min_transcript": "In the video on solubility, I draw little pictures of sodium and chloride ions when sodium chloride dissolves or disassociates into water. This is sodium and this is chloride. And my simple brain, when I looked at it, I said, OK, how should I draw these things? I said, well, they're in the same period, and sodium is a Group 1 element. It's an alkali metal, while chlorine is a halogen, so chlorine's going to have a smaller atomic radius. And the logic there, just to review from the atomic table trends, is that both of their valence electrons are in the third shell. Sodium only has 11 protons pulling in the center. It has 11 in the center, and it has only one electron out there in its valence shell. So the attraction isn't as strong as the case of chlorine, which has 17 protons in the center. Although it has more valence electrons -- it has 7 of them-- these protons are going to have a stronger attraction on them. you'd expect the sodium neutral atom to be bigger than the chlorine neutral atom. Because this guy has more protons pulling everything in. And that's how I drew the ions in that video. I said, oh, when I disassociate in water, I'll have a big sodium ion and a smaller chlorine ion, which is incorrect. Because think about -- and this was pointed out to me by one of the viewers, and they're correct, and I should have realized it. What happens when you ionize these things? This guy will lose an electron, right? He gives the electron to this guy. So his electron configuration is actually going to look a lot more like neon. He now will have no electrons in that third shell, in the third energy state. So now he's going to have an atomic radius that's actually much more similar to neon here, right? Because he's going to have filled up the second shelf. So actually, the sodium ion, this is completely incorrect. The sodium ion is going to have Actually, it will be even a little smaller than neon because it has the same electron configuration, but it has one more proton. So the sodium ion is actually going to be smaller. Because it gets rid of the electron in that third shell, and the chlorine cation, gained an electron, so it has completely completed its third shell. So here you have where the chlorine ion is going to be bigger. So in that solubility video, I should've actually switched the places between the sodium and the chlorine, at least in size-wise. And, of course, i showed how they disassociate in water, and this would be attracted to the oxygen end of the water, and you have the hydrogen end and all that. But you can watch the solubility video for that. It doesn't change the real takeaway from the video. But I think this is a really interesting point that it brings up, that when you ionize these neutral atoms, it can significantly change, especially significantly change their relative atomic sizes. Anyway, hopefully, you found that interesting."
    },
    {
        "Q": "\nSal says that fog is an example of a colloid. And he says at 5:38 that fog is water molecules(H2O) that are inside air. Isn't H2O's molecular mass much less than 2nm. I think it has to be because Cs is 0.26 nm. If the H2O molecule is less than 2nm, shouldn't it also be a solution ?",
        "A": "Eventhough, the size of H2O particles is less than 2nm, fog is a colloid as it exhibits tyndall effect(scattering of light), a distinguishing traits among colloids.",
        "video_name": "3ROWXs3jtQU",
        "timestamps": [
            338
        ],
        "3min_transcript": "where we're at 2 to 500 nanometers, we're dealing with a colloid. That word, I remember in seventh grade, I think you learned it in science class: the colloid. And a friend and I, we thought it was a more appropriate word for some type of gastrointestinal problem. But it's not a gastrointestinal problem. It's a type of homogeneous mixture. And it's a homogeneous mixture where the particles are small enough that they stay suspended. So maybe they could call it a better suspension or a permanent suspension. So here the molecules are -- so let's say that's my mixture. So water, maybe it's water. It doesn't have to be water. It could be air or whatever. Now the molecules are small enough that they stay suspended. So the forces, either their buoyancy or the force -- actually, more important, the forces between the particles and the intermolecular forces kind of outweigh to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles and they'll never come out of the air. They're small enough that they'll always just float around with the air. Now, if you get below 2 nanometers -- maybe I should eliminate my homogenized milk. If you get below 2 nanometers -- I'm trying to draw in black. If you're less than 2 nanometers, you're now in the realm of the solution. And although this is very interesting in the everyday world, a lot of things that we-- and this is a fun thing to think about in your house, or when you encounter things, is this a suspension? Well, first, you should just think is it homogeneous? And then think is it a suspension? Is it eventually going to not be in the state it's in and then I'll have to shake it? Is it a colloid where it will stay in this kind of nice, thick state in the case of Jell-O or fog or smoke where it will really just stay in the state Or is it a solution? And solution is probably the most important in chemistry."
    },
    {
        "Q": "\nAt around 4:45 Sal draws a cup with liquid in it and says \"this could be air, whatever...\" . Funny cause AIR isn't liquid.",
        "A": "Gases are quite able to make solutions.",
        "video_name": "3ROWXs3jtQU",
        "timestamps": [
            285
        ],
        "3min_transcript": "you've got to make sure that the can is well shaken. Otherwise, you're going to get an inconsistent coat. The other, that's close to my heart, is chocolate milk. Because when you mix it up, it's nice and it seems homogeneous, right? It's nice. And I already have milk here. So right at first when you stir it nice, you have all the little chocolate clumps in there, at least the chocolate when I make it is like that. But then if you let it sit around for a long time, eventually all the chocolate is going to collect at the bottom of the glass. Actually, different parts of it. I've seen situations where the sugar all collects at the bottom and then you have these little clumps at the top. But you get the idea, that the mixture separates. And that's because the particles in either the paint or the chocolate milk are greater than 500 nanometers. Now, if we get to a range that's a little bit smaller than that, where we're at 2 to 500 nanometers, we're dealing with a colloid. That word, I remember in seventh grade, I think you learned it in science class: the colloid. And a friend and I, we thought it was a more appropriate word for some type of gastrointestinal problem. But it's not a gastrointestinal problem. It's a type of homogeneous mixture. And it's a homogeneous mixture where the particles are small enough that they stay suspended. So maybe they could call it a better suspension or a permanent suspension. So here the molecules are -- so let's say that's my mixture. So water, maybe it's water. It doesn't have to be water. It could be air or whatever. Now the molecules are small enough that they stay suspended. So the forces, either their buoyancy or the force -- actually, more important, the forces between the particles and the intermolecular forces kind of outweigh to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles"
    },
    {
        "Q": "at 7:20 what does aq stand for?\n",
        "A": "aq is the state description for an aqueous solution.",
        "video_name": "3ROWXs3jtQU",
        "timestamps": [
            440
        ],
        "3min_transcript": "to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles and they'll never come out of the air. They're small enough that they'll always just float around with the air. Now, if you get below 2 nanometers -- maybe I should eliminate my homogenized milk. If you get below 2 nanometers -- I'm trying to draw in black. If you're less than 2 nanometers, you're now in the realm of the solution. And although this is very interesting in the everyday world, a lot of things that we-- and this is a fun thing to think about in your house, or when you encounter things, is this a suspension? Well, first, you should just think is it homogeneous? And then think is it a suspension? Is it eventually going to not be in the state it's in and then I'll have to shake it? Is it a colloid where it will stay in this kind of nice, thick state in the case of Jell-O or fog or smoke where it will really just stay in the state Or is it a solution? And solution is probably the most important in chemistry. 99% of everything we'll talk about in chemistry involves solutions. And in general, it's an aqueous solution, when you stick something in water. So sometimes you'll see something like this. You'll see some compound x in a reaction and right next to it they'll write this aq. They mean that x is dissolved in water. It's a solute with water as the solvent. So actually, let me put that terminology here, just because I used it just now. So you have a solute. This is the thing that's usually whatever you have a smaller amount of, so thing dissolved. And then you have the solvent. This is often water or it's the thing that's in larger quantity. Or you can think of it as the thing that's all around or the thing that's doing the dissolving. Thing dissolving."
    },
    {
        "Q": "\nAt 2:38 you talked about the amplitude, but if I have a higher amplitude does it make the frequency low or high?",
        "A": "Amplitude is independent of frequency.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            158
        ],
        "3min_transcript": "This is a pulse wave because we only have, essentially, one perturbation of the string. Now if I kept doing that-- if I kept going up and down, and up and down, essentially, if I periodically did it at regular intervals, then my string would look something like this. Doing my best to draw it neatly. It might look something like this, where once again, the perturbations are going towards-- the disturbances are going to move to-- the right. They're going to move to the right with some velocity. And what I want to do in this video is really focus on this type of wave. This type of wave right here, which you can imagine, since I'm periodically moving this left side up and down, up and down, and creating these periodic movements in the wave, we call this a periodic wave. This is So what I want to talk about is some of the properties of a periodic wave. Now, the first thing you might say is, hey, how far are you jerking it up and down? How far are these movements from rest? So if this is the resting position right there, how far are these movements above the resting position and below the resting position? And we call that the amplitude of the wave. So that distance right there- I'll do it in magenta-- that distance right there is the amplitude. Sometimes mariners will have an idea of wave height. Wave height, they normally refer to from the bottom-- from the trough-- of a wave to its peak. Amplitude, we're talking about from the resting position to it's peak. So let me label peak. I think you know what peak means. Peak is the highest point on the wave. That's the peak. If you're in a fishing boat and you wanted to see how big a wave is, you'd probably care about the wave height-- not so much if your boat's sitting down here, you have to care about this whole distance. But anyway, we won't talk too much about that. So that's the first interesting idea behind a wave. And not all waves are being generated by Sal jiggling a string on the left-hand side. But I think you get the idea that these waves can represent many different-- this graph can represent many different types of wave forms. And this, essentially, displacement, from the resting position, or from the zero position, that is your amplitude. Now the next question you might ask is, OK, I know how far you're jiggling this string up and down, but how quickly are you doing it? So how long does it take for you to go all the way up, all the way down, and back again? So how long for each cycle?"
    },
    {
        "Q": "\nat 6:48 in the video i still don't get the 10 cycles over second what does that equal",
        "A": "A cycle is the same as a period. That is, one full wave from peak to peak. The number of cycles per second is the frequency of the wave. So 10 cycles per second means there are 10 full waves every second, and this means the frequency is 10 Hz (Hertz is a unit which just means cycles per second ). It also means that the period is T = .1 s. The frequency and the period are always reciprocals of each other.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            408
        ],
        "3min_transcript": "seconds per cycle. Or maybe it's 2 seconds per cycle. But what if we're asked how many cycles per second? So we're asking the opposite question. It's not how long, how many seconds does it take for me to go up, down, and back again. We're saying in each second, how many times am I going up, down, back again? So how many cycles per second? That's the inverse of period. So period, the notation is normally a big capital T for period. This is frequency. It's normally denoted by an F. And this, you're going to say cycles per second. So if you're going 5 seconds per cycle, that means you're doing 1/5 of a cycle or, 1/5 of a cycle per second. And that make sense. Because the period and the frequency are inversions of This is how many seconds per cycle. How long does one up, down, back again take? And this is how many up, down, back agains are there in a second? So they are inverses of each other. So we could say that frequency is equal to 1 over the period. Or you could say that period is equal to 1 over the frequency. So if I told you that I'm vibrating the left end of this rope at 10 cycles per second-- and, by the way, the unit of cycles per second, this is a hertz, so I could have also written this down as 10 hertz, which you've probably heard before. 10 hertz just means 10 cycles per second. If my frequency is 10 cycles per second, my period is going So 1 over 10 seconds per cycle, which makes sense. If in 10 times, I can go up and down, a whole up, down, back again, if I can do that 10 times in a second, it's going to take me 1/10 of a second to do it each time. Now another question we might ask ourselves is, how quickly is this wave moving, in this case, to the right? Since I'm jiggling the left end of the string. How quickly is it moving to the right? So the velocity. So to do that, we need to figure out how far did the wave go after one cycle? Or after one period? So after I jiggled this once, how far did the wave go? What is this distance from this resting point to this resting point there? And we call that a wavelength."
    },
    {
        "Q": "\nAt 8:54, Doesn't the wave length get reduced or shortened as time progresses?",
        "A": "actually the amplitude of the wave decreases which slowly brings it to end the wavelength remains constant.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            534
        ],
        "3min_transcript": "So 1 over 10 seconds per cycle, which makes sense. If in 10 times, I can go up and down, a whole up, down, back again, if I can do that 10 times in a second, it's going to take me 1/10 of a second to do it each time. Now another question we might ask ourselves is, how quickly is this wave moving, in this case, to the right? Since I'm jiggling the left end of the string. How quickly is it moving to the right? So the velocity. So to do that, we need to figure out how far did the wave go after one cycle? Or after one period? So after I jiggled this once, how far did the wave go? What is this distance from this resting point to this resting point there? And we call that a wavelength. wavelength. You could view a wavelength as how far the initial pulse went after completing exactly one cycle. Or you could view it as the distance from one peak to another peak. That is also going to be the wavelength. Or you could view it as a distance from one trough to the other trough. That's also the wavelength. Or in general, you could view the wavelength as one exactly equal point on the wave. From that distance to that distance. That is also one wavelength. Where you're completing, between that point and that point, you're completing one entire cycle to get exactly back to that same point. And when I say exactly back to that same point, this point doesn't count. Because this point, although we're in the same position, we're now going down. We want to go to the point where we're in the same position. And notice over here, we're going up. We want to be going up again. So distance is not one wavelength. back to the same position. And we're moving in the same direction. So this is also one wavelength. So if we know how far we've travelled after one period-- let me write it this way; wavelength is equal to how far the wave has traveled after one period. Or you could say after one cycle. Because remember, a period is how long does it take to complete one cycle. One to complete up, down, and back again notion. So if we know how far we've traveled, and we know how long it took us, it took us one period, how can we figure out the velocity? Well, the velocity is equal to distance divided by time."
    },
    {
        "Q": "at 1:04 you said that waves don't really need a medium to travel in. Can you give an example of a type of wave that doesn't need a medium(other than light)\n",
        "A": "Any other electromagnetic wave. Microwaves, for example. Mechanic waves always require a medium.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            64
        ],
        "3min_transcript": "In the last video we talked about the idea that if I start with some type of a string there, and if I were to take the left end of the string-- I could just have equally have done the right, but if I take the left end of the string and jerk it up, then all the way down, and then back to its resting position, it'll generate this disturbance in the string. And the disturbance might initially look like this after I've done that jerking up and down once. And that disturbance is going to propagate down the string. It's going to move down the string like that. Let me color this in black. So this is right after I do that first cycle-- that first jerking up and down. The string might look something like that. And if we wait a little while, the string might look something like this, assuming that I only did that once. The string might look something like this, where that pulse has actually propagated down the string. That pulse has propagated down the string. And in the last video, we said, hey, this disturbance that's propagating down the string, or propagating down this medium-- although it doesn't necessarily have to have a medium- we called this a wave. And in particular, This is a pulse wave because we only have, essentially, one perturbation of the string. Now if I kept doing that-- if I kept going up and down, and up and down, essentially, if I periodically did it at regular intervals, then my string would look something like this. Doing my best to draw it neatly. It might look something like this, where once again, the perturbations are going towards-- the disturbances are going to move to-- the right. They're going to move to the right with some velocity. And what I want to do in this video is really focus on this type of wave. This type of wave right here, which you can imagine, since I'm periodically moving this left side up and down, up and down, and creating these periodic movements in the wave, we call this a periodic wave. This is So what I want to talk about is some of the properties of a periodic wave. Now, the first thing you might say is, hey, how far are you jerking it up and down? How far are these movements from rest? So if this is the resting position right there, how far are these movements above the resting position and below the resting position? And we call that the amplitude of the wave. So that distance right there- I'll do it in magenta-- that distance right there is the amplitude. Sometimes mariners will have an idea of wave height. Wave height, they normally refer to from the bottom-- from the trough-- of a wave to its peak. Amplitude, we're talking about from the resting position to it's peak. So let me label peak. I think you know what peak means. Peak is the highest point on the wave. That's the peak."
    },
    {
        "Q": "Is Sal saying 'Lamda' at 12:55 or is he saying 'Landa'?\n",
        "A": "Lambda (the Greek letter that looks like an upside down Y ).",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            775
        ],
        "3min_transcript": "wavelength over period. Which is the same thing as wavelength times 1 over my period. And we just said that 1 over the period, this is the same thing is your frequency. So velocity is equal to wavelength times your frequency. And if you know this, you can pretty much solve all of the basic problems that you might encounter in waves. So for example, if someone tells you that I have a velocity of-- I don't know-- 100 meters per second to the right, so in that direction-- velocity you have to give a direction-- and they were to tell you that my frequency is equal to-- let's say my frequency is 20 cycles per second, which is the same thing as 20 hertz. only able to observe this part of your wave, you'd only observe that part of my string. If we're talking about 20 hertz, then in 1 second, you would see this go up and down twenty times. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. In exactly one second, you would see this go up and down 20 times. That's what we mean by the frequency being 20 hertz, or 20 cycles per second. So, they gave you the velocity. They gave you the frequency. What's the wavelength here? So the wavelength, in this situation-- you would say the velocity-- your velocity is equal to your wavelength times your frequency. Divide both sides by 20. And actually, let me make sure I get the units right. So this is meters per second, is equal to lambda times 20 cycles per second. 100 meters per second times 1/20 seconds per cycle. And then this becomes a 5. This becomes a 1. So you get 5, and then the seconds cancel out. So you get 5 meters per cycle. So this is equal to 5 meters per cycle, which would be your wavelength in this situation. So it's 5 meters. You could say 5 meters per cycle, but wavelength implies that you're talking about the distance per cycle. So in this situation, if this is moving to the right at 100 meters per second and this frequency-- I see this moving up and down 20 times in a second-- then this distance, right here, must be 5 meters. Likewise, we can figure out the period very easily. The period here is just going to be 1 over the frequency. It's going to be 1/20 seconds per cycle."
    },
    {
        "Q": "At 10:15 when Sal says that he could write it as a Vector, What does he mean? What is a vector?\n",
        "A": "A vector is a value that also has a direction. Since velocity has a direction, Sal is saying he could write down not only the value of the velocity (the magnitude ), but also the direction. Doing math with vectors is little more complicated and the subject of later material. Sal didn t bother to do that in this video.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            615
        ],
        "3min_transcript": "wavelength. You could view a wavelength as how far the initial pulse went after completing exactly one cycle. Or you could view it as the distance from one peak to another peak. That is also going to be the wavelength. Or you could view it as a distance from one trough to the other trough. That's also the wavelength. Or in general, you could view the wavelength as one exactly equal point on the wave. From that distance to that distance. That is also one wavelength. Where you're completing, between that point and that point, you're completing one entire cycle to get exactly back to that same point. And when I say exactly back to that same point, this point doesn't count. Because this point, although we're in the same position, we're now going down. We want to go to the point where we're in the same position. And notice over here, we're going up. We want to be going up again. So distance is not one wavelength. back to the same position. And we're moving in the same direction. So this is also one wavelength. So if we know how far we've travelled after one period-- let me write it this way; wavelength is equal to how far the wave has traveled after one period. Or you could say after one cycle. Because remember, a period is how long does it take to complete one cycle. One to complete up, down, and back again notion. So if we know how far we've traveled, and we know how long it took us, it took us one period, how can we figure out the velocity? Well, the velocity is equal to distance divided by time. vector, but I think you get the general idea. Your velocity-- what's the distance you travel in a period? Well, the distance you travel in a period is your wavelength after one up, down, back again. The wave pulse would have traveled exactly that far. That would be my wavelength. So I've traveled the distance of a wavelength, and how long did it take me to travel that distance? Well, it took me a period to travel that distance. So it's wavelength divided by period. Now I just said that 1 over the period is the same thing as the frequency. So I could rewrite this as wavelength. And actually, I should be clear here. The notation for wavelength tends to be the Greek letter lambda."
    },
    {
        "Q": "why does nitrogen not 'like to react' with other chemicals? 8:25\n",
        "A": "In its gaseous form (N2) it is triple bonded which is quite strong and thus doesn t want to react.",
        "video_name": "lzWUG4H5QBo",
        "timestamps": [
            505
        ],
        "3min_transcript": "to go both to us, or to Jack and to be used by itself. So it actually has an excess of oxygen that it's making. So this is actually kind of interesting and good to know. Now if you think about it, if I was to, let's say sketch out a planet. Let's draw a little planet over here, and ask you the question. If this was your planet Earth, and you've got thousands, instead of just one Jack, let's say now you have thousands of Jacks and thousands of beanstalks, in fact, not even thousands. Let's say billions, because really, that's what we have. We have a planet full of humans, and full of other animals, and full of plants. What would the atmosphere look like? This is the atmosphere. What would the atmosphere look like? Well you'd guess that the atmosphere is gas. And so what would those gases be? Well, the way I've drawn things again, of carbon dioxide. So I would say well, I guess there must be, I don't know, maybe 50-50 carbon dioxide and oxygen based on what we know so far. And the truth is, that's actually not true. That if you actually look at air, if you actually break down the atmosphere or air-- I'm just going to write \"air\" here-- if you actually break it down, turns out that the ratios are actually a little different. So for example, oxygen makes up about 21% of our air. This is our air breakdown. And carbon dioxide makes up about less than 1%. So, that leaves you wondering, what the heck is making up all that other parts of air? What is it made of? And in truth, it's about 78% nitrogen. Now you know, you've got nitrogen in your proteins, we've got nitrogen in our DNA. things. But nitrogen gas, specifically, is actually N2. And N2, this nitrogen gas, really is not too reactive. It kind of just hangs out by itself, does not like to react with things. So, looking at our little atmosphere graph, if you want to now think about it, knowing that we've got very little carbon dioxide and about 21% oxygen-- you could think of oxygen being, let's say, something like that-- well then relative to that nitrogen would be much more. You have much more nitrogen hanging out. And so this is really what our atmosphere looks like. It looks more filled with nitrogen than anything else. And in terms of carbon dioxide, it's just got a little smidge of carbon dioxide. Maybe right there. That could be carbon dioxide, maybe even less than that."
    },
    {
        "Q": "at 4:18 what is ATP?\n",
        "A": "Adenosine Tri-Phosphate (ATP) is a small molecule that cells use as a coenzyme. It s one of the most basic signal transporting molecules in living things. It does many biochemical processes: From activating membrane transporters to regulating transcription.",
        "video_name": "lzWUG4H5QBo",
        "timestamps": [
            258
        ],
        "3min_transcript": "kind of relying on each other to really work. So you kind of need both of them to work well. And so let's actually take a moment to write out these processes that are happening between Jack and the beanstalk. So let's start with the process of photosynthesis, the beanstalk. So on the one hand, you've got what? You've got water because, of course, the beanstalk needs water, and you've got carbon dioxide. And I'm going to do carbon oxide in orange. So it's taking in water and carbon dioxide. And it's going to put out, it's going to actually take these ingredients if you want to think of it as kind of cooking, it's going to take these ingredients and it's going to put out. It's going to put out what? Oxygen and glucose. So I'll put glucose up top and oxygen down below. So these are the inputs and outputs of photosynthesis. You've got inputs. You've got glucose and oxygen going in. You're going to start seeing some serious similarities here. You've got glucose and oxygen going in. So Jack is taking in those two things. And he is again, of course, processing them. And he's putting out water and carbon dioxide. So this looks really, really nice, right? Looks perfect actually. Because everything is nice and balanced. And you can see how it makes perfect sense that, not only did Jack need the beanstalk, but actually it sounds like the beanstalk needed Jack, based on how I've drawn it. Now remember, none of this would even happen if there was no sunlight. So we actually need light energy. In fact, that's the whole purpose of this, right? Getting energy. So you have to have some light energy. I'm going to put a big plus sign, and I might even circle it because it's so important. And on the other side, of course, Jack is getting something as well. He's getting chemical energy. In fact, he's using the chemical energy to help him climb the beanstalk. And so the chemical energy comes in the form of what we call ATP, which is just a molecule of high energy. And so Jack and the beanstalk are basically going from light energy to chemical energy using these two Now here's the part that people don't always appreciate. And I'm actually going to take just a moment to show you that this isn't the full story. There's actually something else going on as well. And that is that there's actually some cellular respiration happening on the plant's side. So remember, not only does the human, or the Jack, need energy, but so does the plant. The plant needs energy as well. And in fact, if it takes in light energy right here,"
    },
    {
        "Q": "\nAt 3:53, can someone clarify what temperature has to go with the energy in the molecule?",
        "A": "Temperature IS kinetic energy of molecules.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            233
        ],
        "3min_transcript": "Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there. If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant."
    },
    {
        "Q": "\nWhen the equation P=nRT is formed at 4:56 I tried to find the units for R. I got it as m(superscript) -3 i.e. volume. So if the equation is correct then why do we need the equation PV=nRT?",
        "A": "R is a constant for accuracy. R s units should be atm*L/moles*K. That allows all the units to cancel. The reason why you need Volume in the equation is because volume also affects pressure.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            296
        ],
        "3min_transcript": "If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant. We gave these two examples. Obviously, it is dependent on the temperature; the faster each of these particles move, the higher pressure we'll have. It's also dependent on the number of particles, the more particles we have, the more pressure we'll have. What about the size of the container? The volume of the container. If we took this example, but we shrunk the container somehow, maybe by pressing on the outside. So if this container looked like this, but we still had the same four particles in it, with the same average kinetic energy, or the same temperature. So the number of particles stays the same, the temperature is the same, but the volume has gone down. Now, these guys are going to bump into the sides of the container more frequently and there's less area. So at any given moment, you have more force and less area. So when you have more force and less area, your pressure is going to go up. So when the volume went down, your pressure went up."
    },
    {
        "Q": "\nAt 2:20, can you please explain how is temperature related to average energy?\nCan you also tell me which video explains pressure,density,volume for a substance are explained ?",
        "A": "It comes from the video on the Kinetic Molecular Theory of gases. The result at the end of the video is that KE(avg) = \u00c2\u00bdkT. So KE(avg) \u00e2\u0088\u009d T and T \u00e2\u0088\u009d KE(avg). I m don t think there is a video explaining how P, V, and \u00cf\u0081 are related for substances in general, but the video on Boyle s Law explains the relationship between P and V for an ideal gas.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's say I have a balloon. And in that balloon I have a bunch of particles bouncing around. They're gas particles, so they're floating freely. And they each have some velocity, some kinetic energy. And what I care about, let me just draw a few more, what I care about is the pressure that is exerted on the surface of the balloon. So I care about the pressure. And what's pressure? It's force per area. So the area here, you can think of it as the inside surface of the balloon. And what's going to apply force to that? Well any given moment-- I only drew six particles here, but in a real balloon you would have gazillions of particles, and we could talk about how large, but more particles than you can really probably imagine-- but at any given moment, some of those particles are bouncing off the wall of the container. bouncing there, this guy's bouncing like that. And when they bounce, they apply force to the container. An outward force, that's what keeps the balloon blown up. So think about what the pressure is going to be dependent on. So first of all, the faster these particles move, the higher the pressure. Slower particles, you're going to be bouncing into the container less, and when you do bounce into the container, it's going to be less of a ricochet, or less of a change in momentum. So slower particles, you're going to have pressure go down. Now, it's practically impossible to measure the kinetic energy, or the velocity, or the direction of each individual particle. Especially when you have gazillions of them in a balloon. So we do is we think of the average energy of the particles. Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there."
    },
    {
        "Q": "1:45 does that mean that helium particles \"whiz around\" faster then air particles?\n",
        "A": "yes. its for the same reason when you inhale helium gas and try to speak your voice seems to be at a very high pitch",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            105
        ],
        "3min_transcript": "Let's say I have a balloon. And in that balloon I have a bunch of particles bouncing around. They're gas particles, so they're floating freely. And they each have some velocity, some kinetic energy. And what I care about, let me just draw a few more, what I care about is the pressure that is exerted on the surface of the balloon. So I care about the pressure. And what's pressure? It's force per area. So the area here, you can think of it as the inside surface of the balloon. And what's going to apply force to that? Well any given moment-- I only drew six particles here, but in a real balloon you would have gazillions of particles, and we could talk about how large, but more particles than you can really probably imagine-- but at any given moment, some of those particles are bouncing off the wall of the container. bouncing there, this guy's bouncing like that. And when they bounce, they apply force to the container. An outward force, that's what keeps the balloon blown up. So think about what the pressure is going to be dependent on. So first of all, the faster these particles move, the higher the pressure. Slower particles, you're going to be bouncing into the container less, and when you do bounce into the container, it's going to be less of a ricochet, or less of a change in momentum. So slower particles, you're going to have pressure go down. Now, it's practically impossible to measure the kinetic energy, or the velocity, or the direction of each individual particle. Especially when you have gazillions of them in a balloon. So we do is we think of the average energy of the particles. Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there."
    },
    {
        "Q": "At 3:42, the example compares two balloons, one with many particles and one with only one particle. The particles are the gas particles, how can we have only 1 gas particle in a balloon?\n",
        "A": "The example is a simplification of the real world. Each particle in the example represents many billions of trillions of particles in a real balloon.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            222
        ],
        "3min_transcript": "Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there. If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant."
    },
    {
        "Q": "at 6:50 or any other equations why are there constant?\nlike PV=nRT <- what is that \"R\"?\nand they don't explain what's it for.\nare those things so complicated that we can't understand ?\n",
        "A": "The R is just a conversion factor that takes account of the types of units you chose to use for P, V, and T. Nature does not arrange herself to match Pascals, cubic meters, and Kelvin, so we have to adjust to her.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            410
        ],
        "3min_transcript": "We gave these two examples. Obviously, it is dependent on the temperature; the faster each of these particles move, the higher pressure we'll have. It's also dependent on the number of particles, the more particles we have, the more pressure we'll have. What about the size of the container? The volume of the container. If we took this example, but we shrunk the container somehow, maybe by pressing on the outside. So if this container looked like this, but we still had the same four particles in it, with the same average kinetic energy, or the same temperature. So the number of particles stays the same, the temperature is the same, but the volume has gone down. Now, these guys are going to bump into the sides of the container more frequently and there's less area. So at any given moment, you have more force and less area. So when you have more force and less area, your pressure is going to go up. So when the volume went down, your pressure went up. proportional to volume. So let's think about that. Let's put that into our equation. We said that pressure is proportional-- and I'm just saying some proportionality constant, let's call that R, to the number of particles times the temperature, this gives us the total energy. And it's inversely proportional to the volume. And if we multiply both sides of this times the volume, we get the pressure times the volume is proportional to the number of particles times the temperature. So PV is equal to RnT. And just to switch this around a little bit, so it's in a form that you're more likely to see in your chemistry book, if we just switch the n and the R term. You get pressure times volume is equal to n, the number of And this right here is the ideal gas equation. Hopefully, it makes some sense to you. When they say ideal gas, it's based on this little mental exercise I did to come up with this. I made some implicit assumptions when I did this. One is I assumed that we're dealing with an ideal gas. And so you say what, Sal, is an ideal gas? An ideal gas is one where the molecules are not too concerned with each other. They're just concerned with their own kinetic energy and bouncing off the wall. So they don't attract or repel each other. Let's say they attracted each other, then as you increased the number of particles maybe they'd want to"
    },
    {
        "Q": "\nAt 4:39, he changes one of the hydrogens to an R group for the H3O+. Why was this necessary? I understood that that could stay H3O+.",
        "A": "I believe the reaction will be taking place with alcohol as the solvent (not in water), so there really wouldn t be any H3O+ ...",
        "video_name": "rL0lDESwwv8",
        "timestamps": [
            279
        ],
        "3min_transcript": "those pi electrons in here, in our carbonyl, had kicked off onto our oxygen, like that. So, that gives us our intermediate, and we're almost to the formation of a hydrate, the last step would be to deprotonate: So a molecule of water comes along, this time, it's gonna function as a base, and take this proton away, leaving these two electrons behind, on this oxygen. So let's go ahead, and show our product, formation of a hydrate, so we have a carbon bonded to two OH groups now, and so we form our hydrate, or our gem-diol. Let's follow those electrons now. So, let's say that these electrons right in here, moved off onto this top oxygen, doesn't really matter which lone pair you say they are, and that's formation of our hydrate. We could have followed through as a ketone, so instead of an aldehyde, we could have started off here as a ketone, and so, this would've been our prime, and then, this would've been our prime, and then, finally, this would've been our prime; so, that's still formation of a hydrate. If we wanted to form a hemiacetal, we wouldn't have started with water; we would have started with an alcohol, so let's go ahead and show that. So let's make this R double prime OH, and, of course, an acid-catalyzed version, so R double prime OH and H plus, instead of forming a hydrate, this is gonna form a hemiacetal, so instead of H three O plus, which is what we had in our acid-catalyzed hydration, we're gonna protonate the alcohol, so let me go ahead and change this, this would no longer gonna be an \"H,\" here, this would be \"R double prime,\" like that, and so, when we protonate our carbonyl oxygen here, and make our carbon more electrophilic, this wouldn't be water as our nucleophile; this would be alcohol. So our alcohol molecule would be a nucleophile, and attack our carbonyl carbon, So this wouldn't be an \"H,\" this would be \"R double prime,\" and therefore, in our final product, this wouldn't be an \"H,\" this would be \"R double prime,\" and hopefully you can see, that this is now a hemiacetal. It's actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually, the hemiacetal is going to move on to form a full acetal. So let's go ahead, and show the structure of a full acetal. And, this isn't just one step, and I won't go through the steps in this video, that'll be in the next video, but I just wanted to show you the structure of a full acetal here. So we have R double prime, and lone pairs of electrons on our oxygen; and we have, over here, R double prime and lone pairs of electrons on our oxygen, so this would be an acetal, and let's, once again, compare structures. So, with an acetal, we have this OR double prime group, and another OR double prime group"
    },
    {
        "Q": "\nAt 9:12, we learn that the oxygen was reduced by the carbon. Wasn't it also reduced by the hydrogen, as 2 out of the 4 oxygens gained their negative reduction because of the addition of 2 positive hydrogens each?",
        "A": "The hydrogens don t change oxidation states through the reaction though so they didn t take part in the reduction/oxidation part of this reaction",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            552
        ],
        "3min_transcript": "On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen? So oxygen reduced by the carbon. And this type of reaction-- where you have both oxidation and reduction taking place, and really they're two sides of the same coin. One thing is going to be oxidized if another thing is being reduced, and vice versa. We call these oxidation reduction reactions. Or sometimes \"redox\" for short. Take the \"red\" from \"reduction\" and the \"ox\" from \"oxidation,\" and you've got \"redox.\" This is a redox reaction. Something is being oxidized. Something else is being reduced. Not everything is being oxidized or reduced, and we can see that very clearly when we depict it in these half reactions. And one way to check that your half reactions actually makes sense is you can actually sum up the two sides."
    },
    {
        "Q": "Hello! I'm a beginner in chemistry and at 8:20 I lost it. Why does the oxygen gain 8 electrons and not 2? It's oxidation number is 2- Can someone explain? What happens to the other 6 gained electrons?\n",
        "A": "There s four oxygen atoms because of the coefficients. Four of them each gaining 2 electrons = 8",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            500
        ],
        "3min_transcript": "Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen?"
    },
    {
        "Q": "\nAt around 8:00, is hydrogen also reducing oxygen by giving oxygen electrons?",
        "A": "No. Hydrogen doesn t change its oxidation state in this reaction. Thus, it cant oxidise or reduce anything. Hope this helps :)",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            480
        ],
        "3min_transcript": "Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen?"
    },
    {
        "Q": "at 2:28, does it matter if the equation isn't balanced?\n",
        "A": "In most cases yes! If you have an unbalanced redox equation involving charged ions, then if the equation is not balanced, the oxidation numbers will be wrong and therefore your half equations will be wrong. In the case of the video, balancing would not matter, but it s safest to balance first and then work out oxidation numbers. Hope that helps! :)",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            148
        ],
        "3min_transcript": "What we have depicted right over here is a combustion reaction. We have the hydrocarbon methane right over here. You take that. You take some molecular oxygen. You give them enough heat. And then, they are going to combust. And they're going to produce carbon dioxide, water, and then more energy than you put in. This is an exothermic reaction. More energy comes out than you put in. This is why fires keep spreading. This is why combustion is used to power things. But that's not what we're going to focus on in this video. In this video, we want to think about-- which of these components, of these molecules, what's being oxidized and what is being reduced? And to do that, let's first think about the oxidation states of the input atoms and the oxidation states of the outputs-- of the different constituents of these molecules. So I encourage you to now pause this video and try to figure that out on your own. So I'm assuming you've given a go at it. Let's first think about the methane. And I have a bunch of electronegativities here based on the Pauling scale listed out here, but let's just visualize methane. Methane is a carbon bonded to 4 hydrogens. In our oxidation state world-- even though, this, in reality, is a covalent bond-- we pretend like they're hypothetically ionic bonds. So we have to give the electron pair to one of the parties to the bond. If we look between carbon and hydrogen, carbon is more electronegative than hydrogen. So we will assume that carbon will be hogging the electrons and that hydrogen will be giving away the electrons. So carbon's going to take-- in this hypothetical world-- hypothetically take an electron from each of these hydrogens. It is going to have an oxidation state of negative 4, an electron from each of 4 hydrogens. Negative 4. And once again, we write the sign after the number probably so that we don't get these confused with exponents. Each of those hydrogens is having-- in this hypothetical world-- an electron taken away from it. We could say it has an oxidation state of plus 1, which we could write as 1 plus. Or we could just write a positive right over here-- a plus sign. Now, we have molecular oxygen-- oxygen bonded to oxygen. Well, all oxygens are created equal. Or we'll assume that these oxygens are created equal, that they're not different isotopes or anything like that. In this reality, there's no reason why one oxygen would hog any electrons from the other oxygen. In this world, the oxygen has an oxidation state-- when it's in this molecular oxygen form-- it has an oxidation state of 0 or an oxidation number of 0. Now, let's think about this side-- the products. Now, what's happening here with carbon dioxide? Carbon dioxide is a carbon double"
    },
    {
        "Q": "\nAt 9:15 , why don't both the oxygen's make double bonds with sulfur? I know this breaks the octet rule, but sulfur is in the 3rd period, so isn't it more important to have no formal charge than obey the octet rule?",
        "A": "It is because Oxygen is more electronegative, and therefore will consume more of the electrons. Sulfur, being less electronegative, cannot have more electrons around it than the more electronegative oxygen",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            555
        ],
        "3min_transcript": "So I could take, let's say-- let me go ahead and make these blue here. So I could take a lone pair of electrons from either oxygen. I'm just going to say that we take a lone pair of electrons from that oxygen and move them in here to form a double bond between the sulfur and the oxygen. So if I do that, now I have a double bond between the sulfur and the oxygen. The oxygen on the right now has only two lone pairs of electrons around it. The oxygen on the left still has three lone pairs of electrons around it like that. And the sulfur still has a lone pair of electrons here in the center. All right, so if we assign formal charges now-- let's go ahead and do that really fast. So we know that we have electrons in these bonds here. And so if we assign a formal charge to the oxygen on the left-- let's do that one first, all right, so this oxygen on the left here. All right, oxygen normally has six valence electrons And in our dot structure, we give to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail We talked about those in earlier videos. So make sure that you watch them. But this is our final dot structure here. So let me go ahead and redraw it here. This is one of the possible dot structures you can draw sulfur dioxide that fulfills our rules for drawing dot structures. So let me go ahead and put in these valence electrons here so we can see it a little bit better. All right, so I'll leave out formal charges now because we're just focusing in on geometry. We're concerned about VSEPR theory. So we have our dot structure. We go back up to check our steps for predicting the shape. And now we are going to count the number of electron clouds that surround our central atoms, the regions of electron density, which could be valence electrons in bonds, so bonding electrons, or non-bonding electrons, like lone pairs of electrons. So if we look at our central atom, which is sulfur, and let's see if we can count up our electron clouds. So this is an electron cloud over here."
    },
    {
        "Q": "Isn't boron trifluoride tetrahedral, not trigonal planar as the video states at 4:25?\n",
        "A": "No, BF\u00e2\u0082\u0083 is trigonal planar.",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            265
        ],
        "3min_transcript": "of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron So I'm going to draw sheet of paper here, a plane. And we're going to put our boron atom in the center. And here's one of our electron clouds. And then here's another one, and here's our third one here. So they're going to get as far away from each other as they possibly can. We call this shape trigonal planar. So let me go ahead and write that here. So this is a trigonal planar geometry of my electron clouds surrounding my central atom. And since we don't have any lone pairs of electrons to worry about on our central atom, we can go ahead and predict the geometry of the molecule as being the same here as the geometry of the electron cloud. So the molecule has a trigonal planar shape, as well. Now, when we think about bond angles, right? So the easiest way to think about what's the bond angle for trigonal planar or what would we predict the bond angle to be, you think about a circle. And since the electrons repel each other equally, you want to divide your circle into three equivalent angles."
    },
    {
        "Q": "At 10:15, why wouldn't the 2 non-bonded electrons be in different electron clouds, or different orbitals? It seems that Hund's rule would cause the 2 electrons to join different oribitals.\n",
        "A": "Which orbital do you suppose the second one would go in to? Something to keep in mind: VSEPR theory predicts a bond angle of 120 degrees for SO2, and that s very close to what we actually observe (119 degrees)",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            615
        ],
        "3min_transcript": "to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail We talked about those in earlier videos. So make sure that you watch them. But this is our final dot structure here. So let me go ahead and redraw it here. This is one of the possible dot structures you can draw sulfur dioxide that fulfills our rules for drawing dot structures. So let me go ahead and put in these valence electrons here so we can see it a little bit better. All right, so I'll leave out formal charges now because we're just focusing in on geometry. We're concerned about VSEPR theory. So we have our dot structure. We go back up to check our steps for predicting the shape. And now we are going to count the number of electron clouds that surround our central atoms, the regions of electron density, which could be valence electrons in bonds, so bonding electrons, or non-bonding electrons, like lone pairs of electrons. So if we look at our central atom, which is sulfur, and let's see if we can count up our electron clouds. So this is an electron cloud over here. So we have one electron cloud. Over here on the right, this double bond, we can consider it as an electron cloud. We're not worried about numbers of electrons, just regions of them. And then for the first time, we now have a lone pair of electrons, right? And this, we can also think about as occupying an electron cloud. So we have three electron clouds. And we saw in the previous example that when you have three electron clouds, the electron clouds are going to try to adopt a trigonal planar shape. So I could redraw this dot structure and attempt to show it in more of a trigonal planar shape here. So let's go ahead and show it looking like this-- once again, not worried about drawing informal charges here-- so something like this for the structure. Let me go ahead and put those electrons in our orbital here so we can see that electron cloud a little bit better. And so, once again, our electron clouds are in a trigonal planar geometry."
    },
    {
        "Q": "At 2:50 he said that the completed-octet version of the dot structure and the dot structure with minimized charge are acceptable representations of the molecule. But which ones do molecules generally \"pick\" in nature?\n",
        "A": "The actual structure is a hybrid of all possible structures, but the ones with minimized charge separation are closest to the actual structure.",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            170
        ],
        "3min_transcript": "We're going to put those leftover electrons on a terminal atoms, which are our fluorines in this case. Fluorine follows the octet rule. So each fluorine is now surrounded by two valence electrons. So each fluorine needs six more to have an octet of electrons. So I'll go ahead and put six more valence electrons on each of the three fluorines. And 6 times 3 is 18. So we just represented 18 more valence electrons. And so now we are all set. We've represented all 24 valence electrons in our dot structure. And some of you might think, well, boron is not following the octet rule here. And that is true. It's OK for boron not to follow the octet rule. And to think about why, let's assign a formal charge to our boron atom here. And so, remember, each covalent bond consists of two electrons. Let me go ahead and draw in those electrons in blue here. And when you're assigning formal charge, remember how to do that. You take the number of valence electrons in the free atoms, which is three. of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron"
    },
    {
        "Q": "\nAt 2:00, Khan resembles a ray to a marching band, however, I don't see how one part of a single ray would leave the water before its other part. Also, if (in any way), a piece of a ray does leave he water first, why doesn't the speed differentiation cause the photon (ray= photon?) to start spinning instead of bending??",
        "A": "A ray is just a way to draw light. The light is really a wave, right? Not an arrow. To understand refraction, you really have to use the wave behavior of light, not think of it as a photon.",
        "video_name": "jxptCXHLxKQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "Before doing more examples with Snell's Law which essentially amounts to math problems what I do is give you an intuitive understanding for why this straw looks bent in this picture right over here To do that, let me just do a simplified version of that picture This is the side profile of the cup, or glass right over here The best I can draw it And then let me draw the actual straw. I'll first draw the straw where it actually is coming in off the side of the cup and the straw is actually not bending goes to the bottom of the cup just like that and then it goes up like that and then it goes slightly above. Then it actually does bent up here It's irrelevant to what we want to talk about What I want to do in this video is talk about when we look over here why does it look like the straw got bent? As the light from the straw down here changes as it go from one medium to another Now we know from refraction indices or just in general the light moves slower in water than it does in air slower in water; faster in air Let's think about what's going to happen Let me draw 2 rays that are coming from this point on the straw right over here I draw one ray right over here. I'm gonna take an arbitrary direction. Like that Now when it goes from the slower medium to the faster medium, what's going to happen to it? Until this light go here so the left side of the ray is going to end up in the air before the right side and I'm using the car example to think about which way this light's going to bend The left side of the marching band is gonna get out before the right side and start moving faster So this is going to turn to the right Let me do another ray Let the ray come from the same point Right along the straw, so another ray just like that It will also turn to the right Now if someone's eye is right over here-- Draw their nose and all the rest If they're looking down where does it look like this 2 light rays? Let's say his eye's big enough to capture both of these rays Where does it look like this 2 light rays are coming from? So if you trace both of these rays back if you just assume that there's a line here--that's what our eyes and brains do-- if you assume whatever direction this ray is currently going it's the direction it came from"
    },
    {
        "Q": "\nat 1:02, why is refractive index useful and what is it used to measure",
        "A": "Refractive index is used to measure how much the speed of light travels faster in a medium than that of air. For example, refractive index of diamond is 2.41. So, in air, the light travels 2.41 times faster than in diamond. I hope this helped!",
        "video_name": "jxptCXHLxKQ",
        "timestamps": [
            62
        ],
        "3min_transcript": "Before doing more examples with Snell's Law which essentially amounts to math problems what I do is give you an intuitive understanding for why this straw looks bent in this picture right over here To do that, let me just do a simplified version of that picture This is the side profile of the cup, or glass right over here The best I can draw it And then let me draw the actual straw. I'll first draw the straw where it actually is coming in off the side of the cup and the straw is actually not bending goes to the bottom of the cup just like that and then it goes up like that and then it goes slightly above. Then it actually does bent up here It's irrelevant to what we want to talk about What I want to do in this video is talk about when we look over here why does it look like the straw got bent? As the light from the straw down here changes as it go from one medium to another Now we know from refraction indices or just in general the light moves slower in water than it does in air slower in water; faster in air Let's think about what's going to happen Let me draw 2 rays that are coming from this point on the straw right over here I draw one ray right over here. I'm gonna take an arbitrary direction. Like that Now when it goes from the slower medium to the faster medium, what's going to happen to it? Until this light go here so the left side of the ray is going to end up in the air before the right side and I'm using the car example to think about which way this light's going to bend The left side of the marching band is gonna get out before the right side and start moving faster So this is going to turn to the right Let me do another ray Let the ray come from the same point Right along the straw, so another ray just like that It will also turn to the right Now if someone's eye is right over here-- Draw their nose and all the rest If they're looking down where does it look like this 2 light rays? Let's say his eye's big enough to capture both of these rays Where does it look like this 2 light rays are coming from? So if you trace both of these rays back if you just assume that there's a line here--that's what our eyes and brains do-- if you assume whatever direction this ray is currently going it's the direction it came from"
    },
    {
        "Q": "4:08 I think this is hept-2,4-diene.\n",
        "A": "ya it is",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            248
        ],
        "3min_transcript": "So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon."
    },
    {
        "Q": "\nAt 3:30 why is the answer hepta-2,4-diene and not hept-2,4-diene. Where does the letter a come from? Does the letter a indicate something or is it just spelling?",
        "A": "It s just how you name molecules with multiple double bonds. Also makes it flow a bit better off the tongue.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            210
        ],
        "3min_transcript": "So this tells us right here that we're dealing with an alkene, not an alkane. If you have a double bond, it's an alkene. Triple bond, alkyne. We'll talk about that in future videos. This is hept, and we'll put an ene here, but we haven't specified where the double bond is and we haven't numbered our carbons. When you see an alkene like this, you start numbering closest to the double bond, just like as if it was a alkyl group, as if it was a side chain of carbons. So this side is closest to the double bond, so let's start numbering there. One, two, three, four, five, six, seven. The double bond is between two and three, and to specify its location, you start at the lowest of these numbers. So this double bond is at two. So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct"
    },
    {
        "Q": "at \"4:17\", why did the first C of the chain named H3C and why not CH3?\n",
        "A": "It was written as H3C to highlight that it is the carbon of the methyl group (CH3) that bonds the the neighboring carbon in the chain, not a hydrogen. The methyl group can be written as either H3C or CH3, it does not matter; it s just clearer if written as H3C in this case. H3C is usually used at the ends of molecules.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            257
        ],
        "3min_transcript": "So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon."
    },
    {
        "Q": "Brother Sal, I don't understand this> At 4:35 u said that \"double bond will take presidence....\". What does it mean?\n",
        "A": "well answered =]",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            275
        ],
        "3min_transcript": "One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon."
    },
    {
        "Q": "at 6:22, if you are trying to find the largest carbon chain, shouldn't it be 8 if you start counting from the carbon on the far left methyl group? Is it different if its a cycle?\n",
        "A": "If there is a ring structure that has more carbons than any chains connected to the ring, then use the number of carbons in the ring as the longest chain. In this case, 6 carbons = hex , and since it is in a ring use the prefix cyclo- . The methyl group then becomes a substituent.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            382
        ],
        "3min_transcript": "It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon. So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon."
    },
    {
        "Q": "\nAt 2:47 Sal says 'if we are stationary relative to the ether right now'. How can earth be stationary relative to the ether at a point in its orbit, if the earth is always moving at 30 km/s.",
        "A": "Exactly! That s Sal s point.",
        "video_name": "88hs5LcCoX4",
        "timestamps": [
            167
        ],
        "3min_transcript": "the Earth is rotating, but not only is it rotating on its own axis, but it's rotating around the sun. So if this is the sun right over here, this is the Earth. The Earth is rotating, and these are all rough figures, the Earth is moving around the Sun at approximately 30 kilometers per second. 30 kilometers per second! By our everyday standards, that's quite fast, but we're not done yet. 'Cuz the Sun is also moving around the center of the galaxy. And this isn't an actual picture of the Milky Way; obviously we haven't gotten this far from our own galaxy to actually get this type of a vantage point, but if the Sun were right over there, the Sun, estimates are, are moving with a speed of 200, roughly, 200, let me write that in a better color so you can actually see it, 200 kilometers per second. around the center of the Milky Way, and then the Milky Way itself could be moving. So we don't know our actual, kind of, our orientation relative to the Ether, but we are, we're constantly changing our orientation, we're moving in these orbital patterns. If there is some type of luminiferous Ether, if there is some type of luminiferous Ether, and I'm just gonna draw these lines over here to kind of show our luminiferous Ether, we must be moving relative to it if we orient ourselves just the right way. In fact, the odds of us being stationary relative to the Ether are pretty close to zero, especially if we wait a little. If we're stationary relative to the Ether right now, let's say at this point, since we're changing our direction, we're not going to be stationary relative to the Ether at that point. And that's just when you consider the Earth's orbit around the Sun. It's even more true when you think about the solar system's orbit around the center of the galaxy, So, we should be moving relative to the Ether, or the Ether should be moving relative to us. So we should be able to detect some type of, some type of what's called an \"Ether wind\". 'Cuz it should be moving relative... Ether wind. Now how would you detect an Ether wind? Well, let's think about some other type of medium moving relative to us. Let's say that we are sitting on an island, let me do this in a better color for an island. So let's say that we're on an island that's in the middle of a stream. So these are the shores of the stream. These are the shores of the stream. And there is some type of a current. So the water is moving in that direction. So that's the medium. And now let's start a wave propagating through this. So if I were to just take a pebble and drop it"
    },
    {
        "Q": "\n1:30 why Na+ cant react with water ?",
        "A": "Na+ is a cation, having a positive charge. Needing a negative charge from water, it could potentially react with either H+ or OH-. H+ doesn t work since it has a positive charge. OH- does not work because the supposedly formed substance would be NaOH, a strong base. Strong bases are not formed since they dissociate to near completion.",
        "video_name": "HwkEQfsJenk",
        "timestamps": [
            90
        ],
        "3min_transcript": "- Let's say we have some hydrochloric acid, and a solution of sodium hydroxide. We know that hydrochloric acid is a strong acid so we can think about it as consisting of H+ and Cl-. Sodium hydroxide is a strong base, so in solution we're going to have sodium ions and hydroxide anions. Let's think about the products for this reaction. One product would be H+, and OH-. If you put H+ and OH- together, you form H20. So water is one of our products, and the other product would be what's leftover. We have Na+ and Cl-, so that gives us NaCl which is sodium chloride. This is an example of an acid base neutralization reaction where an acid reacts with the base to give you water and a salt. In this case, our salt is sodium chloride. Let's think about an aqueous solution of sodium chloride. and you dissolve your sodium chloride in water to make your solution. In solution, you're gonna have sodium cations and chloride anions. Let's think about what those would do with water. We know that the pH of water is seven. The pH of water is equal to seven. Sodium ions don't react with water so they're not going to affect the pH of our solution. You might think that the chloride anion could function as a weak base, and take a proton from water. However, that's not really gonna happen very well because the chloride anion, Cl-, this is the conjugate base to HCl. We know that HCl is a strong acid, and the stronger the acid, the weaker the conjugate base. With a very strong acid, we're gonna get a very weak conjugate base from water very well so the pH is unaffected. The pH of our solution of sodium chloride is equal to seven. When you have a salt that was formed from a strong acid and a strong base, so sodium chloride was formed from a strong acid and a strong base, these salts form neutral solutions. So your pH should be equal to seven. Let's compare this situation to the salt that's formed from a weak acid, and a strong base. Over here we have acetic acid which we know is a weak acid. Then we have sodium hydroxide which is our strong base. In solution, we would have Na+ and OH-. Hydroxide is going to take the acidic proton on acetic acid, and this is the acidic proton on acetic acid. Once again, H+ and OH- give us H20."
    },
    {
        "Q": "\nAt 1:18 , how is pH related to the formation of salts?",
        "A": "Salt of strong acid + strong base \u00e2\u0086\u0092 pH = 7. Salt of strong acid + weak base \u00e2\u0086\u0092 pH < 7. Salt of weak acid + strong base \u00e2\u0086\u0092 pH > 7. Salt of weak acid + weak base \u00e2\u0086\u0092 pH depends on which is stronger acid or base.",
        "video_name": "HwkEQfsJenk",
        "timestamps": [
            78
        ],
        "3min_transcript": "- Let's say we have some hydrochloric acid, and a solution of sodium hydroxide. We know that hydrochloric acid is a strong acid so we can think about it as consisting of H+ and Cl-. Sodium hydroxide is a strong base, so in solution we're going to have sodium ions and hydroxide anions. Let's think about the products for this reaction. One product would be H+, and OH-. If you put H+ and OH- together, you form H20. So water is one of our products, and the other product would be what's leftover. We have Na+ and Cl-, so that gives us NaCl which is sodium chloride. This is an example of an acid base neutralization reaction where an acid reacts with the base to give you water and a salt. In this case, our salt is sodium chloride. Let's think about an aqueous solution of sodium chloride. and you dissolve your sodium chloride in water to make your solution. In solution, you're gonna have sodium cations and chloride anions. Let's think about what those would do with water. We know that the pH of water is seven. The pH of water is equal to seven. Sodium ions don't react with water so they're not going to affect the pH of our solution. You might think that the chloride anion could function as a weak base, and take a proton from water. However, that's not really gonna happen very well because the chloride anion, Cl-, this is the conjugate base to HCl. We know that HCl is a strong acid, and the stronger the acid, the weaker the conjugate base. With a very strong acid, we're gonna get a very weak conjugate base from water very well so the pH is unaffected. The pH of our solution of sodium chloride is equal to seven. When you have a salt that was formed from a strong acid and a strong base, so sodium chloride was formed from a strong acid and a strong base, these salts form neutral solutions. So your pH should be equal to seven. Let's compare this situation to the salt that's formed from a weak acid, and a strong base. Over here we have acetic acid which we know is a weak acid. Then we have sodium hydroxide which is our strong base. In solution, we would have Na+ and OH-. Hydroxide is going to take the acidic proton on acetic acid, and this is the acidic proton on acetic acid. Once again, H+ and OH- give us H20."
    },
    {
        "Q": "at 1:25 what is an oxidation state?\n",
        "A": "An oxidation state is a tool to write the electronic charge(s) of a molecule without resorting to partial charges (e.g. 1/2-).",
        "video_name": "M7PnxSQedkM",
        "timestamps": [
            85
        ],
        "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has"
    },
    {
        "Q": "\nAt 2:27 what is the reducing agent?",
        "A": "a substance that tends to bring about reduction by being oxidized and losing electrons.",
        "video_name": "M7PnxSQedkM",
        "timestamps": [
            147
        ],
        "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has"
    },
    {
        "Q": "3:36 so if the 2 was positive it would not be neutral ?\n",
        "A": "Right it would be positive, and vice versa. But in both of these compounds the positive and negative charges cancel each other out.",
        "video_name": "M7PnxSQedkM",
        "timestamps": [
            216
        ],
        "3min_transcript": "And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has So it could have an oxidation state of positive 2, which would imply that the entire hydroxide anion-- And let's just say hydroxide for now. Well I'll say hydroxide anion. I kind of gave it away a little bit-- that this hydroxide, or this part of the molecule, the right-hand part of what I've written here, for this whole thing to be neutral, it should have a negative 2 oxidation state. Now how does that make sense? Well we have two hydroxides here. Notice this subscript right over here. So if each of those hydroxides has a negative 1 charge, or a negative 1, I guess you could say, total oxidation state, then when you take two of them together, they would net out against the magnesium. And that does seem to make sense. If oxygen has a negative 2 oxidation state, hydrogen has a positive 1 oxidation state. is going to have a net oxidation state of negative 1. But then you have two of them. So the net oxidation for this part of the molecule or the compound is going to be negative 2 nets out with the positive 2 from magnesium. So once again, it makes sense."
    },
    {
        "Q": "At 1:19,why does magnesium loose two electrons?\n",
        "A": "It loses two electrons because it strives as any element to become stable, which is to say to have either 8 or 2 electrons on its outer shell. So to become stable, it only needs to lose the 2 electrons on its outer shell for the shell under the initial shell to become the outer shell.",
        "video_name": "M7PnxSQedkM",
        "timestamps": [
            79
        ],
        "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has"
    },
    {
        "Q": "9:21 Why aren't children vaccinated against meningococcus earlier than the age o two?\n\nEdit: I found an answer to this question in an Infectious diseases textbook. Apparently the immune system of children less then two years old is not mature enough to respond to a polysacharide antigen of N. meningitidis, thus vaccination before the age of two would be futile.\n",
        "A": "I disagree with Amelia on one point. Babies are given a Heb B vaccine at birth. I d assume that there would be just as much of a stigma regarding that. Also, babies are now given a whole slew of vaccines at age 2 months old. If it were safe and effective in neonates, it would be on the schedule starting at 2 months.",
        "video_name": "CnXuSCaCNBo",
        "timestamps": [
            561
        ],
        "3min_transcript": "And then, most importantly, the fluid is examined with something called a Gram stain, a special kind of stain. It's Gram with a capital G, named after Dr. Gram. And a Gram stain can determine whether or not there are bacteria present. MORGAN THEIS: OK. So you're actually staining the bacteria. DR. CHARLES PROBER: Exactly. And if there are sufficient bacteria present, the Gram stain will reveal those bacteria. And so with bacterial meningitis, the second prudent principle is to know the usual pathogens. So if a spinal fluid is obtained, there are lots of white cells, the glucose is low. Even with a negative Gram stain, one can guess the usual pathogens, because the list is short in normal children. And those bacteria, the short list includes a bacteria called Haemophilus influenzae, type B. I mean, an E at the end? DR. CHARLES PROBER: There is an E on the end. A second bacteria is the new pnuemococcus. MORGAN THEIS: Now, that's funny. It sounds like it causes pneumonia. DR. CHARLES PROBER: And it does indeed cause pneumonia as well. But it also causes bacterial meningitis. And a third bacteria is called meningicoccus. And those are the prominent bacteria in normal children with bacterial meningitis. The reason though, we're not seen as much bacterial meningitis in 2011 as we were seeing 10 and 20 years ago is, we now have vaccination against each of those three different pathogens. MORGAN THEIS: All of them? DR. CHARLES PROBER: We do. We vaccinate against Haemophilus influenzae, type B, starting at two months of age. And by the time the child is about a year and a half, they're completely protected against The pneumococcus, we also vaccinate against. And it's very successful. The vaccine is very successful at reducing the frequency of pneumococcal meningitis. It also starts at two months of age. And meningicoccus, the vaccine is relatively new and is used in children who are a bit older. They're over two years of age, under special circumstances. So that means that we still can see, and do see, cases of meningicoccal meningitis, because it occurs in children under two years of age. MORGAN THEIS: I see. DR. CHARLES PROBER: But those are the usual pathogens. And when you go to other parts of the world who don't use vaccines, those are the pathogens that will be prominent in causing bacterial meningitis. And knowing those pathogens, we go to the third principle of antibiotic prescribing, which is knowing what antibiotics typically kill those bacteria. MORGAN THEIS: OK. So what should I call that category, like matching?"
    },
    {
        "Q": "I've noted that Jay writes ionic compounds like this: Na+-OH (7:32)\nIs this the standard way of writing it? I'm used to seeing Na+OH-, with the sign after the anion... (both of course in superscript).\n",
        "A": "see the things are same........but specifically in the hydroxide anion , due to the more electronegative character of the oxygen atom the bonded pair of electrons in the molecule rest near the oxygen atom.....hence jay has made the negative charge over the oxygen atom...",
        "video_name": "My5SpT9E37c",
        "timestamps": [
            452
        ],
        "3min_transcript": "So we are gonna use DMSO. And we know in an SN2 mechanism the nucleophile attacks our alkyl halide at the same time our leaving group leaves. So our nucleophile is the hydroxide ion. It is going to attack this carbon and these electrons are gonna come off on to the bromide to form our bromide anion. So our OH replaces our bromine and we can see that over here in our product. In an SN2 mechanism we need a strong nucleophile to attack our alkyl halide. And DMSO is gonna help us increase the effectiveness of our nucleophile which is our hydroxide ion. So let's look at some pictures of how it helps us. So we have sodium hydroxide here. So first let's focus in on the sodium, our cation. So here is the sodium cation. DMSO is a good solvator of cations and that's because oxygen has a partial negative charge. The sulfur has a partial positive charge help to stabilize the positive charge on our sodium. So same thing over here. Partial negative, partial positive and again we are able to solvate our cation. So the fact that our polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile. That increases the effectiveness of the hydroxide ion. The hydroxide ion itself is not solvated by a polar aprotic solvent. So you might think, okay well if the oxygen is partially negative and the sulfur is partially positive. The partially positive sulfur could interact with our negatively charged nucleophile. But remember we have these bulky methyl groups here. And because of steric hindrance that prevents our hydroxide ion from interacting with DMSO. So the hydroxide ion is all by itself which of course increases its effectiveness as a nucleophile. If we had used something like water, we know that water is a polar protic solvent with the oxygen being partially negative and the hydrogens being partially positive and a polar protic solvent would interact with our nucleophile solvating it and essentially decreasing the effectiveness of our nucleophile. So that's why polar protic solvents don't work as well if you want an SN2 mechanism. A polar aprotic solvent increases the effectiveness of our nucleophile therefore favoring our SN2 mechanism."
    },
    {
        "Q": "\nat 10:06 what does he mean when he says increases the strength of the nucleophile?",
        "A": "In a polar protic solvent, the nucleophile is solvated by the solvent molecules. The nucleophile has to push these solvating molecules out of the way in order to attack the \u00ce\u00b1 carbon. A polar aprotic solvent preferentially solvates the metal cation. That means that the nucleophile is relatively unsolvated or \u00e2\u0080\u009cbare\u00e2\u0080\u009d. It can then much more easily attack the \u00ce\u00b1 carbon. This makes it a stronger nucleophile.",
        "video_name": "My5SpT9E37c",
        "timestamps": [
            606
        ],
        "3min_transcript": "If we had used something like water, we know that water is a polar protic solvent with the oxygen being partially negative and the hydrogens being partially positive and a polar protic solvent would interact with our nucleophile solvating it and essentially decreasing the effectiveness of our nucleophile. So that's why polar protic solvents don't work as well if you want an SN2 mechanism. A polar aprotic solvent increases the effectiveness of our nucleophile therefore favoring our SN2 mechanism."
    },
    {
        "Q": "At 4:43 it shows water molecules moving through protein channels. I am aware that the cell membrane does not allow watersoluble/hydrophillic/lipidphobic substances to pass straight through but I thought small molecules such as oxygen gas, carbon dioxide and water were able to pass through easily because of their size, without needing to move through channel proteins. Is this incorrect? Thanks\n",
        "A": "Channel proteins allow passive transport of the small molecules like oxygen gas and water, meaning these molecules can flow through from the higher concentration to the lower one without using energy. This is what the video means by easily passing through the membrane, since no ATP is required. The water and oxygen cannot pass through anywhere they want, otherwise the membrane would be so full of holes that it would be useless! Hope this helps!",
        "video_name": "TyZODv-UqvU",
        "timestamps": [
            283
        ],
        "3min_transcript": "We want the in-the-middle Charlie Sheen who can just make us laugh and be happy; and that is the state that water concentrations are constantly seeking, it's called isotonic when the concentration is the same on both sides, outside and in. And this works in real life, we can actually show it to you. This vase if full of fresh water, and we also have a sausage casing, which is actually made out of cellulose, and inside of that we have salt water. We've died it so that you can see it move through the casing, which is acting as our membrane. This time lapse shows how over a few hours, the salt water diffuses into the pure water. It'll keep diffusing until the concentration of salt in the water is the same inside the membrane as outside. When water does this, attempting to become isotonic, it's called moving across its concentration gradient. Most of my cells right now are bathed in a solution that has the same concentration as inside of them, and this is important. For example, if you took one of my red blood cells and put in a glass of pure water, it would be so hypertonic, so much stuff would be in the cell compared to outside the cell, that water would rush into the red blood cell, then it would literally explode. But if the concentration of my blood plasma were too high, all the water would rush out of my cell and it would shrivel up and be useless. And that's why your kidneys are constantly on the job regulating the concentration of your blood plasma to keep it isotonic. No, water can permeate a cell membrane without any help, but it's not actually particularly easy. As we discussed in the last episode, cell membranes are made out of phospholipids; and the phospholipid bilayer is hydrophilic, or water-loving on the outside, and hydrophobic, or water-hating, on the inside. So, water molecules have a hard time passing through these layers because they get stuck at the non-polar hydrophobic core. That is where the channel proteins come in. They allow passage of stuff like water and ions without using any energy. They straddle the width of the membrane, and inside they have channels that are hydrophilic, which draws the water through. The proteins that are specifically for channeling water are called aquaporins. Each one can pass 3 billion water molecules a second. Make me have to pee just thinking about it. Things like oxygen and water that cells need constantly, but most chemicals, they use what's called active transport. This is especially useful if you want to move something in the opposite direction of it concentration gradient from low concentration to a high concentration. So, say we're back at that show, and I'm keeping company with John, who's being all anti-social in his polite and charming way, but after half a beer and an argument about who's the best Doctor Who, I want to get back to my friends across the crowded bar. So, I transport myself against the concentration gradient of humans, spending a lot of energy dodging stomping feet, throwing an elbow to get to them. That is high energy transport. In a cell, getting the energy necessary to do pretty much anything, including moving something the wrong direction across its concentration gradient, requires ATP. ATP, adenosine tri-phosphate. You just wanna replay that over and over again, until it just rolls off the tongue because it's one of the most important chemicals that you will ever, ever, ever hear about."
    },
    {
        "Q": "Sorry I'm confused because at 4:00 it was mentioned, I thought Hypotonic was when water would rush in and explode while hypertonic was when it would shrivel up.\n",
        "A": "That is correct - however, this terminology is very tricky. See the example below: You have a situation with less water in cell, more in envirnoment The cell is hypertonic to the solution, and the solution is hypotonic to the cell. The cell is in a hypotonic solution.",
        "video_name": "TyZODv-UqvU",
        "timestamps": [
            240
        ],
        "3min_transcript": "until it was a uniform mass of John Greens throughout the club. When oxygen gets crowded, it finds places that are less crowded and moves into those spaces. When water gets crowded, it does the same thing. It moves to where there is less water. When water goes across a membrane, it's a kind of diffusion called osmosis. This is how your cells regulate their water content. Not only does this apply to water itself, which as we've discussed is the world's best solvent, you're gonna learn more about water in our water episode, it also works with water that contains dissolved materials or solutions, like solutions of salt water or solutions of sugar water, or booze, which is just a solution of ethanol and water. If the concentration of a solution is higher inside of a cell than it is outside of the cell, then that solution is called hypyertonic, like power thirst, it's got everything packed into it. And if the concentration inside of the cell is lower than outside of the cell, it's called hypotonic. Which is sort of a sad version of hypertonic. So, like with Charlie Sheen, we don't want the crazy, manic Charlie Sheen, We want the in-the-middle Charlie Sheen who can just make us laugh and be happy; and that is the state that water concentrations are constantly seeking, it's called isotonic when the concentration is the same on both sides, outside and in. And this works in real life, we can actually show it to you. This vase if full of fresh water, and we also have a sausage casing, which is actually made out of cellulose, and inside of that we have salt water. We've died it so that you can see it move through the casing, which is acting as our membrane. This time lapse shows how over a few hours, the salt water diffuses into the pure water. It'll keep diffusing until the concentration of salt in the water is the same inside the membrane as outside. When water does this, attempting to become isotonic, it's called moving across its concentration gradient. Most of my cells right now are bathed in a solution that has the same concentration as inside of them, and this is important. For example, if you took one of my red blood cells and put in a glass of pure water, it would be so hypertonic, so much stuff would be in the cell compared to outside the cell, that water would rush into the red blood cell, then it would literally explode. But if the concentration of my blood plasma were too high, all the water would rush out of my cell and it would shrivel up and be useless. And that's why your kidneys are constantly on the job regulating the concentration of your blood plasma to keep it isotonic. No, water can permeate a cell membrane without any help, but it's not actually particularly easy. As we discussed in the last episode, cell membranes are made out of phospholipids; and the phospholipid bilayer is hydrophilic, or water-loving on the outside, and hydrophobic, or water-hating, on the inside. So, water molecules have a hard time passing through these layers because they get stuck at the non-polar hydrophobic core. That is where the channel proteins come in. They allow passage of stuff like water and ions without using any energy. They straddle the width of the membrane, and inside they have channels that are hydrophilic, which draws the water through. The proteins that are specifically for channeling water are called aquaporins. Each one can pass 3 billion water molecules a second. Make me have to pee just thinking about it. Things like oxygen and water that cells need constantly,"
    },
    {
        "Q": "\nat 8:34 Sal said that if we cut a magnet into half then two more magnets will appear. But notice that the left side (from the viewer's point of view) of the 1st magnet is north pole but the left side of the second magnet is south. But why?",
        "A": "because a magnet is made of a bunch of smaller magnets that stick together, north to south and south to north",
        "video_name": "8Y4JSp5U82I",
        "timestamps": [
            514
        ],
        "3min_transcript": "something in that field that can be affected by it, it'll be some net force acting on it. So actually, before I go into magnetic field, I actually want to make one huge distinction between magnetism and electrostatics. Magnetism always comes in the form of a dipole. It means that we have two poles. A north and a south. In electrostatics, you do have two charges. You have a positive charge and a negative charge. So you do have two charges. But they could be by themselves. You could just have a proton. You don't have to have an electron there right next to it. You could just have a proton and it would create a positive electrostatic field. And our field lines are what a positive point charge would do. And it would be repelled. So you don't always have to have a negative charge there. And you don't have to have a proton there. So you could have monopoles. These are called monopoles, when you just have one charge when you're talking about electrostatics. But with magnetism you always have a dipole. If I were to take this magnet, this one right here, and if I were to cut it in half, somehow miraculously each of those halves of that magnet will turn into two more magnets. Where this will be the south, this'll be the north, this'll be the south, this will be the north. And actually, theoretically, I've read-- my own abilities don't go this far-- there could be such a thing as a magnetic monopole, although it has not been observed yet in nature. So everything we've seen in nature has been a dipole. So you could just keep cutting this up, all the way down to if it's just one electron left. And it actually turns out that even one electron is still a magnetic dipole. It still is generating, it still has a north pole and a And actually it turns out, all magnets, the magnetic field is actually generated by the electrons within it. By the spin of electrons and that-- you know, when we talk about electron spin we imagine some little ball of charge spinning. But electrons are-- you know, it's hard to-- they do have mass. But it starts to get fuzzy whether they are energy or mass. And then how does a ball of energy spin? Et cetera, et cetera. So it gets very almost metaphysical. So I don't want to go too far into it. And frankly, I don't think you really can get an intuition. It is almost-- it is a realm that we don't normally operate in. But even these large magnets you deal with, the magnetic field is generated by the electron spins inside of it and by the actual magnetic fields generated by the electron motion around the protons. Well, I hope I'm not overwhelming you. And you might say, well, how come sometimes a metal bar can be magnetized and sometimes it won't be? Well, when all of the electrons are doing random"
    },
    {
        "Q": "In 6:39, what is the nearest star? I just want to know.\n",
        "A": "Proxima Centauri",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            399
        ],
        "3min_transcript": "So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star--"
    },
    {
        "Q": "I'm not sure where I heard this (probably early elementary school), but I remember learning that the red spot on Jupiter is 3x as big as Earth. If that's true, I feel like Jupiter can't be 10x as big as Earth, like Sal says at 4:12, because the red spot isn't 30 percent of the planet. Am I crazy in remembering that the red spot is the size of 3 Earths?\n",
        "A": "Jupiter s circumference is 11 times that of Earths. But that gives it a surface area more than 100x larger. The red spot is about 2 or 3 earth diameters wide.",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            252
        ],
        "3min_transcript": "it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States. You Nevada here. You have Arizona and New Mexico-- so a big chunk of a big continent we're already including. And frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it. But if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes goes so fast-- almost unimaginably fast for us-- that it's made it feel like things like continents aren't as big. Because you can fly across them in five or six hours. But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles, and you multiply that by 10. And you get pretty close-- a little bit over-- the diameter of the Earth-- a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. So even though this is, frankly, larger can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun."
    },
    {
        "Q": "at 0:15 Sal says that we can't comprehend thing that are small compared to the size of the universe, is he referring to the galaxies?\n",
        "A": "An example would be galaxies, other examples also include black holes, superclusters, clusters, et.c",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            15
        ],
        "3min_transcript": "The purpose of this video is to just begin to appreciate how vast and enormous the universe is. And frankly, our brains really can't grasp it. What we'll see in this video is that we can't even grasp things that are actually super small compared to the size of the universe. And we actually don't even know what the entire size of the universe is. But with that said, let's actually just try to appreciate how small we are. So this is me right over here. I am 5 foot 9 inches, depending on whether I'm wearing shoes-- maybe 5 foot 10 with shoes. But for the sake of this video, let's just roughly approximate around 6 feet, or around roughly-- I'm not to go into the details of the math-- around 2 meters. Now, if I were to lie down 10 times in a row, you'd get about the length of an 18-wheeler. That's about 60 feet long. So this is times 10. Now, if you were to put an 18-wheeler-- if you were to make it tall, as opposed to long-- somehow stand it up-- and you were to do that 10 times in a row, you'll get to the height of roughly a 60-story skyscraper. you'll get about a 60-story skyscraper. Now, if you took that skyscraper and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers. It's not always going to be exactly 10. But we're now getting to about something that's a little on the order of a mile long. So the Golden Gate Bridge is actually longer than a mile. But if you go within the twin spans, it's roughly about It's actually a little longer than that. But that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here. And when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles just so you appreciate the scale. And what's interesting here-- and this picture's interesting. Because this is the first time we can relate to cities. it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States."
    },
    {
        "Q": "At 6:56 Sal said that the nearest star is 200,000 times the distance from the earth to the sun; but isn't the sun considered a star?\n",
        "A": "Yes. Obviously he meant the nearest star other than the sun, right?",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            416
        ],
        "3min_transcript": "So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star--"
    },
    {
        "Q": "5:37 What does AU stand for?\n",
        "A": "It stands for Astronomical Unit, which is the average distance between the Earth and the Sun, about 93 million miles, or 1.5*10^11 meters.",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            337
        ],
        "3min_transcript": "can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun. So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball--"
    },
    {
        "Q": "\nAt around 7:45, the second equation of CO2 plus HbO2 does not seem to be a balanced equation. Where does an extra H+ on the right hand side of the equation come from? According to the reactants, shouldn't it be CO2+HbO2 = HbCOO- + O2 ??",
        "A": "Hemoglobin is a protein so it s formed of amino acids , carboxylic acid having amide group -NH2, so in this equation CO2 binds to the terminal amide group forming amide linkage and -NH2 loses H+ _-NH2 + CO2 ------> _-NHCOO- + H+",
        "video_name": "QP8ImP6NCk8",
        "timestamps": [
            465
        ],
        "3min_transcript": "Now, if it's an acid, try to keep in mind what acids do. Acids are going to kick off a proton. So this becomes HCO3 minus. And it kicks off a proton. And notice that now you've got bicarb and proton on this side. And this bicarb is actually going to just make its way outside. So the bicarb goes outside the cell. And the proton, what it does is, it meets up with one of these oxyhemoglobins. It kind of finds an oxyhemoglobin. Remember, there are millions of them around. And it literally binds to hemoglobin. And it boots off the oxygen. So it binds to hemoglobin and oxygen falls away. So this is interesting because now this is a second reason for why oxygen gets delivered to the tissues. And that is that, protons compete with hemoglobin. So they're competing for hemoglobin. Now I said there is another thing that happens to the carbon dioxide. So what's the other thing? Turns out that carbon dioxide actually sometimes independently seeks out oxyhemoglobin. Remember, again, there are millions of them. So it'll find one. And it'll do the same thing. It'll say, well, hey, hemoglobin, why don't you just come bind with me and get rid of that oxygen? So it also competes with oxygen. So you've got some competition from protons, some competition from carbon dioxide. And when carbon dioxide actually binds, interesting thing is that it makes a proton. So guess what happens? That proton can go and compete again by itself. It can compete with oxyhemoglobin and try to kick off another one, kick off another oxygen. because now you've got a few reasons why you have oxygen delivery. You've got protons competing. You've got now CO2 competing with oxygen. So you've got a couple of sources of competition. And you've got, of course, just simply the fact that there's just not too much oxygen around. So these are reasons for oxygen delivery. So at this point, you've got oxygen that's delivered to the cells. And these hemoglobin molecules, they're still our cell, of course, inside of a red blood cell. And these hemoglobin molecules have now been bound by different things. So they're no longer bound by oxygen. So you can't really call them oxyhemoglobin anymore. Instead they have protons on them like this. And they might have some COO minus on them. So they might have-- actually, let me do that in the original kind of orangey color. So they basically have different things binding to them."
    },
    {
        "Q": "At 8:01, we see an equation where CO2 bonds with Hb and creates H+. Where does the hydrogen come from? Has it already created carbonic acid by reacting with H2O?\n",
        "A": "Yes you re right CO2 already has reacted with water and then formed carbonic acid, from which the hydrogen dissociates..",
        "video_name": "QP8ImP6NCk8",
        "timestamps": [
            481
        ],
        "3min_transcript": "Now, if it's an acid, try to keep in mind what acids do. Acids are going to kick off a proton. So this becomes HCO3 minus. And it kicks off a proton. And notice that now you've got bicarb and proton on this side. And this bicarb is actually going to just make its way outside. So the bicarb goes outside the cell. And the proton, what it does is, it meets up with one of these oxyhemoglobins. It kind of finds an oxyhemoglobin. Remember, there are millions of them around. And it literally binds to hemoglobin. And it boots off the oxygen. So it binds to hemoglobin and oxygen falls away. So this is interesting because now this is a second reason for why oxygen gets delivered to the tissues. And that is that, protons compete with hemoglobin. So they're competing for hemoglobin. Now I said there is another thing that happens to the carbon dioxide. So what's the other thing? Turns out that carbon dioxide actually sometimes independently seeks out oxyhemoglobin. Remember, again, there are millions of them. So it'll find one. And it'll do the same thing. It'll say, well, hey, hemoglobin, why don't you just come bind with me and get rid of that oxygen? So it also competes with oxygen. So you've got some competition from protons, some competition from carbon dioxide. And when carbon dioxide actually binds, interesting thing is that it makes a proton. So guess what happens? That proton can go and compete again by itself. It can compete with oxyhemoglobin and try to kick off another one, kick off another oxygen. because now you've got a few reasons why you have oxygen delivery. You've got protons competing. You've got now CO2 competing with oxygen. So you've got a couple of sources of competition. And you've got, of course, just simply the fact that there's just not too much oxygen around. So these are reasons for oxygen delivery. So at this point, you've got oxygen that's delivered to the cells. And these hemoglobin molecules, they're still our cell, of course, inside of a red blood cell. And these hemoglobin molecules have now been bound by different things. So they're no longer bound by oxygen. So you can't really call them oxyhemoglobin anymore. Instead they have protons on them like this. And they might have some COO minus on them. So they might have-- actually, let me do that in the original kind of orangey color. So they basically have different things binding to them."
    },
    {
        "Q": "\nAt 1:10, is diffusion kind of like osmosis?",
        "A": "Yes, osmosis is specifically for water, while diffusion is for other solutes :)",
        "video_name": "QP8ImP6NCk8",
        "timestamps": [
            70
        ],
        "3min_transcript": "Let's talk about exactly how oxygen and carbon dioxide come into and out of the lungs. So you know this is our alveolus in the lungs. This is the last little chamber of air where the lungs are going to interface with blood vessels. So this is our blood vessel down here. And oxygen is going to make its way from this alveolus. It's going to go into the blood vessel. And it's going to go from the blood vessel into a little red blood cell. This is my red blood cell here. He's headed out for the first delivery of oxygen that day. And he's going to pick up some oxygen. And it's going to get inside of the red blood cell through diffusion. That's how it gets inside. So the oxygen has made its way into the red blood cell. And where do you think it goes first? Well, this red blood cell is, we sometimes think of it as a bag of hemoglobin. It's got millions and millions and millions of hemoglobin proteins. So this is our hemoglobin protein. It's got four parts to it. So hemoglobin, I can shorten this to Hb. Now, oxygen is going to bump into, quite literally bump into one of these hemoglobins. And it's going to bind, let's say, right here. And initially, it's kind of tricky because oxygen doesn't feel very comfortable sitting on the hemoglobin or binding to hemoglobin. But once a single oxygen is bound, a second one will come and bind as well. And then a third will find it much easier. Because what's happening is that as each oxygen binds, it actually changes the conformation or shape of hemoglobin. And so each subsequent oxygen has an easier time binding. We call that cooperativity. Has the word, almost like cooperation in it. And an easy way to think of cooperativity, the way I think of it, is that if you're at a dinner party, you are much more likely to sit where two or three of your friends are already sitting, if you think of this as a table with four chairs, rather than just sitting at a table by yourself being the first one to sit there. is kind of a friendly molecule. And so it also likes to sit where or bind where other oxygens have already bound. What are the two, then, major ways, based on this diagram, how I've drawn it. What are the two major ways that oxygen is going to be transported in the blood? One is hemoglobin binding oxygen. And we call that HbO2. Just Hb for hemoglobin, O2 for oxygen. And this molecule, or this enzyme, then, is not really called hemoglobin anymore. Technically, it's called oxyhemoglobin. That's the name for it. And another way that you can actually transport oxygen around is, that some of this oxygen-- I actually underlined it there-- is dissolved, O2 is dissolved in plasma. So some of the oxygen actually just gets dissolved right into the plasma. And that's how it gets moved around. Now, the majority, the vast majority of it is actually going to be moved through binding to hemoglobin. So just a little bit is dissolved in the plasma."
    },
    {
        "Q": "At 0:29 when Sal mentions that c3 mixes with RuBis what does that mean.\n",
        "A": "C3 means the number of c in an triose phosphate molecule and in the calvin cycle the after some ofthe triose phosphate is sent to make stuff like glucose and lipids, the remaining triose phosphate does not mix but is converted to Rubp with the help of ATP.",
        "video_name": "xp6Zj24h8uA",
        "timestamps": [
            29
        ],
        "3min_transcript": "A couple of videos ago, we saw that in classic C3 photosynthesis And once again, it's called C3 because the first time that carbon dioxide (CO2) is fixed, it's fixed into a 3-carbon molecule. But we saw the problem with C3 photosynthesis is that the enzyme that does the carbon fixation, it can also react with oxygen. And when oxygen essentially reacts with ribulose biphosphate instead of your carbon, you get an unproductive reaction. Not only is it unproductive, it'll actually suck up your ATP and your NADPH and you'll go nowhere. So every now and then, when oxygen bonds here instead of a carbon dioxide, you get nothing produced, net. Everything becomes less efficient. And so in the last video, we saw that some plants have evolved a way to get around this. And what they do is, they fix their carbon on the outside, on cells that are actually exposed to the air. And then once they fix the carbon they actually fix it that gets turned into malate. Then they pump the malate deeper within the leaf, where you aren't exposed to oxygen. And then they take the carbon dioxide off the malate, and this is where they actually perform the Calvin cycle. And even though you do have your RuBisCO still there, your RuBisCO isn't going to have - the photorespiration is not going to occur. Because it only has access to carbon dioxide. It does not have access to this oxygen out here. Now that's a very efficient way of producing sugars. And that's why some of the plants that we associate with being very strong sugar, or even ethanol producers, all perform C4 photosynthesis. Corn, sugarcane and crab grass. And these are all very, very efficient sugar producers. Because they don't have to worry too much about photorespiration. Now some plants have a slightly different problem. They're not so worried about the efficiency of the process. They're more worried about losing water. These are plants that are in the desert. Because these stomata, these pores that are on the leaves, they let in air, but they can also let out water. I mean, if I'm in the rainforest, I don't care about that. But if I'm in the middle of the desert, I don't want to let out water vapour through my stomata. So the ideal situation is, I would want my stomata closed during the daytime. This is what I want. So I want, if I'm in the desert, let me make this clear. If I'm in the desert, I want stomata closed during the day. For obvious reasons. I don't want all my water to vaporise out of these holes in my leaves. But at the same time, the problem is photosynthesis can only occur during the daytime. And that includes the dark reactions. Remember, I've said multiple times, the dark reactions are"
    },
    {
        "Q": "\n7:00 Wouldn't getting the dam out of the way release the rest of the toxic copper sediments?",
        "A": "They would use bioremediation to remove toxic elements from the water before getting rid of the dam.",
        "video_name": "3BBqL_F9fxQ",
        "timestamps": [
            420
        ],
        "3min_transcript": "extinction vortex. The next step is to figure out how small a population is too small. Ecologists do this by calculating what's called the minimum viable population, which is the smallest size at which a population can survive and sustain itself. To get at this number, you have to know the real breeding population of, say, grizzly bears in Yellowstone National Park. And then you figure out everything you can about a grizzly's life history, how long they live, who gets to breed the most, how often they can have babies, that kind of thing. After all that information is collected, ecologists can run the numbers and figure out that, for the grizzlies in Yellowstone, a population of, say hypothetically, 90 bears would have about a 95% chance of surviving for 100 years. But if there were a population of 100 bears, the population would likely be able to survive for 200 years. Something to note, ecology involves a lot of math. So if you're interested in this, that's just the way it is. So that's the small population approach to conservation. Another way of preserving biodiversity focuses on populations whose numbers are in decline, no matter how large the original population was. and it involves answering a series of related questions that get at the root of what's causing an organism's numbers to nosedive. First, you have to determine whether the population's actually declining. Then you have to figure out how big the population historically was and what its requirements were. And finally, you have to get at what's causing the decline and figure out how to address it. Milltown Dam actually gives us a good example of this process. In the winter of 1996, authorities had to release some of the water behind the dam as an emergency measure because of a big ice floe in the river that was threatening to break the dam. But when they released the water, a bunch of toxic sediment went with it, which raised the copper concentrations downriver to almost 43 times what state standards allowed. As a result, it's estimated that about half of the fish downstream died. Half of the fish, dead. And researchers have been monitoring the decline in populations ever since. This information was really helpful in determining what to do with the dam because we knew what the fish population was like before and after the release It was decided that it would be best to get the dam out as soon as possible rather than risk another 1996 scenario. Which brings me to the place where conservation biology and restoration ecology intersect. Restoration ecology is kind of where the rubber meets the road in conservation biology. It comes up with possible solutions for ecological problems. Now, short of a time machine, which I'm working on, you can't really get a natural environment exactly the way that it used to be. But you can at least get rid of whatever is causing the problem, and help recreate some of the elements that the ecosystem needs to function properly. All of this involves a whole suite of strategies. For instance, what's happening in Milltown is an example of structural restoration, basically the removal and cleanup of whatever human impact was causing the problem, in this case, the dam and the toxic sediments behind it. And then the rebuilding of the historical natural structure, here the meanders of the river channel and the vegetation. Another strategy is bioremediation, which recruits organisms temporarily to help remove toxins, like bacteria that eat wastes or plants that leach out metals from tainted soils. Some kinds of fungi and bacteria are even"
    },
    {
        "Q": "At 1:06, It is great to learn about the restoration of the river system in Montana. Is there other examples (positive or negative) of Restoration Elology? Is there a way to restore Chernobyl nuclear site, since we know exactly what went wrong?\n",
        "A": "Well, one example is the Chesepeake Bay Watershed. The watershed was very polluted at one time but now, laws, regulations, and ordinary efforts from people are making it better! This applies to most polluted places in the US because of the EPA. Oh, and about Chernobyl, we have to wait until the radiation goes away",
        "video_name": "3BBqL_F9fxQ",
        "timestamps": [
            66
        ],
        "3min_transcript": "For the past 12 weeks, we've been investigating our living planet together, learning how it works on many levels, how populations of organisms interact, how communities thrive and ecosystems change, and how humans are wrecking the nice, perfectly functioning systems Earth has been using for hundreds of thousands of years. And now it's graduation day. This here is like the commencement speech where I talk to you about the future and our role in it and how what we're doing to the planet is totally awful, but we're taking steps to undo some of the damage So what better way to wrap up our series on ecology than by taking a look at the growing fields of conservation biology and restoration ecology. These disciplines use all the Kung Fu moves that we've learned about in the past 11 weeks and apply them to protecting ecosystems and cleaning up the messes that we've already made. And one of the main things they teach us is that doing these things is difficult, like in the way that uncooking bacon is difficult. So let's look at what we're doing and try to uncook this unbelievably large pile of bacon we've made. we've got a Superfund site. Not super-fun. Superfund, a hazardous waste site that the government is in charge of cleaning up. The mess here was made more than 100 years ago, when there was a dam in the Clark Fork River behind me called the Milltown Dam. This part of Montana has a long history of copper mining, and back in 1908, there was a humongous flood that washed about 4.5 million cubic meters of mine tailings chock full of arsenic and toxic heavy metals into the Clark Fork River. And most of it washed into the reservoir created by the Milltown Dam. I mean, actually it was lucky that the dam was there-- it had only been completed six months before-- or the whole river system all the way to the Pacific Ocean would have been a toxic mess. As it happened, though, only about 160 kilometers of the river was all toxic, messed up. A lot of it recuperated over time. But all that nasty hazardous waste was still sitting behind Milltown Dam. that started polluting nearby residents' wells. So scientists spent decades studying the extent of the damage caused by the waste and coming up with ways to fix it. And from 2006 to 2010, engineers carefully removed all the toxic sediment as well as the dam itself. Now this stretch of the Clark Fork River runs unimpeded for the first time in over a century, and the restored area where the dam used to be is being turned into a state park. Efforts like this show us conservation biology and restoration ecology in action. Conservation biology involves measuring the biodiversity of an ecosystem and determining how to protect it. In this case, it was used to size up the health of fish populations in the Clark Fork River, which were severely affected by the waste behind the dam, and the damn blocking their access to spawning grounds upstream, and figuring out how to protect them during the dam's removal. Restoration ecology, meanwhile, is the science of restoring broken ecosystems, like taking an interrupted, polluted river and turning it into what you see taking shape here."
    },
    {
        "Q": "At 2:11, Sal says that \"almost nothing\" is completely smooth on an atomic level. Is there any known substance that is completely smooth?\n",
        "A": "A substance that is completely smooth will be frictionless. As far as our knowledge is concerned, there is no substance that is completely frictionless. Thus, till date, we do not know of any substance that is completely smooth, that is, frictionless. This is because the atoms of all substances interact and nudge each other when they are close to each other. Hope this helps :)",
        "video_name": "J9BWNiOSGlc",
        "timestamps": [
            131
        ],
        "3min_transcript": "I mentioned in the last several videos that the coefficient of kinetic friction tends to be less, sometimes it'll be roughly equal to, the coefficient of static friction. But this might lead you to-- at least, a question that I've had in my mind, and I still have to some degree-- is why? Why is the coefficient of kinetic friction lower? Or why can it be lower? And the current best theory-- one I can visualize in my head, and based on the reading that I've done-- is the difference between-- so let's think about it this way. So if we look at it at a kind of a regular human level, maybe we have a block. So this is the static case. So let's think about the static case. Let me draw it like this. So I'll draw the static case over here. So I have a block that is stationary on top of-- let me do the surface in a different color-- on top of some type of surface right over here. And over here, I'm going to have a block moving at a constant velocity relative to some surface, And so let me draw it out. So this is moving at some constant velocity. And so the interesting thing here is, assuming that these are the same masses, that these are the same surfaces, is, why should the coefficient of friction here-- why should the coefficient of static friction-- so here, since this is stationary, what's under play is the coefficient of static friction. Why should that be larger than the coefficient of kinetic friction over here? Why should that be large than the coefficient of kinetic friction? Or another way to think about it is, you would need to apply more force to overcome the static friction here, and start to get this accelerating, than you would need to apply to get this already moving body to accelerate. Because there would be kind of a less of a responsive friction force. So what I'm going to do is zoom in into the atomic level. And so when you zoom in to the atomic level, almost nothing is completely smooth. So the surface over here might look something like this. So I'm going to draw the molecules that make up the surface, the best to my ability. So the molecules, when you zoom up really close for the surface, might look something like this. So we're really zooming into the atomic level, unimaginably small level. Much smaller than that box I just drew. But I'm just trying to look at what's happening with the atoms where they contact, or the molecules where they contact? And the box's molecules might look something like this. They aren't completely smooth. And hopefully this video also emphasizes that all of these forces and all of this contact that we're talking about in these videos-- and it's actually"
    },
    {
        "Q": "At 7:24, wouldn't the liquid just stay inside because the hole is the only way for air to get in?\n",
        "A": "not only that, but a vacuum would have negative pressure wouldn t it? unless outside is a vacuum too. the liquid would just turn into a vapor and fill the container.",
        "video_name": "QX2YLR09Q78",
        "timestamps": [
            444
        ],
        "3min_transcript": "Similarly, if pressure increases, then velocity is going to decrease. That might be a little unintuitive, but the other way, it makes a lot of sense. When velocity increases, this pressure is going to decrease, and that's actually what makes planes fly and all sorts of neat things happen, but we'll get more into that in a second. Let's see if we can use Bernoulli's equation to do something useful. You should memorize this, and it shouldn't be too hard to memorize. It's pressure, and then you have this potential energy term, but instead of mass, you have density. You have this kinetic energy term. It's not kinetic energy anymore, because we manipulated it some, but instead of mass, you have density. With that said, let's do a problem. I'll keep this down here, since you probably haven't memorized it as yet. Let me erase everything else. That's not how I wanted to erase it. That's how I wanted to erase it. I wanted to erase it like that without getting rid of OK, that's good enough. And then let me clean up. Clean up all this stuff. Let's say that I have a cup. I'll just draw a cup. It's easier to draw sometimes then to draw straight lines and all of that. No, that's too dark. Do purple. I'm using a super-wide tool. I have to switch the length. OK, so that's my cup. It has some fluid. Actually, let's say it has a top to it, and I have some fluid in it. Maybe it happens to be red. We haven't been dealing with red fluids as yet,. Let me-- oh, I didn't want to do that. So you know there's a fluid there. Let's say that h-- we don't know what units are, but let's say h meters below the surface of the fluid. This is all fluid here. I poke a hole right there, and fluid starts spurting out. My question to you is, what is this output velocity of the fluid as a function of this height? Let me tell you something else. Let's say that this hole is so small, let's call the area of that hole A2, and let's say that the surface area of the water is A1. Let's say that hole is so small that the surface area the water-- let's say that A2 if equal to 1/1,000 of A1. This is a small hole relative to the surface"
    },
    {
        "Q": "\nAt 2:00, the carbon gaining the negative charge is sp3 hybridized right? The only reason the compound is still aromatic is that the negative charge(lone pair) is delocalised even when that carbon does not have a p orbital. Am I correct? If not, please correct me. Thanks :)",
        "A": "No it s sp2, it doesn t rehybridise itself. When you have a lone pair next to a pi system, that lone pair will be in a p orbital because there is a large energy benefit to the aromatic system.",
        "video_name": "wvVdgGTrh-o",
        "timestamps": [
            120
        ],
        "3min_transcript": "Voiceover: In the last video, we used the criteria for aromaticity to see that heterocycles can be aromatic too. In this video, we're going to look at more aromatic heterocycles, specifically five-membered rings. We'll start with pyrrole right down here. The pyrrole molecule, as you can see, five atoms in the ring, and if we take a look at the carbons in the ring, we can see that those carbons all have a double bond to them. Therefore, each of those carbons is sp two hybridized, meaning each of the carbons has a free p orbital. So I can go ahead and draw a free p orbital on each of those four carbons like that. In terms of the nitrogen on the ring, I need to know the hybridization state of this nitrogen. The best way to do that is to calculate the steric number. We know the steric number is equal to the number of sigma bonds plus number of lone pairs of elections. I can see that here is a sigma bond, here is a sigma bond, and here is a sigma bond, so three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen there, which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria"
    },
    {
        "Q": "at 2:42 although the nitrogen in pyrrole is sp2 hybridised once the lone pair of electrons participate in resonance does the positive charge on nitrogen not mean anything when said that it is sp2 hybridised.\n",
        "A": "the fact that nitrogen has a positive formal charge in some of the resonance structures is simply a consequence of the way formal charge is calculated. it has no bearing on the hybridization of nitrogen. Geometry specifies hybridization.",
        "video_name": "wvVdgGTrh-o",
        "timestamps": [
            162
        ],
        "3min_transcript": "which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria Pyrrole does contain a ring of continuously overlapping p orbitals, and it does have four n plus two pi electrons in that ring, so let's go ahead and highlight those. We had these pi electrons, so that's two, these pi electrons, so that's four, and then these pi electrons here in magenta are actually de-localized in the ring, so that gives us six pi electrons. So if n is equal to one, four times one plus two gives me six pi electrons. Pyrrole has six pi electrons and also has a ring of continuously overlapping p orbitals, so we can say that it is aromatic. Let's go ahead and look at another molecule here so similar to it. This is imidazole. For imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here, so at first, it looks like that nitrogen might be sp three hybridized, but we can draw a resonance structure for it."
    },
    {
        "Q": "At 2:57, Sal uses the resistance formula to combine the 4 parallel resistors. But doesn't the current just go through the path of smallest resistance? If that is the case, then wouldn't all the current go through 3 ohms resistor and then the 1 ohm resistor? Combining these, since they are in series, gives us a 4 ohm resistor. Then by V=IR, we have 5 amps.\n",
        "A": "No, current goes through any path it can go through. There is no such law as current goes through the path of smallest resistance . If you put 5 volts across a 1 ohm resistor, 5 amps will go through the resistor. The same 5 volts across a 10 ohm resistor will give you 0.5 amps through that resistor. Since parallel resistors have the same potential difference across them, if you put the 1 ohm and the 10 ohm both connected to the 5 volt source, that source is going to send out 1.5 amps.",
        "video_name": "3NcIK0s3IwU",
        "timestamps": [
            177
        ],
        "3min_transcript": "that is 3 ohms. And let's say I have a resistor here. Let's just make it simple: 1 ohm. And just to make the numbers reasonably easy-- I am doing this on the fly now-- that's the positive terminal, negative terminal. Let's say that the voltage difference is 20 volts. So what I want us to do is, figure out what is the current flowing through the wire at that point? Obviously, that's going to be different than the current at that point, that point, that point, that point, all of these different points, but it's going to be the same as the current flowing at this point. So the easiest way to do this is try to figure out the equivalent resistance. Because once we know the equivalent resistance of this big hairball, then we can just use Ohm's law and be done. So first of all, let's just start at, I could argue, the simplest part. Let's see if we could figure out the equivalent resistance of these four resistors in parallel. Well, we know that that resistance is going to be equal to 1/4 plus 1/8 plus 1/16 plus 1/16. So that resistance-- and now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1 plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the numbers are working out-- is equal to 1/2, so that equivalent resistance is 2. So that, quickly, we just said, well, all of these resistors combined is equal to 2 ohms. So let me erase that Simplify it. So that whole thing could now be simplified as 2 ohms. I lost some wire here. I want to make sure that circuit can still flow. So that easily, I turned that big, hairy mess into something that is a lot less hairy. Well, what is the equivalent resistance of this resistor and this resistor? Well, they're in series, and series resistors, they just add up together, right? So the combined resistance of this 2-ohm resistor and this 1-ohm resistor is just a 3-ohm resistor. So let's erase and simplify. So then we get that combined resistor, right?"
    },
    {
        "Q": "At 4:20 when there are 4 H is it impossible there would be a lone pair or lone single e- making FC -1 or 0 respectively? Is steric hindrance at play?\n",
        "A": "Yes it is not possible. Nitrogen can only have up to 8 valence electrons around it and in this case there are 4 bonds which is 8 electrons. It can fit no more.",
        "video_name": "5-MM39VCwc0",
        "timestamps": [
            260
        ],
        "3min_transcript": "So three bonds and one lone pair of electrons, the formal charge is equal to zero. So when nitrogen has three bonds and one lone pair of electrons, the formal charge is equal to zero. And sometimes you don't want to draw in lone pairs of electrons, so you could just leave those off. You could just say alright, well if I just draw this and you know the formal charge of nitrogen is zero, then it's assumed you also know there's a lone pair of electrons on that nitrogen. So this is just another way of representing the same molecule, leaving off the lone pair, because you should know it's there. Let's look at other examples where nitrogen has a formal charge of zero. So we'll start with the example on the left here and if we look at this nitrogen and we know it has a formal charge of zero, let's see how many bonds it has. Let's use red here. So here's one bond, two bonds, and then three bonds. we know there should be a lone pair of electrons on that nitrogen. So you could leave it off and just know it's there, or you could draw them in. So I'll go ahead and draw in the lone pair of electrons on the nitrogen. So formal charge of zero. Let's look at the one on the right. So if we assume that nitrogen has a formal charge of zero, let's see how many bonds we have here. So here's one, two, and three. So we have three bonds, so we'd still need one lone pair of electrons. So if you wanted to show the lone pair of electrons you could put them in there like that. Notice this gives nitrogen an octet of electrons around it. So count those up, here's two, four, six, and eight. So nitrogen would have an octet. And remember, you could just leave off that lone pair of electrons and it's assumed if we know nitrogen has a formal charge of zero that there is a lone pair and we just didn't want to take the time to draw them in. Let's assign formal charge to another nitrogen, So what is the formal charge of nitrogen now? Let's draw in our electrons. So each bond is two electrons, so I draw those in there. And the formal charge on nitrogen is equal to the number of valence electrons that nitrogen is supposed to have, which we already know is five, so we put a five in here, and from that we subtract the number of valence electrons that nitrogen actually has in our drawing. So for these bonds, hydrogen gets one electron and nitrogen gets one for each of these bonds. So that allows us to see there are four electrons around nitrogen. So here's one, two, three, and four. So in our drawing, nitrogen only has four electrons around it, so this would be five minus four, which gives us a formal charge of plus one. So it's like nitrogen lost a valence electron. It's supposed to have five and here we see only four around it, so it's as if it lost"
    },
    {
        "Q": "\n0:05 Sal mentioned that \"parsec\" is science fiction...is this true?",
        "A": "It is often used in science fiction shows like Star Trek and Star Wars (Han Solo misuses it in episode IV). But a parsec is a real unit that astronomers use.",
        "video_name": "6zV3JEjLoyE",
        "timestamps": [
            5
        ],
        "3min_transcript": "You've probably heard the word parsec before in science fiction movies or maybe even some things dealing with astronomy. And what I want to do in this video is really just tell you where the word and the definition of the word really come from. And just to kind of cut to the chase, it's just a unit of distance. It's just about 3.26 light years. But what I want to do is just think about where did this weird distance come from, this distance that is roughly 3.26 light years? It comes from the distance of something, probably a star. But let me say \"something\" because there are no stars exactly this far away from us. The distance of something that has a parallax angle of one And the word comes from the \"par\" in parallax and the \"second\" in arc seconds. So it's literally par-- let me do this in a different color. It's literally parsec. You could think of it as kind of the parallax arc second. How far would this thing be? It turns out it's 3.26 light years. So we can actually calculate that. And that's actually what I'm going to do in this video. So let's say there is something. So this is the sun. This is the Earth at some point in time. This is the Earth six months later at the opposite end of the orbit. And we are looking at some distance. We are looking at some object some distance away. We know that this distance right here is one astronomical unit. And what we want to do is figure out the distance of this object. And all we know is that it has parallax angle of one arc If we're looking right at-- remember, we're looking from above the solar system. So the Earth is rotating in this direction in either case. And so in this point in the year-- we don't know when this is. Depends on what star that is. At this point in the year right at sunrise, right when we first catch the first glimpses of the sun's light, if we look straight up, the angle between that object in the night sky and straight up is going to be the parallax angle. So this is going to be one arc second. And just to make it consistent with the last few videos we did on parallax, let's just visualize how that would look in the night sky. So let me draw the night sky over here. Let me do that in purple maybe. Let me draw the night sky over here."
    },
    {
        "Q": "\nat 8:52 its been said that DMSO i.e a polar aprotic solvent increases the nucleophilicity of the ethoxide aninon therefore increasing SN2.... but i dont understand how DMSO increases the nucleophilic strengh of ethoxide anion?? plz help",
        "A": "I don t think he said it s favoring substitution over elimination in the way that you re thinking, which is that the SN2 products are the major products. I think he s just trying to say that in order to increase the % of SN2 products, you can change the solvent to make it happen.",
        "video_name": "vFSZ5PU0dIY",
        "timestamps": [
            532
        ],
        "3min_transcript": "with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So, it's not gonna be SN1 or SN2 and we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons and we're going to protonate this oxygen for our first step. So, let's draw in our ring and we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons and a plus one formal charge on the oxygen. So, this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen and we remove a bond from this carbon in red which would give us a secondary carbocation. and the carbon in red is this one and that carbon would have a plus one formal charge. So, let me draw in a plus one formal charge here. And now we have water which can function as a weak base in our E1 reaction and take a proton from a carbon next to our carbon with a positive charge. So, let's say this carbon right here. It has two hydrogens on it. I'll just draw one hydrogen in and water functions as a base, takes this proton and these electrons move in to form a double bond. So, let's draw our final product here. We would have a ring, we would have a double bond between these two carbons, so our electrons in, let's use magenta, electrons in magenta moved in to form our double bond. So, our product is cyclohexane. So, a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid For this reaction we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism because water as a polar protic solvent can stabilize the formation of a carbocation. So, let's draw the carbocation that would result. These electrons would come off onto our bromine and we're taking a bond away from this carbon in red. So, the carbon in red gets a plus one formal charge and let's draw our carbocation. So, we have our benzine ring here. I'll put in my pi electrons and the carbon in red is this one, so that carbon gets a plus one formal charge. This is a secondary carbocation"
    },
    {
        "Q": "\nAt 2:10, why was the E1 mechanism excluded ?",
        "A": "The strong base favors E2. I think this is because carbocations are energetically unfavorable. So, if you can do E2 or E1, then E2 will win and in the presence of a strong base you can do E2 ...",
        "video_name": "vFSZ5PU0dIY",
        "timestamps": [
            130
        ],
        "3min_transcript": "- [Instructor] Let's look at elimination versus substitution for a secondary substrate. And these are harder than for a primary or tertiary substrate because all four of these are possible to start with. So, if we look at the structure of our substrate and we say it's secondary, we next need to look at the reagent. So, we have NaCl which we know is Na plus and Cl minus and the chloride anion functions only as a nucleophile. So, we would expect a substitution reaction, nucleophilic substitution. So, E1 and E2 are out. Between SN1 and SN2 with the secondary substrate, we're not sure until we look at the solvent and DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So, SN1 is out and we're gonna think about our chloride anion functioning as a nucleophile. So, let me draw it in over here. So, this is with a negative one formal charge. And an SN2 mechanism are nucleophile attacks and our nucleophile is going to attack this carbon in red. So, we're gonna form a bond between the chlorine and this carbon in red and when the nucleophile attacks, we also get loss of our leaving group. So, these electrons come off onto the oxygen and we know that tosylate is a good leaving group. So, when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group. So, we had a wedge here for our leaving groups, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide. So, just looking at our reactions, we can't really rule any out here. So, all four are possible, until we look at our reagent. Now, we saw in an earlier video, it does not act like a nucleophile. So SN1 and SN2 are out. And a strong base means an E2 reaction. So, E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon and the carbons directly bonded to the alpha carbon are the beta carbons. So, I'll just do the beta carbon on the right since they are the same essentially. And we know that our base is gonna take a proton from that beta carbon. So, let me just draw in a hydrogen here. And DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond and these electrons come off to form our bromide anion. So, our final product is an alkyne and our electrons in magenta in here"
    },
    {
        "Q": "at 7:08 as T=>Ta ,then how T can be HOTTER if T=Ta\n",
        "A": "Sal should have written T > Ta. In that case, T will always be hotter than Ta.",
        "video_name": "IICR-w1jYcA",
        "timestamps": [
            428
        ],
        "3min_transcript": "and once again I could put a constant here, but I'm going to end up with a constant on the right hand side too so I'm just going to merge them into the constant on the right hand side. So that is going to be equal to, now here, this is going to be negative kt, and once again we have plus C. And now we can raise e to both of these powers, or another way of interpreting this is if e to this thing is going to be the same as that. So we can write this as, the absolute value, let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to e, something about e I always think of the color green. e to the negative kt plus C. This of course is the same thing as, we've done this multiple times before. Negative kt times e to the C power. And we could just call this another arbitrary constant. If we called this C1, then we could just call this whole thing C. So this we could say is Ce to the negative kt. So at least it's starting to resemble what we did when we were modelling population. We'll see it's a little bit different. Instead of just temperature on this left hand side, we have temperature minus our ambient temperature. And so, we can do a couple of things. If, in a world, say we were dealing with a hot cup of tea, something that's hotter than the ambient temperature. So we could imagine a world where T is So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive or the thing inside the absolute value So we don't need the absolute value. Or the absolute value of it is going to be the same thing as it. And then we can just add T sub a to both sides, and then we would have our temperature, and I can even write this as a function of time, is going to be equal to this business, is going to be equal to Ce, let me do that in that same color. Ce to the negative kt plus T sub a. All I did is I'm assuming that this inside the absolute value is going to be positive, so the absolute value is not going to change the value. And I added T sub a to both sides to get this. So this right over here is going to be our general solution,"
    },
    {
        "Q": "\nA quick, slightly off topic question. At \"11:45\" Sal says g = 9.8 m/s^2, and that the acceleration is 33m/s^2. Why exactly do we square the seconds? Couldn't we do a few more steps and express it as m/s? That would seem to be easier for people to process.",
        "A": "The m/s^2 just means meters per second every second...so if gravity is 9.8 meters per second every second.",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            705
        ],
        "3min_transcript": "And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth, more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands,"
    },
    {
        "Q": "At 3:57, Sal starts to take displacement for the solution. Why did he not use any other formula except for the displacement or distance?\n",
        "A": "Because the problem asks for ddisplacement/",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            237
        ],
        "3min_transcript": "So if we want to convert this into seconds, we have, we'll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I'll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you're doing any type of this dimensional analysis, you really should see whether it makes sense. If I'm going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that's why we're dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that's why we're multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I'll just round it to 72, because that's about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let's think about what that acceleration could be, given that we know the length of the runway, and we're going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let's think a little bit about it. The total displacement, I'll do that in purple, the total displacement is going to be times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It's going to be our final velocity, plus our initial velocity, over 2. It's just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we're trying to get to 10 m/s, or we're trying to get 10 m/s faster, and we're accelerating at 2 m/s squared, it'll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to"
    },
    {
        "Q": "\nAt 7:34, where is 72 coming from?",
        "A": "He took 260 km/h and changed the units to m/s seconds, which means: 260 km/h, 1km =1000m, 1h = 3600 seconds. Hence: (260 * 1000 m) / 36000 s = 72 m/s.",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            454
        ],
        "3min_transcript": "If we just work through this math, and I'll try to write a little bigger, I see my writing is getting smaller, our displacement can be expressed as the product of these two things. And what's cool about this, well let me just write it this way: so this is our final velocity plus our initial velocity, times our final velocity minus our initial velocity, all of that over 2 times our acceleration. Our assumed constant acceleration. And you probably remember from algebra class this takes the form: a plus b times a minus b. And so this equal to -- and you can multiply it out and you can review in our algebra playlist how to multiply out two binomials like this, but this numerator right over here, I'll write it in blue, is going to be equal to our final velocity squared minus our initial velocity squared. into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what's really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, the scalar version, and really we are thinking only in the scalar, we're thinking about the magnitudes of all of these things for the sake of this video. We're only dealing in one dimension. But sometimes you'll see it written like this, sometimes you'll multiply both sides times the 2 a, and you'll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it'll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it's not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you're just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we've just derived ourselves a kind of a"
    },
    {
        "Q": "\nAt 14:00 Sal says that after 2 seconds, the plane would go 66 meters. Shouldn't it be 99 meters because of acceleration?",
        "A": "aceleration is at the rate of 33m/s^2 .so, after 2 seconds its speed will be 33m/s x 2 = 66m/s",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            840
        ],
        "3min_transcript": "more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands, essentially two more people stacked above you, roughly, I'm just giving you the general sense of it, that's how it would feel, a little bit more than two people, that squeezing sensation. So his entire body is going to feel 3 times heavier than it would if he was just laying down on the beach or something like that. So it's very very very interesting, I guess, idea, at least to me. Now the other question that we can ask ourselves is how long will it take to get catapulted off of this carrier? And if he's accelerating at 33 meters per second squared, how long would it take him to get from 0 to 72 meters per second? So after 1 second, he'll be going 33 meters per second, after 2 seconds, he'll be going 66 meters per second, so it's going to take, and so it's a little bit more than 2 seconds. So it's going to take him a little bit more than 2 seconds. 72 meters per second, and you divide it by 33, it'll take him 2.18 seconds, roughly, to be catapulted off of that carrier."
    },
    {
        "Q": "\nAt 9:34, the amino acid is being called a \"Zwitter ion.\" Sal explains that \"zwitter\" means \"hybrid\" in German. Why can't it be called \"polar?\"",
        "A": "It cannot be explained as polar because polar substance is that which has a partial negative charge inside the molecule. While this has a charge which can be explained as a full charge. Polar substances are not either anion or cation.",
        "video_name": "Pk4d9lY48GI",
        "timestamps": [
            574
        ],
        "3min_transcript": "is the term for two or more amino acids connected together, so this would be a dipeptide, and the bond isn't this big, I just, actually let me just, let me draw it a little bit smaller. So... That's serine. This is valine. They can form a peptide bond, and this would be the smallest peptide, this would be a dipeptide right over here. \"Peptide,\" \"peptide bond,\" or sometimes called a peptide linkage. And as this chain forms, that polypeptide, as you add more and more things to it, as you add more and more amino acids, this is going to be, this can be a protein or can be part of a protein that does all of these things. Now one last thing I wanna talk about, this is the way, the way these amino acids have been drawn is a way you'll often see them in a textbook, but at physiological pH's, the pH's inside of your body, which is in that, you know, that low sevens range, What you have is this, the carboxyl group right over here, is likely to be deprotonated, it's likely to have given away its hydrogen, you're gonna find that more likely than when you have... It's gonna be higher concentrations having been deprotonated than being protonated. So, at physiological conditions, it's more likely that this oxygen has taken both of those electrons, and now has a negative charge, so it's given, it's just given away the hydrogen proton but took that hydrogen's electron. So it might be like this, and then the amino group, the amino group at physiological pH's, it's likely to actually grab a proton. So nitrogen has an extra loan pair, so it might use that loan pair to grab a proton, in fact it's physiological pH's, you'll find a higher concentration of it having grabbed a proton than not grabbing a proton. use its loan pairs to grab a proton, and so it is going to have... So it is going to have a... It is going to have a positive charge. And so sometimes you will see amino acids described this way, and this is actually more accurate for what you're likely to find at physiological conditions, and these molecules have an interesting name, a molecule that is neutral even though parts of it have charge, like this, this is called a zwitterion. That's a fun, fun word. Zwitterion. And \"zwitter\" in German means \"hybrid,\" and \"ion\" obviously means that it's going to have charge, and so this has hybrid charge, even though it has charges at these ends, the charges net out to be neutral."
    },
    {
        "Q": "\nhi, at \"4:00\" when mr khan was naming the compound, i was wondering if he made a mistake by saying 2,3-dimethyl instead of saying 2,3-trimethyl coz there are three CH3's.....or maybe someone correct me and explain it to me please? =) thank you x",
        "A": "There are 3 CH3s but only two of them are groups, one of them is part of the main chain",
        "video_name": "peQsBg9P4ms",
        "timestamps": [
            240
        ],
        "3min_transcript": "to use should have as many simple groups attached to it as possible, as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be a bromomethyl group, which is not as simple as maybe it could be. But if we use this carbon in our longest chain, we'll have We'll have a bromo attached, and we'll also have a methyl group. And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that is closest to something being attached to it, and that bromine is right there. So there's going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. figure out what order they should be listed in. So this is a 1-bromo and then this will be a 2-methyl right here. And then just a hydrogen. Then three we have a fluoro, so on a carbon three, we have a fluoro, and then on carbon three, we also have a methyl group right here, so we also have a 3-methyl. So when we name it, we put in alphabetical order. Bromo comes first, so this thing right here is 1-bromo. Then alphabetically, fluoro comes next, 1-bromo-3-fluoro. We have two methyls, so it's going to be 2 comma 3-dimethyl. And remember, the D doesn't count in alphabetical order. chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group,"
    },
    {
        "Q": "@6:10 so even if the carbon next to the chiral center has a F plus 2 CH3 groups attached to it, we still consider the C attached to the Br as the higher one? Thanks\n",
        "A": "oh, I see. Now what if there were more CH3 groups attached to the other C ? And thanks on your answer!",
        "video_name": "peQsBg9P4ms",
        "timestamps": [
            370
        ],
        "3min_transcript": "chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group, And the way you do it, if there's a tie on the three carbons, you then look at what is attached to those carbons, and you compare the highest thing attached to each of those carbons to the highest things attached to the other carbons, and then you do the same ranking. And if that's a tie, then you keep going on and on and on. So on this carbon right here, we have a bromine. Bromine has an atomic number of 35, which is higher than carbon. So this guy has a bromine attached to it. This guy only has hydrogen attached to it. This guy has a fluorine attached to it. That's the highest thing. So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it, so that is number three. The one has the bromine attached to it is going to be number one, and the one that has the fluorine attached to it is number two. have to look at the next highest constituent, and even if this had three fluorines attached to it, the bromine would still trump it. You compare the highest to the highest. So now that we've done that, let me redraw this molecule so it's a little bit easier to visualize. So I'll draw our chiral carbon in the middle. And I'm just doing this for visualization purposes. And right here we have our number one group. I'll literally just call that our number one group. So right there that is our number one group. It's in the plane of the screen. So I'll just call that our number one group. Over here, also in the plane of the screen, I have our number two group. So let me do it like that. So then you have your number two group, just like that."
    },
    {
        "Q": "at 9:15 why does the # 1 group stay in place and the rest of the groups rotate?\n",
        "A": "Group 1 stays in the same place because the rest of the groups need to rotate in order to put the hydrogen behind all other atoms. You need to think about the molecule in three dimension. The hydrogen in this situation is pointing towards you and therefore in order to put it away from you, you would need to keep group 1 in the same spot and rotate the rest of the groups in order to make hydrogen face away from you.",
        "video_name": "peQsBg9P4ms",
        "timestamps": [
            555
        ],
        "3min_transcript": "molecule right now the way it's drawn. I'll do that in magenta. So then you have your number three group. It's behind the molecule, so I'll draw it like this. This is our number three group. And then we have our number four group, which is the hydrogen pointing out right now. And I'll just do that in a yellow. We have our number four group pointing out in front right now. So that is number four, just like that. Actually, let me draw it a little bit clearer, so it looks a little bit more like the tripod structure that it's So let me redraw the number three group. The number three group should look like-- so this is our number three group. Let me draw it a little bit more like this. The number three group is behind us. And then finally, you have your number four group in straight out. So that is coming straight out of-- well, not straight out, but at an angle out of the page. So that's our number four group, I'll just label it It really is just a hydrogen, so I really didn't have to simplify it much there. Now by the R-S system, or by the Cahn-Ingold-Prelog system, we want our number four group to be the one furthest back. So we really want it where the number three position is. And so the easiest way I can think of doing that is you can imagine this is a tripod that's leaning upside down. Or another way to view it is you can view it as an umbrella, where this is the handle of the umbrella and that's the top of the umbrella that would block the rain, I guess. But the easiest way to get the number four group that's actually a hydrogen in the number three position would be to rotate it. You could imagine, rotate it around the axis defined by the number one group. So the number one group is just going to stay where it is. The number four is going to rotate to the Number three is going to rotate around to the number two group, and then the number two group is going to rotate to where the number four group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit. So we have our chiral carbon. I put the little asterisk there to say that that's our chiral carbon. The number four group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number four group is now behind where the number three group used to be, so number four is now there. Number one hasn't changed. That's kind of the axis that we rotated around. So the number one group has not changed. Number one is still there. Number two is now where number four used to be, so number two is now jutting out of the page."
    },
    {
        "Q": "\nAt 3:10 What is mole ratio?",
        "A": "A mole ratio is a ratio between the amount of mols of each substance. In this case, it s the ratio of the amount of mols of Fe2O3, which is 1, and the amount of mols of Al, which is 2. So, the mole ratio is 1:2",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            190
        ],
        "3min_transcript": "we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here."
    },
    {
        "Q": "i didnt get the part at 8:20 that why do you multiply to get the mass of aluminum required.\ncan u pls help me?\n",
        "A": "You multiply avogadro s number by the number of grams in aluminium to get the number of amus in aluminum. I recommend watching the video The mole and Avogadro s number on khan academy. Hope this helps! Excellent question. If you require further clarification, please don t hesitate to comment on my answer.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            500
        ],
        "3min_transcript": "is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium,"
    },
    {
        "Q": "at 7:12 how does Sal get how many moles of Al are there\n",
        "A": "he just multiplyed the Fe2O3 s mole(0.53) by 2. so you get 1.06. in the chemical equation. we have 2 Al for every Fe2O3. it s that easy",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            432
        ],
        "3min_transcript": "So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units."
    },
    {
        "Q": "I have a question that says \"How many liters of oxygen are required to form 6.0 L of water\" and it gives me this balanced equation: 2H2+1O2-->2H20\nWhat I don't understand is what I'm supposed to to with the 6 L and the 2 mol of H20 to get to O. Do I divide the 6 by two so I have a 1:1 ratio or what?\n",
        "A": "You shouldn t divide the 6, because the equation isn t dealing with liters, it s dealing with moles. 2 moles of H(2) and 1 mole of O(2) form 2 moles of H(2)O. 6 L is 6000 grams, so your next step is to find how many moles are in 6000 grams of water. THEN, you find the number of moles of oxygen you need using the 1:2 ratio, and then find the number of grams(then liters) of oxygen that you need, which should be your response.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            61
        ],
        "3min_transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams."
    },
    {
        "Q": "\nAt 4:34, does it make more sense to use the average mass number?",
        "A": "Yes ideally he should not be rounding these. He should have used the number on the periodic table and round to the appropriate significant figures at the end of the calculation.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            274
        ],
        "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?"
    },
    {
        "Q": "At 6:00, what is the relationship between moles and grams?\n",
        "A": "For atoms 1 mole equals the relative atomic mass with units of grams (g). e.g. 1 mole of oxygen atoms has a mass of 16 g. For molecules, 1 mole equals the relative formula mass with units of grams (g). e.g. 1 mole of water (H2O) has a mass of (2(1) + 16) = 18 g.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            360
        ],
        "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum."
    },
    {
        "Q": "\nat 8:04, in the balanced equation, the coefficient is 2 for Al. So why wouldnt you multiply it's atomic mass by two, since 2 Al atoms are necessary?",
        "A": "When balancing, you are initially not looking for mass ratios, but atom ratios. For instance magnesium oxide is MgO, or a 1 to 1 ratio between Mg and O. So the balanced chemical equation would be: 2 Mg + 1 O2 --> 2 MgO Next you could use the masses to find out what mass of Mg would react with what mass of O2 and thus give the product mass formed IF both reactants get used up completely.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            484
        ],
        "3min_transcript": "is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium,"
    },
    {
        "Q": "At 6:05, you said that there was Fe3O3 but before that you said it was Fe2O3, so wouldn't the final answer be different\n",
        "A": "He meant Fe203. If you have the pop ups enabled, a pop up shows up saying Sal wrote Fe303 but meant 1 mole of Fe203",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            365
        ],
        "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum."
    },
    {
        "Q": "\nAt 8:40 Sal said that the Black hole at the center of our galaxy is 4 million times the mass of our sun. How is this possible aren't black holes an object of infinite density",
        "A": "The mass is in an infinitesimally small volume which would make any finite amount of mass have an infinite density.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            520
        ],
        "3min_transcript": "when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes. Maybe they formed near the beginning of the actual universe. When the density of things was a little uniform, things condensed into each other. And what we're going to talk about in the next video is how these supermassive black holes can help generate unbelievable sources of radiation, even though the black holes themselves aren't emitting them. And those are going to be quasars."
    },
    {
        "Q": "\nAt 6:15, Sal mentioned quantum fluctuations. What does that mean?",
        "A": "It is the temporary change in the amount of energy in a point in space.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            375
        ],
        "3min_transcript": "So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it. So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe,"
    },
    {
        "Q": "Wait, at 0:42, he talks about the event horizon saying that if anything gets past it, it cannot go back, but wouldn't gravity be like a constant, no just one line that separates things?\n",
        "A": "The farther away you move from an object, the weaker its gravity gets. This is why the astronauts in orbit don t feel Earth s gravity and can float about. Well the event horizon is the distance where the gravity becomes strong enough to draw in light.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            42
        ],
        "3min_transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form"
    },
    {
        "Q": "At 5:06, when Sal says that the math starts to break down, what does he mean? Is it that different laws of physics apply inside a black hole as opposed to outside of it? How is that possible?\n",
        "A": "The math starts to break down means that you are starting to try to do things like divide by zero, which you generally can t do.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            306
        ],
        "3min_transcript": "in an area that has a lot of matter that it can accrete around it. So I'll draw the-- this is the event horizon around it. The actual black hole is going to be in the center of it, or rather the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole. And then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing and so the event horizon will also keep growing in radius. Now this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or a 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it."
    },
    {
        "Q": "At 1:12 Sal said there are Stellar Blackholes and at 4:18 he said there are Primorial (super massive) Blackholes. Is there any other type of Blackhole?\n",
        "A": "There could be, but those are the only two types we have discovered.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            72,
            258
        ],
        "3min_transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form"
    },
    {
        "Q": "at 8:18 he says the center of the universe (unintentionally, and then fixes it) but it got me thinking .Is there a center of the universe?\n",
        "A": "There is no center of the universe, so far as we can tell. There is a center of portion of the universe that we can observe, which we call the observable universe. By definition, we are at the center of that, because we can observe the same distance in every direction.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            498
        ],
        "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes."
    },
    {
        "Q": "\nat 7:31, he said that there were might be black holes between stellar mass black holes and supermassive black holes. are there?",
        "A": "Yes but not very many",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            451
        ],
        "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes."
    },
    {
        "Q": "\nAt 8:14 the stars could be orbitting something besides a black hole that scientists haven't discovered yet, right?",
        "A": "We aren t just orbiting the black hole (its much to far away to affect us greatly), we are orbiting every other thing in our galaxy that is closer to the center. We orbit the common center of mass for our galaxy. Our galaxy began just orbiting that one black hole but as it got bigger, it needed the gravity of every other thing in it to keep it together.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            494
        ],
        "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes."
    },
    {
        "Q": "4:31... Why would the two groups attached to a carbon be named in order of how big each of the R-groups are, and not alphabetically?\n\nI would have thought that butyl should come before propyl, not the other way round. (and I do realize that this is a common name, not the IUPAC name)\n",
        "A": "Even in common names, the rule is to list the groups alphabetically. The name methyl ethyl ketone is traditional and was assigned long before anyone thought of a system for common names. That name is still used by many people. But the systematic common name puts the names of the groups in alphabetical order.",
        "video_name": "wD15pD5pCt4",
        "timestamps": [
            271
        ],
        "3min_transcript": "Propa-, and instead of calling it propane, we get rid of that \"e\" over there and we would call it propanone. That tells us that this right here is a ketone. And you have to know where this double bond is. And actually, for propanone, you don't have to specify it, because if you know it's a ketone, you know that it has to have a carbon on either side of the carbonyl group, so you actually don't even have to specify where the carbonyl group is. But if you wanted to, you could say, OK, that's going to be on the two carbon. No matter what direction you start counting from, it's going to be on the two carbon. But the two is kind of optional for propanone. Let's do a couple of other ones. So let's say we had a molecule that looks like this. So the traditional way of naming it, you'd say, OK, on So on that end, I have three carbons. That is a propyl group. And on this other side of the ketone right over here, I have only one carbon. That is a methyl group. So then you would just name them. And you name then in order of increasing chain size, molecule size, or group size. So this one you'd write methyl first. Methyl, because it's only one carbon. So this is methyl propyl ketone. This is kind of the traditional or the common way, often kind of the most used way, of naming this molecule. But the systematic way of naming it, you look at the longest carbon chain and you say, OK, I have one, two, three, four, five carbons. So it's going to be pent-. And then you want to start numbering it so that the So you want to start numbering on the right side: one, two, three, four and five. So this right here, so we said the prefix would be pent-. So it's penton, and instead of saying it's pentane, you say it's pentanone. And to specify where the carbonyl group is, you say it's 2. This is 2-pentanone. And you might also see it written like this: pentan-2-one. Either one of these right here would be acceptable. Let's do a slightly more complicated example. Let's say we had something that looked like this. So we have something, a molecule, that looks like this. And let me stick some chlorines over here. So what would this be? Well, our longest chain, once again, is this cyclohexane:"
    },
    {
        "Q": "\nFrom 6:05 to 6:30, Jay mentioned that exactly 10.2 eV is needed to promote the level 1 electron to energy level 2. Does that mean that if x eV of energy is given to the level 1 electron, such that 10.2 < x < 12.9, the electron will not be promoted to level 2?",
        "A": "Yes, photons whose energies don t match the possible jumps are not absorbed.",
        "video_name": "nJ-PtF14EFw",
        "timestamps": [
            365,
            390
        ],
        "3min_transcript": "When N is equal to one, that was negative 13.6 electron volts, that's the energy associated with that electron as it orbits the nucleus. And so if we go over here on the right, and we say this top line here represents energy is equal to zero, then this would be negative 13.6 electron volts. So none of this is drawn perfectly to scale, but this is just to give you an idea about what's happening. So this is when N is equal to one, the electron is at a distance R one away from the nucleus, we're talking about the first energy level, and there's an energy of negative 13.6 electron volts associated with that electron. Alright, let's say the electron was located a distance R two from the nucleus. Alright, so that's N is equal to two, and we just calculated that energy is equal to negative 3.4 electron volts, alright? And then let's say the electron was at R three and once again we just calculated that energy to be equal to negative 1.51 electron volts. And so it's useful to compare these two diagrams together, because we understand this concept of energy much better. For example, let's say we wanted to promote the electron that I drew, so this electron right here I just marked. Let's say we wanted to promote that electron from the lower energy level to a higher energy level. So let's say we wanted to add enough energy to cause that electron to go from the first energy level to the second energy level, so that electron is jumping up here to the second energy level. We would have to give that electron this much energy, so the difference in energy between our two energy levels, so the difference between these two numbers. And if you're thinking about just in terms of magnitude, alright, 13.6 minus 3.4, alright? And so if you gave that electron 10.2 electron volts of energy, right, you could cause that electron to jump from the first energy level all the way here to the second energy level. But you would have to provide the exact right amounts of energy in order to get it to do that. Alright, let's say you wanted to cause the electron to jump, let's say, from the first energy level all the way to the third energy level, alright? So from the first energy level to the third energy level. So that would be, here's our electron in the first energy level, let's say we wanted to cause it to jump all the way up to here, alright? So once again, you would have to provide enough energy in order to do that. So, just thinking about the magnitudes, right? This was negative 1.51, this was negative 13.6, if we just take 13.6 minus 1.51, alright, we would get how much energy we need to put in"
    },
    {
        "Q": "At 1:11, what hormone is being referred to? Also, at 3:43, what neurotransmitter is she talking about?\n",
        "A": "Ghrelin is the hormone that signals hunger. Orexin/hypocretin is missing in narcolepsy patients (in most cases).",
        "video_name": "VBcEz8bVbL0",
        "timestamps": [
            71,
            223
        ],
        "3min_transcript": "Voiceover: I\u2019m sure we\u2019ve all had trouble sleeping at one point or another, maybe trouble falling asleep, staying asleep or waking up or maybe you're forcing yourself to sleep less because you have too much to do to lie in bed. But sleep deprivation can be a serious issue. People who don't get enough sleep are more irritable and perform worse on memory and detention tasks than people who do. So all this can be just a minor annoyance in everyday life, imagine the long-term implications for let's say, airline pilots, firefighters, security officers or the person driving next to you on the freeway. For example, one study in Canada showed that the Monday after the Spring time change, so when people lose an hour of sleep, the number of traffic accidents increases sharply compared to the Monday after the Fall time change when people get an extra hour of sleep, the number of accidents decreases sharply. So that's just one example, but sleep deprivation also makes people more susceptible to obesity. which is a hormone that tells your body to make more fat. You also produce more of the hormone that tells your body you're hungry, so you end up eating more and turning more of what you eat into fat which can contribute to weight gain. And finally sleep deprivation can also increase your risk for depression and one theory about this link is that REM sleep helps your brain process emotional experiences, which in turn helps protect against depression though we're still not entirely sure about this link. Most people, now most people experience sleep deprivation at some points in their lives, but the good news is that most people can get back on track by getting a few nights of good sleep, sort of paying back your sleep debt. Your next question might be then, \"How much sleep is enough sleep?\" That's kind of a hard question to answer, but most adults need about 7-8 hours of sleep, Babies need a lot more sleep, for example, than older adults often sleep less than 10 or 8 hours without severe detriments. Again everyone has trouble falling asleep at some point, but people who have persistent problems in falling or staying asleep have a more serous sleep disorder called insomnia. There are various medications that can help people fall asleep, but taking them for too long can result in dependence and tolerance, which is if a person continues to rely on the medication then their body will get used to it and they'll eventually need more and more to get the same affects. Now, that can often be a bad thing because there are side effects to drugs, but so treatment for insomnia often involves psychological training as well as or sometimes instead of medication and some lifestyle changes might also be necessary. For example, exercise regularly but not right before bed or spend some time just relaxing before bed"
    },
    {
        "Q": "\nWhy Momentum of Ball A and B in y direction at initial position is zero ? 3:07",
        "A": "okay thanks i got it :)",
        "video_name": "CFygKiTB-4A",
        "timestamps": [
            187
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt around 4:52, Sal calculated that the X-component is the square root of 3. I thought since the velocity was 3, you would just move the vector and the x-component would be 3. Could someone explain please?",
        "A": "Actually v=3 is the initial velocity of the ball before the collision.After collision it reduces to \u00e2\u0088\u009a3.Just think of a collision between two carrom pieces or two snooker balls if you still don t understand this. Hope it helps",
        "video_name": "CFygKiTB-4A",
        "timestamps": [
            292
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:41 you write the name of the second molecule and you say\n\"which alphabet comes after b ,whcih is h (for heptyl)\"\nbut must it not be which chain comes is nearer to the first C Atom?\n",
        "A": "No, when naming molecules, the order in which you name the chains is alphabetically. To indicate where each branch goes, you use numbers to put in front of the name of the branch",
        "video_name": "Se-ekDNhCDk",
        "timestamps": [
            461
        ],
        "3min_transcript": "This over here, let's see what we're dealing with. We have one, two, three carbons, so that is just a standard propyl group. Then here, we have one, two, three, four carbons, so we could say this is a butyl group. But this isn't just any butyl group. If we use the common naming, the carbon we immediately touch on or that we immediately get to when we go off of our main ring, that branches off into three other carbons, so this is tert-, tert- for three. The tert- is usually written in italics. It's hard to differentiate that when you see it. I'll write it in cursive. This is a tertbutyl group. Now, the next question is how do we specify where these different groups sit on this main ring? If you just had one group, you wouldn't have to specify it, but when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one. Now, we have an H in heptyl, a P in propyl, and tert-butyl, you might say, well, do I use the T or do I use the B? And this is just the convention, you use the B. If you have sec-butyl or tert-butyl, you ignore the sec- or the tert-. If this was an isobutyl or an isopropyl, you actually would use the I, so it's a little bit-- I guess the best way to think about it is there's a dash here so you can kind of ignore it, but if this was isobutyl it would all be one word, so you would consider the I. So in this situation, we would consider the B, and B comes before P or an H, so that is where we will start numbering one. Then to figure out which direction to keep numbering to encounter the first or where we will encounter the first side chain, so we'll go in this direction because we get right to the propyl group. One, two, three, four, five, six, seven, eight, nine. So this compound, we're going to start with the alphabetically first side chain, so it's 1-tert-butyl. I'll write this in cursive. 1-tert-butyl. Then the next one alphabetically is the heptyl group. That's H for heptyl, so then it is 5-heptyl. And then we have the propyl, and then it is 2-propyl, and then finally cyclononane, and we're done!"
    },
    {
        "Q": "\nAt around 0:48, Sal says to always choose the chain that has the most alkyl groups on it. Why?",
        "A": "It s a rule made to make naming strict and universal.",
        "video_name": "Se-ekDNhCDk",
        "timestamps": [
            48
        ],
        "3min_transcript": "We've got a few more molecular structures to name, so let's look at this first one right here. The first thing you always want to do is identify the longest chain. If we start over here, we have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen carbons, looks pretty long. Now what if we start over here? This looks like it could also be a long chain: one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. So we have two different chains, depending on whether we want to go up here or whether we go over here, that have a length of thirteen. So you're probably asking, which chain do you choose? You should always, if we have two chains of equal lengths and they're the longest chains, you pick the one that will have more branches or more alkyl groups on it. So this group right here, if we pick this from here to here as our chain, we only have one group on it, that group up there. If we pick this chain, starting over here and then We would have this group over here and then we would have this group over here. This is the better chain to use because it has more groups on it. It has more groups, but the groups are smaller and simpler. So let's start counting. And the direction we want to count, we always want to start on the side of the chain where we're going to encounter something first, and everything is closer this end of the chain so we'll start counting here. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, so we have thirteen carbons on our main chain. Let me draw our main chain. So our main chain is this thing in orange I'm drawing right here. That is our main chain. The prefix is tridec- and it's tridecane because we have all single bonds here, so it's tridecane. Tridecane is the main chain. Then we have two groups over here. This one in green, this only has one the carbon branching off of the main chain, so it's prefix will be meth- and it'll be a methyl group, so that is methyl. That is a methyl group right there. Then this one down here, we have one, two, three carbons. the prefix is prop-, so this is a propyl group. And the methyl is sitting on the three carbon of our main chain and the propyl group is sitting on the four carbon: one, two, three, four. Now when we figure out what order to list them in when we"
    },
    {
        "Q": "did you mean displacement near 3:35 or are you talking about distance?\n",
        "A": "Silver Night, That s a good question considering how displacement and distance can be easily mixed up despite having subtle differences. In his example, he is referring to displacement considering he uses it in an equation involving velocity instead of speed, and denoting it has a directional component.",
        "video_name": "Y5cSGxdDHz4",
        "timestamps": [
            215
        ],
        "3min_transcript": "write this in a different color, so we have some variety-- velocity times time, or d equals rate times time. The reason why I have change in distance here, or change in time, is that I'm not assuming necessarily that we're starting off at the point 0 or at time 0. If we do, then it just turns out to the final distance is equal to the average velocity times the final time, but let's stick to this. We want to figure out time given this set of inputs. Let's go from this equation. If we want to solve for time, or the change in time, we could just could divide both sides by the average velocity-- actually, no, let's not do that. Let's just stay in terms of change in distance. this and start again. We're given change in distance, initial velocity, and acceleration, and we want to figure out what the time is-- it's really the change in time, but let's just assume that we start time 0, so it's kind of the final time. Let's just start with the simple formula: distance, or change in distance-- I'll use them interchangeably, with a lower case d this time-- is equal to the average velocity times time. What's the average velocity? The average velocity is just the initial velocity plus the The only reason why we can just average the initial and the final is because we're assuming constant acceleration, and that's very important, but in most projectile problems, we do have constant acceleration-- downwards-- and that's gravity. We can assume, and we can do this-- we can say that the average of the initial and the final velocity is the average velocity, and then we multiply that times time. Can we use this equation directly? No. we know acceleration, but don't know final velocity. If we can write this final velocity in terms of the other things in this equation, then maybe we can solve for time. Let's try to do that: distance is equal to-- let me take a little side here. What do we know about final velocity? We know that the change in velocity is equal to acceleration times time, assuming that time"
    },
    {
        "Q": "At 0:52, (specifically at the definition you gave) could you also say \"Work is the amount of Kinetic Energy that is required for an object to get from its resting position to its current position\"?\n",
        "A": "No. Kinetic energy is energy associated with motion, not position. You could replace KE in your sentence with PE, and add that at the current position the object is also at rest, and then it would be right.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            52
        ],
        "3min_transcript": "Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force."
    },
    {
        "Q": "At 2:05, Is it ok if we make the same approximations or should we work things out accurately?\n",
        "A": "It depends on how the answer is expected. if the answer is expected in whole nos. then you can do approximations to make your given values into whole nos. if answer is needed upto 2 or 3 decimal places then you can t make approximations.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            125
        ],
        "3min_transcript": "Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force."
    },
    {
        "Q": "Hi Sal, in the example \"5:30\", how could the distance between two points in the universe be 30 million light years if the whole universe has only been expanding for 300K light years? Could the universe be expanding faster than the speed of light?\n",
        "A": "Actually, there s nothing preventing the universe from expanding faster than the speed of light.",
        "video_name": "6nVysrZQnOQ",
        "timestamps": [
            330
        ],
        "3min_transcript": "If we were talking about smaller time scales or I guess smaller distances, you could say approximately that. The expansion of the universe itself would not make as much of a difference. And let me make it even more clear. I'm talking about an object over there. But we could even talk about that coordinate in space. And actually, I should say that coordinate in space-time, because we're viewing it at a certain instant as well. But that coordinate is not 13.7 billion light years away from our current coordinate. And there's a couple of reasons to think about it. First of all, think about it, that light was emitted 13.7 billion years ago. When that light was emitted, we were much closer to that coordinate. This coordinate was much closer to that. Where we are in the universe right now was much closer to that point in the universe. as this-- let me actually draw it. So let's go 300,000 years after that initial expansion of that singularity. So we're just 300,000 years into the universe's history right now. So this is roughly 300,000 years into the universe's life. I guess we could view it that way. And first of all, at that point things haven't differentiated in a meaningful way yet right now. And we'll talk more about this when we talk about the cosmic microwave background radiation. But at this point in the universe, it was kind of this almost uniform white-hot plasma of hydrogen. And then we're going to talk about it. It was emitting microwave radiation. And we'll talk more about that in a future video. But let's just think about two points in this early universe. And let's say you have the coordinate where we are right now. You have the coordinate where we are right now. And in fact, I'll just make that roughly-- I won't make it the center just because I think it makes it easier to visualize if it's not the center. And let's say at that very early stage in the universe, if you were able to just take some rulers instantaneously and measure that, you would measure this distance to be 30 million light years. And let's just say right at that point, this object over here-- I'll do it in magenta-- this object over here emits a photon, maybe in the microwave frequency range. And we'll see that that was the range that it was emitting in. But it emits a photon. And that photon is traveling at the speed of light. It is light. And so that photon, says, you know what, I only got 30 million light years to travel. That's not too bad. I'm going to get there in 30 million years. And I'm going to do it discrete."
    },
    {
        "Q": "At 2:26, arc length can also be found using a degree.\nso why do we use radians if we can find arc length by using\n(Theta/360)*2pi*r? I mean why do we have to use radians?\n",
        "A": "radians have no units. This makes them much more convenient than degrees.",
        "video_name": "dVYYh8C80zo",
        "timestamps": [
            146
        ],
        "3min_transcript": "- [Instructor] So in the previous video, we defined all the new angular motion variables and we made an argument that those are more useful in many cases to use than the regular motion variables for things that are rotating in a circle. Since every point on the string in tennis ball, let's say this is a tennis ball you tied a string to and you're whirling around in a circle. Every point on the string including the tennis ball will have the same angular displacement, angular velocity and angular acceleration. But even though using angular motion variables is more convenient for these rotational motion problems, it's also really important to know how to translate those angular motion variables back into the regular motion variables. So that's what I wanna show you in this video how to translate angular motion variables back into regular motion variables. So let's do this. The simplest possible angular motion variable was the angular displacement because that just represented how much angle an object has rotated through. So let's say it rotated through this much. We represented the angular displacement with a delta theta In Physics, we typically choose to measure this in radians for a reason and I'll show you in just a second. Now, how would we convert this into a regular motion variable? What regular motion variable would that be? If I were to come at this for the first time, I'd be like all right, this is the angular displacement. Let's figure out how to relate it to the regular displacement but that would be weird. Because just think about it, the regular displacement for the ball that started over here and made it over here would be from this point to that point, that would be the regular displacement of the ball, the regular linear displacement of the ball. That's a little weird. I don't wanna show you how to find that for one, you have to use the law of cosines. That's a little more in depth than I'd wanna get to in this video. For two, the better reason this isn't all that useful in turns out. There's a much more useful quantity that would tell you how far the ball went. That's the arc length of the ball. So the ball traced out of path through space We call this the arc length that it turns out this is much more useful in a variety of problems. Good news is it's much easier to find than that regular displacement. So this is the arc length. People vary on what letter to use here. I've seen an l but most math books use s so we'll just use s as well. You might think this is hard to find but it's not. In fact, if we use radians and this is why we use radians, it's extremely easy to find. If we wanted to find the arc length of this tennis ball, we're just gonna take the radius of the circular path that tennis ball is tracing out. So in this case it'd be the length of the string. We take that radius. If we're in radians, we just multiply by the angular displacement. This is why the radians are so convenient. We just take that measurement in radians multiplied by the radius of the circular path the object is tracing out. You get the arc length which is the number of meters along this path that the object has traveled."
    },
    {
        "Q": "at 7:13, Sal says \"...solar systems...\", then something says Sal meant \"planetary systems\". way is that? way aren't they called solar systems?\n",
        "A": "Sol is the old name for our sun specifically. so our planetary system is called The Solar system. Other planetary systems would be named after the star they orbit, like the Wolf system.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            433
        ],
        "3min_transcript": "looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms But it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy, because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least-- I mean it blows my mind if you really think about what it's saying-- these unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully, that gives you a sense of how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion stars. to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart"
    },
    {
        "Q": "\nat the picture in 12:23 what is that shiny bright light in the lower left corner",
        "A": "It is a planet. Difficult to say which one without knowing when this picture was taken. Although my guess would most likely be Jupiter.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            743
        ],
        "3min_transcript": "it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there."
    },
    {
        "Q": "At 1:27 he says that the Voyager space craft was the fastest thing that has been made by man and sent out in space. I thaght it was the New Horizens space craft on the way to Pluto that was the fastest. Wich is the faster one?\n",
        "A": "While New Horizons may have started off faster, Voyager 1 is the fastest as it got more of a speed boost from its gravitational assists from Jupiter and Saturn.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            87
        ],
        "3min_transcript": "In the last video, I hinted that things were about to get wacky. And they are. So if we start where we left off in the last video, we started right over here, looking at the distance to the nearest star. And just as a reminder, in this drawing right here, this depiction right here, this circle right here, this solar system circle, it's not the size of the Sun. It's not the size of the orbits of the Earth, or Pluto, or the Kuiper Belt. This is close to the size of the Oort Cloud. And the actual orbit of Earth is about-- well, the diameter of the orbit of Earth is about 1/50,000 of this. So you wouldn't even see it on this. It would not even make up a pixel on this screen right here, much less the actual size of the Sun or something much smaller. And just remember that orbit of the Earth, that was at a huge distance. It takes eight minutes for light to get from the Sun to the Earth, this super long distance. If you shot a bullet at the Sun from Earth, So one thing, this huge distance wouldn't even show up on this picture. Now, what we saw in the last video is that if you travel at unimaginably fast speeds, if you travel at 60,000 kilometers per hour-- and I picked that speed because that's how fast Voyager 1 actually is traveling. That's I think the fastest object we have out there in space right here. And it's actually kind of leaving the solar system as we speak. But even if you were able to get that fast, it would still take 80,000 years, 75,000 or 80,000 years, to travel the 4.2 light years to the Alpha Centauri cluster of stars. To the nearest star, it would take 80,000 years. And that scale of time is already an amount of time that I have trouble comprehending. As you can imagine, all of modern civilization But most of recorded history is in the last 4,000 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance. Another way to think about it is if the Sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball you would have to put in Kiev, Ukraine in order to have a similar scale. So these are basketballs sitting in these cities. And you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video. The Sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So there are any little small planets over here, they would have to be grains of sand"
    },
    {
        "Q": "At 8:54 in the video, Sal mentioned the word supermassive black holes.What is a supermassive black hole and what is it's purpose?\n",
        "A": "Supermassive black holes are giant black holes that are in the center of galaxies (well most of them). The Milky Way has one called Sagittarius A*",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            534
        ],
        "3min_transcript": "to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here."
    },
    {
        "Q": "\nAt 12:14, Khan shows a picture of the Milky Way galaxy. In the bottom (and a little to the left) there is an IMMENSLEY COLOSSAL star or billions of stars that make a huge, bright and perfect star-shaped figure or formation. Please explain all about this form so I will understand.\n-Thanks, Easton",
        "A": "But it is an artist hypothesized sketch?",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            734
        ],
        "3min_transcript": "it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there."
    },
    {
        "Q": "At 9:40, he talks about how we see things in space as they were. How do they come up with these numbers? If we see something 5 billion light years away, how do we know if it's not 3, or 7 billion light years away?\n",
        "A": "Astronomers have several techniques for measuring distance. These include stellar parallax, spectroscopic parallax, the use of Cepheid variable stars as standard candles , and measurement of the spectral red shift of distant galaxies. Sal has videos about parallax and Cepheid variables. He also has vids about the redshift, although I am not sure he explains how it can be used to estimate distance.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            580
        ],
        "3min_transcript": "And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here. it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand,"
    },
    {
        "Q": "This may sound like a stupid question, but if you look in the middle of the galaxy at around 5:30 you see that there is a very bright yellow spot that is immensely long. What is it?\n",
        "A": "That is a very large cluster of densely packed yellow stars. There are probably several clusters like that in the center of the galaxy.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            330
        ],
        "3min_transcript": "So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really wacky is when you start really looking at this. Even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take-- and we're doing rough estimates right here-- it's about 30 light years. I'll just do LY for short. So that's about 30 light years, And once again, you can take pictures of our galaxy from our point of view. But you actually can't take a picture of the whole galaxy from above it. So these are going to be artists' depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly-- and these are all approximations. This is about 1,000 light years. And this is the 1,000 light years of our Sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here-- now, it would take forever to get anywhere over here-- it would be 1/30 of this. So it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here, that spans a 1,000 light years. But then when you start to really put it into perspective-- so now, let's zoom out a little bit-- so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the Sun. And once again, the word \"local\" is used in a very liberal way at this point. So this right here is 1,000 light years. looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms"
    },
    {
        "Q": "@ 6:55, during the discussion of London dispersion forces, it is assumed that the molecules produce opposite, and therefore attractive, momentary charges. Why wouldn't the molecules produce just as many same, and therefore repulsive, momentary charges, thus canceling any net effect?\n",
        "A": "That is very possible!",
        "video_name": "pBZ-RiT5nEE",
        "timestamps": [
            415
        ],
        "3min_transcript": "that you should remember for hydrogen bonding are fluorine, oxygen, and nitrogen. And so the mnemonics that students use is FON. So if you remember FON as the electronegative atoms that can participate in hydrogen bonding, you should be able to remember this intermolecular force. The boiling point of water is, of course, about 100 degrees Celsius, so higher than what we saw for acetone. And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction, and therefore, it takes more energy or more heat to pull these water molecules apart in order to turn them into a gas. And so, of course, water is a liquid at room temperature. Let's look at another intermolecular force. And this one is called London dispersion forces. So these are the weakest intermolecular forces, and they have to do with the electrons that are always And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. And it's hard to tell in how I've drawn the structure here, but if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon, and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar, and, of course, this one's nonpolar. And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, if for a brief transient moment in time you get a little bit of negative charge to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And, of course, it is. So the boiling point for methane is somewhere around negative 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure. Now, if you increase the number of carbons, you're going to increase the number of attractive forces"
    },
    {
        "Q": "\nAt 7:38 is there an omitted hydrogen on the first carbon? I understand that hydrogens aren't always drawn but is the hydrogen that was added bonded to that firs carbon?",
        "A": "At about 5:33 you ll see pi bond picked up the H and it was added to the far left carbon, which formed a secondary carbocation on the carbon next to it. The secondary carbocation is more stable than if the H was added to the second carbon from the left which would result in a primary carbocation.",
        "video_name": "dJhxphep_gY",
        "timestamps": [
            458
        ],
        "3min_transcript": "We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism,"
    },
    {
        "Q": "\nAt 8:30 so there is no cis or trans in this molecule? And why do?",
        "A": "Because one side has a methyl group and the other side does not. Both carbons of the double bond would have to have the same substituents in order for there to be cis/trans isomerism.",
        "video_name": "dJhxphep_gY",
        "timestamps": [
            510
        ],
        "3min_transcript": "We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism,"
    },
    {
        "Q": "At 3:42, Sal gets the average speed. How does he do that? What is the 12pi?\n",
        "A": "12pi is the circumference of the circle. The formula for circumference is C = 2(pi)(r), and the radius of the loop is 6 meters, so the circumference of the loop is 12pi meters. He finds the speed of the car by doing distance divided by time. The time for the loop is 2 and 14/30 seconds.",
        "video_name": "zcnnZz2pCSg",
        "timestamps": [
            222
        ],
        "3min_transcript": "and in the previous video, we figured out the radius was 6 m So it's 2 pi times 6 m which is equal to 12 pi meters If you wanted to figure out its average speed-- the velocity is constantly changing because the direction is changing but the magnitude of the velocity-- if we wanted to figure out the average magnitude of the velocity or the average speed the total distance traveled is 12 pi meters divided by the time required to travel the 12 pi meters so that is 2 and 14/30 seconds Now let's get our calculator out to actual calculate that value So we're going to have the distance 12 pi m divided by And then this gives us in meters per second 15.3 m/s So the average speed is approximately 15.3 m/s which is almost twice as fast as the minimum speed we figured out because you want that margin of safety and you want to be able to have some traction with the road Although you don't want to go too fast, because then the G force is going to be too big then this--maybe we'll talk about that in the future video I'll just relate this into kilometers per hour. Let's figure out what that is I want to use the calculator here So that's in the meters per second So that's how many meters per hour, and divide it by 1000 which you can see right over there that is 55 km/h If you want to do it in miles, it's rough approximation, divide it by 1.6 It's about 35 mph give or take or 55 km/h So this is approximately 55 km/h So the driver here luckily they did the physics ahead of time and he had the margin of safety He was well in excess of the minimum velocity just to maintain the circular motion So he probably has some nice traction with the track up here"
    },
    {
        "Q": "\nat 2:05, aren't there other pieces of evidence besides fossils, such as mountain ranges?",
        "A": "I think there is.",
        "video_name": "axB6uhEx628",
        "timestamps": [
            125
        ],
        "3min_transcript": "We know that new plate material is being formed, and these lithosphere plates on the surface of the Earth are moving around. And that might raise the question in your brain-- what happens if we kind of reverse things? We know the direction they're moving in. What does that tell us about where they came from? So let's just do the thought experiment. Right now, South America and Africa are moving away from each other, because of new plate material being created at the mid-Atlantic rift. Let's rewind it. Let's bring them back together. We know that India is jamming into the Eurasian Plate right now, causing the Himalayas to get higher and higher. What if we rewind that? Let's bring India back down towards Antarctica. Same thing with Australia. We have new plate material being formed between Australia and Antarctica that's making the continents move apart. Let's bring them back together. Let's rewind the clock. but if you actually look at the GPS data, it becomes pretty obvious that North America, right now is moving in a counterclockwise rotation. So let's rewind it into a-- let's go back, moving it in a clockwise direction. Let's, instead of Eurasia going further away from North America, let's bring it back together. And so what you could imagine is a reality where India, Australia are jammed down into South America-- sorry, into Antarctica. South America and Africa are jammed together. North America is jammed in there. And essentially, Eurasia is also jammed in there. So it looks like they all would clump together if you go back a few hundred million years. And based on, literally-- based on just that thought experiment, you could imagine at one point, all of the continents on the world were merged into one supercontinent. And that supercontinent is called Pangaea-- pan for the world. And it turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called-- well, we call it Pangaea, now. Obviously, there probably weren't things on the planet calling it anything back then. Or, there were things back then, but not things that would actually go and try to label continents that we know of. But all of the evidence tells us that Pangaea existed about 200 to 300 million years ago, roughly maybe 250 million, give or take, years ago. And I want to be clear. This was not the first supercontinent. To a large degree, it's kind of the most recent supercontinent. And it's easiest for us to construct because it was the most recent one. But we believe that there were other supercontinents before this. That if you rewind even more that you would have to break up Pangaea and it would reform. But we're now going back in time. Or that there were several supercontinents"
    },
    {
        "Q": "at 6:07 it is easy to understand how the unpaired electrons repel the hydrogens attached to nitrogen so that a trigonal pyramidal shape is achieved. But what forces drive the formation of a tetrahedral shape for methane? Why doesn't it assume a flat plane? Wouldn't that create the greatest distance among the hydrogens?\n",
        "A": "If it assumed a flat plane, then the angles between the Hs would be 360\u00c2\u00b0/4 = 90\u00c2\u00b0. A tetrahedral shape allows the angles between the Hs to be 109.5\u00c2\u00b0, greater than the 90\u00c2\u00b0 that a flat plane would allow.",
        "video_name": "ka8Yt4bTODs",
        "timestamps": [
            367
        ],
        "3min_transcript": "surrounded the central atom. So regions of electron density. And let's go ahead and find those. So I can see that I have these bonding electrons. That's a region of electron density, so that's an electron cloud. Here's another one. Here's another one. So that's three. And this lone pair of electrons, this non-bonding pair of electrons is also going to be counted as an electron cloud. It's a region of electron density too. And so once again we have four regions of electron density. When you're thinking about the geometry of those electron clouds, those four electron clouds are going to, once again, try to point towards the corners of a tetrahedron. So we can kind of sketch out the ammonia molecule. And we can draw the base the same way we did before, with our three hydrogens right here. And then we're going to go ahead and put our lone pair of electrons right up here. And so again, it's an attempt to show the electron clouds Let's go back up here and look at our steps again. So in step three, we predicted the geometry electron clouds are going to attempt to be in a tetrahedron shape around our central atom. But when we're actually talking about the geometry or shape of the molecule, we're going to ignore any lone pairs when we predict the geometry of the molecule. So when we look at the ammonia molecule, we're going to ignore that lone pair of electrons on top of the nitrogen and we're just going to focus in on the bottom part for the shape here. And so when we do that, we get something that looks like a little squat pyramid here. So if I'm [? ignoring ?] that lone pair of electrons up there at the top, it's going to look something like that for the shape. And we call this trigonal pyramidal. So this is a trigonal pyramidal shape. So even though the electron clouds are attempting to be in a tetrahedron fashion, because we ignore any lone pairs of electrons. In terms of a bond angle, this lone pair of electrons on the nitrogen actually occupies more space. These non-bonding electrons occupy a little more space than bonding electrons. And because of that, those non-bonding electrons are going to repel these bonding electrons. I'm going to go ahead and put them in blue here just as an example. Repel these a little bit more than in the previous example that we saw. And that's actually going to make the bond angle a little bit smaller than the ideal bonding angle we saw before for 109.5 for a tetrahedral arrangement of electron clouds. And so it turns out that this bond angle between the atoms, the hydrogen nitrogen hydrogen bond angle gets a little bit smaller than 109.5. So it actually gets smaller to approximately 107 degrees here for a trigonal pyramidal situation."
    },
    {
        "Q": "At 3:50, how is it that 109.5 separates all the clouds better on a single plane compared to 90?\n",
        "A": "But it isn t on a single plane, it s in 3D space. 109.5 degrees just happens to be the maximum angle when you have 4 equal groups. If you need proof then Jay did do a mathy video that proves this on here, might be easier just accept it for what it is.",
        "video_name": "ka8Yt4bTODs",
        "timestamps": [
            230
        ],
        "3min_transcript": "of that tetrahedron, that four sided figure here. And so we've created the geometry of the electron clouds around our central atom. And in step four, we ignore any lone pairs around our central atom, which we have none this time. And so therefore, the geometry the molecule is the same as the geometry of our electron pairs. So we can say that methane is a tetrahedral molecule like that. All right, in terms of bond angles. So our goal now is to figure out what the bond angles are in a tetrahedral molecule. Turns out to be 109.5 degrees in space. So that's having those bonding electrons as far away from each other as they possibly can using VSEPR theory. So 109.5 degrees turns out to be the ideal bond angle for a tetrahedral molecule. Let's go ahead and do another one. Let's look at ammonia. So we have NH3. So we start by finding our valence electrons, nitrogen in group five. So 5 valence electrons. Hydrogen in group one, and I have three of them. So 1 times 3 plus 5 is 8. So once again, we have 8 valence electrons to worry about. We put nitrogen in the center and we know nitrogen has bonded to 3 hydrogens, so we go ahead and put our 3 hydrogens in there like that. Let's see how many valence electrons we've used up so far. 2, 4, and 6. So 8 minus 6 is 2 valence electrons left. We can't put them on our terminal atoms, because the hydrogens are already surrounded by two electrons. So we go ahead and put those two valence electrons on our central atom, which is our nitrogen like that. And so now we've gone ahead and represented those two valence electrons. So we have all eight valence electrons shown for our dot structure. All right, we go back up here to our steps to remind us what to do after we've drawn our dot structure. And we can see that now we're going surrounded the central atom. So regions of electron density. And let's go ahead and find those. So I can see that I have these bonding electrons. That's a region of electron density, so that's an electron cloud. Here's another one. Here's another one. So that's three. And this lone pair of electrons, this non-bonding pair of electrons is also going to be counted as an electron cloud. It's a region of electron density too. And so once again we have four regions of electron density. When you're thinking about the geometry of those electron clouds, those four electron clouds are going to, once again, try to point towards the corners of a tetrahedron. So we can kind of sketch out the ammonia molecule. And we can draw the base the same way we did before, with our three hydrogens right here. And then we're going to go ahead and put our lone pair of electrons right up here. And so again, it's an attempt to show the electron clouds"
    },
    {
        "Q": "\n5:30 how do you apply bond breaking energy? just heat?",
        "A": "Heat is certainly an option because it is a form of energy, in fact, any form of energy will work as long as there is enough quantity of energy to break the bond. In fact, some experiments are so spontaneous that yelling at the reactants (sound energy) will start the reaction",
        "video_name": "Ce4BGV1DVVg",
        "timestamps": [
            330
        ],
        "3min_transcript": "Notice this is different from the previous video where we talked about electronegativity. There we were comparing elements in the same period on the periodic table. So we were moving horizontally across our periodic table this way. And in that video, the fluoride anion was the most stable one because fluorine's our most electronegative element, and therefore, best able to stabilize a negative charge. But as you go down a group on the periodic table, your electronegativity decreases. So that can't be the dominant trend because if your electronegativity decreases as you go down, just thinking about electronegativity, that would predict HF to be the strongest acid, and that's not what we observe. So as you go down a group on the periodic table, it's the size of the anion that determines the stability of the conjugate base. So the larger the anion, and therefore, the more stable the conjugate base. The more stable the conjugate base, the more likely HX is to donate a proton, and therefore, the stronger the acid. Another very important factor to think about is the strength of the bond. We've already said that hydroiodic acid is our strongest acid with the lowest pKa value. So this bond right here must be the easiest to break. If it's easy to break this bond, that makes it easy to donate this proton. So we can get an idea of the bond strengths for our binary acids by looking at bond association energy. So we could also call these bond energies or bond enthalpies. So remember bond dissociation energy measures the amount of energy that's needed to break a bond in the gaseous state. So if we look at our hydrogen halides and we think about our bonds, notice what happens to the bond energy. It's the hardest to break this bond, This takes the most energy to break this bond, and as you go down, we see we decrease in bond dissociation energy. So it only takes 299 kilojoules per mole to break this bond between H and I. I should say these are approximate bond energies and you'll see several different values in different textbooks. So if the HI bond is the easiest to break, that means when you're thinking about the acids, this bond is the easiest to break, therefore, it's the most likely to donate a proton, and therefore, it has the lowest value for the pKa. Hydroiodic acid is the strongest out of these four binary acids."
    },
    {
        "Q": "\nAt 0:45, how can a reaction have a zero order while the molecularity can never be zero? Because the powers raised over the concentration in an elementary reaction while using rate law is equal to the stoichiometric coefficients of reactants and these coefficients can also never be zero.",
        "A": "I think you are gravely mistaken the powers are not the stoichiometric coefficients, they are the order of reaction with respect to the given atom or molecule, and order as we discussed is experimental and molecularity is very different from order they are two different things..i hope you got your error...",
        "video_name": "49LcF9Zf9TI",
        "timestamps": [
            45
        ],
        "3min_transcript": "- Let's see we have a zero order reaction where A turns into our products and when time is equal to zero we're starting with our initial concentration of A and after some time period T, we would have the concentration of A at that time. So we could express the rate of our reaction, one way to do it would be to say the rate of the reaction is equal to the negative change in the concentration of A over the change in time. And another way to do this would be to right the rate law. So the rate of our reaction is equal to the rate constant K times the concentration of A and since I said this is a zero order reaction, this would be A to the zero power. And any number to the zero power is equal to one. Therefore, the rate of the reaction would be equal to K times one or the rate is just equal to the rate constant K. So the rate of a zero order reaction is a constant. Next, we can set these equal. Right we can say that K is equal to the negative change in the concentration of A over the change in time. So we have the negative change in the concentration of A over the change in time is equal to the rate constant K. And next we could think about our calculus, alright, instead of writing change in A over change in time, we're going to write the rate of change of the concentration of A with respect to time. We have our negative sign in here and then we have our K on the side. So we're ready to think about our differential equation and we're gonna multiple by both sides by DT. Let's go ahead and multiple both sides by negative DT, so then we would have DA on the left side, right, and then we would have negative KDT on the right side. And we're ready to integrate. Right, so we're gonna integrate on the left, K is a constant so we can pull it out of our integral on the right. about what would be integrating from. So we'd be going from time is equal to zero to time is equal to T and from our initial concentration to our concentration at time T. So you plug those in we're going from time is equal to zero to time is equal to T and then for our concentration we're going from our initial concentration to our concentration at time T. So we have some easy integrals here, right? What's the integral of DA? That would be of course A or the concentration of A. So we have the concentration of A, right. We are evaluating this from our initial concentration to our concentration of A at time T. On the right side we have another easy integral, integral of DT. That's just T. So we have negative KT from zero to T. Next fundamental theorem of calculus, right, so we would get the concentration of A"
    },
    {
        "Q": "At 6:46 in the diagram of the 1.0 M NaCl solution what are the two blue and 1 green circles representing? and the red ones?\n",
        "A": "The red and white molecule (which forms Mickey Mouse s head) would be H2O, water. The blue and green molecule is NaCl, salt. As he says, And here they visualise Sodium Chloride at the surface. while pointing to the green/blue molecule.",
        "video_name": "z9LxdqYntlU",
        "timestamps": [
            406
        ],
        "3min_transcript": "It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize. You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. So solute also raises the boiling point. solute, when you add something to a solution, it's going to make it want to be in the liquid state more. Whether you lower the temperature, it's going to want to stay in liquid as opposed to ice, and if you raise the temperature, it's going to want to stay in liquid as opposed to gas. I found this neat -- hopefully, it shows up well on this video. I have to give due credit, this is from chem.purdue.edu/gchelp/solutions/eboil.html, but I thought it was a pretty neat graphic, or at least a visualization. This is just the surface of water molecules, and it gives you a sense of just how things vaporize as well. There's some things on the surface that just bounce off. And here's an example where they visualized sodium chloride at the surface. And because the sodium chloride is kind of bouncing around on the surface with the water molecules, fewer of those water molecules kind of have the room to escape, so the boiling point gets elevated. And this is one of the neat things in life is that the answer is actually quite simple. The change in boiling or freezing point, so the change in temperature of vaporization, is equal to some constant times the number of moles, or at least the mole concentration, the molality, times the molality of the solute that you're putting into your solution. So, for example, let's say I have 1 kilogram of -- so let's say my solvent is water. I'll switch colors. And I have 1 kilogram of water, and let's say we're just at atmospheric pressure."
    },
    {
        "Q": "@3:12, did he mean \"lower the freezing point\"?\n",
        "A": "I believe he meant lower the melting point.",
        "video_name": "z9LxdqYntlU",
        "timestamps": [
            192
        ],
        "3min_transcript": "they're just vibrating in place. So you have to get a little bit orderly right there, right? And then, obviously, this lattice structure goes on and on with a gazillion water molecules. But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example. Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same -- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way. So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big. Now, do you think it's going to be easier Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it. They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other. But then when you introduce a solute particle, let's say a solute particle is sitting right here. It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that. and it's making it harder to get into a regular form. And so the intuition is that this should lower the boiling point or make it -- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is. Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil? But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface,"
    },
    {
        "Q": "(1:50) since the dark reactions are independent, why do they require ATP, NADPH, AND CO2, but not the protons, can anyone help me better understand?\n",
        "A": "ATP and NADPH in non-cyclic photophosphorylation are created in the light reactions using the photons. This byproducts from the Light dependent reactions are used in the Calvin cycle. So, ATP and NADPH are created in light reactions because this are needed in the calvin cycle to produce glucose or any other carbohydrate which can be used by the plants later. Hope this helps.",
        "video_name": "slm6D2VEXYs",
        "timestamps": [
            110
        ],
        "3min_transcript": ""
    },
    {
        "Q": "9:30 How do we get glucose (C6H12O6) from 2 G3Ps which is 3 carbon with phosphate? Don't we need Hydrogen as well as more Oxygen?\n",
        "A": "It s because there are materials from the Light Dependent Reactions (Photosynthesis) that go into making glucose",
        "video_name": "slm6D2VEXYs",
        "timestamps": [
            570
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nCan someone refresh my memory here? At 2:10, Sal says the Oxygen wants to give away an electron. I thought Oxygen's goal was to take electrons in order to complete it's valence shell.",
        "A": "Remember that all molecules want a neutral charge. The oxygen atom here has a negative charge. When he says it wants to give away an electron he really means that the oxygen really wants to share its electron with a proton, forming a covalent bond, which in turn will keep its octet, and make the molecule neutral (no charge).",
        "video_name": "J0gXdEAaSiA",
        "timestamps": [
            130
        ],
        "3min_transcript": "Let's try to come up with the reaction of when we have this molecule right here reacting with sodium methoxide in a methanol solution, or with methanol as the solvent. Just so we get a little practice with naming, let's see, this is one, two, three, four carbons. So it has but- as a prefix and no double bond or triple bonds, so it's butane. And we have a chloro group here. So if we start numbering at the side closest to it, one, two. So it's 2-chlorobutane. Let's think about what might happen here. The first thing-- let me just redraw the molecule right there. The first thing you need to realize is this sodium methoxide is a salt. When it's not dissolved, it's made up of a positive sodium cation. Let me draw them right here. negative methoxide anion. Let me draw the methoxide part right here. It normally would have just two pairs, but now we have three pairs here. The oxygen has an extra electron. Actually, let me draw another electron as the extra electron. This'll be useful for our mechanism. It could be any of them. The oxygen has an extra electron. It has a negative charge. In solid form, when they're not dissolved, they had formed an ionic bond and they form a crystal-like structure. It's a salt, but when you dissolve it in something, in a solution, in this case, we have methanol as the solution, they will disassociate from each other. And what you have right here is this methoxide right here. This is a very, very, very strong base. And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. In this situation, that is exactly what it will do. I'll actually give you the most likely reaction to occur here, and we'll talk about other reactions, and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron. Let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron."
    },
    {
        "Q": "At around 7:15, Sal says that the speed is independent of the inertial frame of reference. But if Sal and Sally are traveling at different speeds, how can they observe the speed of light as being the same? Isn't Sally closer to the speed of light, so she has greater relative velocity compared to it?\n",
        "A": "What you are saying makes common sense but that is not how velocities that are a significant fraction of the speed of light work. Any observer that is not accelerating will always measure light in a vacuum traveling at the same speed.",
        "video_name": "OIwp8m3S30c",
        "timestamps": [
            435
        ],
        "3min_transcript": "or it looks like the velocity of that photon is one and a half times 10 to the 8th meters per second in the positive x direction. And this should hopefully makes sense from a Newtonian point of view, or a Galilean point of view. These are called Galilean transformations because if I'm in a car and there's another car and you see this on the highway all the time, if I'm in a car going 60 miles per hour, there's another car going 65 miles per hour, from my point of view, it looks like it's only moving forward at five miles per hour. So that photon will look slower to Sally. Similarly, if we assume this Newtonian, this Galilean world, if she had a flashlight, if she had a flashlight right over here and right at time equals zero she turned it on, and that first photon we were to plot it on her frame of reference, well, it should go the speed of light So it starts here at the origin. And then after one second, in the s prime, in the s prime coordinates, it should have gone three times 10 to the 8th meters. After two seconds, it should've gone six times 10 to the 8th meters. And so it's path on her space-time diagram should look like that. That's her photon, that first photon that was emitted from it. So you might be noticing something interesting. That photon from my point of view is going faster than the speed of light. After one second, its x coordinate is 4.5 times 10 to the 8th meters. It's going 4.5 times 10 to the 8th meters per second. It's going faster than the speed of light. It's going faster than my photon, and that might make intuitive sense except it's not what we actually observe in nature. And anytime we try to make a prediction that's not what's observed in nature, it means that our understanding of the universe is not complete inertial reference frame we are in, the speed of light, regardless of the speed or the relative velocity of the source of that light, is always going three times 10 to the 8th meters per second. So we know from observations of the universe that Sally, when she looked at my photon, she wouldn't see it going half the speed of light, she would see it going three times 10 to the 8th meters per second. And we know from observations of the universe that Sally's photon, I would not observe it as moving at 4.5 times 10 to the 8th meters per second, that it would actually still be moving at three times 10 to the 8th meters per second. So something has got to give. This is breaking down our classical, our Newtonian, our Galilean views of the world. It's very exciting. We need to think of some other way to conceptualize things, some other way to visualize these space-time diagrams for the different frames of reference."
    },
    {
        "Q": "At 3:00, why does the carbonyl reform? I thought oxygen can hold the negative charge?\n",
        "A": "Yes, it can, but a C=O double bond with no charge is more stable than a C-O\u00e2\u0081\u00bb single bond with a charge.",
        "video_name": "rNJPNlgmhbk",
        "timestamps": [
            180
        ],
        "3min_transcript": "will react with that faster. And so once again, our product will depend on what our starting material is. So the mechanism for the reduction of aldehydes or ketones with lithium aluminum hydride is just like the one for sodium borohydride. So we'll move on to a mechanism for the reduction of an ester. So let's go ahead and do that. So let's start with an ester down here. So we have our carbonyl. Like that. So we'll put in our lone pairs. And down here, we have our R prime group. Like that. So there's our ester. And we add lithium aluminum hydride in excess. So in terms of molar equivalence, let's go ahead and put lithium aluminum hydride down here, Li+. And then, we have Al bonded to 4 hydrogens. And that's going to give the aluminum a negative 1 formal charge. Like that. So the first step in the mechanism is just like the one we did in the previous video. All right. in electronegativity between this carbon and this oxygen-- oxygen being more electronegative, pulling these electrons closer towards it and the double bond, giving it a partial negative charge. Whereas the carbon down here is losing some electron density, becoming partially positive. So the carbonyl carbon is an electrophile. It wants electrons. And of course, it's going to get electrons from these 2 electrons in here. So these 2 electrons are going to attack this carbon. So a nucleophilic attack. And kick these electrons off onto our oxygen. So that's our first step-- the nucleophilic attack portion. So now, we have R. And it used to have 2 bonds to carbon and oxygen. Now, it has only 1 bond because those electrons moved off onto the oxygen, giving the oxygen a negative 1 formal charge. Like that. So we added on our hydrogen. Like that. And then, we still have our O and our R prime group In the next step of this reaction, the carbonyl's going to reform. So the electrons in here are going to kick back into here to reform our carbonyl. That would mean 5 bonds to carbon-- which we know never happens-- so that these electrons are going to have to break and come off onto the oxygen. So let's go ahead and draw the results of that. So now, we have R bonded to carbon. Now, we reformed our carbonyls. Now, we have only two lone pairs of electrons on that oxygen now. And then, we still have a hydrogen. Right here. So we lost an oxygen with an R prime group. And that oxygen is going to have three lone pairs around it with a negative 1 formal charge. Oxygen being relatively electronegative, it can handle that formal charge fairly well and be relatively stable. So now we have an aldehyde. And we know that this reaction can occur again with an aldehyde. So since you have extra lithium aluminum hydride floating"
    },
    {
        "Q": "\nat 10:28 ,isn't h2 and Pd going to reduce the aldehyde ester AND the benzene ring also.",
        "A": "no, reducing the benzene would need H2 over a Rhodium/Carbon catalyst. H2 over Palladium catalyst is selective to alkenes and alkynes than aldehydes and ketones if it s in the same molecule.",
        "video_name": "rNJPNlgmhbk",
        "timestamps": [
            628
        ],
        "3min_transcript": "forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride Like that. And then, down here, what used to be our ester functional group, we're going to add two molar equivalents of hydrogen to that carbonyl. And we're going to end up breaking that bond between the carbonyl carbon and that oxygen. All of this over here is going to go away as a leaving group in our mechanism. And we're going to add on two hydrogens to that carbon. And then, that's going to form our alcohol. So we're going to add two hydrogens onto that carbon, forming a primary alcohol down here as well-- just like in the mechanism that we just discussed. So reduction of esters using lithium aluminum hydride. What about if we were to add a hydrogen gas and palladium as our metal catalyst here? Well, this is also a reduction reaction that we talked about earlier. Hydrogenation is an example of a reduction reaction. And it's going to be chemoselective. reaction, the only thing the hydrogenation reaction is going to touch is this double bond. It's going to reduce this double bond. So let's go ahead and draw the product. It's not going to touch the aldehyde. It's not going to touch the ester. And it's not going to touch the benzene ring. So let's go ahead and draw the product. The benzene ring is not hydrogenated under normal conditions, but we're going to add on two hydrogens across that double bond. And the aldehyde is untouched. And down here, the ester is going to be untouched as well. So that would be our product from a hydrogenation. So three different reductions, three different products. Now, hydrogen will reduce carbonyls under the right conditions. Usually, if you have increased pressure and increased temperatures, you actually can reduce those carbonyls. But again, you can control those conditions. So you can control what part of the molecule is reduced."
    },
    {
        "Q": "Near 8:01, Jay said that NaBH4,reduces the aldehyde to alcohol.\n\nIt's correct. I have no objections in that. But, my teacher taught us that NaBH4 also reduces the alpha-beta-conjugated double bonds. So,by my teacher's side, the product should be a cyclic ring with two double bonds and no double bond in the right side... Was my teacher correct or is Jay correct?\n",
        "A": "No, NaBH4 is a mild reducing agent and will only react with more reactive carbonyls like ketones and aldehydes. Alkenes require more reactive metal catalysts like Pd or Pt. There are special cases where the nucleophilicity of the alkene can be adjusted to react with NaBH4, but these are not addressed in an undergraduate class.",
        "video_name": "rNJPNlgmhbk",
        "timestamps": [
            481
        ],
        "3min_transcript": "So this one and this one. So the reaction happened twice. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product. You're going to add on two hydrogens on to that original carbonyl carbon. Like that. So let's look at the chemoselectivity of this reaction. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start here with our reactants-- so let's make it a benzene ring. Like that. And let's put stuff on the benzene ring. So let's go ahead and put a double bond here. And then, we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester. Like that. transform different parts of this molecule using different reagents. So let's say we were to do a reaction wherein we add on sodium borohydride. And then, the proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion. So we are still going to have our double bond here. And the aldehyde's going to go away to form a primary alcohol. So we're going to get primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride"
    },
    {
        "Q": "at 1:18 Sal circles one set of arrows on the Nazca plate and just a little to the right and above there is another set of arrows. These arrows are pointing in a directions almost perpendicular to the set Sal circled. How can a plate be moving in two directions at once? Or is this map from wikipedia inaccurate?\n",
        "A": "plates move in different directions, so in one part a plate will slide under another and in another part of the same plate might be seperating away from another or they can spread or converge at the same time",
        "video_name": "f2BWsPVN7c4",
        "timestamps": [
            78
        ],
        "3min_transcript": "What I want to do in this video is talk a little bit about plate tectonics. And you've probably heard the word before, and are probably, or you might be somewhat familiar with what it discusses. And it's really just the idea that the surface of the Earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates, and these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here I have a picture I got off of Wikipedia of the actual plates. And over here you have the Pacific Plate. Let me do that in a darker color. You have the Pacific Plate. You have a Nazca Plate. You have a South American Plate. I could keep going on. You have an Antarctic Plate. It's actually, obviously whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic Plate, North American Plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here the Nazca Plate and the Pacific Plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here in the middle of the Atlantic Ocean the African Plate and the South American Plate meet each other, and they're moving away from each other, which means that new land, more plate material I guess you could say, is somehow being created right here-- we'll talk about that in future videos-- and pushing these two plates apart. Now, before we go into the evidence for plate tectonics or even some of the more details about how plates are created and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. And when I talk about chemical layers, I'm talking about what are the constituents of the different layers? So when you talk of it in this term, the top most layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that, you have the mantle. So everything between the blue and the orange line,"
    },
    {
        "Q": "\nHey :)\nAt 13:38 , I dont really understand how can NH3 give NH4 and hydroxide ions?\nand isnt it NOT reversible reaction? :O\n\n-Sahil, Mumbai",
        "A": "Put in the first term of reation water and you can understand better. Amonia is a weak base and water, in this case, is a weak acid. There is a Ka in this case.",
        "video_name": "3Gm4nAAc3zc",
        "timestamps": [
            818
        ],
        "3min_transcript": "Well, if you know the pKa for a weak acid-- For example, let's say we have NH4 plus. This is a weak acid, right? It can donate an H, but it's not an irreversible reaction. That H can be gained back. So this is a weak acid. If you look it up on Wikipedia, it'll say, hey, the pKa of NH4 is equal to 9.25. Right? So this is 9.25 for NH4. For ammonium. So what is going to be the pKb for ammonia? Right? Let me write that reaction down. So this is NH4. Is in equilibrium. This is plus. It can get rid of one of its hydrogen protons, and you're just left with ammonia. So this is what the pKa is associated with. So the equilibrium constant for this reaction is is equal to 9.25. And if I had the reverse reaction, the conjugate base reaction, so ammonia converts to ammonium. Plus it grabbed that hydrogen proton from a water molecule. If I wanted to figure out the pKb, or the equilibrium constant, or the negative log of the equilibrium constant for this reaction, what is it? Well, this one's 9.25. And 9.25 plus this pKb have to be equal to 14. So what's 14 minus 9.25? It's what, 4.75. So we immediately know the equilibrium constant for the conjugate base reaction. So it's a useful thing to know. And always remember, you see these pKa and pKb, and you say, what is that? Well, if you see a p, it's a negative log of something. And in this case, it's the negative log of the equilibrium constant for an acidic reaction. Plus the negative log of the equilibrium basic reaction, where this is the conjugate base of this acid. It's always going to be equal to 14 if we're dealing with an aqueous solution at 25 degrees Celsius, which is essentially room temperature, which is usually going to be the case."
    },
    {
        "Q": "At 1:40, Sal says that is a conjugate-base. How do we know that is not a conjugate-acid?\n",
        "A": "Watch the conjugate acids and bases video. Because HA is an acid in the first place, A- has to be the conjugate base.",
        "video_name": "3Gm4nAAc3zc",
        "timestamps": [
            100
        ],
        "3min_transcript": "In the last video we learned that if I had some -- let's say I have some weak acid. So it's hydrogen plus some-- the rest of whatever the molecule was called. We'll just call it A. And I think this tends to be the standard convention for the rest of the acid. They can disassociate, or it's in equilibrium because it's a weak acid. So it can be in equilibrium with -- since it's a weak acid, it's going to produce some hydrogen proton. And then the rest of the molecule is going to keep the electrons. So it's going to be plus -- oh, this is in an aqueous solution. Let me do that. Aqueous solution. And then you're going to have the rest of the acid, whatever it might be. A minus, and that's also going to be in an aqueous solution. And that's the general pattern. We've seen the case where A could be an NH3, right? If A is an NH3, then when you have this, you have an NH4 plus, and this would be ammonia. A could be a fluorine molecule right there, because then this would be hydrogen fluoride or hydrofluoric acid. And this would just be the negative ion of fluorine. Or a fluorine with an extra electron. So it could be a bunch of stuff. You can just throw in anything there, and it'll work. Especially for the weak acids. So we learned a last video that if this is the acid, then this is the conjugate base. And we could write the same reactions, essentially, as kind of more of a basic reaction. So we could say, if I start with A minus -- it's in an aqueous solution -- that's in equilibrium with-- This thing could grab a hydrogen It's still in an aqueous solution. And then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxide molecule. Right? Because it's hydrogen. Remember, whenever I say pluck the hydrogen, just the proton, not the electron for the hydrogen. So the electron stays on that water molecule, so it has a negative charge. It's in an aqueous solution. So we could write the same reaction both ways. And we can write equilibrium constants for both of these reactions. So let's do that. Let me erase this, just because I can erase this stuff right there, and then use that space. So an equilibrium reaction for this first one. I could call this the K sub a, because the equilibrium reaction for an acid. And so this is going to be equal to its products. So the concentration of my hydrogen times"
    },
    {
        "Q": "\nAt 4:13, why do we know that Si has 4 bonds to F atoms?",
        "A": "Si has 4 valence electrons so it needs to form 4 bonds to complete its octet.",
        "video_name": "p7Fsb21B2Xg",
        "timestamps": [
            253
        ],
        "3min_transcript": "But it is not possible for elements to have more than eight electrons. Always check your dot structures, and make sure that if you have an element in the second period, you do not exceed eight electrons. Once you get to the third period, you have even more orbitals available to you now. So in the first energy level, you have only one s orbital. In the second energy level you have s and p orbitals, and in the third energy level you have s, p, and d orbitals. So you can fit more than eight electrons. And so therefore it's possible to exceed the octet rule for elements in the third period and beyond. And we will see a few examples of that in this video, and some of the ones to come here. So getting back to our molecule, silicon tetrafluoride, if I wanted to find out how many total valence electrons are in this molecule, I need to find these elements on my periodic table. So I go over here and I find silicon, and I see it's in group four. So therefore one atom of silicon has four valence electrons. Fluorine is over here in a group of seven, will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover"
    },
    {
        "Q": "\n5:05 You mention the term terminal atoms. What is that ?",
        "A": "The word Terminal Means End so a terminal atom is just the outermost atom of the molecule.",
        "video_name": "p7Fsb21B2Xg",
        "timestamps": [
            305
        ],
        "3min_transcript": "will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover So in this case, the terminal atoms would be the fluorines. So let's go back down here and look at our dot structure. So fluorine would be the terminal atoms, and we're going to assign electrons to those fluorines. But how many do we need to assign? Well, going back to our periodic table over here, so fluorine is in the second period. So pretty good bet it's going to follow the octet rule here. So we need to surround each fluorine atom with eight electrons. Each fluorine already has two electrons around it, so I'm going to go ahead and put six more around each fluorine, like that. So each fluorine get six more valence electrons. And since I'm assigning six valence electrons to four fluorines, 6 times 4 gives me 24. And so therefore we've now accounted for all of the valence electrons. And so this should be the should be the final dot structure here. And so we don't even need to go on to step four for this molecule. This is a very simple molecule to draw."
    },
    {
        "Q": "\nWhat is \"lumen\" mentioned at 7:08 ?",
        "A": "Lumen is the inside space of a tubular structure, such as an artery or intestine. Source: Wikipedia",
        "video_name": "6UqtgH_Zy1Y",
        "timestamps": [
            428
        ],
        "3min_transcript": "a separate organelle. So you get this thing that looks like this, and I'll just do it the best that I can draw it. And this right over here is called the endoplasmic reticulum. So this right here is endoplasmic reticulum, which I've always thought would be a good name for a band. And the endoplasmic reticulum is key for starting to produce and then later on package proteins that are either embedded in the cellular membrane or used outside of the cell itself. So how does that happen? Well, the endoplasmic reticulum really has two regions. It has the rough endoplasmic reticulum. And the rough endoplasmic reticulum has a bunch of ribosomes. So that's a free ribosome right over here. These are ribosomes that are attached to the membrane of the endoplasmic reticulum. So this region where you have attached ribosomes right over here, that is the rough endoplasmic reticulum. I'll call it the rough ER for short. Perhaps an even better name for a band. And then there's another region, which is the smooth endoplasmic reticulum. And the role that this plays in protein synthesis, or at least getting proteins ready for the outside of the cell, is you can have messenger RNA-- let me do that in that lighter green color-- you can have messenger RNA find one of these ribosomes associated with the rough endoplasmic reticulum. And as the protein is translated, it won't be translated inside the cytosol. It'll be translated on the other side of the rough endoplasmic reticulum. in the lumen of the rough endoplasmic reticulum. Let me make that a little bit-- let me draw that a little bit better. So let's say that this right over here, that right over here is the membrane of the endoplasmic reticulum. And then as a protein, or as a mRNA is being translated into protein, the ribosome can attach. And let's say that this right over here is the mRNA that is being translated. Let's say it's going in that direction right over here. Here is the membrane of the ER. So ER membrane. This right over here-- and actually, the way I've drawn it right over here, this is just one bilipid layer. So let me just draw it like this. I could do it like this. And this is actually, this bilipid layer is continuous."
    },
    {
        "Q": "\nAt 2:51, wouldn't the magnetic field be flowing out of the page instead of into on the right side of wire 2? Sal has it going into the page, but with my right thumb down my fingers go out at that position?",
        "A": "The field he has drawn there is the field from wire 1, that is, the field that produces a magnetic force on wire 2. There will also be a (larger) contribution to the total magnetic field at that point from wire 2 pointing into the page, but since the field from wire 2 is symmetric around wire 2, it will produce no net force on wire 2.",
        "video_name": "4tctB1wZNiI",
        "timestamps": [
            171
        ],
        "3min_transcript": "And of course, it's going into the page, into the video screen, all the way out to infinity. It gets weaker and weaker. It's inversely proportional to the radius away from the wire, But even here, this magnetic field is going into the page. Now we know, just as a little bit of review, the force created by current 1 on current 2-- that's just the convention I'm using, you wouldn't always put the 1 first-- is equal to current 2 times some length-- let's call that length 2-- along the wire. This is going to be a vector because it's a magnitude of length and a direction. And it goes in the same direction as the current. So let's say that that is L2. So we're talking about from here to here. That's the current. Cross product that with the magnetic field. I'll switch back to that. Now it all seems pretty complicated, but you can just take your right hand rule and figure out the direction. So we put our index finger-- I'm doing it right now, you can't see it-- you put your index finger in the direction of L2. You can write the 2 down here, instead of writing a big 2 up there. Put your index finger in the direction of L2. I keep redoing it just to make sure I'm drawing it right. Put your middle finger in the direction of-- so this is L2. This is this. Goes in the direction of the middle finger. Sorry, the index finger. Your middle finger is going to go in the direction of the field. So it's going to be pointing downwards, because the field is going into the page, on this side of this wire. And then your other hands are going to do what they will. And then your thumb is going to go in the direction of the net force. So your thumb is going to go like that. So there you have it. You have your little veins or tendons, whatever those are, that's your nail. So in this situation, when the current is going in opposite direction, the net force is actually going to be outward on this wire. The net force is outward. And then if you don't believe me, you might want to try it yourself, but the force on current 1 or on wire 1, or some length of wire 1, caused by the magnetic field due to current 2, is also going to be outwards. So here, if you want to think about it little bit, or have a little bit of intuition, if the current's going in the same direction they will attract, and if currents are going in opposite directions they will repel each other. So anyway, let's apply some numbers. Let's apply some numbers to this problem. Let's do it with the opposite current direction. So let's say that current 1-- I'm just going to make up some numbers-- is 2 amperes. Current 2 is, I don't know, 3 amperes."
    },
    {
        "Q": "\nAt 1:18, why is the magnetic field of the right wire going into the page? When I use the right hand rule, I flip my thumb upside down which makes my fingers go clockwise, as opposed to counter-clockwise like the left wire.",
        "A": "Yup, that s wrong :).",
        "video_name": "4tctB1wZNiI",
        "timestamps": [
            78
        ],
        "3min_transcript": "In the last video, we saw that if we have two currents, or two wires carrying current, and the current is going in the same direction, that they'll attract each other. Now what would happen-- before we break into the numbers-- what would happen if the two currents are going in opposite directions? Would they attract or repel each other? And you can probably guess that, but let's go through the exercise. Because I realize that last time I did it, I got And I'll do it a little bit cleaner this-- I don't have to draw as many magnetic field lines. So let's say that's wire 1. That's wire 2. And I'll just make the currents go in opposite So this is I1. And this is I2. So what would the magnetic field created by current 1 look like? Well, let's do the wrap around rule. Put our thumb in the direction of the current, and then the magnetic field will wrap around. It'll go into the page here and it'll go out of the page here. If you put your thumb up like that. Your right hand, always use your right hand. And of course, it's going into the page, into the video screen, all the way out to infinity. It gets weaker and weaker. It's inversely proportional to the radius away from the wire, But even here, this magnetic field is going into the page. Now we know, just as a little bit of review, the force created by current 1 on current 2-- that's just the convention I'm using, you wouldn't always put the 1 first-- is equal to current 2 times some length-- let's call that length 2-- along the wire. This is going to be a vector because it's a magnitude of length and a direction. And it goes in the same direction as the current. So let's say that that is L2. So we're talking about from here to here. That's the current. Cross product that with the magnetic field. I'll switch back to that. Now it all seems pretty complicated, but you can just take your right hand rule and figure out the direction. So we put our index finger-- I'm doing it right now, you can't see it-- you put your index finger in the direction of L2. You can write the 2 down here, instead of writing a big 2 up there. Put your index finger in the direction of L2. I keep redoing it just to make sure I'm drawing it right. Put your middle finger in the direction of-- so this is L2. This is this. Goes in the direction of the middle finger. Sorry, the index finger. Your middle finger is going to go in the direction of the field. So it's going to be pointing downwards, because the field is going into the page, on this side of this wire. And then your other hands are going to do what they will. And then your thumb is going to go in the direction of the net force. So your thumb is going to go like that. So there you have it."
    },
    {
        "Q": "Why did you multiply by area 4:18?\n",
        "A": "He multiplied for 1 teorically, [1 = a/a = b/b =A/A] so you can multiply and divide by the area whenever you want, he did that so that it lets him take [F/A = P]",
        "video_name": "uqyLOuAzbvo",
        "timestamps": [
            258
        ],
        "3min_transcript": "that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i--"
    },
    {
        "Q": "\nAt 6:33 when Sal is substituting for the potential and kinetic energies for the energy equation he substitutes with mass times gravity times height and mass times velocity squared divided by 2 respectively. In subsequent videos he uses rho instead of mass. I was wondering which version is correct.",
        "A": "rho is density or mass per unit volume in this case",
        "video_name": "uqyLOuAzbvo",
        "timestamps": [
            393
        ],
        "3min_transcript": "We know that density is just mass per volume, or that volume times density is equal to mass, or we know that volume is equal to mass divided by density. The work that I'm putting into the system-- I know I'm doing a lot of crazy things, but it'll make sense so far-- is equal to the input pressure times the amount of volume of fluid that moved over that period of time. That volume of fluid is equal to the mass of the fluid that went in at that period of time, and we'll call that the input mass, divided by the density. Hopefully, that makes a little bit of sense. As we know, the input volume is going to be equal to the doesn't change-- is equal to the output mass, so we don't have to write an input and output for the mass. The mass is going to be constant; in any given amount of time, the mass that enters the system will be equivalent to the mass that exits the system. There we go: we have an expression, an interesting expression, for the work being put into the system. What is the potential energy of the system on the left-hand side? The potential energy of the system is going to be equal to that same mass of fluid that I talked about times gravity times this input height-- the initial height-- times h1. The initial kinetic energy of the fluid equals the mass of the fluid-- this mass right here, of that same cylinder volume that I keep pointing to-- times the velocity of the We remember this from kinetic energy divided by 2. So what's the total energy at this point in the system over this period of time? How much energy has gone into the system? It's going to be the work done, which is the input pressure-- I'm running out of space, so let me erase all of this. I'll probably have to run out of time, too, but that's OK-- it's better than being confused. Back to what we were doing. So, the total energy going into the system is the work being done into the system, and I rewrote it in this"
    },
    {
        "Q": "At \"3:38\" SAL says that Vi*t= D in the formula work=fd. what does distance have to do with time?\n",
        "A": "distance is the amount something travels in a given time. So, if distance is unknown, you would need time to solve for it- requiring either solving mathematically or measuring. If it needs to be solved mathematically you would need the time to figure out how far something traveled in a certain amount of time. Additionally, Force= mass * acceleration- where again time could be important to know as acceleration is the change in velocity over a given time.",
        "video_name": "uqyLOuAzbvo",
        "timestamps": [
            218
        ],
        "3min_transcript": "that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i--"
    },
    {
        "Q": "\nAt 1:50, why is there 1% of carbon dioxide in the alveoli? I thought we breathe all the carbon dioxide out.",
        "A": "because we are not perfect human beings and 1% will remain",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            110
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 14:12, how can red blood cells have NO nucleus or DNA. How do blood tests work if they don't?",
        "A": "When they blood type they are looking for the presence of cell surface antigens. A antigens, B antigens, both of them together (AB), or no antigens at all (O type).",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            852
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:02 you said that Artery's go away from the heart but don't they come from the heart?",
        "A": "Going away from the heart and coming from the heart are the same thing. Going away from the heart and coming to the heart are opposites. Arteries carry blood away from the heart This can also be said as blood comes from the heart through arteries.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            182
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 9:08, Sal says that the N2 does diffuse in the blood, so what do we do with it? And can we make amino acids out of nitrogen?",
        "A": "simply no. Nitrogen is removed as a waste product. Nitrogen is either expelled out of our body when we breathe, or removed from our waste products.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            548
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 12:35, Sal stated that from the outside of the body, veins appear a blue-green color. I looked more closely and saw that little purple colored vessels. Is this a capalary or an artery and if a capalary then what is contained in side of it?",
        "A": "No, the technical term for them is veins. Veins are the vessels that carry blood towards the lungs so this blood is not oxygenated which makes it blue. When oxygen binds to the iron in the blood, the iron rusts so the blood is red. When the iron is not bonded to oxygen, then it is gray and the other colors in the cell are more prominent making it blue.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            755
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 15:16 you said \"They get rid of their nucleus.\" Does the dent in the center have anything to do with that, or is that just a structure created by evolution?",
        "A": "The dent increases surface area allowing for more oxygen to be absorbed, yes, you could say it s evolution that imposed its structural development.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            916
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At around 3:00, when he's talking about arteries, he said earlier in the video that it was capillaries. Are arteries and capillaries the same thing?\n",
        "A": "arteries are blood vessels that take blood aay from the heart and lungs. Capillaries are tiny, very tiny blood vessels that help with diffusion etc.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            180
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 1:40 he says there is less than 1 percent of Carbon dioxide in the atmosphere but their are trillions of humans breathing it every second how is that possible?",
        "A": "Photosynthetic organisms (plants, bacteria etc.) take CO2 and transform it to O2 in the process of photosynthesis. That s why there is more O2 than CO2 in the atmosphere. Interestingly, before photosynthetic organisms appeared there was less O2 in the atmosphere. About 300 million years ago, the amount of Oxygen peaked at about 30%. Also, (as of 2012.) there are about 7 billion humans on Earth not trillions.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            100
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 2:06 sal says pulmonary capillaries.\nwhat are pulmonary capillaries?",
        "A": "Pulmonary capillaries surround the alveoli and are where gas exchange takes place between the pulmonary arteries and pulmonary veins.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            126
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:56, why is this membrane so thin, and why is it there at all?",
        "A": "The membrane is thin because you need to get O2 and CO2 across it. The thicker a membrane is, the harder it would be for small molecules like this to diffuse through. And it is there in the first place to keep everything else in the right place. Without a membrane there, your blood would pour into the alveoli and you would cough it up. Not very good for oxygen delivery to your tissues! :)",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            236
        ],
        "3min_transcript": ""
    },
    {
        "Q": "5:16 to 5:30 does that mean time/2 * acceleration of gravity = initial velocity. I want to now if in this equation does it have to be acceleration of gravity or can it be just acceration.\n",
        "A": "I think it can just be acceleration.",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            316,
            330
        ],
        "3min_transcript": "We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time, So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand"
    },
    {
        "Q": "Why at around 7:10 Sal divides by 2?\n",
        "A": "He finding the average velocity. To get the average of 2 number you add them and devide by 2. Since one of the numbers is 0 you end up dividing the other by two. For example the average of 1 and 3 is (1 + 3)/2 = 4/2 = 2 but the average of 0 and 2 is (0 + 2)/2 = 2/2 = 1",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            430
        ],
        "3min_transcript": "So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here And we just gonna divide that by 2 This will give us the average velocity And we wanna multiply that by 2.5 s, times 2.5 s So we get this part right over here 24.5 divided by 2 When you go with the 0, it is still 24.5 It gives us 12.25 times 2.5, and remember this right over here is in seconds let me write the units down So this is 12.25 m/s times 2.5 seconds And just to remind ourselves We are calculating the displacement over the first 2.5 seconds So this gives us, I get the calculator out once again We have 12.25 times 2.5 seconds gives us 30.625"
    },
    {
        "Q": "\n@8:00 Sal is talking about dividing by 2 then multiplying by 2.5, why cant you multiply by 1.5 instead of complicating it? I did the math and it doesn't add up the same. Why is that?",
        "A": "I am not sure how you got the factor of 1.5. Sal is talking about taking the average of the initial velocity which is 24.5 m/s and the final velocity in the upward trajectory of the ball, which is 0 m/s. In constant acceleration motion, to find the average velocity, you just take the average of final and initial velocities: (u+v)/2. Displacement \u00ce\u0094S = average velocity * \u00ce\u0094t. = {(u+v)/2} * \u00ce\u0094t",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            480
        ],
        "3min_transcript": "We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here And we just gonna divide that by 2 This will give us the average velocity And we wanna multiply that by 2.5 s, times 2.5 s So we get this part right over here 24.5 divided by 2 When you go with the 0, it is still 24.5 It gives us 12.25 times 2.5, and remember this right over here is in seconds let me write the units down So this is 12.25 m/s times 2.5 seconds And just to remind ourselves We are calculating the displacement over the first 2.5 seconds So this gives us, I get the calculator out once again We have 12.25 times 2.5 seconds gives us 30.625 The seconds cancel out This is actually a ton, you know, roughly give or take about 90 feet throw into the air, this looks like a nine stories building And I frankly do not have the arm for that But if someone is able to throw the ball for 5 seconds in the air They have thrown 30 meters in the air Hopefully you will find that entertaining In my next video I'll generalize this maybe we can get a little bit of formula so maybe you can generalize it So regardless of the measurement of time you can get the displacement in the air Or even better, try to derive it yourself And you will see how, at least how I tackle it in the next video"
    },
    {
        "Q": "At 3:24, Sal showed that -9.8 was the velocity. I am confused about how he got that number. Was it from the previous video? Because i remember seeing it there.\n",
        "A": "-9.8 is the acceleration not the velocity maybe if you ask again i can clarify more",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            204
        ],
        "3min_transcript": "We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time,"
    },
    {
        "Q": "\nHi! At 1:13, Sal says that the time it takes the object to go up is the same as the time it takes for the object to come down. I understand that we are not taking air resistance into account, but wouldn't gravity make the ball come down faster? The ball is working against gravity on it's way up and falling with gravity on the way down so it seems like the ball's decent would be faster than its ascent. Thanks for the help!",
        "A": "The acceleration due to gravity is the same throughout the flight. Acceleration is rate of change in velocity. Won t it take exactly the same amount of time to go from 20 m/s to 0 as it will take to go from 0 to -20 m/s?",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            73
        ],
        "3min_transcript": "Let's say you and I are playing a game or I'm trying to figure out how high a ball is being thrown in the air How fast would we throwing that ball in the air? And what we do is one of us has a ball and the other one has a stop watch over here So this is my best attempt to make it more like a cat than a stop watch but I think you get the idea And what we do is one of us throw the ball the other one times how long the ball is in the air And what we do is gonna use that time in the air to figure out how fast the ball was thrown straight up and how long it was in the air or how high it got And there is going to be one assumption I make here frankly that's an assumption we are gonna make in all of these projectile motion type problem is that air resistance is negligible And for something like it, this is a baseball or something like that That's a pretty good approximation So when can I get the exact answer, I encourage you experiment it on your own or even to see what air resistance does to your calculations We gonna assume for this projectile motion in future one We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds"
    },
    {
        "Q": "how does he know (19:34) that there is more heat going in the systm than heat going out?\n",
        "A": "Since work=heat, and from step C to A (that the system is receiving heat) the work is less than from step A to C. And this is because when the temperature of a system is at lower state you need less force to do work to the system.",
        "video_name": "aAfBSJObd6Y",
        "timestamps": [
            1174
        ],
        "3min_transcript": "And in the process, it was also doing some work. And what was the work that it did? Well, it's the area under this curve, or the area inside of this cycle. So this is the work done by our Carnot engine. And the way you think about it is, when you're going in the rightward direction with increasing volume, it's the area under the curve is the work done by the system. And then when you move in the leftward direction with decreasing volume, you subtract out the work done to the system, and then you're left with just the area in the curve. So we can write this Carnot engine like this. It's taking, it's starting-- so you have a reservoir at T1. And then you have your engine, right here. And then it's connected-- so this takes Q1 in from this reservoir. The work is represented by the amount of-- the work right here is the area inside of our cycle. And then it transfers Q2, or essentially the remainder from Q1, into our cold reservoir. So T2. So it transfers Q2 there. So the work we did is really the difference between Q1 and Q2, right? You say, hey. If I have more heat coming in than I'm letting out, where did the rest of that heat go? It went to work. Literally. So Q1 minus Q2 is equal to the amount of work we did. And actually, this is a good time to emphasize again that heat and work are not a state variable. A state variable has to be the exact same value when we complete a cycle. Now, we see here that we completed a cycle, and we had a net amount of work done, or a net amount of heat added to the system. So we could just keep going around the cycle, and keep having heat added to the system. So there is no inherent heat state variable right here. this point in time. All you could say is what amount of heat was added or taken away from the system, or you can only say the amount of work that was done to, or done by, the system. Anyway, I want to leave you there right now. We're going to study this a lot more. But the real important thing is, and if you never want to get confused in a thermodynamics class, I encourage you to even go off on your own, and do this yourself. Kind of-- you can almost take a pencil and paper, and redo this video that I just did. Because it's essential that you understand the Carnot engine, understand this adiabatic process, understand what isotherms are. Because if you understand that, then a lot of what we're about to do in the next few videos with regard to entropy will be a little bit more intuitive, and not too confusing."
    },
    {
        "Q": "At 4:30 you said you want the triple bond to have the lowest # of Carbon placement possible for its naming. But what if there is a double bond within the structure too, and they happen to tie for this \"lowest # of Carbon placement possible.\" Which bond would get priority for naming?\n",
        "A": "To answer your question, alkenes have priority over alkynes. Thus the molecule C-C\u00e2\u0089\u00a1C-C=C-C would be named hex-2-en-4-yne by IUPAC.",
        "video_name": "qZTeyhR1akA",
        "timestamps": [
            270
        ],
        "3min_transcript": "So when we talked about two carbons for nomenclature, our root was eth-, and since we're dealing with alkynes, our ending is going to be -yne here. So I stick the root together with the ending, so I take eth- and I add the -yne to it, so it's called ethyne. So ethyne would be the IUPAC name for this molecule. That's probably not the name that's used most frequently. This is the most famous alkyne, also called acetylene. So you've heard of acetylene torches before, and you can do a very cool demonstration called underwater fireworks where you use acetylene in there. So this is acetylene, which is the common name or ethyne, which is more the IUPAC name, and this is the simplest alkyne. So let's name another one here. Let's get an actual chain, a carbon chain in here, and let's see how to name this one like that. Well, the first thing to notice is the straight line And remember, that's because that portion of the molecule is linear, because those carbons are sp hybridized. So when you're drawing Lewis Dot Structures, you'll notice that almost everyone will draw that portion of the molecule straight to better reflect the actual geometry of the molecule. All right, so I need to figure out how many carbons are in this alkyne. So I find my longest carbon chain including my triple bond. So let's see, here is one carbon, two carbons, three carbons, four, and five. So a five carbon alkyne. So remember your root for five carbons would be pent-. So it would be pentane if it was an alkane, but since it's an alkyne, it would be pentyne. So this is pentyne. So I have that much so far. Next, I need to figure out how to number my chain. So I could number it from the left, My goal is to give my triple bond the lowest number possible, so it's a lot like double bonds here. So I want to give my triple bond the lowest number possible, and that means, of course, I need to start from the right here. So if I start from the right, and I say this is carbon one, then my triple bond starts at carbon two. And then this would be carbon three, carbon four, and carbon five, like that. So the triple bond starts at carbon two, so all I have to do is put a two in here and say 2-pentyne. So in that respect, it's just like naming alkenes here. So that's 2-pentyne. Let's do another one, one that's a little bit more complicated than that. So let's look at this molecule here. So there's my triple bond, and then I have-- let's see, I'll put in a chain and some methyl groups there-- and so this is my molecule. All right, so once again find the longest carbon chain that includes your triple bond."
    },
    {
        "Q": "At 2:24 sal says halogens are group 7 elements but they are group 17\n",
        "A": "In the new system halogens occupy group 17... Older system puts halogens into SevenA group. Also keep in mind that halogens have seven valence electrons. he may have been referring to that...",
        "video_name": "__zy-oOLPug",
        "timestamps": [
            144
        ],
        "3min_transcript": "- [Voiceover] When you're studying chemistry you'll often see reactions, in fact you always see reactions. For example if you have hydrogen gas it's a diatomic molecule, 'cause hydrogen bonds with itself in the gassy state, plus iodine gas, I2, that's also in the gassy state, it's very easy to just sort of, oh you know, if you put 'em together they're going to react and form the product, if you have two moles of, hydrogen, two moles of iodine, so it's gonna form two moles of hydrogen iodide. That's all nice and neat and it makes it seem like it's a very clean thing that happens without much fuss. But we know that that isn't the reality and we also know that this doesn't happen just instantly, it's not like you can just take some hydrogen, put it with some iodine, and it just magically turns into hydrogen iodide. That there's some process going on, that these gaseous state particles are bouncing around, and somehow they must bounce into each other and break bonds that they were in before, and that's what we're going to study now. This whole study of how the reaction progresses, and the rates of the reactions is called kinetics. Which is a very fancy word, but you're probably familiar with it because we've talked a lot about kinetic energy. Kinetics. Which is just the study of the rate of reactions. How fast do they happen, and how do they happen? So let's just in our minds, come up with a intuitive way that hydrogen and iodine can combine. So let's think about what hydrogen looks like. So if we get our periodic table out, hydrogen's got one valence electron so if they have two hydrogen atoms they can share them with each other. And then iodine, iodine has seven valence electrons, so if they each share one they get complete as well. So let's just review that right now. So hydrogen this hydrogen might have one, well, will have one electron out there. And then you can have another hydrogen that has another electron out there, this hydrogen can pretend like he has this electron, this hydrogen can pretend like she has that electron, and then they're happy. They both feel like they've completed their 1S shell. Same thing on the iodine side. Where you have two iodines, they both have seven valence electrons. They're halogens, you know that already. Halogens are the group seven elements, so they have seven electrons this guy's got one here, this guy's got one here, if this guy can pretend like he's got that electron, he's happy, he has eight valence electrons. If this guy can pretend like he's got that one, same thing. So there's a bond right here, and this is why hydrogen is a diatomic molecular gas, and this is why iodine is the same. Now, when they're in the gaseous state, you have a bunch of these things that are moving around bumping into each other, I'll do it like this. So the hydrogen might look something like this, the hydrogen is these two atomic spheres that are bonded together,"
    },
    {
        "Q": "\nAt 0:44, Why do sal put O2 instead of just O?",
        "A": "Because oxygen doesn t hang around as single atoms on the surface of earth, it comes as a diatomic gas with 2 oxygen atoms bonded together, so O2",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            44
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 10:14, why is he converting moles to moles?",
        "A": "You can also think about it this way grams A -> moles A -> moles B -> grams B",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            614
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 12:00 why did Sal multiply 0.833 with 32? Where did the 32 come from?",
        "A": "For every mol of O2 there are 32g O2, so you can multiply something by 32g O2/1mol O2. It s like multiplying by 1/1. He did that so that you can cancel out the mol O2 on top and bottom and solve for g of O2.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            720
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 0:25 why does glucose react with oxygen?\n",
        "A": "It s the equation for respiration and is how the body acquires energy.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            25
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n@10:50, is it 0.833 of O2 or O? Does it make a difference?",
        "A": "Its moles of O2. This makes a difference in calculating grams as there are 2 oxygen atoms per mole. Hope this helps :)",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            650
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 10:11, so the coefficient symbolizes the amount of moles?",
        "A": "Mostly, but not exact. The coefficients represent the ratios of the numbers of molecules of the chemicals involved in the reaction. As such, something like 2H\u00e2\u0082\u0082 (g) + O\u00e2\u0082\u0082(g) \u00e2\u0086\u0092 2H\u00e2\u0082\u0082O (g) The coefficients could mean 2 molecules of H\u00e2\u0082\u0082 and 1 molecule of O\u00e2\u0082\u0082 reacting. Or it could be 2 moles of H\u00e2\u0082\u0082 and 1 molecule of O\u00e2\u0082\u0082 reacting. Or it could be 0.0150 moles of H\u00e2\u0082\u0082 and 0.0075 moles of O\u00e2\u0082\u0082 reacting. So, it doesn t actually specify the quantity you have, it is just the ratio of the numbers of molecules,",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            611
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n4:35 oxygen is spelled incorrectly",
        "A": "Yes, Sal has missed the y in the spelling of oxygen.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            275
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 12:51, you used every one mole of glucose we produce 6 mole of carbon dioxide\" and found the moles of co2 first, does it make any difference if i take the moles of glucose and find out the number of h2o moles first, or i cant do that because there is no carbon in h2o so i have to find the moles of carbons first?\n",
        "A": "He didn t even have to write the balanced equation. He knew that i mol glucose contains 6 mol f C, and 6 mol of C must produce 6 mol of CO\u00e2\u0082\u0082. You can start anywhere you want when balancing an equation, but usually the procedure that works best is to start with the most complicated formula (glucose); balance all atoms other than O and H (i.e., balance C); balance O; balance H last.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            771
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 0:56, is there a way to check your stoichiometry problem to make sure its right?\n",
        "A": "You can google the reaction after trying it out for yourself to see if you are right :) read the language in the question carefully to work out what is a reactant (left side) and what is a product (right side). Most gases like Oxygen and Hydrogen will exist as O2 and H2, not just a single atom O Fun fact! This reaction is known as respiration.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            56
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 1:42 sal uses the term notional charge\nwhat does it mean?\n",
        "A": "Physicists do not use the term notional charge. It is probably because Sal has a finance background and is bringing terminology from that world into play. In this case, he is using the term notional charge to imply a test charge.",
        "video_name": "0YOGrTNgGhE",
        "timestamps": [
            102
        ],
        "3min_transcript": "Let's imagine that instead of having two charges, we just have one charge by itself, sitting in a vacuum, sitting in space. So that's this charge here, and let's say its charge is Q. That's some number, whatever it is. That's it's charge. And I want to know, if I were to place another charge close to this Q, within its sphere of influence, what's going to happen to that other charge? What's going to be the net impact on it? And we know if this has some charge, if we put another charge here, if this is 1 coulomb and we put another charge here that's 1 coulomb, that they're both positive, they're going to repel each other, so there will be some force that pushes the next charge away. If it's a negative charge and I put it here, it'll be even a stronger force that pulls it in because it'll be closer. So in general, there's this notion of what we can call an electric field around this charge. And what's an electric field? We can debate whether it really exists, but what it affecting the space around it in some way that whenever I put-- it's creating a field that whenever I put another charge in that field, I can predict how the field will affect that charge. So let's put it in a little more quantitative term so I stop confusing you. So Coulomb's Law told us that the force between two charges is going to be equal to Coulomb's constant times-- and in this case, the first charge is big Q. And let's say that the second notional charge that I eventually put in this field is small q, and then you divide by the distance between them. Sometimes it's called r because you can kind of view the distance as the radial distance between the two charges. So sometimes it says r squared, but it's the distance between them. So what we want to do if we want to calculate the field, we want to figure out how much force is there placed per charge at any point around this Q, so, say, at a given At this distance, we want to know, for a given Q, what is the force going to be? So what we can do is we could take this equation up here and divide both sides by this small 1, and say, OK, the force-- and I will arbitrarily switch colors. The force per charge at this point-- let's call that d1-- is equal to Coulomb's constant times the charge of the particle that's creating the field divided by-- well, in this case, it's d1-- d1 squared, right? Or we could say, in general-- and this is the definition of the electric field, right? Well, this is the electric field at the point d1, and if we wanted a more general definition of the electric field, we'll just make this a general variable, so instead of having a particular distance, we'll define the field for all distances away from the point Q."
    },
    {
        "Q": "\nat 8:22 , he said positive test charge . what is a test charge ?",
        "A": "A test charge is a charge that we imagine is so small that it does not affect the field we are trying to test.",
        "video_name": "0YOGrTNgGhE",
        "timestamps": [
            502
        ],
        "3min_transcript": "that circle around it, this is generating a field that if I were to put-- let's say I were to place a 1 coulomb charge here, the force exerted on that 1 coulomb charge is going to be equal to 1 coulomb times the electric fields, times 2,500 newtons per coulomb. So the coulombs cancel out, and you'll have 2,500 newtons, which is a lot, and that's because 1 coulomb is a very, very large charge. And then a question you should ask yourself: If this is 1 times 10 to the negative 6 coulombs and this is 1 coulomb, in which direction will the force be? Well, they're both positive, so the force is going to be outwards, right? So let's take this notion and see if we can somehow draw an electric field around a particle, just to get an intuition of what happens when we later put a charge anywhere So there's a couple of ways to visualize an electric field. One way to visualize it is if I have a-- let's say I have a point charge here Q. What would be the path of a positive charge if I placed it someplace on this Q? Well, if I put a positive charge here and this Q is positive, that positive charge is just going to accelerate outward, right? It's just going to go straight out, but it's going to accelerate at an ever-slowing rate, right? Because here, when you're really close, the outward force is very strong, and then as you get further and further away, the electrostatic force from this charge becomes weaker and weaker, or you could say the field becomes weaker and weaker. But that's the path of a-- it'll just be radially outward-- of a positive test charge. And then if I put it here, well, it would be radially outward that way. It wouldn't curve the way I drew it. It would be a straight line. If I did it here, it would be like that, but then I can't draw the arrows. If I was here, it would out like that. I think you get the picture. At any point, a positive test charge would just go straight out away from our charge Q. And to some degree, one measure of-- and these are called electric field lines. And one measure of how strong the field is, is if you actually took a unit area and you saw how dense So here, they're relatively sparse, while if I did that same area up here-- I know it's not that obvious. I'm getting more field lines in. But actually, that's not a good way to view it because I'm covering so much area. Let me undo both of them. You can imagine if I had a lot more lines, if I did this area, for example, in that area, I'm capturing two of these field lines. Well, if I did that exact same area out here, I'm only"
    },
    {
        "Q": "What is static electric force at 12:10?\n",
        "A": "static electric force is the force acting between two charges Static charge is charge that is not moving (unlike current)",
        "video_name": "0YOGrTNgGhE",
        "timestamps": [
            730
        ],
        "3min_transcript": "shorter and shorter the electric field vectors get. And so, in general, there's all sorts of things you can draw the electric fields for. Let's say that this is a positive charge and that this is a negative charge. Let me switch colors so I don't have to erase things. If I have to draw the path of a positive test charge, it would go out radially from this charge, right? But then as it goes out, it'll start being attracted to this one the closer it gets to the negative, and then it'll curve in to the negative charge and these arrows go like this. And if I went from here, the positive one will be repelled really strong, really strong, it'll accelerate fast and it's rate of acceleration will slow down, but then as it gets closer to the negative one, it'll speed up again, and then that would be its path. Similarly, if there was a positive test charge here, its path would be like that, right? If it was here, its path would be like that. If it was there, maybe its path is like that, and at some point, its path might never get to that-- this out here might just go straight out that way. That one would just go straight out, and here, the field lines would just come in, right? A positive test charge would just be naturally attracted to that negative charge. So that's, in general, what electric field lines show, and we could use our little area method and see that over here, if we picked a given area, the electric field is much weaker than if we picked that same area right here. We're getting more field lines in than we do right there. So that hopefully gives you a little sense for what an electric field is. It's really just a way of visualizing what the impact would be on a test charge if you bring it close to another charge. And hopefully, you know a little bit about Coulomb's constant. And let's just do a very simple-- I'm getting this out of the AP Physics book, but they say-- let's do a little simple problem: Calculate the static electric force between So 6 times-- oh, no, that's not on an electric field. Oh, here it says: What is the force acting on an electron placed in an external electric field where the electric field is-- they're saying it is 100 newtons per coulomb at that point, wherever the electron is. So the force on that, the force in general, is just going to be the charge times the electric field, and they say it's an electron, so what's the Well, we know it's negative, and then in the first video, we learned that its charge is 1.6 times 10 to the negative nineteenth coulombs times 100 newtons per coulomb. The coulombs cancel out. And this is 10 squared, right? This is 10 to the positive 2, so it'll be 10 to the minus 19"
    },
    {
        "Q": "\nWhat is silican? As said in 3:23",
        "A": "Silicon is an element. It has the symbol Si.",
        "video_name": "T2DaaGuKOTo",
        "timestamps": [
            203
        ],
        "3min_transcript": "(lively intro music) I'm a collection of organic compounds called Hank Green. An organic compound is more or less any chemical that contains carbon, and carbon is awesome. Why? Lots of reasons. I'm gonna give you three. First, carbon is small. It doesn't have that many protons and neutrons. Almost always 12, rarely it has some extra neutrons making it C-13 or C-14. Because of that, carbon does not take up a lot of space and can form itself into elegant shapes. It can form rings. It can form double or even triple bonds. It can form spirals and sheets and all kinds of really awesome things that bigger molecules would never manage to do. Basically, carbon is like an olympic gymnast. It can only do the remarkable and beautiful things it can do because it's petite. Second, carbon is kind. It's not like other elements that desperately want to gain or lose or share electrons No, carbon knows what it's like to be lonely, so it's not all, \"I can't live without your electrons.\" Needy, like chlorine or sodium is. This is why chlorine tears apart your insides if you breathe it in gaseous form, and why sodium metal, if ingested, will explode. Carbon, though, eh. It wants more electrons, but it's not going to kill for them. It's easy to work with. It makes and breaks bonds like a 13-year-old mall rat, but it doesn't ever really hold a grudge. Third, carbon loves to bond because it needs 4 extra electrons, so it will bond with whoever happens to be nearby. Usually, it will bond with 2 or 3 or 4 of them at the same time. Carbon can bond with lots of different elements. Hydrogen, oxygen, phosphorus, nitrogen, and other atoms of carbon. It can do this in infinite configurations, allowing it to be the core element of the complicated structures that make living things like ourselves. Because carbon is small, kind, and loves to bond, life is pretty much built around it. Carbon is the foundation of biology. even conceiving of life that is not carbon-based. Silicon, which is analogous to carbon in many ways, is often cited as a potential element for alien life to be based on, but it's bulkier, so it doesn't form the same elegant shapes as carbon. It's also not found in any gases, meaning that life would have to be formed by eating solid silicon, whereas life here on earth is only possible because carbon is constantly floating around in the air in the form of carbon dioxide. Carbon, on its own, is an atom with 6 protons, 6 electrons, and 6 neutrons. Atoms have electron shells, and they need or want to have these shells filled, in order to be happy, fulfilled atoms. The first electron shell called the S-orbital needs 2 electrons to be full. Then there's the 2nd S-orbital, which also needs 2, carbon has this filled as well. Then we have the first P-orbital, which needs 6 to be full. Carbon only has 2 left over, so it wants 4 more. Carbon forms a lot of bonds that we call \"covalent\". These are bonds where the atoms actually share electrons,"
    },
    {
        "Q": "\nAt 8:35, he says that since water has positive and negatively charged sides, the molecules will stick together, hydrogen side to oxygen side, and I am wondering, does that explain the hexagonal patterns that are formed at a molecular level when water freezes?",
        "A": "That is exactly why water freezes the way it does.",
        "video_name": "T2DaaGuKOTo",
        "timestamps": [
            515
        ],
        "3min_transcript": "around the oxygen than around the hydrogens. This creates a slight positive charge around the hydrogens and a slight negative charge around the oxygen. When something has a charge, we say that it's polar. It has a positive and negative pole. This is a polar covalent bond. Ionic bonds occur when instead of sharing electrons, atoms just donate or accept an electron from another atom completely and then live happily as a charged atom or ion. Atoms would, in general, prefer to be neutral but compared with having the full electron shells is not that big of a deal. The most common ionic compound in our daily lives? that would be good old table salt, NaCl, sodium chloride, but don't be fooled by its deliciousness. Sodium chloride, as I previously mentioned, is made of 2 very nasty elements. Chlorine is a halogen, or an element that only needs one proton to fill its octet, while sodium is an alkali metal, an element that only has one electron in its octet. They will happily tear apart any chemical compound searching to satisfy the octet rule. No better outcome could occur than sodium meeting chlorine. They immediately transfer electrons so sodium doesn't have its extra and the chlorine fills its octet. They become Na+ and Cl-, and are so charged that they stick together, and that stickiness is what we call an ionic bond. These chemical changes are a big deal, remember? Sodium and chlorine just went from being deadly to being delicious. They're also hydrogen bonds, which aren't really bonds, so much. So, you remember water? I hope you didn't forget about water. Water is important. Since water is stuck together with a polar covalent bond, the hydrogen bit of it is a little bit positively-charged and the oxygen is a little negatively-charged. When water molecules move around, they actually stick together a little bit, hydrogen side to oxygen side. This kind of bonding happens in all sorts of molecules, particularly in proteins. It plays an extremely important role in how proteins fold up to do their jobs. bonds, even when they're written with dashes or solid lines, Sometimes ionic bonds are stronger than covalent bonds, though that's the exception rather than the rule, and covalent bond strength varies hugely. The way that those bonds get made and broken is intensely important to how life and our lives operate. Making and breaking bonds is the key to life itself. It's also like if you were to swallow some sodium metal, the key to death. Keep all of this in mind as you move forward in biology. Even the hottest person you have ever met is just a bunch of chemicals rambling around in a bag of water. That, among many other things, is what we're gonna talk about next time."
    },
    {
        "Q": "At 0:01 you started out with the static friction and the kinetic friction. What is the formula to be able to figure those amounts out? I need help for my physics class,.. Please help.\n",
        "A": "He gives these equations at 1:24 and 2:01. The equation for static friction is: ||Fb|| / ||Fn|| = \u00ce\u00bcs Where ||Fb|| is the budging force, or the amount of force required to make an object start moving, ||Fn|| is the normal force, and \u00ce\u00bcs is the coefficient of static friction. The equation for kinetic friction is: ||Ff|| / ||Fn|| = \u00ce\u00bck Where ||Ff|| is the force of friction, ||Fn|| is the normal force, and \u00ce\u00bck is the coefficient of kinetic friction.",
        "video_name": "ZA_D4O6l1lo",
        "timestamps": [
            1
        ],
        "3min_transcript": "So I have got this block of wood here that has a mass of 5 kilograms and it is sitting on some dirt and we are near the surface of the earth and the coefficient of static friction between this type of wood and this type of dirt is 0.60 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 This was measured by someone else long ago or you found it in some type of a book someplace And let's say we push on this side of the block with a force of a 100 N What is going to happen? So the first thing you might realize is if there is no friction if this was a completely frictionless boundary and there is no air resistance, we are assuming that there is no air resistance in this example That in this dimension, in the horizontal dimension there would only be one force here, this 100 N force It would be completely unbalanced and that would be the net force and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms Force = Mass times acceleration acceleration and force are vector quantities would give you 20 meters per second of acceleration in the rightward direction That is if there were no friction but there is friction in this situation So let's think about how we'll deal with it So the coefficient of friction tells us So this right here is the ratio between the magnitude of the force that I have called the budging force The amount of force you need to apply to get this thing to budge to get this thing to start moving. So we can start using the coefficient of kinetic friction It's the ratio between that and the magnitude of the force of contact between this block and the floor or ground here And the magnitude of that force of contact is the same thing as the normal force that the ground is applying on the block the magnitude of the normal force the ground is applying on the block Then once its moving then we can say that this is going to be--this will then be equal to this over here will be equal to the force of friction and this over here will be equal to the force of friction The magnitude of the force of friction over the force of contact the contact force between those two, so over the normal force and it makes sense that the larger the contact force the more that these are being pressed together the little at the atomic level, they kind of really get into each others grooves the more budging force you would need or the more friction force would go against your motion And in either situation the force of friction is going against your motion So even if you push it in that way sounds like force of friction is all of a sudden going to help you So let's think about what the necessary force will we need to overcome the force of friction right here in the static situation So the force of gravity on this block is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down"
    },
    {
        "Q": "1:22 I don't understand why he says that coefficient of static friction deals with the force to move the object (budging force). Is it not rather maximum friction force that can be applied BEFORE it moves. Meaning any more force and then it will move (becoming kinetic friction)?\n",
        "A": "The coefficient of static friction tells you how much frictional force you get for a given normal force. Friction force = coeff of sf * normal force. Friction is a reactive force, which means that it only responds to applied forces. When you multiply the coefficient of static friction times the normal force, that tells you the maximum budging force that the friction will resist. Apply more force than that, and the object will start to move.",
        "video_name": "ZA_D4O6l1lo",
        "timestamps": [
            82
        ],
        "3min_transcript": "So I have got this block of wood here that has a mass of 5 kilograms and it is sitting on some dirt and we are near the surface of the earth and the coefficient of static friction between this type of wood and this type of dirt is 0.60 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 This was measured by someone else long ago or you found it in some type of a book someplace And let's say we push on this side of the block with a force of a 100 N What is going to happen? So the first thing you might realize is if there is no friction if this was a completely frictionless boundary and there is no air resistance, we are assuming that there is no air resistance in this example That in this dimension, in the horizontal dimension there would only be one force here, this 100 N force It would be completely unbalanced and that would be the net force and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms Force = Mass times acceleration acceleration and force are vector quantities would give you 20 meters per second of acceleration in the rightward direction That is if there were no friction but there is friction in this situation So let's think about how we'll deal with it So the coefficient of friction tells us So this right here is the ratio between the magnitude of the force that I have called the budging force The amount of force you need to apply to get this thing to budge to get this thing to start moving. So we can start using the coefficient of kinetic friction It's the ratio between that and the magnitude of the force of contact between this block and the floor or ground here And the magnitude of that force of contact is the same thing as the normal force that the ground is applying on the block the magnitude of the normal force the ground is applying on the block Then once its moving then we can say that this is going to be--this will then be equal to this over here will be equal to the force of friction and this over here will be equal to the force of friction The magnitude of the force of friction over the force of contact the contact force between those two, so over the normal force and it makes sense that the larger the contact force the more that these are being pressed together the little at the atomic level, they kind of really get into each others grooves the more budging force you would need or the more friction force would go against your motion And in either situation the force of friction is going against your motion So even if you push it in that way sounds like force of friction is all of a sudden going to help you So let's think about what the necessary force will we need to overcome the force of friction right here in the static situation So the force of gravity on this block is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down"
    },
    {
        "Q": "\nAt 11:40 Sal says that he should have looked up the exact frequency of hemophilia in men. He estimates that is in roughly 1/7000. Does anyone know a more exact frequency?",
        "A": "Hemophilia occurs in about one out of every 7,500 live male births. There are approximately 17,000 people in the United States who have hemophilia. Factor VIII deficiency accounts for about 80% (1 in every 5,000 male births) of the hemophilia population and Factor IX deficiency accounts for about 20% (1 in every 30,000 male births) of the hemophilia population",
        "video_name": "-ROhfKyxgCo",
        "timestamps": [
            700
        ],
        "3min_transcript": "and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has. So this man will have hemophilia. So a natural question should be arising is, hey, you know this guy-- let's just say that this is a relatively infrequent mutation that arises on an X chromosome-- the question is who's more likely to have hemophilia? A male or a female? All else equal,who's more likely to have it? Well if this is a relatively infrequent allele,a female, in order to display it,has to get two versions of it. So let's say that the frequency of it-- and I looked it up before this video-- roughly they say between 1 in 5,000 to 10,000 men exhibit hemophilia. So let's say that the allele frequency of this is 1 in 7,000, the frequency of Xh,the hemophilia version of the X chromosome. because it's completely determined whether-- there's a 1 in 7,000 chance that this X chromosome they get is the hemophilia version. Who cares what the Y chromosome they get is, cause that essentially doesn't code at all for the blood clotting factors and all of the things that drive hemophilia. Now,for a woman to get hemophilia,what has to happen? She has to have two X chromosomes with the mutation. Well the probability of each of them having the mutation is 1 in 7,000 So the probability of her having hemophilia is 1 in 7,000 times 1 in 7,000,or that's 1 in what,49 million. So as you can imagine,the incidence of hemophilia in women is much lower than the incidence of hemophilia in men. And in general for any sex-linked trait, if it's recessive, if it's a recessive sex-linked trait,which means men, if they have it,they're going to show it,"
    },
    {
        "Q": "12:25 If haemophilia is recessive, why would a man who carries only one of the alleles for haemophilia (because he is XY, and so it can only ever be on one gene) have the condition present? If he has two chromosome (the X and Y) and only one can / does carry the allele for haemophilia, and then the condition is present, wouldn't that make it dominant?\n",
        "A": "No, in order for an allele to be dominant, it has to show its phenotype even in the presence of another recessive allele of that gene. When only one gene is present an allele cannot be determined to be dominant or recessive. Haemophilia is recessive as in females (who are XX), there are indeed two copies of the gene and one haemophilia-causing allele is not enough for the disease to be present.",
        "video_name": "-ROhfKyxgCo",
        "timestamps": [
            745
        ],
        "3min_transcript": "and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has. So this man will have hemophilia. So a natural question should be arising is, hey, you know this guy-- let's just say that this is a relatively infrequent mutation that arises on an X chromosome-- the question is who's more likely to have hemophilia? A male or a female? All else equal,who's more likely to have it? Well if this is a relatively infrequent allele,a female, in order to display it,has to get two versions of it. So let's say that the frequency of it-- and I looked it up before this video-- roughly they say between 1 in 5,000 to 10,000 men exhibit hemophilia. So let's say that the allele frequency of this is 1 in 7,000, the frequency of Xh,the hemophilia version of the X chromosome. because it's completely determined whether-- there's a 1 in 7,000 chance that this X chromosome they get is the hemophilia version. Who cares what the Y chromosome they get is, cause that essentially doesn't code at all for the blood clotting factors and all of the things that drive hemophilia. Now,for a woman to get hemophilia,what has to happen? She has to have two X chromosomes with the mutation. Well the probability of each of them having the mutation is 1 in 7,000 So the probability of her having hemophilia is 1 in 7,000 times 1 in 7,000,or that's 1 in what,49 million. So as you can imagine,the incidence of hemophilia in women is much lower than the incidence of hemophilia in men. And in general for any sex-linked trait, if it's recessive, if it's a recessive sex-linked trait,which means men, if they have it,they're going to show it,"
    },
    {
        "Q": "\nIn the video ,10:28 you said a man only one hemophelia cromosome to past it to the off spring. so does that mean that most cancer comes from men?",
        "A": "Cancer is a mutation of genes, not genetically inherited from parents (unless the cancer was present in gametes).",
        "video_name": "-ROhfKyxgCo",
        "timestamps": [
            628
        ],
        "3min_transcript": "But what it tells you is it does very little other than determining what the gender is. And the way it determines that, it does have one gene on it called the SRY gene. You don't have to know that. SRY,that plays a role in the development of testes or the male sexual organ.So if you have this around, this gene right here can start coding for things that will eventually lead to the development of the testicles. And if you don't have that around,that won't happen, so you'll end up with a female. And I'm making gross oversimplifications here. But everything I've dealt with so far,OK, this clearly plays a role in determining sex. But you do have other traits on these genes. And the famous cases all deal with specific disorders. The genes,or the mutations I should say. So the mutations that cause color blindness. Red-green color blindness,which I did in green, which is maybe a little bit inappropriate. Color blindness and also hemophilia. This is an inability of your blood to clot. Actually, there's several types of hemophilia. But hemophilia is an inability for your blood to clot properly. And both of these are mutations on the X chromosome. And they're recessive mutations.So what does that mean? It means both of your X chromosomes have to have-- let's take the case for hemophilia-- both of your X chromosomes have to have the hemophilia mutation in order for you to show the phenotype of having hemophilia. and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has."
    },
    {
        "Q": "is the X chromosome always longer compared to the Y?\njust curious... 07:43\n",
        "A": "Yes it is a huge giant by comparison.",
        "video_name": "-ROhfKyxgCo",
        "timestamps": [
            463
        ],
        "3min_transcript": "I mean it's not just the case with kings. It's probably true,because most of our civilization is male dominated, that you've had these men who are obsessed with producing a male heir to kind of take over the family name. And,in the case of Henry the VIII,take over a country. And they become very disappointed and they tend to blame their wives when the wives keep producing females,but it's all their fault. Henry the VIII,I mean the most famous case was with Ann Boleyn. I'm not an expert here,but the general notion is that he became upset with her that she wasn't producing a male heir. And then he found a reason to get her essentially decapitated, even though it was all his fault. He was maybe producing a lot more sperm He eventually does produce a male heir so he was-- and if we assume that it was his child-- then obviously he was producing some of these,but for the most part, it was all Henry the VIII's fault. So that's why I say there's a little bit of irony here. Is that the people doing the blame are the people to blame for the lack of a male heir? Now one question that might immediately pop up in your head is, Sal,is everything on these chromosomes related to just our sex-determining traits or are there other stuff on them? So let me draw some chromosomes. So let's say that's an X chromosome and this is a Y chromosome. Now the X chromosome,it does code for a lot more things, although it is kind of famously gene poor. It codes for on the order of 1,500 genes. And the Y chromosome,it's the most gene poor of all the chromosomes. It only codes for on the order of 78 genes. But what it tells you is it does very little other than determining what the gender is. And the way it determines that, it does have one gene on it called the SRY gene. You don't have to know that. SRY,that plays a role in the development of testes or the male sexual organ.So if you have this around, this gene right here can start coding for things that will eventually lead to the development of the testicles. And if you don't have that around,that won't happen, so you'll end up with a female. And I'm making gross oversimplifications here. But everything I've dealt with so far,OK, this clearly plays a role in determining sex. But you do have other traits on these genes. And the famous cases all deal with specific disorders."
    },
    {
        "Q": "At around 5:48 what would the hybridization of the transition state be?\n",
        "A": "The hybridization of the carbon atom in the transition state would be sp\u00c2\u00b2.",
        "video_name": "3LiyCxCTrqo",
        "timestamps": [
            348
        ],
        "3min_transcript": "We can see in our final products here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic because of course it is our nucleophile that is doing the substituting. And finally the two here refers to the fact that this is bimolecular which means that the rate depends on the concentration of two things. The substrate and the nucleophile. That's different from an SN1 mechanism where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left we have a methyl halide followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide, the carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. Turns out that the methyl halides and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond and steric hindrance would prevent that from happening. Something like a tertiary alkyl halide has this big bulky methyl groups which prevent the nucleophile for attacking. Let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen which I'm seeing as yellow. And here's our nucleophile which could be the hydroxide ion. for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation. These bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need decreased steric hindrance. So, if we look at this alkyl halide,"
    },
    {
        "Q": "8:40 what represnt those arrows?\n",
        "A": "The arrows represent the electrons. You can have up to two electrons in an orbital but, if there are two electrons, they must have opposite spin. (Spin is defined by the fourth quantum number that you will have seen in earlier videos.) Opposite spin is represented by drawing one arrow pointing up and the other arrow pointing down.",
        "video_name": "649ZlWMp0LE",
        "timestamps": [
            520
        ],
        "3min_transcript": "of the two p electron from the full attraction of the nucleus, right? So, even though we have five protons in the nucleus, and a positive five charge for boron, the fact that these two s electrons add a little bit of extra shielding means it's easier to pull this electron away. So, it turns out to be a little bit easier to pull this electron in the two p orbital away due to these two s electrons. And that's the reason for this slight decrease in ionization energy. As we go from boron to carbon, we see an increase in ionization energy, from carbon to nitrogen, an increase in ionization energy. Again, we attribute that to increased effective nuclear charge, but when we go from nitrogen to oxygen, we see a slight decrease again. From about 1400 kilojoules per mole, down to about 1300 kilojoules per mole for oxygen. So, let's see if we can explain that by writing out Nitrogen has seven electrons to think about. So it's electron configuration is one s two, two s two, and two p three. So that takes care of all seven electrons. For oxygen, we have another electron, so one s two, two s two, two p four is the electron configuration for oxygen. Let's just draw using orbital notation the two s orbital and the two p orbital. So for nitrogen, here's our two s orbital. We have two electrons in there, so let's draw in our two electrons. And for our two p orbitals, we have three electrons. So here are the two p orbitals, and let's draw in our three electrons using orbital notation. Let's do the same thing for oxygen. So there's the two s orbital for oxygen, which is full, so we'll sketch in those two electrons, and we have four electrons in the two p orbitals. There's one electron, there's two, there's three, and notice what happens when we add the fourth electron. We're adding it to an orbital that already has an electron in it, so when I add that fourth electron to the two p orbital, it's repelled by the electron that's already there, which means it's easier to remove one of those electrons, so electrons have like charges, and like charges repel. And so that's the reason for this slight decrease in ionization energy. So, it turns out to be a little bit easier to remove an electron from an oxygen atom, than nitrogen, due to this repulsion in this two p orbital. From there on, we see our general trend again. The ionization energy for fluorine is up to 1681, and then again for neon, we see an increase in the ionization energy due to the increased effective nuclear charge."
    },
    {
        "Q": "1.At 5:24, Sal said isostatic process. What does it mean?\n2. Why did he use natural log instead of common log? Either way is fine?\nThanks!\n",
        "A": "1. By isostatic, he means any one of the isobaric, isochoric, adiabatic, isothermal, etc (if any) processes. 2. Since integral of 1/x = ln x, he used natural log so that the answers match.",
        "video_name": "WLKEVfLFau4",
        "timestamps": [
            324
        ],
        "3min_transcript": "Or maybe if I had a smaller container, I would also have fewer potential states. There would be fewer potential places for our little particles to exist. So I want to create some type of state variable that tells me, well, how many states can my system be in? So this is kind of a macrostate variable. It tells me, how many states can my system be in? And let's call it s for states. For the first time in thermodynamics, we're actually using a letter that in some way is related to what we're actually trying to measure. s for states. And since the states, they can grow really large, let's say I like to take the logarithm of the number of states. Now this is just how I'm defining my state variable. I get to define it. So I get to put a logarithm out front. So let me just put a logarithm. So in this case, it would be the logarithm of my number of states-- so it would be x to the n, where this is number of And you know, we need some kind of scaling factor. Maybe I'll change the units eventually. So let me put a little constant out front. Every good formula needs a constant to get our units right. I'll make that a lowercase k. So that's my definition. I call this my state variable. If you give me a system, I should, in theory, be able to tell you how many states the system can take on. So let me close that box right there. Now let's say that I were to take my box that I had-- let me copy and paste it. I take that box. And it just so happens that there was an adjacent box next to it. They share this wall. They're identical in size, although what I just drew But they're close enough. They're identical in size. And what I do, is I blow away this wall. I just evaporate it, all of a sudden. It just disappears. So this wall just disappears. Now, what's going to happen? Well, as soon as I blow away this wall, this is very much not an isostatic process. All hell's going to break loose. I'm going to blow away this wall, and you know, the particles that were about to bounce off the wall are just going to keep going. They're going to keep going until they can maybe bounce off of that wall. So right when I blow away this wall, there's no pressure here, because these guys have nothing to bounce off to. While these guys don't know anything. They don't know anything until they come over here and say, oh, no wall. So the pressure is in flux. Even the volume is in flux, as these guys make their way across the entire expanse of the new volume. So everything is in flux. Right? And so what's our new volume? If we call this volume, what's this?"
    },
    {
        "Q": "\nat around 10:06 you say that the temperature doesnt change because everything is in isolation thus adiabatic. In previous videos, with the cilinder/pebbles, when everything was adiabatic, the temperature DID change. The only time the temperature didnt change back then was when there was a reservoir added (Isotherm).\n\nSo why is it different here? Just because the system doesnt have to do any work?",
        "A": "I guess i found my answer at ~6:55, sorry",
        "video_name": "WLKEVfLFau4",
        "timestamps": [
            606
        ],
        "3min_transcript": "Now, each particle could be in 2x different states. Why do I say 2x? Because I have twice the area to be in. Now, the states aren't just, you know, position in space. But everything else-- so, you know, before here, maybe I had a positions in space times b positions, or b momentums, you know, where those are all the different momentums, and that was equal to x. Now I have 2a positions in volume that I could be in. I have twice the volume to deal with. So I have 2a positions in volume I can be at, but my momentum states are going to still be-- I just have b momentum states-- so this is equal to 2x. I now can be in 2x different states, just because I have 2 times the volume to travel around in, right? Well, each particle can be in 2x states. So this is 2x times 2x times 2x. And I'm going to do that n times. So my new s-- so this is, you know, let's call this s initial-- so my s final, my new way of measuring my states, is going to be equal to that little constant that I threw in there, times the natural log of the new number of states. It's 2x to the n power. So my question to you is, what is my change in s when I blew away the wall? You know, there was this room here the entire time, although these particles really didn't care because this wall was there. So what is the change in s when I blew away this wall? The temperature didn't change, because no kinetic energy was expended. I should have said. It's adiabatic. There's no transfer of heat. So that's also why the temperature didn't change. So what is our change in s? Our change in s is equal to our s final minus our s initial, which is equal to-- what's our s final? It's this expression, right here. It is k times the natural log--and we can write this as 2 to the n, x to the n. That's just exponent rules. And from that, we're going to subtract out our initial s value, which was this. k natural log of x to the n. Now we can use our logarithm properties to say, well, you know, you take the logarithm of a minus the logarithm of b, you can just divide them. So this is equal to k-- you could factor that out-- times the logarithm of 2 to the N-- it's uppercase N, This is uppercase N. I don't want to get confused with Moles."
    },
    {
        "Q": "@02:13,\nshouldn't it be x(x-1)(x-2)(x-3).....\nbecause a already ocupied 1 state so b has only x-1 states to go to?\n",
        "A": "But the states are changing as the molecules have velocity. If they wouldn t have had a velocity then your proposition would have been correct",
        "video_name": "WLKEVfLFau4",
        "timestamps": [
            133
        ],
        "3min_transcript": "If you followed some of the mathematics, and some of the thermodynamic principles in the last several videos, what occurs in this video might just blow your mind. So not to set expectations too high, let's just start off with it. So let's say I have a container. And in that container, I have gas particles. Inside of that container, they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Let me write that down. What do I mean by state? Well, let's say I take particle A. Let me make particle A a different color. have some velocity like that. It could also be in that corner and have a Those would be two different states. It could be up here, and have a velocity like that. It could be there and have a velocity like that. If you were to add up all the different states, and there would be a gazillion, of them, you would get x. That blue particle could have x different states. You don't know. We're just saying, look. I have this container. It's got n particles. So we just know that each of them could be in x different states. Now, if each of them can be in x different states, how many total configurations are there for the system as a whole? Well, particle A could be in x different states, and then particle B could be in x different places. So times x. If we just had two particles, then you would multiply all the different places where X could be times all the different places where the red particle could be, then you'd get all the different configurations for the system. But we don't have just two particles. We have n particles. So for every particle, you'd multiply it times the number a total of n times. And this is really just combinatorics here. You do it n times. This system would have n configurations. For example, if I had two particles, each particle had three different potential states, how many different configurations could there be? Well, for every three that one particle could have, the other one could have three different states, so you'd have nine different states. If you had another particle with three different states, you'd multiply that by three, so you have 27 different states. Here we have n particles. Each of them could be in x different states. So the total number of configurations we have for our system-- x times itself n times is just x to the n. So we have x to the n states in our system. Now, let's say that we like thinking about how many states a system can have. Certain states have less-- for example, if I had fewer particles, I would have fewer"
    },
    {
        "Q": "\nThe 1,3-Diflurocyclopentane.....at 8:58; wouldnt it have an axis of symmetry straight down the center as if folding a sheet of paper hot dog style?",
        "A": "Yes. Sal was just using the same axis of symmetry to compare it to the first example. You would, in fact, have an axis of symmetry if you chose to do it that way; however, it still would not be chiral, because if you started counting from either fluorine there would still be an unequal number of carbon/fluorine groups on each side.",
        "video_name": "QQMZ1ljepWg",
        "timestamps": [
            538
        ],
        "3min_transcript": "If you were to go clockwise, you'd have CH2, then a CH, which happens to be connected to a fluorine. So you're actually going to see something different, depending whether you're going down into that group or into And then it's also bonded to a hydrogen and a fluorine, four different groups. This is also a chiral center. Another way to think about it, and it's actually interesting to compare it to this molecule up here, which was not chiral and did not have a chiral center, this molecule up here-- let me draw it a little different to make it a little bit more clear. So this one, I could draw it like this. If you have the chlorine like that, over here, we thought about this as a potential chiral center, and it's kind of playing the same role as in that example down here, but you see over here, this is not a chiral center because that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, if we try to draw an axis of symmetry, we can make that bond to the fluorine go through our axis of symmetry,, we'll see that that still is not the reflection of this because we have a fluorine up here. We don't have a fluorine over here. Now, we can do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule. center, you had a chiral molecule. But let's take its mirror image. To take its mirror image, let me clear out some real estate over here, So let me clear out this. Let me clear it out. So what's the mirror image going to look like? Let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. That's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this. You connect them. Now, these are mirror images of each other. But they are also the exact same molecule. I could just literally move this guy over to the right, and it would be superimposed. They are exactly the same. So even though we have two chiral atoms, two chiral"
    },
    {
        "Q": "\nhow can a structure have a chiral atom and be not chiral? does it matter? 9:33.",
        "A": "Well a Meso compound would have a chiral atom but not actually be chiral, wouldn t it?",
        "video_name": "QQMZ1ljepWg",
        "timestamps": [
            573
        ],
        "3min_transcript": "that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, if we try to draw an axis of symmetry, we can make that bond to the fluorine go through our axis of symmetry,, we'll see that that still is not the reflection of this because we have a fluorine up here. We don't have a fluorine over here. Now, we can do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule. center, you had a chiral molecule. But let's take its mirror image. To take its mirror image, let me clear out some real estate over here, So let me clear out this. Let me clear it out. So what's the mirror image going to look like? Let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. That's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this. You connect them. Now, these are mirror images of each other. But they are also the exact same molecule. I could just literally move this guy over to the right, and it would be superimposed. They are exactly the same. So even though we have two chiral atoms, two chiral It is a non-chiral molecule."
    },
    {
        "Q": "At around 5:30, the video is talking about chromate being CrO4^2-, but when it is dichromate it is Cr2O7^2-. Why are there 7 oxygen atoms? I thought there would be 8, as 4*2 is 8. Sorry for the notation.\n",
        "A": "The di means that there are two chromium atoms in the ion, not that it consists of two chromate ions. The missing O atom ends up as water. For example: 2K\u00e2\u0082\u0082CrO\u00e2\u0082\u0084 + H\u00e2\u0082\u0082SO\u00e2\u0082\u0084 \u00e2\u0086\u0092 K\u00e2\u0082\u0082Cr\u00e2\u0082\u0082O\u00e2\u0082\u0087 + H\u00e2\u0082\u0082O + K\u00e2\u0082\u0082SO\u00e2\u0082\u0084",
        "video_name": "DpnUrVXSLaQ",
        "timestamps": [
            330
        ],
        "3min_transcript": "And then we have four Oxygens, so if we go to three Oxygens, SO three two minus, this is Sulfite, cause ite means fewer Oxygens. What about if we took Sulfate, SO four two minus, and we added on an H plus. So H plus and SO four two minus should give us HSO four and then, instead of a negative two here, instead of a two minus, we would just have a one minus, because we added on a positive charge. So one positive charge and two negative charges, give us one negative charge. So HSO four minus is called the Hydrogen Sulfate ion. You might also hear Bisulfate for this one. Next CO three two minus is called Carbonate, so if we add on an H plus to CO three two minus, we'd get HCO three and then we go from minus two or two minus, to minus one, 'cause we're adding on a positive charge here. and you'll also hear Bicarbonate a lot. Next we have PO four three minus, which is called Phosphate. If we add on an H plus to Phosphate, think about what we would get. We would get HPO four and then instead of three minus, we're adding on positive charge, so we get two minus. So we call this Hydrogen Phosphate. Alright, let's add on a proton to Hydrogen Phosphate. So we're adding an H plus onto Hydrogen Phosphate. That would give us two H's. PO four and we'd go from two minus down to one minus. So H two PO four minus is called Dihydrogen Phosphate. Alright, let's continue on. One more set of polyatomic ions to know. So we have CrO four two minus, which is called Chromate. And if we have two Chromiums, so Cr two O seven two minus Next, C two O four two minus is called the Oxalate ion. and we have O two, two minus is called Peroxide. And here we have SCN minus, which we call Thiocyanate. So thio, think about sulfur if you see thio there. So for our next one, we have sulfur present again, is S two O three two minus and this one's called Thiosulfate. So you might see a few additional polyatomic ions in your class, but these are the ones that you see most frequently. So make sure to memorize your polyatomic ions."
    },
    {
        "Q": "At around 4:45, he did not explain where the second hydrogen atom came from to make hydrogen phosphate become dihydrogen phosphate. Can someone explain that to me?\n",
        "A": "It doesn t really matter where it came from, he s just showing you how those phosphate molecules are related to one another. It probably came from water if you have to have an answer to understand.",
        "video_name": "DpnUrVXSLaQ",
        "timestamps": [
            285
        ],
        "3min_transcript": "And then we have four Oxygens, so if we go to three Oxygens, SO three two minus, this is Sulfite, cause ite means fewer Oxygens. What about if we took Sulfate, SO four two minus, and we added on an H plus. So H plus and SO four two minus should give us HSO four and then, instead of a negative two here, instead of a two minus, we would just have a one minus, because we added on a positive charge. So one positive charge and two negative charges, give us one negative charge. So HSO four minus is called the Hydrogen Sulfate ion. You might also hear Bisulfate for this one. Next CO three two minus is called Carbonate, so if we add on an H plus to CO three two minus, we'd get HCO three and then we go from minus two or two minus, to minus one, 'cause we're adding on a positive charge here. and you'll also hear Bicarbonate a lot. Next we have PO four three minus, which is called Phosphate. If we add on an H plus to Phosphate, think about what we would get. We would get HPO four and then instead of three minus, we're adding on positive charge, so we get two minus. So we call this Hydrogen Phosphate. Alright, let's add on a proton to Hydrogen Phosphate. So we're adding an H plus onto Hydrogen Phosphate. That would give us two H's. PO four and we'd go from two minus down to one minus. So H two PO four minus is called Dihydrogen Phosphate. Alright, let's continue on. One more set of polyatomic ions to know. So we have CrO four two minus, which is called Chromate. And if we have two Chromiums, so Cr two O seven two minus Next, C two O four two minus is called the Oxalate ion. and we have O two, two minus is called Peroxide. And here we have SCN minus, which we call Thiocyanate. So thio, think about sulfur if you see thio there. So for our next one, we have sulfur present again, is S two O three two minus and this one's called Thiosulfate. So you might see a few additional polyatomic ions in your class, but these are the ones that you see most frequently. So make sure to memorize your polyatomic ions."
    },
    {
        "Q": "\nBasic question... Why at 3:45, the H is coming out, instead of stay in the sheet planar?",
        "A": "We have to consider what the molecule looks like in three dimensions, in reality the molecules are not flat like the paper we draw them on. There really is no good way to prove this to yourself other than playing around with a molecular modelling kit. They are very handy for learning about stereoisomers. In the example in this video there is a methyl group going away from us, and we can only see 3 bonds to that specific carbon, so there is an implied hydrogen atom coming out of the page towards us.",
        "video_name": "9IYTqlVk_ZI",
        "timestamps": [
            225
        ],
        "3min_transcript": "The carbon on the left is directly bonded to three hydrogens, so we write down here three hydrogens and the carbon on the right is directly bonded to two carbons and one hydrogen, so we write carbon, carbon, hydrogen. Next, we look for the first point of difference and that's the first atom here, so carbon versus hydrogen. Carbon has the higher atomic number, so this group wins, so this group on the right is higher priority which means this must be number two for our groups, and the methyl group gets a number three. Now that we've assigned priority to our four groups, the next step was to orient the molecules so that the lowest priority group is projecting away from us, and that's what we have here because the hydrogen is the lowest priority group so we can ignore this and we can focus in on one, two, and three, so let me label this on the right, the OH was the highest highest priority and the methyl group was the third highest priority. Next we draw a circle and determine whether we're going around clockwise or counterclockwise, so we draw a circle from one to two to three, and obviously, we're going around in a clockwise direction and clockwise is R, so we are R at carbon two, so I write here 2R. Now, let's focus on carbon three, so we know that carbon three is a chiral center, so what is directly bonded to this carbon? Well, there's a carbon attached to three hydrogens, so there's a methyl group, there's a hydrogen that must be coming out at us in space and on the right, there's a carbon bonded to two hydrogens and directly bonded to another carbon. On the left, there's a carbon directly bonded to an oxygen. There must be a hydrogen going away from us, and this carbon is directly bonded to another carbon, so here is our chiral center. We look at the atoms directly bonded to that carbon. There are three carbons, one, two, three, and a hydrogen. We know the hydrogen has the lowest atomic number, so hydrogen is the lowest priority and we call that group four. Now we have three carbons. We have a tie because carbon, of course, has the same atomic number. To break a tie, we look at the atoms that are bonded to those carbons, so let's start with the carbon on the right here, so the carbon on the right is directly bonded to one carbon and then two hydrogens, so we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number. This carbon down here is directly bonded to three hydrogens so one, two, and three. The carbon on the left is directly bonded to an oxygen, a carbon and a hydrogen, so we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen."
    },
    {
        "Q": "Starting around 7:30ish, Sal is using Hess's Law. Correct?\n",
        "A": "Sorry, I think you aren t right Hess s law states that the overall enthalpy change in a reaction is the sum of all the reactions for the process and is independent to the route taken this might look like Hess law at the first glance...",
        "video_name": "fYUwEAPejbY",
        "timestamps": [
            450
        ],
        "3min_transcript": "That's what makes the electrolyte paste alkaline. And you are going to then go to, they're going to react and you're going to form zinc oxide. Zinc oxide in the solid state, plus water in the liquid state, plus two, plus two electrons. And so this reaction that I just described, this is going to be happening right over here. So you could think of it as the hydroxide anions get formed at the cathode, and then they move their way over to the left to the anode where they react with the zinc and they turn, or the zinc reacts and forms zinc oxide, so you have more and more zinc oxide being formed, water, which can then seep its way back into the electrolyte paste and then it can eventually react again at the cathode, and then you have these two electrons. So these electrons, this turns into an electron source right over here, and then the electrons would migrate to react again. And so you can start to see how this will be an energy source that you can tap into this current that'll form to do some useful work. That's why you have an energy source. All right, so let's read that questions now that we have a decent understanding of what's going on. Early forms of metal air cells use zinc as the anode. Well, that's the example we just thought about. Zinc oxide is produced as the cell operates according to the overall equation below. For every two molecules of zinc and one molecule of molecular oxygen, you produce two molecules of zinc oxide. Use the data in the table above, calculate the cell potential for the zinc air cell. So let's think about, let's break down this reaction. Actually, we can just break it down into these two steps here. Because notice, this is, we can't see them both at the same time but this, the things that are reacting, Let me underlie the things that are reacting. You have zinc, you have zinc. You have oxygen, you have oxygen. And so if you return this, if you return the second reaction around like we did before, and let me rewrite it, let me rewrite both of them actually a little bit lower right over here. So you have this top reaction. So you have O2 gaseous state plus two H2O liquid, plus four electrons. Whoops, four electrons, yields four hydroxide anions in aqueous solution. And then this one, we said we're gonna go in the other direction. So you're going to have zinc, you're going to have zinc in the solid state, plus two hydroxide anions. And then that's going to yield zinc,"
    },
    {
        "Q": "at \"8:52\", if the crab nebula is 65,00 light years away, couldn't a star or more than one star formed in the 65,00 years it took for that image to reach earth?\n",
        "A": "It takes more than 6500 (that was the number in the video) years to form a star. It actually takes more than 65,000 years to form a star.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            532
        ],
        "3min_transcript": "it'll be something about maybe two times the mass of the sun, give or take one and a half to three times the mass of the sun. So this is one and a half to three times the mass of the sun in a volume that has a diameter of about 10-- on the order of tens of kilometers. So it's roughly the size of a city, in a diameter of a city. So this is unbelievably dense, diameter of a city. I mean, we know how much larger the sun is relative to the Earth. And we know how much larger the Earth is relative to a city. But this is something large-- more mass than the sun being squeezed into the density, or into the size of a city, so unbelievably dense. Now if the original star is even more massive, if it's more than 20 times the sun-- so let me write it over here. Let me scroll up. If it's greater than 20 times the sun, even the neutrons' inability to squeeze further will give up. And it'll turn into a black hole. And that's-- and I could do many videos on that. And that's actually an open area of research, still, on exactly what's going on inside of a black hole. But then you turn into a black hole, where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, so this will be more mass then even three times the mass So we're talking about an incredibly high amount of mass. Just to kind of visualize things, here is actually a remnant of a supernova. This is the Crab Nebula. This is, right here, is the Crab Nebula. And it's about 6,500 light years away. So it's still, from a galactic scale-- if you think of our galaxy as being 100,000 light years But it's an enormous distance. The closest star to us is four light years away. And it would take Voyager travelling at 60,000 kilometers an hour, 80,000 years to get there. So this is a very, very-- that's only four light years. Now this is 6,500 light years. But this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of this sphere of material, is six light years. So we could say this distance right here is six light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, started to condense because of a shock wave created"
    },
    {
        "Q": "\nWhat does exothermic mean? At around 00:50",
        "A": "Exothermic reactions are one of the two types of chemical reactions. An exothermic reaction releases energy, while endothermic reactions absorb energy.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            50
        ],
        "3min_transcript": "Where we left off in the last video, we had a mature massive star, a star that had started forming a core of iron. It has enormous pressure, enormous inward pressure on this core. Because as we form heavier and heavier elements in the core, the core gets denser and denser and denser. And so we keep fusing more and more elements into iron. This iron core becomes more and more massive, more and more dense. It's squeezing in on itself. And it's not fusing. That is not exothermic anymore. If iron were to fuse, it would not even be an exothermic process. It would require energy. So it wouldn't be even something that could be helped to fend off this squeezing, to fend off this increasing density of the core. So we have this iron here, and it just gets more and more massive, more and more dense. And so at some mass, already a reasonably high mass, the only thing that's keeping this from just completely So let me write this here, electron degeneracy pressure. And all this means is we have all of these iron atoms getting really, really, really close to each other. And the only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether, is that they have these electrons. You have these electrons, and these are being squeezed together, now. I mean, we're talking about unbelievably dense states of matter. And electron degeneracy pressure is, essentially-- it's saying these electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it. But they cannot be squeezed into each other any more. So that, at least temporarily, holds this thing from collapsing even further. in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons."
    },
    {
        "Q": "\nAt 2:51 he talks about \"Beta negative decay\". What is it?",
        "A": "It s the conversion of a neutron into a proton in the nucleus of an atom. When that happens, an electron is emitted. That electron is called a beta particle.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            171
        ],
        "3min_transcript": "So let me write this here, electron degeneracy pressure. And all this means is we have all of these iron atoms getting really, really, really close to each other. And the only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether, is that they have these electrons. You have these electrons, and these are being squeezed together, now. I mean, we're talking about unbelievably dense states of matter. And electron degeneracy pressure is, essentially-- it's saying these electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it. But they cannot be squeezed into each other any more. So that, at least temporarily, holds this thing from collapsing even further. in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons. But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles."
    },
    {
        "Q": "\nHey Sal, at 9:56 you mention that the current model for the formation of our solar system is that it was created by a supernova. That should have left behind a neutron star. Do we have a good candidate for the particular neutron star that is the remnant of the supernova that created our solar system?",
        "A": "The objects in our galaxy orbit around the center at different speed. After 4.5 billion years, it could be anywhere. Currently, any neutron star could be from the supernova that stimulated the nebula collapse.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            596
        ],
        "3min_transcript": "even the neutrons' inability to squeeze further will give up. And it'll turn into a black hole. And that's-- and I could do many videos on that. And that's actually an open area of research, still, on exactly what's going on inside of a black hole. But then you turn into a black hole, where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, so this will be more mass then even three times the mass So we're talking about an incredibly high amount of mass. Just to kind of visualize things, here is actually a remnant of a supernova. This is the Crab Nebula. This is, right here, is the Crab Nebula. And it's about 6,500 light years away. So it's still, from a galactic scale-- if you think of our galaxy as being 100,000 light years But it's an enormous distance. The closest star to us is four light years away. And it would take Voyager travelling at 60,000 kilometers an hour, 80,000 years to get there. So this is a very, very-- that's only four light years. Now this is 6,500 light years. But this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of this sphere of material, is six light years. So we could say this distance right here is six light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, started to condense because of a shock wave created And just to answer another question that was kind of jumping up, probably, in the last video-- and this is still not really, really well understood. We talk about how elements up to iron, or maybe nickel, can be formed inside of the cores of massive stars. So you could imagine when the star explodes, a lot of that material is released into the universe. And so that's why we have a lot of these materials in our own bodies. In fact, we could not exist if these heavier elements were not formed inside of the cores of primitive stars, stars that have supernova-ed a long time ago. Now the question is, how do these heavier elements form? How do we get all of this other stuff on the periodic table? How do we get all these other heavier elements? And they're formed during the supernova itself. It's so energetic. You have all sorts of particles streaming out and streaming in, streaming out because of the force of the shock wave, streaming in because of the gravity. But you have all sorts of kind of a mishmash of elements"
    },
    {
        "Q": "\n\"During the eclipse Sal talks about at 5:28, why is the moon the exact size of the sun, offering complete 1 for 1 coverage?\"",
        "A": "Happy coincidence.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            328
        ],
        "3min_transcript": "But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles. But it's an enormous amount of energy. And this actually is not really, really well understood, of all of the dynamics here. Because at the same time that this iron core is undergoing through this-- at first it kind of pauses due to the electron degeneracy pressure. And then it finally gives in because it's so massive. And then it collapses into this dense ball of neutrons. But when it does it, all of this energy's released. And it's not clear how-- because it has to be a lot of energy. Because remember, this is a massive star. So you have a lot of mass in this area over here. But it's so much energy that it causes the rest of the star to explode outward in an unbelievable, I guess, unbelievably bright or energetic explosion. And that's called a supernova. it comes from, I believe-- I'm not an expert here-- Latin for \"new.\" And the first time people observed a nova, they thought it was a new star. Because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared. Because maybe it wasn't bright enough for us to observe it before. But then when the nova occurred, it did become bright enough. So it comes from the idea of new. But a supernova is when you have a pretty massive star's core collapsing. And that energy is being released to explode the rest of the star out at unbelievable velocities. And just to kind of fathom the amount of energy that's being released in a supernova, it can temporarily outshine an entire galaxy. And in a galaxy, we're talking about hundreds of billions of stars. Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime. So these are unbelievably energetic events. And so you actually have the material that's not in the core being shot out of the star"
    },
    {
        "Q": "\nAt 7:45, is killing itself the process of apoptosis? or is it another mechanism specific to the immune system?",
        "A": "Apoptosis is where a cell receives signals in order for it to kill itself. In the immune system, the cytotoxic T cells release granzymes and other chemicals that kill the cell. For example, spores may be punctured in the cell causing an increase in water intake and eventual lysing (bursting).",
        "video_name": "YdBXHm3edL8",
        "timestamps": [
            465
        ],
        "3min_transcript": "so it's presenting that same antigen on this MHC one complex. Remember, the variable portions need to match up. Let's say that this is an effector cytotoxic T cell and actually let me draw in a little bit different. Let me draw it like this. This is effector cytotoxic T cell. Its receptor, its variable portion is the one that's compatible with this antigen that's being presented right over here. Let me just label this again. This is the MHC one complex. This is an effector cytotoxic T cell. We'll put the C there for cytotoxic and what it does is essentially kind of latches on to the cell that needs to die and it does it, not only have this receptor interfacing with the MHC one complex but actually has a whole series of proteins This would be much smaller or relative to the scale cell. It essentially latches on between the two and I'm not going to go in detail but essentially forms what you can call an immuno synapse which is kind of where the two things are interacting with each other. When it identifies this, it says, okay, I need to kill this thing or essentially I need to make this thing kill itself. It starts releasing all of these molecules. It can release molecules like preference so to release this preference which will essentially cause gaps or holes to form in the membrane of the cell that needs to die and it could release other things like granzymes that can go in and essentially cause this thing to kill itself. The whole point of this video is to appreciate I guess what we haven't talked about yet. We had already talked about what happens when you identify shady things outside of the cells and then how you can kind of bring them in and then present them and then use that to further activate Now we're talking about identifying shady things inside the cells. Those get presented by MHC one complexes and then the cytotoxic T cells recognize them and then force the cell to kill themselves. This wouldn't just be cancer cells, this could also apply to a cell that has already been affected by a virus. For example a cell like this all ready, so that's its nucleus. It's already been infected by some virus so the virus has hijacked the cell's replication machinery in order to replicate itself. The proper immune response is hey look, I'm a virus making machine. I should kill myself. It will wreck some of the antigens that are being produced inside by the viruses. They're going to bind to MHC one complexes. Pieces of the virus are going to bind to MHC one complexes"
    },
    {
        "Q": "at 8:00 why is the o positive , didn't it make a covalent bond ? so shouldn't it be partially positive ?\n",
        "A": "For oxidation number purposes, the O lost one of its lone pair electrons when it bonded with the H, so it gained a full positive charge.",
        "video_name": "U9dGHwsewNk",
        "timestamps": [
            480
        ],
        "3min_transcript": "In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that That electron right here is now over here, and now this bond right over here, is this bond. We formed an alkene and now, what was an ethanol took a hydrogen proton and now becomes a positive cation. Let me draw that. So this electron ends up being given. It's no longer with the ethanol. It gets given to this hydrogen right here. That hydrogen right there. And now they have formed a new bond and since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in The bromide anion is floating around with its eight valence electrons, one, two, three, four, five, six, seven, and then it has this one right over here. That makes it negative. Then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule, and that allowed this molecule to become an alkene, formed a double bond. This is called, and I already told you, an E1 reaction. E for elimination, in this case of the halide. One, because the rate-determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have the different types of E1"
    },
    {
        "Q": "\nAt 5:53 why will the O of ethanol donate its electron to the bonded H atom ?? Why not directly to the positively charged carbocation ? Well I think O can donate an electron to C (+) ......",
        "A": "This is because of the steric hindrance of the ethanol. Due to its larger/bulkier size it is more apt to react with the hydrogen that is further away and less crowded by the overall molecule . This is ultimately why an E1 reaction occurred and why and S1 reaction didn t occur.",
        "video_name": "U9dGHwsewNk",
        "timestamps": [
            353
        ],
        "3min_transcript": "doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that"
    },
    {
        "Q": "6:20 - Would EtOH take the H if the C were not a cation? In other words, is if necessary for the carbocation to exist to enable EtOH to take the H? Does the presence of the carbocation enable EtOH to take the H by reducing the strength of the H-C bond by pulling electrons towards itself? Do the electrons in the H-C bond move towards the carbocation before the H is taken, or only afterwards?\n",
        "A": "No. The EtOH would not take the H if the C were not a cation. The cation must be there to make the H atom acidic enough to be removed by the relatively weak base EtOH. The cation does this by pulling the electrons in the C-H bond towards itself. The electrons in the C-H bond have already moved towards the carbon before the EtOH attacks, But they move entirely to C atom after the EtOH has attacked.",
        "video_name": "U9dGHwsewNk",
        "timestamps": [
            380
        ],
        "3min_transcript": "doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that"
    },
    {
        "Q": "\nat \"7:40\" could you also write 350. feet (with a decimal at the end) to show that you meausred it to the nearest one? Or does nobody do that? Because I remember you doing that in the first significant figures video.",
        "A": "You could also write 350. feet, although that has three significant didgits when the building was 350 feet with only two significant didgits, meaning that the measurement of the building may have been a guess or may have been exact. We just cannot tell without the decimal point, which is what the guy was trying to express.",
        "video_name": "xHgPtFUbAeU",
        "timestamps": [
            460
        ],
        "3min_transcript": "and so we can only legitimately say, if we want to represent what we did properly that the tower is 3.99 meters. And I also want to make it clear that this doesn't just apply to when there is a decimal point. If I were tell you that... Let's say that I were to measure... I want to measure a building. I was only able to measure the building to the nearest 10 feet. So I tell you that that building is 350 feet tall. So this is the building. This is a building. And let's say there is a manufacturer of radio antennas, so... or radio towers. And the manufacturers has measured their tower to the nearest foot. And they say, their tower is 8 feet tall. So notice: here they measure to the nearest 10 feet, here they measure to the nearest foot. And actually to make it clear, because once again, as I said, this is ambiguous, Maybe it was exactly 350 feet or maybe they just rounded it to the nearest 10 feet. So a better way to represent this, they... would be to say instead of writing it 350, a better way to write it would be 3.5 times 10 to the second feet tall. And when you are writing in scientific notation, that makes it very clear that there is only 2 significant digits here, you are only measuring to the nearest 10 feet. Other way to represent it: you could write 350 this notation has done less, but sometimes the last significant digit has a line on the top of it, or the last significant digit has a line below it. Either of those are ways to specify it, this is probably the least ambiguous, but assuming that they only make measure to the nearest 10 feet, If someone were ask you: \"How tall is the building plus the tower?\" Well, your first reaction were, let's just add the 350 plus 8, you get 358. You'd get 358 feet. So this is the building plus the tower. 358 feet. We are making it look like we were able to measure the combination to the nearest foot. But we were able to measure only the tower to the nearest foot. So in order to represent our measurement at the level of precision at we really did, we really have to round this to the nearest 10 feet. Because that was our least precise measurement. So we would really have to round this up to, 8 is greater-than-or-equal to 5, so we round this up to 360 feet. So once again, whatever is... Just to make it clear, even this ambiguous, maybe we put a line over to show, that is our level of precision, that we have 2 significant digits. Or we could write this as 3.6 times 10 to the second. Which is times 100. 3.6 times 10 to the second feet in scientific notation. And this makes it very clear that we only have 2 significant digits here."
    },
    {
        "Q": "\nat 7:23, 350 ft is 2 significant figures but when you multiply out 3.5 e^2, that would make it 350. which is 3 significant figures, not 2, right?",
        "A": "No, the significant digits are still 2. You are not adding precision by multiplying. If you want to represent 3 significant digits, you will right 3.50*10^2",
        "video_name": "xHgPtFUbAeU",
        "timestamps": [
            443
        ],
        "3min_transcript": "0 plus 9 is 9, 9 plus 0 is 9, you get the decimal point, 1 plus 2 is 3. So you get 3.991. And the problem with this, the reason why this is a little bit... it's kind of misrepresenting how precise you measurement is. You don't know, if I told you that the tower is 3.991 meters tall, I'm implying that I somehow was able to measure the entire tower to the nearest millimeter. The reality is that I was only be able to measure the part of the tower to the millimeter. This part of the tower I was able to measure to the nearest centimeter. So to make it clear the our measurement is only good to the nearest centimeter, because there is more error here, then... it might overwhelm or whatever the precision we had on the millimeters there. To make that clear, we have to make this only as precise as the least precise thing that we are adding up. So over here, the least precise thing was, we went to the hundredths, so over here we have to round to the hundredths. and so we can only legitimately say, if we want to represent what we did properly that the tower is 3.99 meters. And I also want to make it clear that this doesn't just apply to when there is a decimal point. If I were tell you that... Let's say that I were to measure... I want to measure a building. I was only able to measure the building to the nearest 10 feet. So I tell you that that building is 350 feet tall. So this is the building. This is a building. And let's say there is a manufacturer of radio antennas, so... or radio towers. And the manufacturers has measured their tower to the nearest foot. And they say, their tower is 8 feet tall. So notice: here they measure to the nearest 10 feet, here they measure to the nearest foot. And actually to make it clear, because once again, as I said, this is ambiguous, Maybe it was exactly 350 feet or maybe they just rounded it to the nearest 10 feet. So a better way to represent this, they... would be to say instead of writing it 350, a better way to write it would be 3.5 times 10 to the second feet tall. And when you are writing in scientific notation, that makes it very clear that there is only 2 significant digits here, you are only measuring to the nearest 10 feet. Other way to represent it: you could write 350 this notation has done less, but sometimes the last significant digit has a line on the top of it, or the last significant digit has a line below it. Either of those are ways to specify it, this is probably the least ambiguous, but assuming that they only make measure to the nearest 10 feet, If someone were ask you: \"How tall is the building plus the tower?\" Well, your first reaction were, let's just add the 350 plus 8, you get 358. You'd get 358 feet. So this is the building plus the tower. 358 feet."
    },
    {
        "Q": "\n5:30 What would drive the reformation of the carbonyl?",
        "A": "Delocalization of electrons leads to increased stability. Before 5:30, the negative charge is localized on the O atom. After 5:30. there is no charge and a lone pair on O is now delocalized in the \u00cf\u0080 orbital of the C=O bond.",
        "video_name": "CafRuKs7EfE",
        "timestamps": [
            330
        ],
        "3min_transcript": "have formed this bond now, and this carbon is bonded to this nitrogen, bonded to this, and our R double prime group, and if we showed, let's make these electrons in here green, these electrons move off onto our nitrogen, like that. And so our carbon is also double-bonded to this nitrogen, with R double prime group. So in the next step our amine is gonna function as a nucleophile. So let's go ahead and draw in our amine. So we have our nitrogen with two hydrogens, and an R prime group, lone pair of electrons on our nitrogen, makes this amine able to function as a nucleophile, so let me go ahead and put those electrons in magenta. So right here, this oxygen is partially negative, It's gonna withdraw some electron density from this carbon, so partially positive. going to attack our electrophile. So our amine is going to attack this carbon, push these electrons off onto our oxygen. So let's go ahead and show the result of this nucleophilic attack. Alright, so let's get some room down here. We would have our carbon bonded to this oxygen. So let's go ahead and draw in all of those electrons, so, -1 formal charge on this oxygen, so if these electrons in here in green move off onto our oxygen we get a -1 formal charge. This carbon is bonded to an R group, it's also bonded to this nitrogen, this nitrogen now has a +1 formal charge. So +1 formal charge on our nitrogen, after the electrons in magenta move in here to form this bond. So we still have our carbon bonded to this oxygen, and draw in our lone pairs of electrons, and then we have this carbon, we have a nitrogen, we have our R double prime group, we have a hydrogen, lone pair of electrons double-bonded to group, so a lot of stuff going on here. So, the use of DCC gives you a good leaving group. So if we think about all this stuff over here, this is an excellent leaving group. So if we reform our carbonyl, let's go ahead and show that, so if these electrons in here move in to reform our carbonyl, these electrons could come off onto our oxygen, and we could even show them moving over to here, to save some time, and if there's a proton out here, these electrons could pick up that proton, and we have an excellent leaving group, so this actually forms dicyclohexylurea over here on the right. So if I circle all of this stuff, we're gonna get dicyclohexylurea, and let's go ahead and show what would happen. So if we reform our carbonyl, let's use those electrons in here in red, so if those electrons in red move in, now we would have our R group,"
    },
    {
        "Q": "\n10:00 I'm not sure I understand why it's important to know what components of the vectors in the 2-d plane are perpendicular to the other... I mean as long as your 3rd vector is sticking in or out of the page it will be perpendicular to both a & b so how does finding their components help with this ?",
        "A": "This will become important with torque, which will most likely be seen in the next video. If they don t cover it I would google it.",
        "video_name": "o_puKe_lTKk",
        "timestamps": [
            600
        ],
        "3min_transcript": "magnitude of vector a that is perpendicular to vector b, and those are the two numbers that I want to multiply and then give it that direction as specified by the right hand rule. And I'll show you some applications. This is especially important-- well, we'll use it in torque and we'll also use it in magnetic fields, but it's important in both of those applications to figure out the components of the vector that are perpendicular to either a force or a radius in question. So that's why this cross product has the sine theta because we're taking-- so in this, if you view it as magnitude of a sine theta times b, this is kind of saying this is the magnitude of the component of a perpendicular to b, or you could interpret it the other way. You could interpret it as a times b sine theta, right? Put a parentheses here. And then you could view it the other way. that is perpendicular to a. Let me draw that, just to hit the point home. So that's my a, that's my b. This is a, this is b. So b has some component of it that is perpendicular to a, and that is going to look something like-- well, I've run out of space. Let me draw it here. If that's a, that's b, the component of b that is perpendicular to a is going to look like this. It's going to be perpendicular to a, and it's going to go that far, right? And then you could go back to SOH CAH TOA and you could prove to yourself that the magnitude of this vector is b sine theta. So that is where the sine theta comes from. It makes sure we're multiplying the components of the vectors that are perpendicular to each other to get a third vector that is perpendicular to both of them. And then the people who invented the cross product said, well, it's still ambiguous because it doesn't tell us-- there's always two vectors that are perpendicular to these two. One goes in, one goes out. They're in opposite directions. And that's where the right hand rule comes in. They'll say, OK, well, we're just going to say a convention that you use your right hand, point it like a gun, make all your fingers perpendicular, and then you know what direction that vector points in. Anyway, hopefully, you're not confused. Now I want you to watch the next video. This is actually going to be some physics on electricity, magnetism and torque, and that's essentially the applications of the cross product, and it'll give you a little bit more intuition of how to use it. See you soon."
    },
    {
        "Q": "Can you please explain that what does Sal means by distances are proportional? (at 5:35)\n",
        "A": "It means that if you increase the distance to the edge, the distance the edge travels down also increases. When he says that the distance increases proportionally, he means that if the distance to the 7N force increases by 10%, the travel distance will also increase by 10%.",
        "video_name": "DiBXxWBrV24",
        "timestamps": [
            335
        ],
        "3min_transcript": "compound it further and et cetera, et cetera, using some of the other concepts we've learned. But I won't worry about that right now. So let's say that I'm going to push up here. Well no let me see what I want to do. I want to push down here with a force of-- let's say that this distance right here is 35 meters, this distance is 5 meters-- and let's say I'm going to push down with the force of 7 newtons, and what I want to figure out is how heavy of an object can I lift here. How heavy of an object. Well, all we have to do is use the same formula. But the moments-- and I know I used that word once before, so you might not know what it is-- but the moments on both sides of the fulcrum have to be the same. So what's the moment again? Well, the moment is just the force times the distance from the force to the fulcrum. So the input moment is 7 newtons times 35 meters. And realize that that does not work, because the distance this force is traveling is not 35 meters. The distance this force is traveling is something like, here. But this 35 meters is going to be proportional to the distance that this is traveling when you compare it to this other side. So this quantity, 7 newtons times 35 meters, is the moment. And that is going to be equal to the moment on this side, the output moment. So that is equal to 5 meters times the force that I'm lifting, or the lifting force of the machine, times let's say the force out. So we can figure out the force out by just dividing both sides by 5. So let's see, 35 divided by 5 is 7, so you get 7 times 7 And you can see that, because you can see that the length of this side of the lever is 7 times the length of this side of the lever. So when you input a force of 7, you output a force of 7 times that. And of course, in order to move the block 1 meter up in this direction, you're going to have to push down for 7 meters. And that's where we know that the input work is equal to the output work. Well anyway hopefully I didn't confuse you and you have a reasonable sense of how levers work. In the next couple of videos, I'll introduce you to other machines, simple machines like a wedge-- I've always had trouble calling a wedge a machine, but it is one-- and pulleys. I'll see you in the next video."
    },
    {
        "Q": "So basically the Na-K pump exchanges Na+ for K+ ions? (1:50)\nWhy would the cell exchange one ion for the other?\nIf the answer is concentration gradients then another question.\nSuppose one side of the wall contains a higher concentration of both Na+ and K+ than the other side. In that case what will happen?\n",
        "A": "Yes the answer is concentration gradient. If there is hihger conc at one side then a potential difference will be established that causes polarization of membrane that propagates. This happens in the case of impulse conduction by nerves.",
        "video_name": "vh166DKxYiM",
        "timestamps": [
            110
        ],
        "3min_transcript": "- [Voiceover] What I hope to do in this video is give ourselves an appreciation for the sodium-potassium pump, and as the name implies, it pumps sodium and potassium, but it does it in different directions. So this little depiction right over here, this is my drawing, my rendition of the sodium-potassium pump, it's a trans-membrane, I guess you could say protein complex right over here. And in this resting state, it is open to the inside of the cell and it has an affinity for sodium ions. And so the sodium ions, you see three sodium ions depicted here in blue. They're going to bind to the pump. And once they bind to it, then it's going to want to be phosphorylated by an ATP, and we see that right over here. This is ATP, adenosine triphosphate. And when it gets phosphorylated, it's a release of energy and it allows the confirmation of the actual protein to change. So the new confirmation of the protein, it's now going to open up to the outside, close off to the inside, have an affinity for sodium ions, but an affinity for potassium ions, and this is fascinating, that release of energy, change of confirmation, that these proteins really are these molecular machines, these fascinating molecular machines. But once that happens, this change of confirmation, the sodium ions are going to be released outside of the cell. And then you're going to have potassium ions that are going to bind from the outside. And then once that happens, the change in confirmation, it's going to get dephosphorylated and then you're going to go back to your original confirmation, your original confirmation right over here. Where you no longer have an affinity for potassium ions, they're going to be released, and then you're going to be back in the original phase. So this is fascinating. By using ATP, by using energy, this is active transport, it takes energy to do this. Let me write this down. This is active, this is active transport that we are talking about right over here. We're able to pump, using an ATP, we're able to pump three sodium ions out, so let me write that down. Three sodium ions out. And in the process, we pump two potassium ions in. So we pump two potassium ions in. Now you might say, okay, the outside, since these both have positive charge, but I have three sodium going out, two potassium going in. That must make the outside more positive than the inside, and that actually is true. But that by itself isn't fully responsible. It's actually only partially responsible for the electric potential difference between the inside of the membrane and the outside of the membrane. What really sets that up is that you actually have channel proteins that allow potassium ions to move down, to diffuse down their concentration gradient. So let's think about what happens before I even talk about these channel proteins right over here. Because of the sodium-potassium pump, what is sodium's concentration gradient?"
    },
    {
        "Q": "At 4:00, aren't the three dot structures the same, just rotated? Why are they considered different?\n",
        "A": "I wondered the same thing initially. I think that an important thing to consider is that the diagrams are only the-same-but-rotated if you don t care which oxygen atom is which. If you labeled the oxygen atoms, then it wouldn t be the same. So in real life, if you were somehow able to hold the molecule still and look at just one oxygen atom, the three structures would not be the same (it could have either a single or double bond to the nitrogen).",
        "video_name": "bUCu7bPkZeI",
        "timestamps": [
            240
        ],
        "3min_transcript": "And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here, would have been equally valid. We could have shown this oxygen on the bottom left now bonded to this nitrogen, and it used to have three lone pairs. Now it has only two. And now this top oxygen is still a single bond with three lone pairs around it. And this bottom right oxygen is still a single bond with three lone pairs around it. So this is a valid dot structure as well. So let's go ahead and put our brackets with a negative charge. And then, of course, we could have taken a lone pair of electrons from the oxygen on the bottom right. So I could have moved these in here to form a double bond. And so now, we would have our nitrogen double bonded to an oxygen on the bottom right. The oxygen on the bottom right now has only two lone pairs of electrons. The oxygen at the top, single bond with three lone pairs. And then the same situation for this oxygen on the bottom left. And so this is, once again, another possible dot structure. And so these are considered to be And the way to represent that would be this double-headed resonance arrow here. And I think when students first see resonance structures, the name implies that, in this case, the ion is resonating back and forth between these three different possible, equally valid dot structures. And that's not quite what's going on here. Each of these dot structures is an attempt to represent the structure of the ion. But they're really not the best way of doing that. You need to think about combining these three dot structures in a resonance hybrid of each other. And so let's go ahead and draw just a simple representation of a way of thinking about a resonance hybrid. So if I combined all three of my dot structures here into one picture, I had a double bond to one oxygen in each of my three resonance structures And so the top oxygen had a double bond in one of them, the bottom left in the middle one, and then the bottom right"
    },
    {
        "Q": "At 1:41 doesn't each O atom should have a negative charge and at 2:49 N atom should have a possitive charge?\n",
        "A": "Yes to both of these",
        "video_name": "bUCu7bPkZeI",
        "timestamps": [
            101,
            169
        ],
        "3min_transcript": "Now that we know how to draw dot structures, let's apply our rules to the nitrate anion. And we're going to see that we can draw a few different dot structures for this anion. And we're going to call those resonance structures of each other. But first, we need to calculate the total number of valence electrons. And so nitrogen is in Group 5 in the period table, therefore, five valence electrons. Oxygen is in Group 6, therefore, six valence electrons for each oxygen. I have three of them. So 6 times 3 is 18 valence electrons, plus the 5 from the nitrogen gives me 23. And I have a negative charge. This is an anion here. So we have to add one electron to that. So 23 plus 1 gives us a total of 24 valence electrons that we need to represent in our dot structure. So we know that nitrogen is going to go in the center, because oxygen is more electronegative. So nitrogen goes in the center. Nitrogen is bonded to three oxygens. So I can go ahead and put them in there like that. And let's see. How many valence electrons have we represented so far? Therefore, 24 minus 6 gives us 18 valence electrons left over. We're going to put those leftover valence electrons on our terminal atoms, which are our oxygens. And oxygen's going to follow the octet role. Currently, each oxygen has two valence electrons around it, the ones in magenta. So if each oxygen has two, each oxygen needs six more to complete the octet. And so I go ahead and put six more valence electrons on each one of my oxygens. Now each oxygen is surrounded by eight electrons. So the oxygens are happy. We added a total of six valence electrons to three oxygens. So 6 times 3 is 18. So we've used up all of the electrons that we need to represent. And so this dot structure, so far, it has all of our valence electrons here. Oxygen has an octet. So oxygen is happy. But nitrogen does not have an octet. If you look at the electrons in magenta, there are only six electrons around the nitrogen. And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here,"
    },
    {
        "Q": "\nI don't quite understand when at 6:28, Sal says that two particles \"buzz past each other.\" Can two particles collide too quickly? I thought that as long as they had energy greater than that of the reaction's activation energy, they would react.",
        "A": "Water and carbon dioxide can form carbonic acid . But in high temperature they can t form a lot because it will breakdown (They did not just buzz past each other , they did react ) (positive delta S & H, high T , negative delta G , not spontaneously reactions ). In low temperature they can form a lot of carbonic acid because it is stable. (positive delta S & H, low T , positive delta G). It will not spontaneously breaking down (not enough reaction s activation energy)",
        "video_name": "CHHu-iTwHjg",
        "timestamps": [
            388
        ],
        "3min_transcript": "and whether they're spontaneous depends on the temperature. So, over here, if we are dealing, our Delta H is less than zero. So, we're going to have a release of energy here, but our entropy decreases. What's gonna happen? Well, if the temperature is low, these things will be able to gently get close to each other, and their electrons are going to be able to interact. Maybe they get to a lower energy state, and they can release energy. They're releasing energy, and the electrons will spontaneously do this. But, the entropy has gone down. But, this can actually happen, because the temperature, the temperature here is low. \"Wait, doesn't that violate \"The Second Law of Thermodynamics?\" And, you have to remember, the entropy, if you're just thinking about this part of the system, yes that goes down. But, you have heat being released. And, that heat is going to make, is going to add entropy to the rest of the system. So, still, The Second Law of Thermodynamics holds because of this released heat. But, if you just look at the constituents here, the entropy went down. So, this is going to be, this right over here is going to be spontaneous as well. And, we're always wanting to back to the formula. If this is negative and this is negative, well, this is going to be a positive term. But, if 'T' low enough, this term isn't going to matter. 'T' is, you confuse it as the weighing factor on entropy. So, if 'T' is low, the entropy doesn't matter as much. Then, enthalpy really takes over. So, in this situation, Delta G, we're assuming 'T' is low enough to make Delta G negative. And, this is going to be spontaneous. Now, if you took that same scenario, but you had a high temperature, well now, you have these same two molecules. Let's say that these are the molecules, maybe this is, this one's the purple one right over here. You have the same two molecules here. Hey, they could get to a more kind of a, they could release energy. But over here, you're saying, The change in enthalpy is negative. But, they're buzzing past each other so fast that they're not gonna have a chance. Their electrons aren't gonna have a chance to actually interact in the right way for the reaction to actually go on. And so, this is a situation where it won't be spontaneous, because they're just gonna buzz past each other. They're not gonna have a chance to interact properly. And so, you can imagine if 'T' is high, if 'T' is high, this term's going to matter a lot. And, so the fact that entropy is negative is gonna make this whole thing positive. And, this is gonna be more positive than this is going to be negative. So, this is a situation where our Delta G is greater than zero. So, once again, not spontaneous. And, everything I'm doing is just to get an intuition for why this formula for Gibbs Free Energy makes sense. And, remember, this is true under constant pressure and temperature. But, those are reasonable assumptions if we're dealing with, you know, things in a test tube, or if we're dealing with a lot of biological systems. Now, let's go over here. So, our enthalpy,"
    },
    {
        "Q": "\nAt 3:03 mitosis goes in to a bunch of what?",
        "A": "Into a bunch of me . In other words, the bunch of cells turns into a human s organs.",
        "video_name": "PvoigrzODdE",
        "timestamps": [
            183
        ],
        "3min_transcript": "is going to keep replicating. So it will, you know, after one, so after one, and we're going to go into the details of the mechanics of mitosis, but after one round of mitosis, it is now two cells. It is now two cells. And I'm going to draw it, once again, I'm not going to draw it at scale. It's now two cells. I want to make sure I have enough space on my little chalkboard here. It has two cells. Instead of drawing all of the chromosomes, let me just say that each of these, in my nucleus, I still have 2N. I have the diploid number. So each of these two cells that it has differentiated to still have the full contingency. That's what mitosis does. It essentially replicates the entire cells. You have the same number of chromosomes. And then this process is just going to keep happening. These two characters are going to replicate, are going to replicate, and so then you're going to have, through mitosis, Mitsosis. So they just keep duplicating themselves. And each of these cells have the full contingency. 2N, the diploid number of chromosomes for, well, in this case it's going to be 46 And then this process is just going to keep happening. So this process is going to keep happening. I'll do dot dot dot to show that, you know, a lot of this has been going on. So mitosis is just going to keep happening. And so eventually you're going to have thousands of these cells, and eventually as we'll see, you're going to have millions and ten millions of them. So let me draw them really really really small. There's a bunch of them there. And each of them, each of them are going to have the diploid number of chromosomes. They're going to have 46 chromosomes. 23 pair of homologous chromosomes. So we now have a big ball of these here. they're going to differentiate into me. They're going to differentiate into the different parts of my body. So for example, these cells right over here might eventually, they'll keep replicating, but then it's them and their offspring might eventually differentiate into my brain cells. These cells here will keep replicating, and them and their offspring, I guess you could say, or the things that they replicate into, might differentiate into my heart. These right over here might differentiate into my lungs, and of course all of these eventually will differentiate into all the different, and they and their offspring will differentiate into all of the things that make me me. And so you have a lot more of this mitosis. You're eventually going to have a human being. So let me just say this is more mitosis going on. Mitosis."
    },
    {
        "Q": "\nOk so just to clarify, can mutations happen in both the female and male cells? Sal only sued sperm cells in his example at 8:45 so I just wanted to make sure. Also when Sal is talking about how the somatic cells differentiate into all the different body parts at 2:31 aren't they called stem cells?",
        "A": "the mutation occurs in both cells they somatic cells can be called as stem cells",
        "video_name": "PvoigrzODdE",
        "timestamps": [
            525,
            151
        ],
        "3min_transcript": "So if you're female you're going to produce eggs. If you're male you're going to produce sperm. But this is through the process of meiosis. Meiosis you're going to produce sperm in the case of a man, and you're going to produce ova in the case of a female. And this brings up a really interesting thing, because throughout biology we talk about mutations and natural selection and whatever else. And it's important to realize how mutations may affect you and your offspring. So if you have a mutation in one of the somatic cells here, let's say in a skin cell, or in you brain, or in the heart, that may affect your ability to, you know, especially if God forbid it's a really dangerous thing like cancer, and it happens when you're young, before you've had a chance to reproduce and you're not able to survive, But if this is happening in a somatic cell, it's not going to affect the DNA make-up of what you pass on. The DNA make-up of what you pass on, that's determined by what's going on in the gametes. So a mutation, if on the way to differentiating into gametes a mutation happens, so if one of these mutate and then keep replicating, so let's say there's a mutation here, and they keep replicating and they differentiate into the germ cells the mutation is right over there, then through meiosis that produces some mutated sperm. Then that would pass on to your, well, that has a chance of passing on to your children. Because once again, it might not be all of the sperm cells that have that mutation. It could be only a handful of the two to three hundred million of the sperm cells, and so if that mutation somehow makes it harder to kind of function, either fuse with an egg or even potentailly develop and kind of swim through fluid, then it still might not be the thing that makes it. So mutations only affect your offspring in the situation where those, the cells in which they happen are eventually differentiated into things, into gametes that you will pass on to your children."
    },
    {
        "Q": "\nIt talks about Theia colliding with Earth in 4:51 , how come the Earth is a perfect sphere? Did the collision do anything?",
        "A": "The collision would have released so much energy that the surface of the Earth would have been turned into molten rock. Gravity acts on the molten rock, and the molten rock fills the crater and leaves a relatively smooth surface.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            291
        ],
        "3min_transcript": "because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is"
    },
    {
        "Q": "\nAt 8:18, Sal mentions we don't have any rocks or artifacts of any sort from the Hadean Eon. So what makes us sure we had a Hadeon Eon?",
        "A": "Actually, in the last few decades of the 20th century, some rocks from this period were found in several countries. They were found in Greenland, Canada and west Australia. They were apparently altered due to volcanic dike. But, you see, there is proof that the Hadean Eon existed.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            498
        ],
        "3min_transcript": "kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned"
    },
    {
        "Q": "\nAt 9:50, during the late Hadean, Do these asteroid impacts contribute to the elemental makeup of our planet? I understand some metals are so heavy that they can only form in supernovae. But do these impacts provide enough energy to form new elements?",
        "A": "Yes, it added compounds (like most of our water) and elements to the planet. But it DID NOT create new elements.. Elements are only created inside a sun or a supernova, An asteroid impact will never create a new element.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            590
        ],
        "3min_transcript": "well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned the Earth still is a giant molten ball, it's just we live on the super-thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma, and if you go dig deeper, you'll have liquid iron. I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself a volcanic, molten ball, it began to harden as you get into the late Hadean eon, but we also had stuff falling from the sky and constantly colliding with Earth, and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there, and, as you can imagine, at this point there was no, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could've formed in the late Hadean eon, but for the most part this was just completely inhospitable for any life forming. we'll talk a little bit about the Archean eon."
    },
    {
        "Q": "At 5:03 What was the evidence that was used to come up with the idea that collision occurred between Earth and Theia? How strong is the evidence?\n",
        "A": "We have noticed that the Moon has a lot less heavy elements than the Earth and that the composition of the surface of the Moon is very similar to that of the Earth except that sometime in the past, it experienced extreme heat. This could mean a few things but it would seem to suggest that the Moon used to be part of the Earths mantle or crust and was launched into orbit by a powerful event.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            303
        ],
        "3min_transcript": "because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is"
    },
    {
        "Q": "at 06:43 so Thea formed into a ball fusing with earth, is that why earth is egg shaped?\n",
        "A": "Earth is not egg shaped. It is very close to a sphere, with a small bulge near the equator. The bulge is the result of the rotation of the earth.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            403
        ],
        "3min_transcript": "It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years"
    },
    {
        "Q": "\n@3:47 Sal draws a double bond? what is this? is this just showing they are all attached together? If they weren't drawn would this be incorrect?",
        "A": "It s complicated. Benzene doesn t really have alternating single and double bonds as depicted but there isn t a very good way to show its true structure. Another common way is the hexagon but with a circle in the middle.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            227
        ],
        "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have"
    },
    {
        "Q": "At 6:29, Sal draws a water molecule. I've always seen water molecules drawn wit the hydrogen atoms at the same angle. what is that angle and why is it at that angle\n",
        "A": "this is to do with bonding angles. as the water molecules have two lone pairs and two bonding pairs, the electrons are trying to get far away as possible which is 107.5 degrees which is why Sal had drawn them that way. hope this helps",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            389
        ],
        "3min_transcript": "but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen. It just so happens to be, what I just wrote down I kind of thought of in terms of empirical formula, in terms of ratios, but that's actually the case. A molecule of hydrogen, sorry, a molecule of water has exactly two hydrogens and, and one oxygen. If you want to see the structural formula, you're probably familiar with it or you might be familiar with it. Each of those oxygens in a water molecule are bonded to two hydrogens, are bonded to two hydrogens. So hopefully this at least begins to appreciate different ways of referring to or representing a molecule."
    },
    {
        "Q": "At 3:45, Sal said that you have a double bond every other bond. Could someone explain why that is and what do double bonds mean?\n",
        "A": "A double bond is just 2 bonds to the same atom, it s shown by 2 lines between the atoms instead of 1. With benzene it s a bit more complicated than this but it isn t worth worrying about yet.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            225
        ],
        "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have"
    },
    {
        "Q": "\nAt 3:44 why we have a double bond in every other carbon atom",
        "A": "That is a way of drawing the structure of benzene, which has what we call an aromatic ring. You will learn more about this when you start to study organic chemistry.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            224
        ],
        "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have"
    },
    {
        "Q": "At 3:46, why are there three double bonds in the carbon hexagon ? What do they signify ?\n",
        "A": "That is the structural formula for benzene. In a double bond, four electrons are shared between the carbon two atoms, compared with only two in a single bond. In fact, the structure of benzene is a little more complicated than shown and you will learn more about single and double bonds, and the benzene structure, in later videos. Benzene is not the best example that Sal could have chosen to explain the different types of formulae.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            226
        ],
        "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have"
    },
    {
        "Q": "at 3:47, he says that every other bond is a double bond, but he doesn't explain why that is. How do we know when something has a double bond and in this case would be every other one?\n",
        "A": "Yeah I think using this isn t the best example as this exact question seems to come up often. You know what a single covalent bond is right? 1 bond (2 shared electrons) between 2 atoms. Well a double bond is 2 covalent bonds (4 shared electrons) between atoms. You can t really know where to place those bonds simply from the formula once the number of carbon atoms gets this high, but if you continue with organic chemistry you are going to become very familiar with that benzene molecule.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            227
        ],
        "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have"
    },
    {
        "Q": "\nAt 4:40 ; What is organic chemistry?",
        "A": "It also involves the shape!",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            280
        ],
        "3min_transcript": "That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen."
    },
    {
        "Q": "2:35, How much evidence have scientists found for this theory? If that theory turns out true, that would be amazing!\n",
        "A": "First thing I will say is that a theory is never proven. We can t test all conceivable circumstances that could happen and with historical events like the incorporation of ancient ancestor of the mitochondria into a cell we were not there to observe it. As for the mitochondria having been a separate bacteria like entity it is very well supported by what we know about bacteria and the mitochondria.",
        "video_name": "i1dAnpSFbyI",
        "timestamps": [
            155
        ],
        "3min_transcript": "and this is actually a little bit more of a textbook visualization, as we'll learn in a few minutes or seconds that we now have more sophisticated visualizations of what's actually going on inside of a mitochondria, but we haven't actually answered all of our questions, but you might have already learned that, so let me make it clear, these are mitochondria. That's the plural. If we're just talking about one of them, we're talking about a mitochondrion. That's the singular of mitochondria. But you might have already learned, some time in your past or in another Khan Academy video, that these are viewed as the ATP factories for cells. So let me right it this way. So ATP factories. A-T-P factories and if you watched the videos on ATP or cellular respiration or other videos, I'd repeatedly talk about how ATP is really the currency for energy in the cell that when it's in its ATP form you have adenosine triphosphate. If you pop one of the phosphate groups off, you pop one of the P's off, it release energy from movement to thinking to all sorts of things that actually go on in your bodies, so you can imagine mitochondria are really important for energy, for when the cell has to do things. And that's why you'll find more mitochondria in things like muscle cells, things that have to use a lot of energy. Now before I get into the structure of mitochondria, I wanna talk a little bit about its fascinating past because we think of cells as the most basic unit of life and that is true, that comes straight out of cell theory, but it turns out the most prevalent theory of how mitochondria got into our cells is that at one time the predecessors, the ancestors to our mitochondria, were free, independent organisms, microorganisms. So they're descendent from bacterial-like microorganisms that might have been living on their own and they were maybe really good at processing energy or maybe they were even good at other things, they got ingested by what the ancestors of our cells and instead of just being engulfed and being torn to shreds and kind of being digested and eaten, it was like, \"Hey, wait, if these things stick around, \"those cells are more likely to survive \"because they're able to help process glucose \"or help generate more energy out of things.\" And so the cells that were able to kind of live in symbiosis have them kind of give a place for the mitochondria to live or the pre-mitochondria, the ancestor mitochondria, those survived and then through kind of the processes of natural selection, this is what we now associate, we now associate eukaryotic cells as having mitochondria, so I find this whole idea of one organism being inside of another organism in symbiosis even at the cellular level, that's kind of mind-boggling, but anyway, I'll stop talking about that and now let's just talk about the present, let's talk about what the actual structure of mitochondria are."
    },
    {
        "Q": "\nIn 9:14, Sal mentions Glycolysis. What is that?",
        "A": "Glycolysis is the process by which glucose or any other food substance is broken down for use in the creation of ATP",
        "video_name": "i1dAnpSFbyI",
        "timestamps": [
            554
        ],
        "3min_transcript": "if you're talking about one of these folds, you're talking about a crista, but if you're talking about more than one of them, you would call that a cristae, cristae. Sometimes I've seen the pronunciation of this as cristae, cristae or cristae, that's plural for crista. These are just folds in the inner membrane and once again the inner membrane is also a phospholipid bilayer. Now inside of the inner membranes, so between the outer membrane and the inner membrane you could imagine what this is gonna be called. That space is called the intermembrane space, not too creative of a name, intermembrane space and because of the porins, the small molecule concentration of the intermembrane space and then outside of the mitochondria, those concentrations are gonna be similar, but then the inner membrane does not have the porins in it and so you can actually have a different concentration on either side and that is essential for the electron transport chain. The electron transport chain really culminates with hydrogen, a hydrogen ion gradient being built between the two sides and then they flow down that gradient through a protein called ATP synthase which helps us synthesize ATP, but we'll talk more about that maybe in this video or in a future video, but let's finish talking about the different parts of a mitochondrion. So inside the inner membrane you have this area right over here is called the matrix. It's called, let me use this in a different color, this is the matrix and it's called the matrix 'cause it actually has a much higher protein concentration, it's actually more viscous than the cytosol that would be outside of the So this right over here is the matrix. When we we talk about cellular respiration, cellular respiration has many phases in it. We talk about glycolysis. Glycolysis is actually occurring in the cytosol. So glycolysis can occur in the cytosol. Glycolysis. But the other major phases of cellular respiration. Remember we talk about the citric acid cycle also known as the Krebs cycle, that is occuring in the matrix. So Krebs cycle is occuring in the matrix and then I said the electron transport chain which is really what's responsible for producing the bulk of the ATP, that is happening through proteins that are straddling the inner membrane or straddling the cristae right over here. Now we're just done. Probably one of the most fascinating parts of mitochondria, we said that we think that they are descendent"
    },
    {
        "Q": "at around 1:55 he says that there are 29 protons and electrons. are the amount of electrons and protons equal in every atom?\n",
        "A": "The electrons and protons are equal in every neutral atom. If an atom loses or gains an electron, we call them ions .",
        "video_name": "ZRLXDiiUv8Q",
        "timestamps": [
            115
        ],
        "3min_transcript": "- [Voiceover] All right, now we're gonna talk about the idea of an electric current. The story about currents starts with the idea of charge. We've learned that we have two kinds of charges, positive and negative charge. We'll just make up two little charges like that. And we know if they're the opposite sign, that there'll be a force of attraction between them. And if they have two like signs, here's two charges that are both positive, and these charges are gonna repel each other. So this is the basic electrostatics idea, and the same thing for two minus charges. They also repel. So like charges repel, and unlike charges attract. That's one idea. We have the idea of charge. And now we need a place to get some charge. One of the places we like to get charge from is copper, copper wires. A copper atom looks like this. Copper atom has a nucleus and it also has electrons flying around the outside, electrons in orbits around the outside. So we'll draw the electrons like this. There'll be orbits around this nucleus. Pretty good circles. And there'll be electrons in these. Little minus signs. There's electrons stacked up in this. And even farther out, there's electrons. So there's kind of a interesting looking copper atom. Copper, the symbol for copper is Cu, and its atomic number is 29. That means there's 29 protons inside here, and there's 29 electrons outside. It turns out, just as a coincidence for copper, that the last orbital out here that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms."
    },
    {
        "Q": "At 1:10 he talks about electrons in orbits; I thought electrons are believed to flow around the nucleus in \"shells\" (delineated by mathematical probability) as opposed to defined orbits...\n",
        "A": "Hello Chaba, You are correct about the mathematical probability. However, for a first introduction it is preferable to use the simplified model that defines orderly fixed orbits. It is easier to explain and visualize. Regards, APD",
        "video_name": "ZRLXDiiUv8Q",
        "timestamps": [
            70
        ],
        "3min_transcript": "- [Voiceover] All right, now we're gonna talk about the idea of an electric current. The story about currents starts with the idea of charge. We've learned that we have two kinds of charges, positive and negative charge. We'll just make up two little charges like that. And we know if they're the opposite sign, that there'll be a force of attraction between them. And if they have two like signs, here's two charges that are both positive, and these charges are gonna repel each other. So this is the basic electrostatics idea, and the same thing for two minus charges. They also repel. So like charges repel, and unlike charges attract. That's one idea. We have the idea of charge. And now we need a place to get some charge. One of the places we like to get charge from is copper, copper wires. A copper atom looks like this. Copper atom has a nucleus and it also has electrons flying around the outside, electrons in orbits around the outside. So we'll draw the electrons like this. There'll be orbits around this nucleus. Pretty good circles. And there'll be electrons in these. Little minus signs. There's electrons stacked up in this. And even farther out, there's electrons. So there's kind of a interesting looking copper atom. Copper, the symbol for copper is Cu, and its atomic number is 29. That means there's 29 protons inside here, and there's 29 electrons outside. It turns out, just as a coincidence for copper, that the last orbital out here that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms."
    },
    {
        "Q": "at 3:30 Hank says that all organelles are suspended in cytoplasm, but isn't cytosol the correct term for the fluid inside of the cell?\n",
        "A": "No cytoplasm is the correct term.",
        "video_name": "d9GkH4vpK3w",
        "timestamps": [
            210
        ],
        "3min_transcript": "to more than 400 million years ago. These plants were lycophytes, which are still around today and which reproduce through making a bunch of spores, shedding them, saying a couple of Hail Marys, and hoping for the best. Some of these lycophytes went on to evolve into scale trees, which are now extinct, but huge, swampy forests of them used to cover the earth. Some people call these scale tree forests coal forest because there were so many of them, they were so dense and they covered the whole earth, that they eventually fossilized into giant seams of coal, which are very important to our lifestyles today. This is now called the Carboniferous Period. See what we did there? Because coal is made of carbon, so they named the epoch of geological history over how face-meltingly intense and productive these forests were. I would give my left eyeball, three fingers on my left hand, the middle ones so that I could hang loose, and my pinky toe, if I were able to go back and see these scale forests, because they would be freaking awesome. Anyway, angiosperms, or plants that used flowers the Cretaceous Period, about 65 million years ago, just as the dinosaurs were dying out. Which makes you wonder if, in fact, the first angiosperms assassinated all the dinosaurs. I'm not saying that's definitely what happened. I'm just, it's a little bit suspicious. Anyway, on a cellular level, plant and animal cells are actually pretty similar. They're called eukaryotic cells, which means they have a good kernel, and that kernel is the nucleus. Not \"nuculus\". The nucleus can be found in all sorts of cells: animal cells, plant cells, algae cells. Basically, all the popular kids. Eukaryotic cells are way more advanced than prokaryotic cells. We have the eukaryotic cell and we have the prokaryotic cell. Prokaryotic basically means before the kernel, pro-kernel. Then we have the eukaryotic, which means good kernel. The prokaryotes include your bacteria and your archea, which you've probably met before in your lifetime. Every time you've had strep throat, for example. Or if you've ever been in a hot spring, or an oil well or something, they're everywhere: Like I said, eukaryotes have that separately enclosed nucleus, that is, that all-important nucleus that contains its DNA and is enclosed by a separate membrane, because a eukaryotic cell is a busy place. There's chemical reactions going on in all different parts of the cell. It's important to keep those places divided up. Eukaryotic cells also have these little stuff-doing factories called organelles, because we decided that we'd name everything something weird. But organelles, and they're suspended in cytoplasm, continuing with the really esoteric terminology that you're going to have to know. Cytoplasm is mostly just water, but it's some other stuff, too. Well, basically, if you want to know about the structure of the eukaryotic cell, you should watch my video on animal cells, which let's just link to it right here. Plant and animal cells are very similar environments. They control themselves in very similar ways. But obviously plants and animals are very different things, so what are the differences in a plant cell that makes it so different from an animal? That's what we're going to go over now. First, plants are thought to have evolved from green algae,"
    },
    {
        "Q": "At 5:59, Hank says that animals who can digest cellulose and lignin have a certain type of bacteria in their stomachs that break down these complex molecules into glucose molecules. What exactly are these bacteria, and how are they able to do this?\n",
        "A": "Fibrobacter succinogenes, Ruminococci, etc They have the enzyme cellulase that can break down cellulose to get the nutrients they need.",
        "video_name": "d9GkH4vpK3w",
        "timestamps": [
            359
        ],
        "3min_transcript": "Something plants inherited from their ancestors was a rigid cell wall surrounding the plasma membrane of each cell. This cell wall of plants is mainly made out of cellulose and lignin, which are two really tough compounds. Cellulose is by far the most common and easy to find complex carbohydrate in nature, though if you were to include simple carbohydrates as well, glucose would win that one. This is because, fascinating fact, cellulose is in fact just a chain of glucose molecules. You're welcome. If you want to jog your memory about carbohydrates and other organic molecules, you can watch this episode right here. Anyway, as it happens, you know who needs carbohydrates to live? Animals. But you know what's a real pain in the ass to digest? Cellulose. Plants weren't born yesterday. Cellulose is a far more complex structure than you'll generally find in a prokaryotic cell, and it's also one of the main things that differentiates a plant cell from an animal cell. Animals do not have this rigid cell wall. They have a flexible membrane that frees them up to move around and eat plants and stuff. However, the cell wall gives structure and also protects it, to a degree. Which is why trees aren't squishy, and they don't giggle when you poke them. The combination of lignin and cellulose is what makes trees, for example, able to grow really, really freaking tall. Both of these compounds are extremely strong and resistant to deterioration. When we eat food, lignin and cellulose is what we call roughage because we can't digest it. It's still useful for us on certain aspects of our digestive system, but it's not nutritious. Which is why eating a stick is really unappetizing and like your shirt. This is a 100% plant shirt, but it doesn't taste good. We can't go around eating wood like a beaver, or grass like a cow, because our digestive systems just aren't set up for that. A lot of other animals that don't have access to delicious donut burgers have either developed gigantic stomachs, like sloths, or multipe stomachs, like goats, in order to make a living eating cellulose. These animals have a kind of bacteria in their stomachs that actually does the digestion of the cellulose for it. glucose molecules which can then be used for food. Other animals, like humans, mostly carnivores, don't have any of that kind of bacteria, which is why it's so difficult for us to digest sticks. But there is another reason why cellulose and lignin are very, very useful to us as humans. It burns, my friends. This is basically what would happen in our stomachs. It's oxidizing. It's producing the energy that we would get out of it if we were able to, except it's doing it very, very quickly. This is the kind of energy, like this energy that's coming out of it right now, is the energy that would be useful to us if we were cows. But we're not, so instead, we just use it to keep ourselves warm on the cold winter nights. (blows air) Ow; it's on me; ow. Anyway, while we animals are walking around, spending our lives searching for ever more digestable plant materials, plants don't have to do any of that. They just sit there and they make their own food. We know how they do that."
    },
    {
        "Q": "At 4:08 why are zeros that are in between other digits significant? Aren't those in-between zeros just placeholders?\n",
        "A": "because there not at the very beginning they represent numbers in 705.001 the two zeros represent tenths and hundredths.",
        "video_name": "eCJ76hz7jPM",
        "timestamps": [
            248
        ],
        "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two."
    },
    {
        "Q": "At 3:51 I don't get how 10.0 is 3 significant digits because I thought zeros after the decimal or \"leading zeros\" is consider non-signifcant\n",
        "A": "The 1 is a significant digit because it is not a zero. The 3rd zero is a significant digit because it is after a decimal point AND after a significant digit (the 1). Also, the 2nd zero is significant because this zero is between two other significant digits.",
        "video_name": "eCJ76hz7jPM",
        "timestamps": [
            231
        ],
        "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two."
    },
    {
        "Q": "\n0:50. He shows that there are only three significant digits in that number. How do you discern significant digits from insignificant digits?",
        "A": "Significant digits are the numbers that actually give you the precision. The 0 s after the 7 tell you that the measurement is more precise, but the 0 s before are just basically placeholders. It is the first non-zero number then all the numbers, including zeros, after that which are significant figures.",
        "video_name": "eCJ76hz7jPM",
        "timestamps": [
            50
        ],
        "3min_transcript": "Let's see if we can learn a thing or two about significant figures, sometimes called significant digits. And the idea behind significant figures is just to make sure that when you do a big computation and you have a bunch of digits there, that you're not over-representing the amount of precision that you had, that the result isn't more precise than the things that you actually measured, that you used to get that result. Before we go into the depths of it and how you use it with computation, let's just do a bunch of examples of identifying significant figures. Then we'll try to come up with some rules of thumb. But the general way to think about it is, which digits are really giving me information about how precise my measurement is? So on this first thing right over here, the significant figures are this 7, 0, 0. So over here, you have three significant figures. And it might make you a little uncomfortable that we're not including these 0's that are after the decimal point and before this 7, that we're not including those. Because you're just like, that does help define the number. how precise our measurement is. And to try to understand this a little bit better, imagine if this right over here was a measurement of kilometers, so if we measured 0.00700 kilometers. This would be the exact same thing as 7.00 meters. Maybe, in fact, we just used a meter stick. And we said it's exactly 7.00 meters. So we measured to the nearest centimeter. And we just felt like writing it in kilometers. These two numbers are the exact same thing. They're just different units. But I think when you look over here, it makes a lot more sense why you only have three significant figures. These 0's are just shifting it based on the units of measurement that you're using. But the numbers that are really giving you the precision are the 7, the 0, and the 0. And the reason why we're counting these trailing 0's is that whoever wrote this number didn't have to write them down. look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved."
    },
    {
        "Q": "\nAt 3:40, why was all three numbers in the number 10.0 significant figures when the number was rounded to the nearest tenth?",
        "A": "When counting significant numbers, you count all the numbers, counting from the right of the first non-zero number. so as 1 is the first non-zero number and there are two zeros to the right of it you have 3 significant numbers. As opposed to the first example, which has 3 leading zeros and seven being the first non-zero number you would count 3 significant numbers in the number 0.00700",
        "video_name": "eCJ76hz7jPM",
        "timestamps": [
            220
        ],
        "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two."
    },
    {
        "Q": "\n2:20--How is it that there can be any kind of \"wind\" in space? Wind is movement between atoms, right? In space there's nothing to move....",
        "A": "Why thank you, Andrew M.",
        "video_name": "jEeJkkMXt6c",
        "timestamps": [
            140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 7:27, he says voyager 1 has a speed about 17 km/s. So how can be New Horizons the fastest spacecraft with a speed of about 15 km/s?",
        "A": "New Horizons has the fastest launch velocity of any spacecraft. However, Voyager 1 got several gravity assist boosts from the gas giant planets to its velocity, giving it a faster overall velocity.",
        "video_name": "jEeJkkMXt6c",
        "timestamps": [
            447
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 1:25, it was stated high energy electrons were spewed out at 400km/s. isnt that faster than light? i thought nothing travels faster than light?",
        "A": "That is a lot slower than light. Light speed in a vacuum is 299,792,458 m/s or 299,792.458 km/s. Empty space can expand faster than light, in the first few Planck times after the Big Bang, space was expanding at around 100-1000 light years per Planck time (light travels one Planck length per Planck Time, there are 5.85356655 \u00c3\u0097 10^50 planck lengths in a light year). However, space expanding faster than light does not violate Einstein s Special Relativity.",
        "video_name": "jEeJkkMXt6c",
        "timestamps": [
            85
        ],
        "3min_transcript": ""
    },
    {
        "Q": "6:42 If Alpha Centauri is closest, what's that really close light ball diagonal from the sun.\n",
        "A": "This is a 2D picture of a 3D mapping of stars. That star would simply be within the line of sight of the Sun, but closer or further from the viewer than the Sun.",
        "video_name": "jEeJkkMXt6c",
        "timestamps": [
            402
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:17, Sal says that light is able to pass through the half-silvered mirror. How can light pass through a mirror?",
        "A": "Same way it passes through other things it passes through, like windows.",
        "video_name": "3G_Q6AggQF8",
        "timestamps": [
            197
        ],
        "3min_transcript": "and then the whole solar system is orbiting around the center of the galaxy at a nice clip. The galaxy itself might be moving, so if you have some absolute frame of reference that's defined by the ether, well we are going to be moving relative to it. And if we're moving relative to it well maybe you just measure the speed of light in different directions and see whether the speed of light is faster or slower in a certain direction and then that might help you identify-- well, one, validate that the ether exists-- but also think about what our velocity is relative to the ether, relative to that absolute frame of reference. But the problem in the 19th century is that we didn't have any precise way of actually measuring--or a precise enough way of measuring--the speed of light where we could detect the relative difference due to the light going for or against, or into or away from, the actual direction of the ether wind. with first kind of breaking things open, starting to really make a dent in this whole idea of a luminiferous ether, is the Michelson-Morley Experiment. Michelson-Morley Experiment. They recognized, okay, we can't measure the speed of light with enough precision to detect has it gotten slowed down by the ether wind or sped up by the ether wind, but what we could do, and this is what Michelson and Morley did do, and I'm gonna do an oversimplification of the experiment, is that, okay, you have a light source, you have a light source right over here. So, you have a light source. And so that's going to send light in this direction. It's going to send light just like that. And what you do is you have a half-silvered mirror that allows half the light to pass directly through it and half of it to be reflected. So let's put a half-silvered mirror right over here. So, there's a half-silvered mirror. and this is just a simplification of it. Let me do it a little neater than that. So half will bounce off like that. And then the other half will be able to go through it. Will be able to go through it. It's a half-silvered mirror. And then we make each of those light rays-- we've essentially taken our original light ray and split it into two-- well then we'll then bounce those off mirrors. Bounce those off mirrors that are equidistant. And there are some adjustments when you actually have to factor in everything, but just as a simple notion, these things are just now going to bounce back. So, this one is now going to bounce back. It's half-silvered, it can go through, or part of it can go through, that mirror. So that's that ray. And then this one is going to bounce back. This one's going to go bounce back. And part of it is going to bounce into this direction. And then you can detect what you see."
    },
    {
        "Q": "At 1:57, why is the information of carbon-13 being added to carbon-12's information?\n",
        "A": "Because that s how the atomic weight of an element is calculated. (mass of isotope 1 x abundance of isotope 1) + (mass of isotope 2 x abundance of isotope 2) + ...",
        "video_name": "EPvd-3712U8",
        "timestamps": [
            117
        ],
        "3min_transcript": "- [Instructor] We have, listed here... We know that carbon 12 is the most common isotope of carbon on Earth. 98.89% of the carbon on Earth in carbon 12. And we know that, by definition, its mass is exactly 12 atomic mass units. Now that's not the only isotope of carbon on Earth. There are other isotopes. The next most frequent one is carbon 13. 1.11% of the carbon on Earth is carbon 13. And we can experimentally find that its mass is 13.0034 atomic mass units. So, these numbers that we have here, just as a review, these are atomic mass. These are atomic mass. And so, what we're gonna think about, in this video, is how do they come up with the atomic weight number that they'll give you on a periodic table like that? So, atomic weight. Well, in the video on atomic weight and on atomic mass, we see that the atomic weight is the weighted average of the atomic masses of the various isotopes of that element. So, to find this roughly 12.01, we take the weighted average of these two things. And what do we weight it by? We weight it by how common that isotope actually is. So, what we wanna do is, we could take 98.89% and multiply it by 12. And I'll rewrite this percentage as a decimal. So it'll be 0.9889 times 12. And, to that, we are going to add... We are going to add 1.11% times 13.0034. So, as a decimal, that's going to be 0.011. That's 1.11% is 0.011, oh, 111. So, what does that give us? Let's get our calculator out here. So, we are going to have 0.9889 times 12 is equal to 11.8668. And, to that, we are going to add... We are going to add 0.0111 times 13.0034. And I know it's going to do this multiplication first because it's a calculator knows about order of operations. And so, that's all going to be, as you can see, 12.01113774, which, if you were to round to the hundredths place, is how this atomic weight was gotten."
    },
    {
        "Q": "\nAt 9:00 and shortly after in the video, he refers to the snowball period in which the earth kind of iced over. My question is how did life survive this? Wouldn't that cold have negatively affected (rather, killed) both the prokaryotic and the eukaryotic organisms?",
        "A": "could be but some could survive underground where it is always 64 degrees Fahrenheit so they would be able to survive the whole snowball timeline and when everything thaws out they can come on the surface again",
        "video_name": "E1P79uFLCMc",
        "timestamps": [
            540
        ],
        "3min_transcript": "by with ultraviolet radiation from the Sun, which is very inhospitable to DNA and to life. And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation. The land was just open to it. Anything on the land would have just gotten irradiated. It's DNA would get mutated. It just would not be able to live. So what happened, and what I guess has to happen, and the reason why we are able to live on land now is that we have an ozone layer. We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the Sun. And now that oxygen began to accumulate, we have the Oxygen Catastrophe. Oxygen accumulates in the atmosphere. Some of that oxygen goes into the upper atmosphere. So we're now in this time period right over here. It goes into the upper atmosphere. to turn into ozone, which then can help actually block the UV light. And I'll do another video maybe on the ozone/oxygen cycle. So this oxygen production, it's crucial, one, to having an ozone layer so that eventually life can exist on the land. And it's also crucial because eukaryotic organisms need that oxygen. Now, the third thing that happened, and this is also pretty significant event, we believe that that oxygen that started to accumulate in the atmosphere, reacted with methane in the atmosphere. So it reacted with methane. And methane is an ozone-- not an ozone. It's a greenhouse gas. It helps retain heat in the atmosphere. And once it reacts with the oxygen and starts dropping out of the atmosphere as methane, we believe the Earth cooled down. And it entered its first, and some people believe it's longest, snowball period. So that's what they talk about right here It's sometimes called the Huronic glaciation. And that happened because we weren't able to retain our heat, if that theory is correct. And so the whole-- as the theory goes-- the whole Earth essentially just iced over. So as we go through the Proterozoic Eon, I guess the big markers of it is it's the first time that we now have an oxygen-rich atmosphere. It's the first time that eukaryotes can now come into existence because they now have oxygen to, I guess we could say, breathe. And the other big thing is now this is where the ozone forms. So this kind of sets the stage for in the next eon, for animals or living things, to eventually get on to the land. And we'll talk about that in the next video."
    },
    {
        "Q": "At 6:11, why is tension considered an internal force? What is the difference between internal and external forces?\n",
        "A": "My teacher taught me to just draw a big circle around the whole system you re trying to deal with. Anything outside of that circle is external, and anything inside is internal. If you drew a circle around both of the boxes and the string attaching them, the tension force is inside of the circle and thus internal.",
        "video_name": "_0nDUXO0k7o",
        "timestamps": [
            371
        ],
        "3min_transcript": "it's equal to the force of kinetic friction \"mu\" \"k\" times \"Fn\" and the \"mu\" \"k\" is going to be 0.2, you have to be careful because the \"Fn\" is not just equal to \"m\" \"g\" the reason is that on an incline the normal force points this way so the normal force doesn't have to counteract all of gravity on an incline it just has to counteract that component of gravity that's directed perpendicular to the incline and that happens to be \"m\" \"g\" \"cos(theta)\" for an object on an incline and if that makes no sense go back and look at the video on inclines or look at the article on inclines and you'll see that this component of gravity pointing into the surface is \"m\" \"g\" cosine that means that normal force is \"m\" \"g\" cosine. Because there's no acceleration in this perpendicular direction and I have to multiply by 0.2 because I'm not really plugging in the in this perpendicular direction. I'm plugging in the kinetic frictional force this 0.2 turns this perpendicular force into this parallel force, so I'm plugging in the force of kinetic friction and it just so happens that it depends on the normal force. That's why I'm plugging that in, I'm gonna need a negative 0.2 times 4 kg times 9.8 meters per second squared. And then I need to multiply by cosine of the angle in this case the angle is 30 degrees. Alright, now finally I divide by my total mass because I have no other forces trying to propel this system or to make it stop and my total mass is going to be 13 kg. You might object and think wait a minute, there's other forces here like this tension going this way, why don't we include that? benefit and appeal of treating this two-mass system as if it were a single mass is that we don't have to worry about these internal forces, it's there but that tension is also over here and on this side it's resisting the motion because it's pointing opposite the directional motion. On this side it's helping the motion, it's an internal force the internal force is canceled that's why we don't care about them, that's what this trick allows us to do by treating this two-mass system as a single object we get to neglect any internal forces because internal forces always cancel on that object. So if we just solve this now and calculate, we get 4.75 meters per second squared is the acceleration of this system. So this 4 kg mass will accelerate up the incline parallel to it with an acceleration of 4.75 meters per second squared. This 9 kg mass will accelerate downward"
    },
    {
        "Q": "At 1:42 he says hat you shouldn't open a plug without a professional, why not? what might happen?\n",
        "A": "I would say only worry about that if you plan to plug it back in. if certain parts are messed up inside the plug it can be a fire hazard, parts may explode, or it may ruin some of the wiring in your house. If you make sure it is unplugged and you won t be plugging it back in you wouldn t need a professional.",
        "video_name": "gFFvaLzhYew",
        "timestamps": [
            102
        ],
        "3min_transcript": "Alright today we're going to take a look at the Conair 1875 hair dryer.We're going to look at the different systems and functions inside of it, how it was made and how it works. And we're also going to take a look at how they were able to produce a hair dryer for less than $8.00 and still make a profit and still stay in business as a company because that's a very low price and the way low price. And the way they've done that is they've reduced a lot of cost and complexity and we'll take a look at how they've done that. So the first thing I want to take a look at is the plug here. This is called a \"ground fault interruptor circuit plug\" and it has two different sized prongs right here. There's a larger prong and a smaller prong. And that's very important. The larger prong is the neutral prong and that means you can't plug this in incorrectly, it has to go in in only one way. And that means that the power is grounded properly. So the power always goes to ground and that's a critical thing in a circuit like this. So what this plug does it's actually pretty smart it can tell if there's a power difference might occur when the hair dryer was, say, dropped in water or there was some sort of short that happened. Inside the hair dryer there are open electrical contacts that if they're put into water or some other conductive fluid they'll short out. And it will cause the, you know, it'll electrify the fluid. And in the past that was a huge problem because people would get schocked or electrocuted and now it's not as big a deal because we have these ground fault interruptor circuits. So let's take a look at what's inside of that. And I've already popped this apart to some degree. I'm going to see if I can get it the rest of the way here. Now I want to say one thing really quick here from a safety stand point: It's absolutely critical that you DON'T take apart any plugs ever without a professionnal! And if you do have a professional and you do end up taking apart a plug like this make sure that you never ever moulded housing it was injection moulded. There were two pieces of steel that came together and the molten plastic was injected and you can see there are little pin marks right here. And pins came in inside the mold and pushed this part out. And then there's a little plastic piece here with a spring and that's for the test switch. So the test switch pushes on this part right here on the printed circuit board and the reset switch is right here. So you push on the reset switch and it will reset it so if it gets triggered you can still use your hair dryer again later. So one thing I want to take a look at here is the printed circuit board here. So we've got a lot of really cool things happening on this printed circuit board. It is made out of fiberglass. It's got a thin layer of copper applied to it. And then on top of the copper is a layer of lacquer. (The copper) Before they put the copper layer down they actually etch away parts of the copper. So"
    },
    {
        "Q": "\nHow would you name tert-butyl chloride in the UPAC system please? (7:41)",
        "A": "First, find the longest carbon chain: 3 carbons, so it is a propane. Next, find the substituents: one methyl group at carbon 2 and one chloro group also at carbon 2. We name the substituents in alphabetical order, so the overall name is: 2-chloro-2-methylpropane.",
        "video_name": "aaZ-isZs4ko",
        "timestamps": [
            461
        ],
        "3min_transcript": "So let me write that in here, so dimethyl. This time we have a methyl group at two and five. So I'll write in here two and five. And then a bromine at five, so it'd be 5-bromo. So our full name would be 5-bromo-2,5-dimethylheptane. Now let's name this compound. So it's the same approach so let's count up our carbons here in our longest carbon chain. One, two, three, four, and five. So a five carbon alkane is pentane so I write in here pentane. Next I think about how do I number my carbon chain to give the lowest number possible to my first substituent? If I number it from the left, I give the bromine here a one. So that's the best way to number it. So that'd be one, two, three, four, and five. So I have a methyl group at four, a chlorine at three, and a bromine at one. So it's gonna be the bromine first, then the chlorine and then the methyl. So I'll go ahead and put the methyl in here. So this is at carbon four so 4-methyl. Next I have my chlorine at three so that would be 3-chloro. And finally my bromine at one, so 1-bromo. So the full name is 1-bromo-3-chloro-4-methylpentane. Now let's look at this compound. So we have one, two, three, four, five and six. So six carbons in our chain, a six carbon alkane is hexane. So I write in here hexane. Now I have two substituents, I have a bromine and I have a methyl group. So let's number from the left and see what happens. One, two, three, four, five and six. That gives me a bromine at two and a methyl at five. So this would be one, two, three, four, five, six. This gives me a methyl at two and a bromine at five. So just looking at numbers, we can't decide who wins here. So we have two versus two which is a tie, five versus five so that's a tie. So the way to break the tie is to think about the alphabet for your substituents. So we have bromine versus methyl so b versus m. Obviously b comes before m in the alphabet so the bromine's gonna win. We're gonna give the bromine the lower number. And that of course is the example on the left where the bromine is coming off of carbon two. So we're gonna choose the system on the left or the way of numbering the carbon chain from the left. Which means we have a methyl group at five. So 5-methylhexane. And then bromine at two, so 2-bromo."
    },
    {
        "Q": "The compound at 2:05 is a haloalkane so shouldn't it be named 3-Bromo-4,6-dimethylheptane? I was taught that the functional group is given first preference?\n",
        "A": "You were taught wrong. A halogen substituent and an alkyl substituent have equal priority, so you number from the end that is closer to either one. IN A TIE, halogens take priority over alkyl groups. Hence, 2-bromo-4-methylhexane is correct, while 4-bromo-2-methylhexane is not.",
        "video_name": "aaZ-isZs4ko",
        "timestamps": [
            125
        ],
        "3min_transcript": "- [Lecturer] You often see two different ways to name alkele halides. And so we'll start with the common way first. So think about alkele halides. First you wanna think about an alkele group, and this alkele group is an ethyl group, there are two carbons on it. So we write in here ethyl. And then since it's alkele halides, you wanna think about the halogen you have So this is chlorine so it's gonna end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case it's gonna be named as a halo alkane. So for a two carbon alkane that would be ethane. So I write in here ethane. And of course our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro, notice the spelling on that. If I had a bromine instead of the chlorine, And finally if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here ioto. Let's name this compound using our common system. So again think about the alkyl group that is present. So we saw in earlier videos this alkyl group is isopropyl. So I write in here isopropyl. And again we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain, so that'd be one, two and three. I know that is propane so I write in here propane. And we have a chlorine attached to carbon two. So that would be 2-chloro, 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen and we see how many alkyl groups There's only one alkyl group, this methyl group here, attached to this carbon so that's called primary. So ethyl chloride is an example of a primary alkyl halide. If you look at isopropyl chloride down here. This is the carbon that's bonded to our halogen and that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound the carbon that is bonded to our halogen is bonded to three alkyl groups. So three methyl groups here. So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time. For larger molecules it's usually easier to use"
    },
    {
        "Q": "At 7:22 I got lost at the finding the ration of H2C2O4 AND NaOH how did he get this result of 1;2\n",
        "A": "H2C2O4 + 2NaOH ----> Na2C2O4+2H20 The chemical equation.",
        "video_name": "XjFNmfLv9_Q",
        "timestamps": [
            442
        ],
        "3min_transcript": "And we know its concentration, 0.485 molar-- so let me do that in a different color-- 0.485 molar, this information allows us to figure out the actual molecules of sodium hydroxide. So we want to multiply this by-- we have 0.485 moles of sodium hydroxide for every 1 liter of this solution. That's what the molarity tells us. We have 0.485 moles per liter. So the liters cancel out, and then now we're going to actually have to get a calculator out. And this'll tell us how many moles of sodium hydroxide we have in this solution. So let me get my calculator. There we go. All right, let me just multiply these two numbers. So we have 0.03447 times 0.485 is equal to-- let me put this And we only have three significant digits here, so we're going to round to three significant digits. So we'll just go with 0.0167. So let me move that over off the screen. So this is going to be equal to 0.0167, and all we have left here are moles of sodium hydroxide. Now we know that this many moles of sodium hydroxide are going to completely react with however many moles of oxalic acid we have. Now we know that we need two moles of this for Or for every mole of oxalic acid that completely reacts, we need two moles of this. So let's write that down. And then you color. So we need two moles of sodium hydroxide, we got that from our balanced equation right there, and it's obvious it needs one mole, or one molecule will take this proton, and then you need another molecule to take that proton. So we need two moles of sodium hydroxide for every one mole of oxalic acid. For every one mole of H2C2O4. So essentially, we are just going to divide this number by 2. Let me get the calculator back. So we're just going to divide 0.0167 divided by 2. Once again, three significant digits 0.00835."
    },
    {
        "Q": "\nat 2:09, why do they pump blood into the vein not the artery?",
        "A": "It because the Artery is under very high pressure, and the vein has very low. Does this help?",
        "video_name": "Nnqp_3HMlDU",
        "timestamps": [
            129
        ],
        "3min_transcript": "- [Voiceover] Let's say that this is a red blood cell. What makes up the outer layer of this red blood cell? Since it's a cell, it has a cell membrane, and that's made up of lipids. But embedded in those lipids, there's all kinds of proteins and molecules, some of which I'm drawing here, that have all kinds of different functions. There are two of them that are sort of more important than the others, at least for this topic that we're going to talk about, and those are the A molecule and the B molecule. I think you'll be pretty happy with those names, not too hard to remember. I'm actually calling them molecules and not proteins because they're actually not proteins. They are actually something called glycolipids, which I actually didn't realize at first. Glycolipids, glyco meaning a sugar group, lipid meaning a fat group, so it's some kind of mix of a sugar and fat. You can look it up if you're interested. have both of these molecules, these glycolipids, on their red blood cells. Some people do have both, but some have only one. For example, some people might have only the A, or, as you can imagine, some people would have only the B, and some people, can you figure out the last possibility? Some people have neither A nor B. Of course, all these people have all kinds of other proteins and molecules embedded in their red blood cells. The reason why we care about this and why I'm talking about these As and these Bs, is that in medicine, we often have to give blood transfusions. Let's say you got in a car accident and you lost a lot of blood. You're rushed to the hospital. If you've lost enough blood, they'll give you a transfusion of blood. Transfusion just means they'll put a needle in your vein and pump blood into your veins. and it has to do with these A and B groups. For example, it turns out, and we'll explain this, but for example, it turns out that if you are the kind of person who has this kind of blood that only has As on your blood cells, then it turns out that you can't get a blood transfusion from someone who has this kind of blood, As and Bs. Let's learn why. If you remember from the immune system, there's something called an antibody. We usually draw it in this shape here. If you remember from the immune system, your body has something called antibodies, and it uses these antibodies to fight things that it doesn't want in the body. For example, if you have a bacterium, I'm drawing one here, you'll have an antibody that will bind to that bacterium. The purpose of that is that now your body knows that is should destroy this bacterium. This antibody is kind of like a tag"
    },
    {
        "Q": "\non 4:23, shouldn't there also be a H2O at the left side of the equation?",
        "A": "No, because the H20 (2 Hydrogens + 1 Oxygen) bonded with one more Hydrogen. Thus it s now H3O+ (positive balance since H is a proton). An equation features the same molecules but rearranged.",
        "video_name": "Y4HzGldIAss",
        "timestamps": [
            263
        ],
        "3min_transcript": "so the hydrogen is just going to be left as a hydrogen proton. And then the chlorine, the chlorine has just nabbed that electron. It had the electrons it had before, and then it just nabbed an electron from the hydrogen, and so it now has a negative charge, and these are both in aqueous solution still. It's still, they're still both dissolved in the water. And so you see very clearly here, you put this in an aqueous solution, you're going to increase the amount of, you're going to increase the amount of hydrogen ions, the amount of protons in the solution. And we've talked about this before, you'll often see a reaction written like this, but the hydrogen protons, they just don't sit there by themselves in the water. They are going to bond with the water molecules to actually form hydronium. So another way that you'll often see this is like this. You have the hydrochloric acid, hydrochloric acid. It's in an aqueous solution, and then you have the H2O. You have the water molecules, H2O, and you'll sometimes see written, okay, it's in its liquid form, and it's going to yield. Instead of just saying that you have a hydrogen ion right over here, you'll say, \"Okay, that thing, \"the hydrogen is actually gonna get bonded \"to a water molecule.\" And so what you're gonna be left with is actually H3O. Now this thing, this was a water molecule, and all it got was a hydrogen ion. All that is is a proton. It didn't come with any electrons, so now this is going to have a positive charge. It's going to have a positive charge, and we could now say that this is going to be in an aqueous solution, hydronium is going to be in an aqueous solution, and you're going to have plus, now you're going to still have the chloride ion, Chloride, chloride anion, and this is still in an aqueous solution. It is dissolved in water, and remember all that happened here is that the chlorine here took all of the electrons, leaving hydrogen with none. Then that hydrogen proton gets nabbed by a water molecule and becomes hydronium. So even by this definition you might say it increases the concentration of hydrogen protons. You could say it increase the concentration of hydronium, of hydronium right over here. Hydronium ions. So that makes, by the Arrhenius definition, that makes hydrochloric acid a strong acid. That makes it a strong acid. Now what would be a strong base by the Arrhenius definition of acids and bases? Well one would be sodium hydroxide. So let me write that down, so if I have sodium hydroxide,"
    },
    {
        "Q": "\nAt 6:40 if the water splits the sodium hydroxide to OH- and Na +\nthen why the sodium ion dont react with the water? its an alkaline metal",
        "A": "The polar nature of water causes water to form intermolecular (between molecule) bonds with the sodium, which is what causes it to split from hydroxide ion (OH-). The sodium ion doesn t react with water to form a new chemical substance because it is happy with eight outer shell electrons.",
        "video_name": "Y4HzGldIAss",
        "timestamps": [
            400
        ],
        "3min_transcript": "and then I have the hydroxide. That's an oxygen bonded to a hydrogen. So that's sodium hydroxide, and actually if you wanted to see what this molecule looked like you have a oxygen having a covalent bond to a hydrogen. Let me do these in different colors. Oxygen has a covalent bond to a hydrogen. to a hydrogen right over here. And it actually has three alone pairs. It actually has three alone pairs right over here. It's actually nabbed the electron from, from somebody some place, and so it's going to have a negative charge. It is going to have a negative charge. Actually I could write it both, It has a negative charge, and then you have a sodium ion that has lost its electron somehow. So you have a sodium ion that has lost an electron somehow, so it has a positive charge, to the oxygen right over here, making the oxygen negative and making the sodium positive, and so this is now positive, this is negative, they're going to be attracted to each other, and they form an ionic bond, so sodium hydroxide, they have an ionic bond because the sodium is actually positive, and the hydroxide part right over here. That is negative, and that's what draws them together, but anyway, you put this in an aqueous solution. You throw some sodium hydroxide into an aqueous solution, it will disassociate into, into sodium with its positive charge, the sodium ions, and actually you know the sodium ion is still part of this. That's what makes it attracted to the hydroxide anion, but it's still going to be in an aqueous solution, and then you're going to have the hydroxide. You're going to have the hydroxide anion, This has a negative charge, and it's still going to be dissolved in the water, so aqueous solution. So you throw sodium hydroxide in water, it's going to increase the concentration. It's going to increase the concentration of hydroxide in the water. It's going to increase the hydroxide concentration, and so by the Arrhenius definition of acids and bases, this would be a strong Arrhenius base. This would be a strong, a strong base by the Arrhenius definition. Now, and I encourage you to look at that relative to the other definitions, the Bronsted-Lowry definition of acids and bases and the Lewis definition of acids and bases, and see how you would think about categorizing things."
    },
    {
        "Q": "how does sal say in 5:03 that hcl is a strong acid?\n",
        "A": "Arrhenius acids are proton donors, they completely ionize to give H+ and conjugate base. HCl - H+ Cl- H2SO4 - 2H+ SO4-2 HNO3- H+ - NO3-1 The key word is completely, they do not form an equilibrium.",
        "video_name": "Y4HzGldIAss",
        "timestamps": [
            303
        ],
        "3min_transcript": "and then you have the H2O. You have the water molecules, H2O, and you'll sometimes see written, okay, it's in its liquid form, and it's going to yield. Instead of just saying that you have a hydrogen ion right over here, you'll say, \"Okay, that thing, \"the hydrogen is actually gonna get bonded \"to a water molecule.\" And so what you're gonna be left with is actually H3O. Now this thing, this was a water molecule, and all it got was a hydrogen ion. All that is is a proton. It didn't come with any electrons, so now this is going to have a positive charge. It's going to have a positive charge, and we could now say that this is going to be in an aqueous solution, hydronium is going to be in an aqueous solution, and you're going to have plus, now you're going to still have the chloride ion, Chloride, chloride anion, and this is still in an aqueous solution. It is dissolved in water, and remember all that happened here is that the chlorine here took all of the electrons, leaving hydrogen with none. Then that hydrogen proton gets nabbed by a water molecule and becomes hydronium. So even by this definition you might say it increases the concentration of hydrogen protons. You could say it increase the concentration of hydronium, of hydronium right over here. Hydronium ions. So that makes, by the Arrhenius definition, that makes hydrochloric acid a strong acid. That makes it a strong acid. Now what would be a strong base by the Arrhenius definition of acids and bases? Well one would be sodium hydroxide. So let me write that down, so if I have sodium hydroxide, and then I have the hydroxide. That's an oxygen bonded to a hydrogen. So that's sodium hydroxide, and actually if you wanted to see what this molecule looked like you have a oxygen having a covalent bond to a hydrogen. Let me do these in different colors. Oxygen has a covalent bond to a hydrogen. to a hydrogen right over here. And it actually has three alone pairs. It actually has three alone pairs right over here. It's actually nabbed the electron from, from somebody some place, and so it's going to have a negative charge. It is going to have a negative charge. Actually I could write it both, It has a negative charge, and then you have a sodium ion that has lost its electron somehow. So you have a sodium ion that has lost an electron somehow, so it has a positive charge,"
    },
    {
        "Q": "at 2:41 Shouldn't the 2-propanol be named as 1-methyl-1-ethanol since it would give the hydroxyl group the lowest number ie. 1?\n",
        "A": "No, you pick the longest chain that has the OH attached and then give the lowest possible number to the C bearing the OH group. The longest chain has three C atoms, and the OH group is on C-2.",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            161
        ],
        "3min_transcript": "Now, if that O-H weren't there, then we'd have just a three carbon alkane, which we would call propane. But since we have our O-H there, this is actually an alcohol. Alcohol is going to have the -ol ending. So this is called propanol. So let's go ahead and write propanol here. And the O-H group is coming off of carbon one. So we're going to say that's one propanol like that. How would we classify this alcohol? Well, the carbon right here that is bonded to the O-H, that carbon is bonded to one other carbon right here. So this would be a primary alcohol. So one propanol is a primary alcohol in terms of its classification. Let's look at a similar-looking molecule. Still three carbons, but this time we put the O-H on the carbon in the middle there. So once again, you're going to go ahead and number it. This is carbon one, this carbon two, this is carbon three. So it's also called propanol. The difference is the hydroxl group is on a different carbon, It's now on carbon two. So we're going to write two-propanol here, which is the IUPAC name. This is also called isopropanol, rubbing alcohol, it's all the same stuff. But two-propanol would be the proper IUPAC nomenclature. How would you classify two-propanol? So once again, we find the carbon attached to the O-H. That's this one. How many carbons is that carbon attached to? It's attached to one and two other carbons. So therefore, this is a secondary alcohol. So we have an example of a primary alcohol, and an example of a secondary alcohol here. Let's do a little bit more complicated nomenclature question. And so let's go ahead and draw out a larger molecule with more substituents. So let's put an O-H here. And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring?"
    },
    {
        "Q": "At 4:52 as by the longest chain rule i think the name would be 6-chloro-6-methyl-4-octane-1-ethyl-ol or maybe 6-chloro-6-methyl-1-ethyl-ol-4-octane instead of what you said in the video : 5-chloro-5-methyl-3-propyl-2-heptanol ?? I am not sure if i am saying right or wrong but i have learned from previous videos that the chain should be longest.\n",
        "A": "You took into consideration only the longest carbon chain rule and forgot about other rules. we can never have ethyl-ol it should be ethan- ol and according to you, that should be octan-1-ol. that s true that the chain should be longest but it also has been mentioned in the video that we have to provide the lowest locant rule to the -OH and other groups.",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            292
        ],
        "3min_transcript": "And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring? So that would be propyl. So we have three-propyl. So go ahead and write three-propyl in here. And what else do we have? At carbon five, we have two substituents. So we have a chloro group right here. And we have a methyl group right here. And remember your alphabet. Right, so C comes before M. So we can go ahead and put our methyl in there. All right, coming off of carbon five, so that would be five-methyl, like that. And then also coming off five, we have chloro. So five-chloro. Right in here. And that should do it. Everything follows the alphabet rule. So we have five-chloro, five-methyl, three-propyl, two-heptanol for this molecule. What about a problem that includes some stereochemistry? So let's say they give us one where we have to worry about stereochemistry. So let's see, something like that. And let's make an O-H group going away from us. And then let's go ahead and make this one coming out at us like that. So give the full IUPAC name for this molecule, and you have to include stereochemistry. So once again, find your longest carbon chain that includes your O-H group. And you want to give that O-H the lowest number possible so it takes precedence over things like alkyl groups, and halogens, and double bonds. So we're going to start from the left. So one, two, three, four, five, six, seven, eight, nine like that. So we have a nine-carbon alcohol. So that would be nonanol. And the alcohol is coming off of carbon three. The O-H is coming off of carbon three. So we have three-nonanol. Like that. So three-nonanol."
    },
    {
        "Q": "\nWhat greek character is Sal using at 7:30 when he's explaining the polarity of water and what does it mean? For instance, a theta usually means angle and delta means change.",
        "A": "It is another form of the Greek letter delta and in this case means slightly or partially .",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            450
        ],
        "3min_transcript": "their electrons to roam, they all become slightly positive. And so they're kind of embedded in this mesh or this sea of electrons. And so the metallic crystals, depending on what cases you look at, sometimes they're harder than the ionic crystals, sometimes not. Obviously, we could list a lot of very hard metals, but we could list a lot of very soft metals. Gold, for example. If you take a screwdriver and a hammer, you know, pure gold, 24-carat gold, if you take a screwdriver and hit it onto the gold, it'll dent it, right? So this one isn't as brittle as the ionic crystal. It'll often mold to what you want to do with it. It's a little bit softer. Even if you talk about very hard metals, they tend to not be as brittle, because the sea of electrons kind of gives you a little give when you're moving around the metal. But that's not to say that it's not hard. In fact, sometimes that give that a metal has, or that ability to bend or flex, is what actually gives it its strength So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water,"
    },
    {
        "Q": "Sal you mention at 7:53 that the Na+ is attracted to the o- of the water and Cl- is attracted towards the H+ side of water. My question is that why they get attracted to the o- and h+ poles of the water? Is this due to the more positive charge on the h+ than Na+ and cause Na have remaining 10 electrons in its outer shell? and the same case in Cl-.\n",
        "A": "The Na+ is much more positive than the partial H+ in water, BUT many H2O molecules work together to seperate each Ion from the crystal lattice. So the attraction between 1 Na+ and 1 Cl- is stronger than 1 Cl- and one partial H+, many partial H+ polar molecules can pull the ions apart. This works the same for the Na+ and partial O- attraction as well.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            473
        ],
        "3min_transcript": "So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water, not a lot of charge that is really movable in this state. But here, all of a sudden, we have these charged particles that can move. And because they can move, all of a sudden, when you put salt, sodium chloride, in water, that does become conductive. So anyway, I wanted you to be at least exposed to all of these different forms of matter. And now, you should at least get a sense when you look at something and you should at least be able to give a pretty good guess at how likely it is to have a high boiling point, a low boiling point, or is it strong or not. And the general way to look at it is just how strong are the intermolecular bonds. Obviously, if the entire structure is all one molecule, it's going to be super-duper strong. And on the other hand, if you're just talking about neon, a bunch of neon molecules, and all they have are the London dispersion forces, this thing's going to have ultra-weak bonds."
    },
    {
        "Q": "\nAt 1:37, what does covalent network mean?",
        "A": "It means the atoms form a rigid network as Sal explains.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            97
        ],
        "3min_transcript": "In the last video, I talked about some of the weaker intermolecular forces or structures of elements. The weakest, of course, was the London dispersion force. In this video, I'll start with the strongest structure, and that's the covalent network. So if you have a covalent network crystal and let me actually define the word crystal. Crystal is just when you have a solid, where the molecules that make up the solid are in a regular, relatively consistent pattern, and this is versus an amorphous solid, where everything is kind of just a hodge-podge and there's different concentrations of different things, of different ions, and different molecules, and different parts of the solid. So crystal is just a very regular structure. Ice is a crystal, because once you get the temperature low enough in water, the hydrogen bonds form a crystal, a regular structure. And we've talked about that a bunch. But the strongest of all crystal structures And the biggest, or the prime, example of that is carbon when it forms a diamond. So in the covalent network, carbon has four valence electrons, so it always wants four more. So when carbon shares with itself, it's very happy. So what it can do is it can form four bonds to four more carbons, and then each of those carbons can form four more bonds to four more carbons. And this one, 1, 2, 3, and it just keeps going on. This is the structure of a diamond. And the reason why this is such a strong structure is because you can almost view the entire -- in fact, you should view the entire diamond as one molecule, because they all have covalent bonds. These are actual sharing of electrons, and these are actually the strongest of all molecular bonds. So you can imagine if the entire solid you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron"
    },
    {
        "Q": "At 2:15, what is the melting point of carbon?\n",
        "A": "That depends on what allotrope of carbon it is and what the atmospheric pressure is. Graphite normally sublimes rather than boils. The temperature at which this happens depends on the pressure but it is usually above 3900 K. Diamond can be made to boil at very, very extreme pressures and temperatures. If you can get the pressure high enough, diamond will boil at over 5100 K, which is close to the temperature of the surface of the sun.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            135
        ],
        "3min_transcript": "In the last video, I talked about some of the weaker intermolecular forces or structures of elements. The weakest, of course, was the London dispersion force. In this video, I'll start with the strongest structure, and that's the covalent network. So if you have a covalent network crystal and let me actually define the word crystal. Crystal is just when you have a solid, where the molecules that make up the solid are in a regular, relatively consistent pattern, and this is versus an amorphous solid, where everything is kind of just a hodge-podge and there's different concentrations of different things, of different ions, and different molecules, and different parts of the solid. So crystal is just a very regular structure. Ice is a crystal, because once you get the temperature low enough in water, the hydrogen bonds form a crystal, a regular structure. And we've talked about that a bunch. But the strongest of all crystal structures And the biggest, or the prime, example of that is carbon when it forms a diamond. So in the covalent network, carbon has four valence electrons, so it always wants four more. So when carbon shares with itself, it's very happy. So what it can do is it can form four bonds to four more carbons, and then each of those carbons can form four more bonds to four more carbons. And this one, 1, 2, 3, and it just keeps going on. This is the structure of a diamond. And the reason why this is such a strong structure is because you can almost view the entire -- in fact, you should view the entire diamond as one molecule, because they all have covalent bonds. These are actual sharing of electrons, and these are actually the strongest of all molecular bonds. So you can imagine if the entire solid you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron"
    },
    {
        "Q": "Since the element would change if the amount of protons in an atom changed (7:31), then why is the subscript necessary? If all carbon atoms have 6 protons, why do we need to make a subscript that says it has 6 protons, if by definition carbon has 6 protons? Why doesn't just noting the element suffice?\n",
        "A": "Yes it s redundant when you know the identity of the element. The only time you re likely to use atomic number as a subscript is in nuclear chemistry reactions.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            451
        ],
        "3min_transcript": "electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons. So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium?"
    },
    {
        "Q": "\nAre there non-neutral atoms?At 6:21, Jay said,\"if it is a neutral atom.... It got me confused",
        "A": "Of course. Atoms that aren t neutral are called ions. They are still atoms, they just do not have the same number of electrons and protons.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            381
        ],
        "3min_transcript": "And for protium, let's look at protium here. So in the nucleus there's only one proton and zero neutrons, so one plus zero gives us a mass number of one. And I'll use red here for mass number so we can distinguish. Alright, so mass number is red and let me use a different color here for the atomic number. Let me use magenta here. So the subscript is the atomic number and that's Z, and the superscript is the mass number and that's A. So this symbol represents the protium isotope. Let's draw one for deuterium. So it's hydrogen so we put an H here. There is still one proton in the nucleus, right one proton in the nucleus, so we put an atomic number of one. The mass number is the superscript, So we look in the nucleus here. There's one proton and one neutron. So one plus one is equal to two. So we put a two here for the superscript. And finally for tritium, it's still hydrogen. So we put hydrogen here. There's one proton in the nucleus, atomic number of one, so we put a one here. And then the combined numbers of protons and neutrons, that would be three. So one proton plus two neutrons gives us three. So there's the symbol for tritium. So here are the isotopes of hydrogen and using these symbols allows us to differentiate between them. So let's take what we've learned and do a few more practice problems here. So let's look at a symbol for carbon. So here we have carbon with subscript six, superscript 12. electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons."
    },
    {
        "Q": "At 2:41 the third isotope is called tritium, I'm wondering if there are specialized names for each element's isotopes or if all elements use the same names. For carbon to have more/less neutrons are its isotopes also called deuterium and tritium?\n",
        "A": "Only hydrogen isotopes had the privilege to be called with their own names. All other elements isotopes are simply called by the element s name and the atomic mass (for instance carbon fourteen, uranium twohundred and thirtyfive, and so on).",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            161
        ],
        "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons."
    },
    {
        "Q": "At 2:10, how do a neutrons have mass? Isn't neutrons part of an atom?\n",
        "A": "Yes, neutrons have mass. An isolated neutron is about 1.008665 u. And neutrons are found in all atoms beyond Hydrogen-1.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            130
        ],
        "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen."
    },
    {
        "Q": "At 1:17, why do the isotopes of hydrogen all have different names (protium, deuterium, tritium), and where do they come from?\n",
        "A": "They have names for convenience. Pro means 1 Deu means 2 Tri means 3 See?",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            77
        ],
        "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen."
    },
    {
        "Q": "\nAt 3:20, does he mean that protium and deuterium, and tritium are specific words used only for NITROGEN atoms, or does he mean you can use these words on any element?\nThank you!",
        "A": "They are specific terminology for Hydrogen only. Hydrogen with 1 proton and 0 neutrons= Protium, Hydrogen with 1 proton and 1 neutron = Deuterium, Hydrogen with 1 proton and 2 neutrons =tritium.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            200
        ],
        "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons."
    },
    {
        "Q": "\nAt 8:57, he says the atomic mass is 235, but isn't the atomic mass 238.03?",
        "A": "He says 235 is the mass number of this isotope The relative atomic mass or atomic weight of uranium is 238.03. These are not the same thing!",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            537
        ],
        "3min_transcript": "So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium? at the atomic number, that's 92. So there must be 92 protons. In a neutral atom, the number of electrons is equal to the number of protons. So there are 92 electrons and then finally, to figure out the number of neutrons, we subtract this number from the mass number. So we just need to do 235 minus 92. And that gives us 143. So there are 143 neutrons."
    },
    {
        "Q": "What is meant by superscript ? said at 3:27\n",
        "A": "A superscript is just text set above the line and smaller than usual, like the 2 in x\u00c2\u00b2. Similarly, a subscript is like the 0 in x\u00e2\u0082\u0080. I used special characters to make this work, so you can copy the \u00c2\u00b2 if you want to use it elsewhere.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            207
        ],
        "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons."
    },
    {
        "Q": "At 0:07, he talks about the Atomic Number (z), why do scientists use letters to symbolize numbers or elements? Just curious.\n",
        "A": "i think its because it helps in writing chemical formula and equation ..its easier to use them symbolically coz there are so many elements and compounds formed from them(elements)",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            7
        ],
        "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen."
    },
    {
        "Q": "At 5:41 why can you not have more than three isotopes or can you and you just did not show it?\n",
        "A": "There is no set number of isotopes per element. Each element has whatever number of naturally occurring isotopes it just happens to have. With hydrogen, there are three naturally occurring isotopes. There are a few more that have been made artificially, but they are so radioactive they only exist for a tiny fraction of a second. On the other hand, mercury has seven stable, naturally occurring isotopes and dozens of radioactive isotopes (mostly artificially made isotopes).",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            341
        ],
        "3min_transcript": "And for protium, let's look at protium here. So in the nucleus there's only one proton and zero neutrons, so one plus zero gives us a mass number of one. And I'll use red here for mass number so we can distinguish. Alright, so mass number is red and let me use a different color here for the atomic number. Let me use magenta here. So the subscript is the atomic number and that's Z, and the superscript is the mass number and that's A. So this symbol represents the protium isotope. Let's draw one for deuterium. So it's hydrogen so we put an H here. There is still one proton in the nucleus, right one proton in the nucleus, so we put an atomic number of one. The mass number is the superscript, So we look in the nucleus here. There's one proton and one neutron. So one plus one is equal to two. So we put a two here for the superscript. And finally for tritium, it's still hydrogen. So we put hydrogen here. There's one proton in the nucleus, atomic number of one, so we put a one here. And then the combined numbers of protons and neutrons, that would be three. So one proton plus two neutrons gives us three. So there's the symbol for tritium. So here are the isotopes of hydrogen and using these symbols allows us to differentiate between them. So let's take what we've learned and do a few more practice problems here. So let's look at a symbol for carbon. So here we have carbon with subscript six, superscript 12. electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons."
    },
    {
        "Q": "\nAt 8:15 we are shown another way to depict an isotope is by writing it like, carbon-14, for instance. Would it not be simplest to just use the atomic symbol and keep the mass number in the superscript since the atomic number in subscript will never change?",
        "A": "You can use whatever method is most convenient. It is simpler to avoid superscripts in typing, but you can t avoid them when writing nuclear equations.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            495
        ],
        "3min_transcript": "electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons. So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium?"
    },
    {
        "Q": "Just to make sure I did not misunderstood this: At 1:19, we have four covalent bond for ammonium but calculated that we need 9 valence electrons.\nIs the reason that there are only 8 electrons (covalent bonds) because we have a cation and therefore took one electron \"away\", as in 9-1?\nI know it is explained pretty clearly but I want to make sure that I understood the reasoning behind it.\n",
        "A": "Yes that is it. It is a +1 cation so there is 1 less electron than protons, so there s 8 valence electrons.",
        "video_name": "dNPs-cr_6Bk",
        "timestamps": [
            79
        ],
        "3min_transcript": "The previous video, we saw some steps for drawing dot structures. In this video we're going to use the same steps to draw a few more structures. But we're also going to talk about how formal charge relates to dot structure. So we'll get back to this definition in a minute. Right now, let's draw a quick dot structure for the ammonium cation. So NH4 plus. The first thing you do is find the total number of valence electrons. And so to do that you look at a periodic table and find nitrogen, which is in group five. Therefore, nitrogen has five valence electrons. Right, each hydrogen has one. And we have four of them. So we have 5 plus 4, giving us 9 electrons. However, there's a plus 1 charge. Meaning this is a cation, meaning we're going to lose an electron here. So instead of representing nine in our dot structure, going to represent eight electrons. And so let's go ahead and put the nitrogen at the center. Remember you put the least electronegative atom at the center, except for hydrogen. So nitrogen is going to go in the center here. And we know it's going to have bonds to four hydrogens, so we go ahead and put in those hydrogens right here. we've used up in our dot structure. Two, four, six, and eight. So that takes care of all eight valence electrons that we were supposed to represent. So this is the structure. And I can go ahead and put some brackets around it here. And also a plus 1 charge to indicate that this is an ion. And so that's the dot structure for the ammonium cation here. Let's see if we can assign formal charges to the nitrogen and the hydrogen. So I'm going to go ahead and redraw our dot structure here. And I'm also going to draw in the electrons, right? We know that each of those covalent bonds consists of two electrons. I'm going to go ahead and put in those two electrons right here. And if I want to find a formal charge for, let's say, the central nitrogen. What I would do is think about the number of valence electrons in the free atoms. So if you had a nitrogen all by itself, right? You look at the periodic table, it's in group five. And so therefore we're talking about five valence electrons From that number we're going to subtract the number of valence electrons in the bonded atom. And the way to approach that is to look at your dot structure here, and think about those two electrons in those covalent bonds. One of them we're going to assign to the hydrogen, and one of them are going to assign to the nitrogen. And so we go around, we do that for each one of our covalent bonds like that. And so now, we can see that nitrogen is surrounded by four valence electrons in the bonded atom. So let me go ahead and write that. So it's 5 minus 4. And so 5 minus 4 is of course plus 1. So we have a plus 1 of formal charge on the nitrogen. So this nitrogen as a plus 1 formal charge. Now let's do it for hydrogen here. So hydrogen's in group one on the periodic table. So let me just point this out. This is for . Nitrogen and then for hydrogen. It's in group one. So one valence electron in the free atom. And from that we're going to subtract a number of valence electrons in the bonded atom."
    },
    {
        "Q": "\nAt 9:49, he mentions significant figures. What are significant figures? My science teacher didn't exactly explain them properly. And she explained them differently.",
        "A": "At 9:45, when he is trying to round the pKa to 2 significant figures, he actually rounds to 4 significant figures (10.57 has 4 significant figures). The pKa correctly rounded to 2 significant figures would be 11 (I rounded 10.5686362358 up to 11).",
        "video_name": "DGMs81-Rp1o",
        "timestamps": [
            589
        ],
        "3min_transcript": "as the log of Ka plus the log of Kb equal to the log of Kw. If we take the negative of everything, let's go ahead and do that, the negative of everything, negative log of Ka. I'll put that in parenthesis, plus the negative log of Kb is equal to the negative log of Kw. The negative log of Ka, we know that this is equal to the pKa. The negative log of Ka was our definition for our pKa, and the negative log of Kb was our definition for pKb. pKa plus pKb is equal to finally the negative log of Kw. That would give you 14. 14.00. The negative log of 1.0 times 10 to the negative 14 Now we have something else that we can work with, so let me go ahead and box this right here. Let's take the Ka value that we just found. Let's find the pKa. The pKa would be equal to the negative log of 2.7 times 10 to the negative 11. Let's do that on our calculator here. Let's get some room. The negative log of 2.7 times 10 to the negative 11 gives us 10.57. We have to round that. 10.57. These are our two significant figures because we have two significant figures here. Let's go back up to our problem here, so that's the pKa for the methylammonium ion. Let's say you're given the pKa for the methylammonium ion What is the pKb for methylamine? All we have to do is plug in to our equation. 10.57 was our pKa value. Let's go ahead and write that in here. We have 10.57 plus pkb is equal to 14. When we solve for the pKb that would give us 3.43. So the pKb is equal to 3.43. We could double-check that. Let's go back up here and we could double-check that, because if we took the negative log of this number that's what we should get. Let's go ahead and do that. Let's take the negative log of 3.7 times 10 to the negative 4."
    },
    {
        "Q": "At 2:26, what does he mean that we multiply them?\n",
        "A": "He means that we multiply the given values of Ka and Kb respectively. We do it in order to equate it to 1 x 10^-14 which is actually Kw..... Hope this helps.... : )",
        "video_name": "DGMs81-Rp1o",
        "timestamps": [
            146
        ],
        "3min_transcript": "- We've already seen that NH4 plus and NH3 are a conjugate acid-base pair. Let's look at NH4 plus. The ammonium ion would function as an acid and donate a proton to water to form H3O plus. If NH4 plus donates a proton you're left with NH3. The Ka for this reaction is 5.6 times 10 to the negative 10. Now let's look at NH3 which we know is a weak base, and it's going to take a proton from water, therefore forming NH4 plus. If we take a proton from water we're left with OH minus. Since we talked about a base here we're gonna use Kb, and Kb for this reaction is 1.8 times 10 to the negative 5. What would happen if we add these two reactions together? We have two water molecules for our reactants, so let me go ahead and write H2O What about ammonium? We have ammonium on the left side for reactant, we also have ammonium over here for our product. That cancels out. Same thing happens with ammonia, NH3. We have NH3 on the left. We have NH3 on the right. We have NH3 as a reactant, NH3 as a product. We can cancel those out too. Our only reactants would be two H2O. For our products we would get H3O plus and OH minus, so H3O plus, hydronium, and hydroxide. This reaction should sound familiar to you. This net reaction is the auto-ionization of water where one water molecule acts as an acid, one water molecule acts as a base. We get H3O plus and OH minus. The equilibrium constant for the auto-ionization of water you've already seen that Kw is equal to 1.0 We added these two reactions together and we got this for our net reaction. What would we do with Ka and Kb to get Kw? It turns out that you multiply them, Ka times Kb for a conjugate acid-base pair is equal to Kw. Let's do that math. Ka is 5.6 times 10 to the negative 10. So 5.6 times 10 to the negative 10. Kb is equal to 1.8 times 10 to the negative 5, 1.8 times 10 to the negative 5, and let's get out the calculator and let's go ahead and do that math. We have 5.6 times 10 to the negative 10. We're going to multiply that by 1.8 times 10 to the negative 5."
    },
    {
        "Q": "\nWhat did he mean by ''overcome the heat of fusion'' at 0:37? there's no fusion occurring there.",
        "A": "by definition heat of fusion is: The energy required to change a gram of a substance from the solid to the liquid state without changing its temperature is commonly called it s heat of fusion . So this is the energy to break the bonds in the solid state to be able to flow in the liquid state. In this case fusion is not about two atoms fusing to form a heavier atom as in a physics way, but that the atoms are in a giant solid lattice that must be broken to change the state.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            37
        ],
        "3min_transcript": "We know that when we have some substance in a liquid state, it has enough kinetic energy for the molecules to move past each other, but still not enough energy for the molecules to completely move away from each other. So, for example, this is a liquid. Maybe they're moving in that direction. These guys are moving a little bit slower in that direction so there's a bit of this flow going on, but still there are bonds between them. They kind of switch between different molecules, but they want to stay close to each other. There are these little bonds between them and they want to If you increase the average kinetic energy enough, or essentially increase the temperature enough and then overcome the heat of fusion, we know that, all of a sudden, even these bonds aren't strong enough to even keep them close, and the molecules separate and they get into a gaseous phase. And there they have a lot of kinetic energy, and they're bouncing around, and they take the shape of their container. But there's an interesting thing to think about. Which implies, and it's true, that all of the molecules do not have the same kinetic energy. Let's say even they did. Then these guys would bump into this guy, and you could think of them as billiard balls, and they transfer all of the momentum to this guy. Now this guy has a ton of kinetic energy. These guys have a lot less. This guy has a ton. These guys have a lot less. There's a huge distribution of kinetic energy. If you look at the surface atoms or the surface molecules, and I care about the surface molecules because those are the first ones to vaporize or-- I shouldn't jump the gun. They're the ones capable of leaving if they had enough kinetic energy. If I were to draw a distribution of the surface molecules-- let me draw a little graph here. So in this dimension, I have kinetic energy, and on this And this is just my best estimate, but it should give you the idea. So there's some average kinetic energy at some temperature, right? This is the average kinetic energy. And then the kinetic energy of all the parts, it's going to be a distribution around that, so maybe it looks something like this: a bell curve. You could watch the statistics videos to learn more about the normal distribution, but I think the normal distribution-- this is supposed to be a normal, so it just gets smaller and smaller as you go there. And so at any given time, although the average is here, there's some molecules that have a very low kinetic energy. They're moving slowly or maybe they have-- well, let's just say they're moving slowly. And at any given time, you have some molecules that have a very high kinetic energy, maybe just because of the random bumps that it gets from other molecules. It's accrued a lot of velocity or at least a lot of momentum."
    },
    {
        "Q": "\nAt 12:08, Is the atmosphere part of the closed container as well? Otherwise, I am not sure how the molecules that comprise the atmosphere are interacting with the gaseous water molecules.\nAlso, when we speak of partial pressure does that indicate that gaseous water molecules within the container experience different amount of pressure than the liquid water molecules?",
        "A": "The closed container is assumed to contain air which will exert pressure on the surface of the liquid water. Both the molecules in the air and the molecules in the gaseous water are exerting pressure on the surface of the liquid water.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            728
        ],
        "3min_transcript": "state in order for the equilibrium to be reached. Let me do it all in the same color. So the pressure created by its evaporated molecules is going to be higher for it to get to that equilibrium state, so it has high vapor pressure. And on the other side, if you're at a low temperature or you have strong intermolecular forces or you have a heavy molecule, then you're going to have a low vapor pressure. For example, iron has a very low vapor pressure because it's not vaporizing while-- let me think of something. Carbon dioxide has a relatively much higher vapor pressure. Much more of carbon dioxide is going to evaporate when you have it. straight from the liquid to the solid state, but I think you get the idea. And something that has a high vapor pressure, that wants to evaporate really bad, we say it has a high volatility. You've probably heard that word before. So, for example, gasoline has a higher-- it's more volatile than water, and that's why it evaporates, and it also has a higher vapor pressure. Because if you were to put it in a closed container, more gasoline at the same temperature and the same atmospheric pressure, will enter into the vapor state. And so that vapor state will generate more pressure to offset the natural inclination of the gasoline to want to escape than in the case with water. Now, an interesting thing happens when this vapor pressure is equal to the atmospheric pressure. the atmosphere here at a certain pressure. Let's say until now, we've assumed that the atmosphere was at a higher pressure, for the most part keeping these molecules contained. Maybe some atmosphere molecules are coming in here, and maybe some of the vapor molecules are escaping a bit, but it's keeping it contained because this is at a higher pressure out here than this vapor pressure. And of course the pressure right here, at the surface of the molecule, is going to be the combination of the partial pressure due to the few atmospheric molecules that come in, plus the vapor pressure. But once that vapor pressure becomes equal to that atmospheric pressure, so it can press out with the same amount of force-- you can kind of view it as force per area-- so then the molecules can start to escape. It can push the atmosphere back. And so you start having a gap here. You start having a vacuum. I don't want to use exactly a vacuum, but since the"
    },
    {
        "Q": "at 7:50 ,does given temperature mean that the temperature has to be kept constant while this process goes on?\n",
        "A": "Yes. The vapor pressure depends on the temperature of the liquid. As you raise the temperature, you raise the average kinetic energy of the molecules. More of these energetic molecules will be able to escape from the liquid, and the vapor pressure will increase. This means that you must always state the temperature at which you measured the vapor pressure. For example, the vapor pressure of water is 0.6 kPa at 0\u00c2\u00b0C, 2.3 kPa at 20\u00c2\u00b0C, 12.3 kPa at 50\u00c2\u00b0C, and 101.3 kPa at 100\u00c2\u00b0C.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            470
        ],
        "3min_transcript": "Now something interesting happens. This is the distribution of the molecules in the liquid state. Well, there's also a distribution of the kinetic energies of the molecules in the gaseous state. Just like different things are bumping into each other and gaining and losing kinetic energy down here, the same thing is happening up here. So maybe this guy has a lot of kinetic energy, but he bumps into stuff and he loses it. And then he'll come back down. So there's some set of molecules. I'll do it in another set of blue. These are still the water-- or whatever the fluid we're talking about-- that come back from the vapor state back into the liquid state. And so what happens is, there's always a bit of evaporation and there's always a bit of condensation because you always have this distribution of kinetic energies. At any given moment in time, out of the vapor above the liquid, some of the vapor loses its kinetic energy and then it goes back into the liquid state. Some of the surface liquid gains kinetic energy by random And the vapor state will continue to happen until you get to some type of equilibrium. And when you get that equilibrium, we're at some pressure up here. So let me see, some pressure. And the pressure is caused by these vapor particles over here, and that pressure is called the vapor pressure. I want to make sure you understand this. So the vapor pressure is the pressure created, and this is at a given temperature for a given molecule, right? Every molecule or every type of substance will have a different vapor pressure at different temperatures, and obviously every different type of substance will also have different vapor pressures. For a given temperature and a given molecule, it's the pressure at which you have a pressure created by the vapor molecules where you have an equilibrium. going back into the liquid state. And we learned before that the more pressure you have, the harder it is to vaporize even more, right? We learned in the phase state things that if you are at 100 degrees at ultra-high pressure, and you were dealing with water, you would still be in the liquid state. So the vapor creates some pressure and it'll keep happening, depending on how badly this liquid wants to evaporate. But it keeps vaporizing until the point that you have just as much-- I guess you could kind of view it as density up here, but I don't want to think-- you have just as many molecules here converting into this state as molecules here converting into this state. So just to get an intuition of what vapor pressure is or how it goes with different molecules, molecules that really want to evaporate-- and so why would a molecule want to evaporate? It could have high kinetic energy, so this would be at a"
    },
    {
        "Q": "Can anyone define equilibrium for me to understand? Sal says it at 10:11.\n",
        "A": "The word equilibrium in general means balance. The situation Sal is talking about in the video is balanced because in a certain amount of time the -same number- of particles will evaporate and leave the liquid to become gas as -the same number -condense and return to the liquid state. So the number of particles in each state does not change. It is in equilibrium.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            611
        ],
        "3min_transcript": "going back into the liquid state. And we learned before that the more pressure you have, the harder it is to vaporize even more, right? We learned in the phase state things that if you are at 100 degrees at ultra-high pressure, and you were dealing with water, you would still be in the liquid state. So the vapor creates some pressure and it'll keep happening, depending on how badly this liquid wants to evaporate. But it keeps vaporizing until the point that you have just as much-- I guess you could kind of view it as density up here, but I don't want to think-- you have just as many molecules here converting into this state as molecules here converting into this state. So just to get an intuition of what vapor pressure is or how it goes with different molecules, molecules that really want to evaporate-- and so why would a molecule want to evaporate? It could have high kinetic energy, so this would be at a It could have low intermolecular forces, right? It could be molecular. Obviously, the noble gases have very low molecular forces, but in general, most hydrocarbons or gasoline or methane or all of these things, they really want to evaporate because they have much lower intermolecular forces than, say, water. Or they could just be light molecules. You could look at the physics lectures, but kinetic energy it's a function of mass and velocity. So you could have a pretty respectable kinetic energy because you have a high mass and a low velocity. So if you have a light mass and the same kinetic energy, you're more likely to have a higher velocity. You could watch the kinetic energy videos for that. But something that wants to evaporate, a lot of its molecules-- let me do it in a different color. Something that wants to evaporate really bad, a lot state in order for the equilibrium to be reached. Let me do it all in the same color. So the pressure created by its evaporated molecules is going to be higher for it to get to that equilibrium state, so it has high vapor pressure. And on the other side, if you're at a low temperature or you have strong intermolecular forces or you have a heavy molecule, then you're going to have a low vapor pressure. For example, iron has a very low vapor pressure because it's not vaporizing while-- let me think of something. Carbon dioxide has a relatively much higher vapor pressure. Much more of carbon dioxide is going to evaporate when you have it."
    },
    {
        "Q": "\nAt 4:47, we calculate the total moles of NH4 given off by the reaction assuming it starts with 0 moles of NH4. Although, wouldn't we start with more than 0 moles because the NH3 molecule would have been disassociating before we even added the HCl?",
        "A": "You start with NH3 and you have 0.004mol of NH3. Afterward, it reacts with HCl to form NH4. Before the reaction, NH3 remains as NH3 and does not start dissociating.",
        "video_name": "kWucfgOkCIQ",
        "timestamps": [
            287
        ],
        "3min_transcript": "NH three plus HCl. So let's write it that way too. So ammonia plus HCl, HCl donates a proton to NH three giving us NH four plus and once HCl donates a proton, you have Cl minus, the chloride anion, the conjugate base. So this would be NH four plus Cl minus. Alright, let's write down how many moles of acid that we have. So we added 0.004 moles of our acid. So let's write down 0.004 here, for moles of acid that we added. And for moles of base that we started with, that's also equal to 0.004. Let's get some more room down here. So we started with 0.004 moles of our base. If you look at your molar ratios there, one to one mole ratio. to completely react with the base that we originally had present. So this represents the equivalence point of our titration. This is the equivalence point for our titration, because we've now added enough acid to completely react with our base. So all of this acid is going to react, right? So we use up all of that and it uses up all of the base that we had present. So all this base is going to react with the acid we're gonna lose, all 0.004 moles of NH three. So we're left with zero of that. Ammonia turns into ammonium, NH four plus. And so, if we start with zero for NH four plus, whatever we lose for ammonia, was what we gain for NH four plus. So for losing 0.004 moles of ammonia, that's how many moles of ammonium we are gaining. So we gain 0.004 moles of NH four plus. we're left with 0.004 moles of NH four plus. So this is at our equivalence point. So we have some ammonium present. What's the concentration of ammonium that we have present? Well, the concentration, once again, is moles over liters. So we have 0.004 moles of NH four plus, so let's write 0.004 moles of NH four plus. What is the volume? What's the total volume now? Well let's go back up to the problem. We started with 40 milliliters of our... We started with 40 milliliters and we added another 40. So 40 plus 40 is 80 milliliters. Or one, two, three, 0.08 liters. So our total volume now is 0.08 liters. So let's go ahead and write that in here."
    },
    {
        "Q": "At 8:23 when you start the last reaction mechanism, I know it is E1 elimination, but the first two were dehydration of alcohol, is there a specific name for this third reaction?\n",
        "A": "Dehydrohalogenation - loss of a hydro(gen) and a halogen.",
        "video_name": "l-g2xEV-z7o",
        "timestamps": [
            503
        ],
        "3min_transcript": "would be ethanol, so let me go ahead and draw in lone pairs of electrons on the oxygen, so notice we're also heating this reaction, so the ethanol is gonna function as a base, so ethanol's not a strong base, but it can take a proton, so let me go ahead and draw in a proton right here, and a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene, so let me go ahead and draw our product, let me put that in here, and let me highlight some electrons, so the electrons in blue moved in here to form our double bond. So a couple of points about this reaction, one point is, when you're looking at SN1 mechanisms, the first step is loss of a leaving group to form your carbocation, so when you get to this carbocation, you might think, well, why is ethanol acting as a base here? Why couldn't it act as a nucleophile? act as a nucleophile, and it would attack the positively-charged carbon, and you would definitely get a substitution product for this reaction as well, so if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism. If the ethanol acts as a base, you're gonna get an E1 elimination mechanism, so here, we're just gonna focus on the elimination product, and we won't worry about the substitution product, but we will talk about this stuff in a later video, 'cause that would definitely happen. Alright, something else I wanna talk about is we had three beta carbons over here, and if I look at these three beta carbons, and I just picked one of them, I just said that this carbon right here, let me highlight it, I just took a proton from this carbon, but it doesn't matter which of those carbons that we take a proton off because of symmetry, let me go ahead and draw this in over here, so this is my carbocation, let's say, so our weak base comes along, and takes a proton from here, and these electrons have moved into here, that would give us the same product, right? So this would be, let me go and highlight those electrons, so these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon, so reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene, and sometimes, phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group, but if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group, and the hydroxide anion is a poor leaving group, and we know that by looking at pKa values."
    },
    {
        "Q": "\nAt 4:00 the guy says that the water will take the the proton from hydrogen attached to the beta carbon...\n\nwhy would the water behave in such a way? we just said that oxygen is very electro negative and does not like to have a positive formal charge.\n\nso what's the deal?",
        "A": "In solution there will be a ton of water and a small quantity of acid......since water is amphoteric (it is both an acid and a base) it can act as a Bronsted base....",
        "video_name": "l-g2xEV-z7o",
        "timestamps": [
            240
        ],
        "3min_transcript": "of only your substrate, this over here on the left, so it's first order with respect to the substrate. And that's because of this rate determining step. The loss of the leaving group is the rate determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction, so it's the concentration of the substrate only, and since it's only dependent on the concentration of the substrate, that's where the one comes from in E1, so I'm gonna go ahead and write this out here, so in E1 mechanism, the one comes from the fact this is a unimolecular, a unimolecular rate law here, and the E comes from the fact that this is an elimination reaction, so when you see E1, it's an elimination reaction, and it's unimolecular, the overall rate of the reaction only depends on the concentration of your substrate, so if you increase, let's say you have, let's say this was your substrate right here, and you increase the concentration of your substrate, let me just write this down, so if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two, so it's first order with respect to the substrate, so this is some general chemistry here. If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction. So let's talk about one more point here in the mechanism, and that is the formation of this carbocation. Since we have a carbocation in this mechanism, we need to think about the possibility of rearrangements in the mechanism, and you need to think what would form, what substrate would form a stable carbocation, forming a tertiary carbocation would be favorable for an E1 mechanism. Here we have a tertiary alkyl halides, and let's say this tertiary alkyl halide undergoes an E1 elimination reaction. So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons, so that's a beta carbon, that's a beta carbon, and that's a beta carbon. So the first step in an E1 mechanism is loss of our leaving group, so if I draw the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anion, so let me draw that in here, so we would make the iodide anion, and let me highlight our electron, so the electrons in this bond come off onto the iodine to form the iodide anion. And this is an excellent leaving group."
    },
    {
        "Q": "At 1:06 I Noticed that elements 58-71 are missing? Was this intentional or a bad Periodic Table?\n",
        "A": "At 2:10 it shows why.",
        "video_name": "t_f8bB1kf6M",
        "timestamps": [
            66
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 1:09 when you're numbering the groups with A's after them, why do you skip groups 3 through 12?",
        "A": "Because some chemists decided a long time ago that that is how they are to be numbered. It is an old system and both the European and US ways to number groups used the same numbers and letters to refer to different groups. That is why it has been depreciated and have officially been replaced with 1-18. Still some people continue to use it today.",
        "video_name": "t_f8bB1kf6M",
        "timestamps": [
            69
        ],
        "3min_transcript": ""
    },
    {
        "Q": "What are valence electrons? (Mentioned at 1:29.)\n",
        "A": "With sal s vids about electron configurations",
        "video_name": "t_f8bB1kf6M",
        "timestamps": [
            89
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 00:12, the Speaker explains the definition of acceleration as \"the change in velocity over time\" and that got me to rethink about its meaning. Does his definition mean the velocity increases and decreases? Before, my thoughts were acceleration only meant the increase in velocity.",
        "A": "Acceleration does not only mean increase in velocity, but also decrease.",
        "video_name": "FOkQszg1-j8",
        "timestamps": [
            12
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 4:23 if acceleration is 20 miles per hour per second then how can we multiply seconds with hour ?",
        "A": "You aren t multiplying seconds and hours. by saying 20 miles per hour per second you are saying that every second you are going to go 20 miles per hour faster. if you start at 0 miles per hour then 1 second later you would be going 20 miles per hour and the next second you would be going 40 miles per hour. so it is the change in speed over the amount of time it takes to change the speed.",
        "video_name": "FOkQszg1-j8",
        "timestamps": [
            263
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:20 Sal said that \"Acceleration is the change in velocity over time.\" but shouldn't the definition be 'The rate of the velocity INCREASING (over time' because if the velocity decreases it is DECELERATION.",
        "A": "We define acceleration as increasing or decreasing. All we do is change the direction of the acceleration. For example, you are moving east at 10m/s but the acceleration is 5m/s^2 to the west. So you are still accelerating. Deceleration can be used at that point but the scientific and more correct term would be what I described above.",
        "video_name": "FOkQszg1-j8",
        "timestamps": [
            20
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:20 Sal says he is going to use a logarithmic scale. What is that and how does it proportionally work?\n",
        "A": "A logarithmic scale is a nonlinear scale used when there is a large range of quantities. Common uses include the earthquake strength, sound loudness, light intensity, and pH of solutions.",
        "video_name": "BWs-ONRDDG4",
        "timestamps": [
            320
        ],
        "3min_transcript": "But you could see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable. That they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star-- and this is just kind of a simulation, a very cheap simulation-- it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again. So it's actual intensity is going up and down with a well-defined period. So if this takes three days and then this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity, is going to be six days. So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she are roughly the same distance away. Everything in the Large Magellanic Cloud is roughly the same distance away. And it's obviously not exact. This is an entire galaxy. So you have obviously things further away in that galaxy and things closer up. You have stars here and here. And their distance isn't going to be exactly the same to us, even though we're sitting maybe over here someplace. But it's going to be close. It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. If she plotted-- so let me plot this right over here. So she plotted on the horizontal axis, if she plotted the relative luminosity. So really, the only way that she can measure this is just how bright did they look to her? And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, how bright it is at the actual star. So she plotted relative luminosity of the star on one axis. And on the other axis, she plotted the period of these variable stars. She plotted the period. And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say this is in days. So this is one day. This is 10 days. This is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10. If we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale. So I'm using the log of the period or I'm just marking them as 1, 10, 100. But I'm giving each of these factors of 10 an equal spacing. But when you plot it on this scale, the relative luminosity versus the period, she got a plot that looked something like this. And this is obviously not exact. She got a plot that looks something like this."
    },
    {
        "Q": "\nAt around 7:09 Sal says that the heat added to the system is equal to the work we did. but it is the system (the piston and pebbles) that do the work by expanding right?",
        "A": "Work is the transfer of energy. There are two instances where energy is transferred. First from the thermal reservoir to the thermodynamic system (gas enclosed by cylinder walls and piston). This energy is then transferred from this volume of gas to the piston and pebbles. The piston and pebbles gain (kinetic) energy as their speed increases.",
        "video_name": "M_5KYncYNyc",
        "timestamps": [
            429
        ],
        "3min_transcript": "This is the heat we put into it. So it's the net heat we applied to the system divided by the heat we put into it. So this is equal to-- Q1 divided by Q1 is 1, minus Q2 over Q1. So once again, this is another interesting definition of They're all algebraic manipulations using the definition of internal energy, and whatever else. Now let's see if we can somehow relate efficiency to our temperatures. Now, let me-- this is Q1 right there. So what were Q2 and Q1? What were they? What were their absolute values, not looking at the signs of them? I mean, we know that Q2 was transferred out of the system, so if we said Q2 in terms of the energy applied to the system, it would be a negative number. But if we just wanted to know the magnitude of Q2, what would it be, and the magnitude of Q1? Well, the magnitude of Q1-- let me draw a new Carnot cycle, just for cleanliness. That's my volume axis. That's my pressure axis. PV. I start here at some state, and then I go isothermally. What's a good color for an isothermal expansion? Maybe purple. It's kind of-- let's see. An isothermal expansion. I'm on an isotherm here. So I go down there, and I go to state, and I went down to state B. So this is A to B. And we know we are an isotherm. This is when Q1 was added. This is an isotherm. If your temperature didn't change, your internal energy didn't change. And like I said before, if your internal energy is 0, then the heat applied to the system is the same as the work you did. They cancel each other out. That's why you got to 0. So this Q1 that we apply to the system, it must be equal to the work we did. And the work we did is just the area under this curve. And why is it the area under the curve? Because it's a bunch of rectangles of pressure times volume. And then you just add up all the rectangles, an infinite number of infinitely narrow rectangles, and you get the area. And what is that? Just a review-- pressure times volume was the work, right? Because we're expanding the cylinder. We're moving up that piston. We're doing force times distance. So the amount Q1 is equal to that integral-- the amount of work we did as well-- is equal to the integral from V final-- I shouldn't say V final. from VB, from our volume at B-- oh sorry. From our volume at A to our volume at B. We're starting here, and we're going to our volume at B. And we're taking the integral of pressure-- and I've done this multiple times, but I'll do it again. So the pressure, our height, times our change in volume, dv. We go back to our ideal gas formula, PV is equal to nRT,"
    },
    {
        "Q": "At 3:00, would the kinetic energy in the same gas be transferred?\n",
        "A": "Yes of course, just like there are some parts of a thing that are colder or hotter, I suppose that every molecule has a different escalar value of kinetic energy, but due to the collisions in a very small area they are approximately equal.",
        "video_name": "PA-T6lMxCBI",
        "timestamps": [
            180
        ],
        "3min_transcript": "They're gonna have, each of them are gonna have kinetic energy and if you average them, that's gonna be proportional to temperature. So let me depict each of this individual molecule's kinetic energy. Maybe this one is doing that. Maybe this one is doing that. Maybe this one is going in this direction. Maybe that one is going that direction. That one is going in that direction. This is going in that direction. That is going in that direction. So, notice, they all have different directions. And the magnitude of their velocity can be different. They all have different speeds. They all might have different speeds. So they have different speeds right over here. And they're all bumping into each other. Transferring their kinetic energy, transferring their momentum to from one particle to another. But when we talk about temperature we talk about the average kinetic energy or what's proportional to the average kinetic energy of the system. Well this one, each of these molecules are also gonna have some kinetic energy. But on average it's gonna be lower. Maybe this one is doing something like this. This is doing something like this. This is doing something like this. So they're different, but on average they're gonna be lower. So, hopefully you see that this magenta arrows are bigger than these blue arrows that I'm doing. And they don't all have to be, for example this one might have a lot of kinetic energy. But when you average it out, the average here is gonna be lower than the average here. So, just like that. Now, if this is our initial state, what do we think is going to start happening? Well, before our different groups of gases were colliding with itself or the magenta was colliding with the magenta. The blue is colliding with the blue. But now they're gonna start colliding with each other. And so you can imagine when this molecule right over here, collides with this molecule, it's gonna transfer some kinetic energy to it. So after the collision, this one might be going. After the collision. So lets just say they just bounced in to. So this is right before, So right after they finish bouncing into each other, this one might ricochet off. So this one is going to go this way. Let me do this in a different color. So it might hit this one, bounce off, and then transfer some of its kinetic energy and then it bounces off in this direction. While this one, after the collision, after the collision is going to is maybe going to move much faster in this direction. And so notice, you have a transfer of energy. Just with that one collision you had a transfer of kinetic energy from this molecule to that molecule. And this is going to happen throughout the system. That the faster molecules, the ones with more kinetic energy as they collide you're gonna have transfer of energy. So you're gonna have a transfer of energy from the higher temperature to the lower temperature. Transfer, transfer of energy. And this transfer of energy."
    },
    {
        "Q": "At 2:10, why is there a kelvin scale? where did it come from?\n",
        "A": "The kelvin is a measure for temperature. It is an absolute scale for thermodynamics as it has 0 in the middle instead of at the beginning like the celcius scale ie. The scale of Celsius scale starts with 0 degrees but in the Kelvin scale, 0 is at the centre. To convert some temperature from Celsius to Kelvin, we add 273.15 from the temperature. For example, 45 degrees Celsius= 45 + 273.15 = 318.15 kelvin. Hope this helps :)",
        "video_name": "tvO0358YUYM",
        "timestamps": [
            130
        ],
        "3min_transcript": "I have this problem here from chapter five of the Kotz, Treichel, and Townsend Chemistry and Chemical Reactivity book, and I'm doing this with their permission. So they tell us that ethanol, C2H5OH, boils at-- let me do this in orange-- it boils at 78.29 degrees Celsius. How much energy, in joules, is required to raise the temperature of 1 kilogram of ethanol from 20 degrees Celsius to the boiling point and then change the liquid to vapor at that temperature? So there's really two parts of this problem. How much energy, in joules, to take the ethanol from 20 degrees to 78.29 degrees Celsius? That's the first part. And then once we're there, we're going to have 78.29 degrees Celsius liquid ethanol. But then we also need the energy to turn it into vapor. So those are going to be the two parts. So let's just think about just raising the liquid Let's figure out how we're going to do that. Just the liquid temperature. So the first thing I looked at is well how many degrees are we raising the temperature? Well we're going from 20 degrees Celsius-- let me write Celsius there just so it's clear-- 20 degrees Celsius to 78.29 degrees Celsius. So how much did we raise it? Well, 78.29 minus 20 is 58.29. So our change in temperature is equal to 58.29 degrees Celsius or this could even be 58.29 kelvin. the Celsius scale and the kelvin scale are the same thing. The kelvin scale is just a shifted version of the Celsius scale. If you added 273 to each of these numbers you would have the kelvin temperature, but then if you take the difference, it's going to be the exact same difference. Either way you do it, 78.29 minus 20. So that's how much we have to raise the temperature. So let's figure out how much energy is required to raise that temperature. So we want a delta T. We want to raise the temperature 58.29. I'll stay in Celsius. Actually let me just change it to kelvin because that looks like what our units are given in terms of specific heat. So let me write that down. 58.29 kelvin is our change in temperature. I could have converted either of these to kelvin first, then found the difference, and gotten the exact same number. Because the Celsius scale and the kelvin scale, the increments are the same amount. Now, that's our change in temperature."
    },
    {
        "Q": "at 6:43 why didn't sal multiply it with \u00ce\u0094T?\n",
        "A": "Because there was no change in temperature at that point (liquid turning into vapour). Heat of vaporisation is just mass times latent heat.",
        "video_name": "tvO0358YUYM",
        "timestamps": [
            403
        ],
        "3min_transcript": "So this is going to be 142-- we'll just round down-- 142,000 kelvin. So this is 142,000. Sorry 142,000 joules. Joules is our units. We want energy. So this right here is the amount of energy to take our ethanol, our 1 kilogram of ethanol, from 20 degrees Celsius to 78.29 degrees Celsius. Or you could view this as from 293 kelvin to whatever this number is plus 273, that temperature in kelvin. Either way, we've raised its temperature by 58.29 kelvin. Now, the next step is, it's just a lot warmer ethanol, liquid ethanol. We now have to vaporize it. It has to become vapor at that temperature. So now we have to add the heat of vaporization. So that's right here. We should call it the enthalpy of vaporization. The enthalpy of vaporization, they tell us, is And this is how much energy you have to do to vaporize a certain amount per gram of ethanol. Assuming that it's already at the temperature of vaporization, assuming that it's already at its boiling point, how much extra energy per gram do you have to add to actually make it vaporize? So we have this much. And we know we have 1,000 grams of enthanol. The grams cancel out. 855 times 1,000 is 855,000 joules. So it actually took a lot less energy to make the ethanol go from 20 degrees Celsius to 78.29 degrees Celsius than it took it to stay at 78.29, but go from the liquid form to the This took the bulk of the energy. But if we want to know the total amount of energy, let's 855,000 plus 842,000. 800 plus 100 is 900. That's 900,000. 50 plus 40 is 90. 5 plus 2 is 7. So it's 997,000 joules or 997 kilojoules. Or we could say it's almost 1 megajoule, if we wanted to speak in those terms. But that's what it will take for us to vaporize that 1 kilogram of ethanol."
    },
    {
        "Q": "\nAt 3:25 Ronald says \"It's called rhodopsin because it's in a rod\" but I feel like that's not actually true. Wikipedia says it's derived from the greek word \"rhodos\" which sounds much more likely. And the Opsines in cones are called Iodopsines.",
        "A": "True. But Ron s reasoning serves as at least a mnemonic to remember the protein name in rods. :)",
        "video_name": "CqN-XIPhMpo",
        "timestamps": [
            205
        ],
        "3min_transcript": "So normally, he is turned on. When there's no light, the rod is turned on. But when light is present, it actually turns him off. So with light, he's turned off. So how does this happen? So this occurs through this thing called the phototransduction cascade. So the phototransduction cascade is the set of steps that occurs at the molecular level that basically takes this rod and turns him off. And in turning him off, he's actually able to turn on a bunch of other cells that eventually lets the brain know, hey, there's light here. So it allows the brain to realize that there's light and allows it to comprehend what's going on make sense of the world. So let's go ahead and examine this phototransduction cascade. So let's give ourselves a little bit of space here, just look at it in a little bit more detail. So I'm going to redraw the rod over here. And let's just look at this part of the rod, just this very top little bit of the rod. And I'm going to draw it a lot bigger, so that we can really make sense of what's going on here. So inside the rod, which we just made much bigger, there are a bunch of these little disks. So they are really thin little disks, and they're just stacked on top of one another, and there are hundreds of them. And basically they are like this. They just go stacked on top of one another, and they fill up the entire rod. Same thing with cones but we're just going to look at the rods here. So inside of these little disks, there are a whole bunch of different proteins all interspersed throughout the disks. So this protein that I'm drawing in red-- let me just go ahead and draw it a lot bigger, so we're going to go ahead and blow this protein up. so it consists of a bunch of different subunits. There are actually seven subunits, so there are 5, 6, and 7. So there are all these subunits that make up this little protein right here. So this protein as a whole is called rhodopsin. And it's called rhodopsin because it's in a rod. If it were in a cone, it would be called cone opsin, but it's basically the same protein. And sitting inside of this protein is a small molecule, and it sits just inside. Let's go ahead and just zoom in a little bit, so we can make a little bit more sense of what we're looking at here. Let's go ahead and zoom in over here. So this little molecule is just sitting inside of rhodopsin, and it's kind of bent. You can see here how I drew it just a little bit bent. So this little molecule is called retinal."
    },
    {
        "Q": "At 8:24, it is said \"cells are hyperpolarized and turn off\". why is it OFF, not ON? hyperpolarization activates the cell and ON seems to be better to name the event.\n",
        "A": "Hyperpolarization is inhibitory. It opposes the depolarization necessary to fire an action potential that allows glutamate to be released into the synapse. This glutamate is inhibitory in ON bipolar cells and excitatory in OFF bipolar cells. When the the rod is hyperpolarized, it is no longer able to release glutamate and it loses the inhibition of the ON bipolar cells, thus turning them on.",
        "video_name": "CqN-XIPhMpo",
        "timestamps": [
            504
        ],
        "3min_transcript": "shape and causes rhodopsin to change shape, transducin breaks away from rhodopsin. And the alpha subunit actually comes over here to another part of the disk and binds to a protein called phosphodiesterase, which I'll just draw as this little box over here. So this little protein is called cyclic GMP phosphodiesterase, or PDE for short. So PDE, basically what it does-- let's go ahead and zoom out a little bit. What PDE does when it's activated is it takes cyclic GMP, which is floating all and it basically takes the cyclic GMP and converts it into just regular GMP. So this basically reduces the concentration of cyclic GMP and increases the concentration of GMP. And the reason that this is important is because there's another channel over here, so there's a whole bunch of these sodium channels and they're all over the cell. So they're just a whole bunch of them, and basically what they let the cell do is they allow the cell to take in sodium from the outside. So let's just say there's a little sodium ion, and it allows it to come inside the cell. So in order for this sodium channel to be open, it actually needs cyclic GMP to be bound to it. So as long as cyclic GMP is bound, the channel is open. But as the concentration of cyclic GMP decreases because causes sodium channels to close. So now we have a closing of sodium channels, and now we basically have less sodium entering the cell. And as less sodium enters the cell, it actually causes the cell to hyperpolarize and turn off. So as the sodium channels close, it actually causes the rods to turn off. So basically, without light, the rods are on because these sodium channels are open. Sodium is flowing through, and the rods are turned on. They can actually produce an action potential and activate the next cell and so on. But as soon as they're turned off, what happens is very interesting because-- let's just look at this rod over here. So what happens is really interesting, because there's this other cell over here that is called the bipolar cell. And we'll just give it a kind of a neutral base, because it's bipolar."
    },
    {
        "Q": "\nat 12:30,one carbon is oxidised but one carbon is reduced.This implies that there must have been a reducing as well as an oxidising agent simultaneously.How is that possible?",
        "A": "To answer that, you really need to see the chemical reaction ( redox reaction) which created the change in oxidation states. Remember, individual atoms within a molecule are not oxidizing or reducing agents, but the entire molecule ITSELF is the agent.",
        "video_name": "bJMUKNbAsTY",
        "timestamps": [
            750
        ],
        "3min_transcript": "directly bonded to two more carbons. On the right, those carbons now only have a single bond between them and this carbon on the left is bonded to a hydrogen. This carbon on the right is directly bonded to an oxygen and let's put in those other carbons, right? We have another carbon, carbon bonds. I mean, let's put in those carbon, carbon bonds like that. Let's assign our oxidation states. So, we draw in our bonding electrons and we think about electronegativity differences but we're going assume once again that our carbons have the same electronegativity on the left and so when we're assigning electrons let's just pick one of those carbons. So, let's pick the carbon on the left, right? We divide up those two electrons in that bond. Give one electron to one carbon. One electron to the other. We do the same thing here. And for the double bond with four electrons we divide up those four electrons. We give two electrons to each carbon. and around it, this carbon has one, two, three, and four. So, four minus four is an oxidation state of zero. So, this carbon has an oxidation state of zero. Same thing for the carbon on the right side of the double bond, right? So, the same situation. So, it also has an oxidation state of zero. What about for our product, right? Let's examine those two carbons. Let's put in our bonding electrons. So, we draw in our bonding electrons here. So, we put those in. A lot more bonding electrons to draw. And we think about electronegativity differences, right? Things have changed. So, now, now let's focus in on the carbon on the left. So, the carbon on the left is now bonded to a hydrogen and carbon is a little bit more electronegative. So, we give both of those electrons to carbon and then we have carbon bonded to carbon, carbon bonded to carbon, and carbon bonded to carbon. So, what's the oxidation state Carbon is supposed to have four valence electrons and how many do have around it? Let's see, one, two, three, four, five. So, four minus five gives us an oxidation state of minus one. So, this carbon on the left now has an oxidation state of minus one. What about the carbon on the right? Well, we have a tie here, right? So, we divide up those electrons, a tie here, but oxygen is more electronegative than carbon. So, oxygen gets both of those electrons. So, the carbon on the right should have an oxidation state of for minus three. Here are the three electrons around carbon once we've accounted for electronegativity. So, that's an oxidation state of plus one. So, this carbon has an oxidation state of plus one. So, overall, what happened here? Well, let me use red. So, the carbon on the left went from an oxidation state of zero to an oxidation state of minus one. That is a decrease in the oxidation state."
    },
    {
        "Q": "\nIs there a difference between amplitude as described in :47 (which is on the displacement vs time graph) and the green arrows at 4:20 (on the displacement vs 'position x' graph)?",
        "A": "No the amplitude is same",
        "video_name": "-_xZZt99MzY",
        "timestamps": [
            260
        ],
        "3min_transcript": "Typical sounds have frequencies in the 100s or even 1000s of hertz. For instance, this note, which is an A note, is causing air to oscillate back and forth 440 times per second. So, the frequency of this A note is 440 hertz. Higher notes have higher frequencies, and lower notes have lower frequencies. Humans can hear frequencies as low as about 20 hertz and as high as about 20,000 hertz, but if a speaker were to oscillate air back and forth more than about 20,000 times per second, it would create sound waves, but we wouldn't be able to hear them. (sound starts, then stops) For instance, this speaker is still playing a note, but we can't hear it right now. Dogs could hear this note, though. Dogs can hear frequencies up to at least 40,000 hertz. Another key idea in sound waves is the wavelength of the sound wave. The idea of a wavelength is that when this sound the air molecules will be compressed close together in some regions and spread far apart from each other in other regions. If you find the distance between two compressed regions, that would be the wavelength of that sound wave. Since the wavelength is a distance, we measure it in meters. People get wavelength and period mixed up all the time. The period of a sound wave is the time it takes for an air molecule to oscillate back and forth one time. The wavelength of a sound wave is the distance between two compressed regions of air. People get these mixed up because there's an alternate way to create a graph of this sound wave. Consider this. Before the wave moves through the air, each air molecule has some undisturbed position from the speaker that we can measure in meters. This number represents the equilibrium undisturbed position of that air molecule. Then as the sound wave passes by, the air molecules get displaced slightly from that position. would be the displacement of the air molecule versus the undisturbed position or equilibrium position of that air molecule. This graph would let us know for a particular moment in time how displaced is that air molecule at that particular position in space. This graph shows us that in some regions the air is displaced a lot from its equilibrium position, and in other regions, the air is not displaced much at all from its equilibrium position. For this kind of graph, the distance between peaks represents the wavelength of the sound wave, not the period, because it would be measuring the distance between compressed regions in space. So, be careful. For a sound wave, a displacement versus time graph represents what that particular air molecule is doing as a function of time, and on this type of graph, the interval between peaks represents the period of the wave, but a displacement versus position graph represents a snapshot of the displacement of all the air molecules along that wave"
    },
    {
        "Q": "\nI just realized something. At 7:30, Sal says that the length of the vial is 1 centimeter, while the problem says that it is 1.0 centimeter. When he gets to his final answer at 12:15, he reports the answer to three significant digits. Since the length of the vial only had 2 significant digits, shouldn't the answer be 0.010, instead of 0.0998? Thanks for the input.",
        "A": "Yes, Sal should only keep 2 significant figures if the length of the vial is to two significant figures. You are correct in your understanding of this.",
        "video_name": "VqAa_cmZ7OY",
        "timestamps": [
            450,
            735
        ],
        "3min_transcript": "So you see the linear relationship? Let me draw the line. I don't have a line tool here, so I'm just going to try to freehand it. I'll draw a dotted line. Dotted lines are a little bit easier to adjust. I'm doing it in a slight green color, but I think you see this linear This is the Beer-Lambert law in effect. Now let's go back to our problem. We know that a solution, some mystery solution, has an absorbance of 0.539-- let me do our mystery solution in-- well, I've pretty much run out of colors. I'll do it in pink-- of 0.539. So our absorbance is 0.5-- this is 0.55, so 0.539 is going to be right over there. And we want to know the concentration of potassium permanganate. Well, if we just follow the Beer-Lambert law, it's got to sit on that line. this line right over here. And this over here looks like 0.10 molar. So this right here is 0, or at least just estimating it, looking at this, that looks like 0.10 molar, or 0.10 molarity for that solution. So that's the answer to our question just eyeballing it off of this chart. Let's try to get a little bit more exact. We know the Beer-Lambert law, and we can even figure out the constant. The Beer-Lambert law tells us that the absorbance is equal to some constant, times the length, times the concentration, where the length is measured in centimeters. So that is measured in centimeters. And the concentration is measured in moles per liter, So we can figure out-- just based on one of these data points because we know that it's 0-- at 0 concentration the absorbance is going to be 0. So that's our other one. We can figure out what exactly this constant is right here. So we know all of these were measured at the same length, or at least that's what I'm assuming. They're all in a 1 centimeter cell. That's how far the light had to go through the solution. So in this example, our absorbance, our length, is equal to 1 centimeter. So let's see if we can figure out this constant right here for potassium permanganate at-- I guess this is probably standard temperature and pressure right here-- for this frequency of light. Which they told us up here it was 540 nanometers."
    },
    {
        "Q": "At 1:15 he said 'spectrometer', we he soon corrected. I wonder, what is the difference between spectrometer and a spectrophotometer?\n",
        "A": "A spectrometer is An apparatus used for recording and measuring spectra, esp. as a method of analysis. . However, a spectrophotometer is ;An apparatus for measuring the intensity of light in a part of the spectrum, esp. as transmitted or emitted by particular substances.;, Hope that helped!",
        "video_name": "VqAa_cmZ7OY",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's see if we can tackle this spectrophotometry example. I took this from the Kotz, Treichel and Townsend Chemistry & Chemical Reactivity book and did it with their permission. So let's see what the problem is. It says a solution of potassium permanganate-- let me underline that in a darker color-- potassium permanganate has an absorbance of 0.539 when measured at 540 nanometer in a 1 centimeter cell. So this 540 nanometers is the wavelength of light that we're measuring the absorbance of. And so this is probably a special wavelength of light for potassium permanganate, one that it tends to be good at absorbing. So it'll be pretty sensitive to how much solute we have in the solution. OK, and the beaker is 1 centimeter. So that's just the length right there. What is the concentration of potassium permanganate? solution, the following calibration data were collected for the spectrophotometer. The absorbances of these known concentrations were already measured. So what we're going to do is we're going to plot these. And then, essentially, this absorbance is going to sit on the line. We learned from the Beer-Lambert law, that is a linear relationship between absorbance and concentration. So this absorbance is going to sit some place on this line. And we're just going to have to read off where that concentration is. And that will be our unknown concentration. So let's plot this first. Let's plot our concentrations first. So this axis, the horizontal axis, will be our concentration axis. I'll draw the axis in blue right there. I just need to make sure I have all this data here. So this is concentration in molarity. And let's see, it goes from 0.03 all the way to 0.15. So let's make this 0.03, then go three more. This over here is 0.06. One, two, three, then this over here is 0.09. This over here is 0.12. And then this over here is 0.15. And then the absorbances go-- well it's close to 0, or close to 0.1-- all the way up to close to 1. So let's make this right here 0.1. Let's make this 0.2, 0.3, 0.4, 0.5-- almost done-- 0.6, 0.7,"
    },
    {
        "Q": "At 0:08, What is a Spectrophotometer?\n",
        "A": "A spectrophotometer is a photometer that can measure intensity as a function of the light source wavelength. Important features of spectrophotometers are spectral bandwidth and linear range of absorption or reflectance measurement. Therefore it is an apparatus that uses light, that is a particular colour (wavelength), and measures the absorption/transmittance or reflected light strength/intensity.",
        "video_name": "VqAa_cmZ7OY",
        "timestamps": [
            8
        ],
        "3min_transcript": "Let's see if we can tackle this spectrophotometry example. I took this from the Kotz, Treichel and Townsend Chemistry & Chemical Reactivity book and did it with their permission. So let's see what the problem is. It says a solution of potassium permanganate-- let me underline that in a darker color-- potassium permanganate has an absorbance of 0.539 when measured at 540 nanometer in a 1 centimeter cell. So this 540 nanometers is the wavelength of light that we're measuring the absorbance of. And so this is probably a special wavelength of light for potassium permanganate, one that it tends to be good at absorbing. So it'll be pretty sensitive to how much solute we have in the solution. OK, and the beaker is 1 centimeter. So that's just the length right there. What is the concentration of potassium permanganate? solution, the following calibration data were collected for the spectrophotometer. The absorbances of these known concentrations were already measured. So what we're going to do is we're going to plot these. And then, essentially, this absorbance is going to sit on the line. We learned from the Beer-Lambert law, that is a linear relationship between absorbance and concentration. So this absorbance is going to sit some place on this line. And we're just going to have to read off where that concentration is. And that will be our unknown concentration. So let's plot this first. Let's plot our concentrations first. So this axis, the horizontal axis, will be our concentration axis. I'll draw the axis in blue right there. I just need to make sure I have all this data here. So this is concentration in molarity. And let's see, it goes from 0.03 all the way to 0.15. So let's make this 0.03, then go three more. This over here is 0.06. One, two, three, then this over here is 0.09. This over here is 0.12. And then this over here is 0.15. And then the absorbances go-- well it's close to 0, or close to 0.1-- all the way up to close to 1. So let's make this right here 0.1. Let's make this 0.2, 0.3, 0.4, 0.5-- almost done-- 0.6, 0.7,"
    },
    {
        "Q": "In 5:57, does 'visible universe' mean how much space we can see with really powerful telescopes or even more than that? If it is more than why is it 'visible'?\n",
        "A": "It is that part of the universe which is within range so that light from there would have had time to reach us since the time of the big bang. Light that originated too far away has not had time to reach us, and objects at that distance are outside the visible (or observable ) universe, regardless of how good our telescopes are.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            357
        ],
        "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago."
    },
    {
        "Q": "\nWhy do some people think that the universe is flat? At 5:55, the visible universe looks like a sphere. So how can it be flat?",
        "A": "The flatness refers to the curvature of space itself, not the geometry of the boundary of the visible universe, which obviously would have to be spherical.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            355
        ],
        "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago."
    },
    {
        "Q": "\n5:55 I thought that the universe was much \"darker\" compared to this picture. Or white is just used to mark superclusters and not actual visible electromagnetic radiation? If we could actually have a look at this universe from somewhere else (impossible, since what we see here is essentially a map of time, the periphery of the circle being the first radiation reaching the Earth that we could detect) I think we would not see much light...",
        "A": "I believe that the white dots represent superclusters, and not electromagnetic radiation. I agree with your idea that if we were to look at the universe in a real time picture of the periphery of the circle we would see a much darker image, new born stars reduced to black dwarfs, etc.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            355
        ],
        "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago."
    },
    {
        "Q": "\nCa. at 9:40 we see the Virgo Supercluster oriented as center. Is there an actual center to the visible universe and where is the Milky Way located in proximity to that center?",
        "A": "The center of the visible universe is Earth. This is because it is from our reference point that we determine if something is visible.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            580
        ],
        "3min_transcript": "where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's about this picture right here. And the really mind numbing thing is, if someone told me that this is the entire universe, this by itself would certainly put things in perspective. But it's unknown what's beyond it. There's some estimates that this might be only be 1 times 10 to the 23rd of the entire universe. And it might even be the reality that the entire universe is smaller than this. And that's an interesting thing to think about. But I'll leave you there because I think no matter how you think about it, it's just-- I don't know. I actually, before doing this video, I stared at some of these photos for half an hour. This is my least productive day just because it's just so awe inspiring to think about what these dots and dots of the dots really are."
    },
    {
        "Q": "12:57 Can you give examples of specific protein receptors on the postsynaptic membrane that the neurotransmitters would bind to? Are they all channel coupled receptors? Thanks:)\n",
        "A": "Most of them are channel coupled receptors. One example would be a ligand-gated sodium channel ( ligand-gated because the presence of a ligand, such as a neurotransmitter, is what opens the gate, and sodium channel because it causes an influx of sodium into the post-synaptic neuron)",
        "video_name": "Tbq-KZaXiL4",
        "timestamps": [
            777
        ],
        "3min_transcript": "Let me do a zoom in of that just to make it clear what's going on. So after they've bonded-- this is kind of before the calcium comes in, bonds to those SNARE proteins, then the SNARE protein will bring the vesicle ultra-close to the presynaptic membrane. So that's the vesicle and then the presynaptic membrane will look like this and then you have your SNARE proteins. And I'm not obviously drawing it exactly how it looks in the cell, but it'll give you the idea of what's going on. Your SNARE proteins have essentially pulled the things together and have pulled them apart so that these two membranes merge. And then the main side effect-- the reason why all this is happening-- is it allows those neurotransmitters to be dumped into the synaptic cleft. So those neurotransmitters that were inside of our vesicle then get dumped into the synaptic cleft. It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's"
    },
    {
        "Q": "\n13:50 What is the deciding factor in neurotransmitters being excitatory (Na channel) vs inhibitors (K channel). Does one neuron hold multiple types of neurotransmitters or are they specific?",
        "A": "First Question: I m not really sure either. Sorry :( Second Question: Neurons often release more than one transmitter. Until relatively recently, it was believed that a given neuron produced only a single type of neurotransmitter. There is now convincing evidence, however, that many types of neurons contain and release two or more different neurotransmitters. Hope that helps! =D",
        "video_name": "Tbq-KZaXiL4",
        "timestamps": [
            830
        ],
        "3min_transcript": "It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's outside of the cell. So positive things are going to leave the cell if it's potassium. Remember, I used triangles for potassium. And so if positive things leave the cell, then if you go further down the neuron, it'll become less positive and so it'll be even harder for the action potential to start up because it'll need even more positive someplace else to make the threshold gradient. I hope I'm not confusing you when I say that. So this connection, the way I first described it, it's exciting. When this guy gets excited from an action potential, calcium floods in. It makes these vesicles dump their contents in the synaptic cleft and then that will make other sodium gates open up and then that will stimulate this neuron, but if it makes potassium gates open up, then it will inhibit it-- and that's how, frankly, these synapses work. I was about to say there's millions of synapses, but"
    },
    {
        "Q": "At around 14:07, Sal points out that if a K+ pump is triggered, K+ ions will flow out. Why would it go out and not in?\n",
        "A": "Sal mentions they move out according to the concentration gradient meaning it moves from the area of high concentration K+ to low concentration K+",
        "video_name": "Tbq-KZaXiL4",
        "timestamps": [
            847
        ],
        "3min_transcript": "It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's outside of the cell. So positive things are going to leave the cell if it's potassium. Remember, I used triangles for potassium. And so if positive things leave the cell, then if you go further down the neuron, it'll become less positive and so it'll be even harder for the action potential to start up because it'll need even more positive someplace else to make the threshold gradient. I hope I'm not confusing you when I say that. So this connection, the way I first described it, it's exciting. When this guy gets excited from an action potential, calcium floods in. It makes these vesicles dump their contents in the synaptic cleft and then that will make other sodium gates open up and then that will stimulate this neuron, but if it makes potassium gates open up, then it will inhibit it-- and that's how, frankly, these synapses work. I was about to say there's millions of synapses, but"
    },
    {
        "Q": "At 4:41 he said water can dissolve more molecules than any liquid. What makes it better than lava?\n",
        "A": "It has to do with the polarity of water, it causes ionic compounds to dissociate because there s an attraction between the molucules. Lava isn t dissolving things, it s burning or melting them because of the intense heat instead of interacting with them on a molecular level like a solvent would. Melting something is not the same as dissolving something.",
        "video_name": "QymONNa5C6s",
        "timestamps": [
            281
        ],
        "3min_transcript": "wax paper or some teflon or something where the water beads up like this. Some leaves of plants do it really well; it's quite cool. Since water adheres weakly to the wax paper or to the plant, but strongly to itself, the water molecules are holding those droplets together in a configuration that creates the least amount of surface area. This is high surface tension that allows some bugs and even I think one lizard and also one Jesus to be able to walk on water. The cohesive force of water does have its limits, of course. There are other substances that water quite likes to stick to. Take glass, for example. This is called adhesion. The water is spreading out here instead of beading up because the adhesive forces between the water and the glass are stronger than the cohesive forces of the individual water molecules in the bead of water. Adhesion is attraction between two different substances, so in this case the water molecules and the glass molecules. These properties lead to one of my favorite things about water; the fact that it can defy gravity. That really cool thing that just happened is called capillary action. Explaining it can be easily done Thanks to adhesion, the water molecules are attracted to the molecules in the straw. As the water molecules adhere to the straw, other molecules are drawn in by cohesion following those fellow water molecules. Thank you cohesion. The surface tension created here causes the water to climb up the straw. It will continue to climb until eventually gravity pulling down on the weight of the water in the straw overpowers the surface tension. The fact that water's a polar molecule also makes it really good at dissolving things, which we call it's a good solvent then. Scratch that. Water isn't a good solvent. It's an amazing solvent. There are more substances that can be dissolved in water than in any other liquid on earth. Yes, that includes the strongest acid that we have ever created. These substances that dissolve in water, sugar or salt being ones that we're familiar with, are called hydrophilic, and they are hydrophilic because they are polar. Their polarity is stronger than the cohesive forces of the water. When you get one of these polar substances in water, it's strong enough that it breaks all the little Instead of hydrogen bonding to each other, the water will hydrogen bond around these polar substances. Table salt is ionic, and right now it's being separated into ions as the poles of our water molecules interact with it. What happens when there is a molecule that cannot break the cohesive forces of water? It can't penetrate and come into it. Basically, what happens when that substance can't overcome the strong cohesive forces of water, can't get inside of the water? That's when we get what we call a hydrophobic substance, or something that is fearful of water. These molecules lack charged poles. They are non-polar and are not dissolving in water because essentially they're being pushed out of the water by water's cohesive forces. Water, we may call it the universal solvent, but that does not mean that it dissolves everything. (boppy music) There have been a lot of eccentric scientists throughout history, but all this talk about water"
    },
    {
        "Q": "\nAt 5:38, since we started with a ketone, shouldn't the compound be called a hemiketal?",
        "A": "It could (and maybe should) be called a hemiketal. Some people use hemiacetal for both types of intermediates. SInce this reaction type works for both aldehydes and ketones, I guess they just used the more general term hemiacetal .",
        "video_name": "8-ccnvn9DxI",
        "timestamps": [
            338
        ],
        "3min_transcript": "we would have, this oxygen over here, would now have two lone pairs of electrons around it, so let's show those, so let's make 'em blue here. So these electrons moved out onto our oxygen, like that. And we just formed a bond between the oxygen on our ethanol, and this carbon, so we have a bond here, like that. And this still had a hydrogen attached to it, an ethyl group, and a plus one formal charge, like that. Alright, so next, let's get a little bit of room down here. The third step would be deprotonation, so let me go ahead and write that. So, step three, we deprotonate. So, another molecule of ethanol could come along and function as a base, and a lone pair of electrons on ethanol could take this proton, which leaves these electrons behind on our oxygen. So let's go ahead, and show that. we would have an OH over here, on the left, let's go ahead and put in those electrons, and then over here, on the right, we would have, this time, two lone pairs of electrons on our oxygen. So let me go ahead, and use green for those. These electrons right in here moved off, onto our oxygen, and so, if you look at that structure closely, that's a hemiacetal. So deprotonation yields our hemiacetal here, which is an intermediate in our reaction. Alright, so next step, next step here is protonations; let me go ahead, and mark this as being step four. We're going to protonate this OH over here, on the left. And so, one of the possibilities would be a protonated ethanol over here, functioning as an acid, so let's go ahead, and draw that. So a plus one formal charge on this oxygen, and a lone pair of electrons picks up a proton, and so let's go ahead and show that. So we protonate the OH, and the reason why protonating the OH would be good, is that would give us water as a leaving group. So, let's once gain show those electrons; let's use magenta again. So these electrons, right here, picked up a proton, and let's show these electrons as being that bond now. And then over here, on the right, we have, once again, our oxygen, and ethyl, and then we have two lone pairs of electrons, and then, let's keep this lone pair green right here. And then, since we protonated the OH, we get a plus one formal charge on this oxygen here, and, if you look closely, let me use red for this, if you look closely over here, you can kinda see water hiding, right? We know water's an excellent leaving group, so, if these electrons in green moved in here,"
    },
    {
        "Q": "I haven't understood why the final velocity is negative (1:24) and neither why the gravitational acceleration is negative too... (2:02)\n",
        "A": "Because we said that up is positive.",
        "video_name": "15zliAL4llE",
        "timestamps": [
            84,
            122
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 1:36, Sal says velocity is a scalar quantity but isn't velocity a vector quantity since it requires both magnitude and direction?",
        "A": "Yeah,they corrected it later,you see?",
        "video_name": "15zliAL4llE",
        "timestamps": [
            96
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:23, I don't get why acceleration due to gravity is negative because acceleration is a scalar, so it does not have direction?!\n",
        "A": "Acceleration is a vector, not a scalar.",
        "video_name": "15zliAL4llE",
        "timestamps": [
            143
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 6:28, I am confused on how you found the momentum of the car.",
        "A": "The momentum of the car is it s mass (1,000kg) multiplied by it s velocity (9m/s eastward). The units of momentum are kg-m/s The momentum of any object is it s mass multiplied by it s velocity vector.",
        "video_name": "XFhntPxow0U",
        "timestamps": [
            388
        ],
        "3min_transcript": "momentum problems. Whoops. Invert colors. OK. So let's say we have a car. Say it's a car. Let me do some more interesting colors. A car with a magenta bottom. And it is, let's see, what does this problem say? It's 1,000 kilograms. So a little over a ton. And it's moving at 9 meters per second east. So its velocity is equal to 9 meters per second east, or to the right in this example. And it strikes a stationary 2, 000 kilogram truck. So here's my truck. Here's my truck and this is a 2,000 kilogram truck. And it's stationary, so the velocity is 0. somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't save space. And then the mass of the truck is 2,000. And what's its velocity? Well, it's 0. It's stationary initially. So the initial momentum of the system-- this is 2,000 times 0-- is 9,000 plus 0, which equals 9,000 kilogram meters per second. That's the momentum before the car hits the back of the truck. Now what happens after the car hits the back of the truck? So let's go to that situation. So we have the truck. I'll draw it a little less neatly. And then you have the car and it's probably a little bit-- well, I won't go into whether it's banged up and whether it released heat and all of that. Let's assume that there was nothing-- if this is a simple problem that we can do. So if we assume that, there would be no change in momentum. Because we're saying that there's no net forces acting on the system. And when I say system, I mean the combination of the car and the truck."
    },
    {
        "Q": "In Right hand rule, doesn't the index finger indicate the direction of magnetic field and the middle finger the direction of current? Then why does the video say otherwise? (at 9:45)\n",
        "A": "Try it your way and try it the way it s described in the video and see what happens.",
        "video_name": "jQ2nD8ZGeEw",
        "timestamps": [
            585
        ],
        "3min_transcript": "magnitude of l. Actually, let me write it. Well, 2 meters times the magnetic field, 1 tesla. And so when you take a cross product of something, this is just a reminder. l cross B. That's equal to the magnitude of l times the magnitude of B times the sine of the angle between them times some unit directional vector that we figure out with the right hand rule. So we already did the magnitude of the distance vector. That was 2 meters. We did the magnitude of the magnetic field. And what's the sine of the angle between them? Well, if the magnetic field is going into the screen, if it's going straight into the screen, you could imagine a bunch of arrows shooting into the screen. Those are the vectors. While our distance vector, or this l is in the screen, they So this angle is 90 degrees. So this actually just becomes 1. So in terms of the magnitude, we're done. The l cross B magnitude is 2 times 1 tesla. And then we multiply that times the current. And then we actually have the magnitude of the force. The magnitude of this force is going to be equal to 5 amperes times 2 meters times 1 tesla. Which is equal to 10 newtons. And then the only question left is, what is the direction of the force that the magnetic field is exerting? And this is where we break out the right hand rule. And it's no different. You could just imagine one of the positive particles moving in that direction, and just use the right hand rule. So let's take our hand out. And if we-- let me draw a hand. A right hand. So this is my right hand. If I have my thumb sticking out like that. So the l is going to be my index finger. And then the B is the magnetic field. That's going into the screen. So you can't see it. All you can take my word for it is that my middle finger is pointed downwards into the screen and then my other fingers are just doing something else. And there you have it. Your thumb is actually the direction of the force. Your index finger is the direction of-- we'll say l for these purposes. And then the magnetic field is going into it, so you can't see my middle finger but it's pointing downwards. I could draw a little x there, to show it's going downwards. And then the force is what my thumb is doing. So the force on this wire, or at least on that section of wire, is going to be perpendicular to the direction of the current. And that direction is going to be a 10 newton force. Anyway, I've run out of time."
    },
    {
        "Q": "\nThe particles in the experiment, shown in 7:51, seems to have an increased radius. How come? The particle is experiencing a constant acceleration due to the electrical force - is it really true that the absolute value of v stays the same? Should the radius in our example really be constant?",
        "A": "If the speed is constant, the radius is constant.",
        "video_name": "b1QFKLZC11U",
        "timestamps": [
            471
        ],
        "3min_transcript": "can actually visualize. And so this whole business of magnetic fields making charged particles go into circles, this is one of the few times that I can actually say has a direct application into things that you've seen. Namely, your TV. Or at least the old-school TVs. The non-plasma or LCD TVs, your cathode ray TVs, take advantage of this. Where you essentially have a beam of not protons but electrons. And a magnet-- if you take apart a TV, which I don't think you should do, because you're more likely to hurt yourself because there's a vacuum in there that can implode, and all that-- but essentially, you have a magnet that deflects this electron beam and does it really fast so it scans your entire screen of different intensities, and that's what forms the image. I won't go into that detail. Maybe one day I'll do a whole video on how TVs work. So that's one application of a magnetic field causing a beam of charged particles to curve. And then the other application, and this is actually one where it's actually useful to make the particle go in a circle, is these cyclotrons that you read in circles really, really fast, and then they smash them together. Well, have you ever wondered, how do they even make a proton go in a circle? It's not like you could hold it and guide it around in a circle. They pass it through an appropriate strength magnetic field, and it curves the path of the proton so that it can keep going through the same field over and over again. And then they can actually use those electric fields. I don't claim to have any expertise in this, but then they can keep speeding it up using the same devices, because it keeps passing through the same part of the collider. And then once it collides, you've probably seen those pictures. You know, that you spend billions of dollars on supercolliders, and you end up with these pictures. And somehow these physicists are able to take these pictures and say, oh, this is some new particle because of the way it moved. Well, what they're actually talking about is these are moving at relativistic speeds. move at different velocities, their mass is changing, and all that. But the basic idea is what we just learned. They move in circles. They move in circles because they're going through a magnetic field. But their radiuses are different because their charges and their velocities are going to be different. And actually some will move to the left and some will move to the right. And that might be because they're positive or negative, and then the radius will be dependent on their masses. Anyway, I don't want to confuse you. But I just wanted to show you that we actually are touching on some physics that a physicist would actually care about. Now with that said, what would have happened if this wasn't a proton but if this was an electron moving at this velocity at 6 times 10 to the seventh meters per second through a 0.5 tesla magnetic field popping out of this video. What would have happened? Well, this formula would have still been safe. The magnitude of the force is the charge-- but it wouldn't"
    },
    {
        "Q": "\nAt 2:13, what's an arduino?",
        "A": "A programmable computer chip.",
        "video_name": "VnfpSf6YxuU",
        "timestamps": [
            133
        ],
        "3min_transcript": "This is all our parts all laid out for the Bit-zee bot, the things that you'll need to make one. Now you can make yours out of a broad variety of things, and we highly recommend that you do that. The only thing that you really have to have is the Arduino. Everything else you can switch out for other things. You can use different types of batteries. You can use different motors, et cetera. I'm going to go through what I've got here and where the products came from, or where the parts came from, I should say. And then we're going to start to put a Bit-zee together on this board so you can see how it's all wired up. But if you don't happen to have two hair dryers that you can take apart, you can either go and buy two electric motors and get some wheels for them. Or there's a variety of things you can do to solve that problem. So again, these are two motors from our hair dryer. You can see the hair dryer blower fan there. And underneath that is a sheet of Lexan. It's a stiff plastic that's really resilient. And that sheet of Lexan, it's easy to machine. So it's going to be used for mounting some of our devices. And you can get that at a hardware store for a few dollars. And this is a universal remote, and it can be gotten at Target for around $8. And we're going to use that to control our Bit-zee bot. And then we've got some electrical tape and different 22-gauge wire. And then we have some solder here. We'll use that to make our solder connections. Just like if you saw the video for the motor controller, it was used to solder that together. And this is a motor controller, which will allow us to control the speed and direction of our motors. And this is our Arduino. It's our microprocessor that we can plug into-- I should say a microcontroller that we can plug into our computer and download code to it to get the motor controller and other things to function the way we want them to. So this is a breadboard, and it's used for prototyping. And we're going to show you how to wire it up And this is our digital recording module, and it's for basically recording sounds and playing them back. And we're going to use the Arduino to trigger that so that when the little bot drives around it can make some sounds. Of course, these are just double-A batteries, and they're going to go in this battery holder. The double-As are 1.5 volts. But when we connect them in series together, they're going to be 12 volts. So that'll be great for powering our motors, because they want to run on a higher voltage than 1.5. And so we have our different transistors here that we're going to use to do some switching in our circuits. And we've got some three-color LED and some screws and nuts and then a bunch of resistors. And these are 330 ohm, 10K ohm, 220 ohm. We'll go into the details on that kind of stuff later. And then I have some of the board of our alarm clock radio, so we're going to use some components off of that board."
    },
    {
        "Q": "At 2:08, I am not getting why he divided the concept into chiral molecules and chiral atoms, while in the end they are chiral molecules and not atoms and have the same concept?\n",
        "A": "Chiral atom is a potentially confusing term - chiral centre is a better expression. It s important to be able to identify a chiral centre because that leads to the identification of a chiral molecule. Understanding the exact configuration of groups around the chiral centre is also necessary for naming the molecule.",
        "video_name": "tk-SNvCPLCE",
        "timestamps": [
            128
        ],
        "3min_transcript": "If I were to draw a hand, and let me just draw a hand really fast, so I'll draw a left hand. It looks something like that. That is a left hand. Now, if I were to take its mirror image, let's say that this is a mirror right there, and I want to take its mirror image, and I'll draw the mirror image in green. So its mirror image would look something like this. Not exact, but you get the idea. The mirror image of a left hand looks a lot like a right hand. Now, no matter how I try to shift or rotate this hand like this, I might try to maybe rotate it 180 degrees, so that the thumb is on the other side like this image right here. But no matter what I do, I will never be able to make this thing look like that thing. I can shift it and rotate it, it'll just never happen. I will never be able to superimpose the blue hand on top of this green hand. When I say superimpose, literally put it exactly on top of the green hand. So whenever something is not superimposable on its mirror So this hand drawing right here is an example of a chiral object. Or I guess the hand is an example of a chiral object. This is not superimposable on its mirror image. And it makes sense that it's called chiral because the word chiral comes from the Greek word for hand. And this definition of not being able to be superimposable on its mirror image, this applies whether you're dealing with chemistry, or mathematics, or I guess, So if we extend this definition to chemistry, because that's what we're talking about, there's two concepts here. There are chiral molecules, and then there are chiral centers or chiral-- well, I call them chiral atoms. They tend to be carbon atoms, so sometimes they call them chiral carbons. So you have these chiral atoms. Now, chiral molecules are literally molecules that are not superimposable on their mirror image. I'm not going to write the whole thing. You know, not superimposable-- I'll just write the whole thing. Not superimposable on mirror image. Now, for chiral atoms, this is essentially true, but when you look for chiral atoms within a molecule, the best way to spot"
    },
    {
        "Q": "\nAt 9:57, what eactly does it mean to exert a repelling chage of Fe=5N/C in this case? Does the charge that's being exerted this force move away from the other positive charge at a rate that we can compare to F=ma?",
        "A": "If the charge is 1 C, it will experience a force of 5N. if there are no other forces on it, it will accelerate accordingly. If it is 2 C it will experience a force of 10 N",
        "video_name": "elJUghWSVh4",
        "timestamps": [
            597
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 2:46, Sal said that we need an equal and opposite force to lift the mass but wouldn't the force we've just applied and the gravitational force cancel each other?",
        "A": "Yes, and that would enable the mass to maintain its velocity, whatever that velocity is.",
        "video_name": "elJUghWSVh4",
        "timestamps": [
            166
        ],
        "3min_transcript": ""
    },
    {
        "Q": "around 9:10 sal says that to get the charge moving downwards, we have to exert a force of 10N. But if we exert that force in the downward direction, seeing that the metal plate is ALSO exerting a force by the same amount, won't the charge just stay stationary over there (like suspended in the electric field)???\nPLEASE HELP ME ON THIS ONE :(\nVERY IMPORTANT\n",
        "A": "The force you actually have to exert is 10.000000000(etc)1 N, which is basically 10, so that s the answer people give. Theoretically, yes it would float if they were both exactly 10N.",
        "video_name": "elJUghWSVh4",
        "timestamps": [
            550
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 11:06 David says that the height is -3m\\s, using the area formula for the second triangle, i-e Area of Triangle = 1/2(Base)(height) . So how is this possible that height can be measured using a negative sign, after all Height ain't no Vector quantity\n",
        "A": "Above the axis is positive, below is negative. Scalars can be positive or negative But in this case the height represents displacement, which is a vector",
        "video_name": "DD58B2siDv0",
        "timestamps": [
            666
        ],
        "3min_transcript": "so we're gonna end with positive 11 meters per second. You might object. You might say, \"Wait a minute, hold on now. \"If we want delta v, \"right, and that's positive two, \"shouldn't delta v be the whole thing \"from like zero to six seconds? \"Shouldn't I say v at six seconds minus v at zero \"is positive two meters per second?\" I can't do that. The reason I can't do that is because look at what I did on the left hand side, my time interval goes from zero to six but on the right hand side, I only included the area from four to six. That's the area, there's a yellow triangle right here. If I wanted to put six and zero on this left hand side, I could do that but from my total area, I wouldn't use that. I have to use the total area. In other words, the total are from zero all the way to six These sides have to agree with each other. So from zero to six, my total area would be, this area here was eight, right? We found that rectangle was eight. This area here was two. So my total area would be 10. I can do that if I want. I could say v at six minus v at zero was, well v at zero we said was one because I just gave you that, equals 10 meters per second. I get that the v at six would be 11 meters per second just like we got it before. So you can still do it mathematically like this but make sure your time intervals agree on those sides. Now let's do the last part here. So we can find this area. This area and the area always represents the area from the curve to the horizontal axis. So in this case it's below the horizontal axis. That means it can negative area. The reason is it's a triangle again. So 1/2 base times height. So 1/2, the base is one, two, three seconds. negative now, negative three meters per second squared. I get that the total area is gonna be negative 4.5 meters per second. All right, now Daisy's gonna have a change in velocity of negative 4.5. If we want to get the velocity at nine, there's a few ways we can do it. Right, just conceptually, we can say that Daisy started at six with a velocity of 11. Her change during this period was negative 4.5. If you just add the two, you add the change to the value she started with. Well you're gonna get positive 6.5 if I add 11 and negative 4.5 meters per second or, if that sounded like mathematical witchcraft, you can say that, all right, delta v equals, what, negative 4.5 meters per second. Delta v would be, all right, you gotta be careful,"
    },
    {
        "Q": "at 6:24. how is it possible to have infinitely small rectangles?\n",
        "A": "It s not, but we can imagine that we can get as many rectangles as we want, as small as we want. And then we can imagine what the limit of that is as the number of rectangles goes up and up and up.",
        "video_name": "DD58B2siDv0",
        "timestamps": [
            384
        ],
        "3min_transcript": "has to just be the change in velocity. So area and change in velocity are representing the exact same thing on this graph. Area is the change in velocity. That's gonna be really useful because when you come over to here the area is still gonna be the change in velocity. That's useful because I know how to easily find the area of a triangle. The area of a triangle is just 1/2 base times height. I don't easily know how to deal with an acceleration that's varying within this formula but I do know how to find the area. For instance the area here, though I have 1/2, the base is two seconds, the height is gonna be positive two meters per second squared. What are we gonna get? One of the halves, cancel. Well, the half cancels one of the twos and I'm gonna get that this is gonna be equal to two meters per second. That's gonna be the area that represents the change in velocity. by two meters per second during this time. Now you might object. You might say, \"Wait a minute. \"I'll buy this over here because height times width \"is just a times delta t, \"but triangle, that has an extra factor of a half in it, \"and there's no half up here. \"How does this, I mean, how can we still make this claim?\" We can make this claim because we'll do the same thing we always do. We can imagine, all right, imagine a rectangle here. We're gonna estimate the area with a bunch of rectangles. Then this rectangle, and this rectangle in your line like that looks horrible. That doesn't look like the area of a triangle at all. It's got all these extra pieces right here, right? You don't want all of that. And okay, I agree. That didn't work so well. Let's make them even smaller, right? Smaller width. So we'll do a rectangle like that. We'll do this one. You see we're getting better. This is definitely closer. This is not as bad as the other one but it's still not exact. so we'll make it even smaller rectangle and an even smaller rectangle here all of these at the same width but they're even smaller than the ones before. Now we're getting really close. This area is really gonna get close to the area of the triangle. The point is if you make them infinite testable small, they'll exactly represent the area of a triangle. Each one of them can be found with this formula. The delta v for each one will be the area, or sorry, the acceleration of the height of that rectangle times the small infinite testable width and you'll get the total delta v which is so gonna be the total area. Long story short, area on a, acceleration versus time graphs represents the change in velocity. This is one you got to remember. this is the most important aspect of an acceleration graph, oftentimes the most useful aspect of it, the way you analyze it. So why do we care about change in velocity? Because it will allow us to find the velocity."
    },
    {
        "Q": "\nAt 9:00 to 10:40 he keeps subtracting the initial (positive) velocity from the change in velocity, yet ends up with a greater number. Why or how is this?",
        "A": "He adds 1m/s to 10m/s on the right side of the equation so that it cancels out on the left side so you are left with the velocity at 6 seconds equals 11 m/s.",
        "video_name": "DD58B2siDv0",
        "timestamps": [
            540,
            640
        ],
        "3min_transcript": "then we can find the velocity at any other point. For instance, let's say I gave you the velocity Daisy had. For some reason I'm gonna stopwatch. I start my stopwatch at right at that moment. At t equals zero, Daisy had a velocity of, let's say positive one meter per second. So Daisy was traveling that fast at t equals zero. That was her velocity at t equals zero seconds. Now I can get the velocity wherever I want. If I want the velocity at four, let's figure this out. To get the velocity at four, I can say that the delta v during this time period right here, this four seconds. I know what that delta v was. That delta v was positive eight. We found that area, height times width. So positive eight is what the delta v is gotta equal. What's delta v? That's v at four seconds minus v at zero seconds. I know what v at zero second was. That was one. So we can get that v at four minus one meter per second is equal to positive eight meters per second. So I get the velocity at four was positive nine meters per second. And you're like, phew, that was hard. I don't wanna do that every time. Yeah, I wouldn't wanna do that every time either so there's a quick way to do it. We can just do this. What's the velocity we had to start with? That was one. What was our change in velocity? That was positive eight. So what's our final velocity? Well, one plus eight gives us our final velocity. It's positive nine. Well it's just gonna take this change in velocity of this area which represents the change in velocity which is gonna add our initial velocity to it when we solve for this final velocity. for instance, if I didn't make sense, for instance, if we want to find the velocity at six, well, we can just say we started at t equals four seconds with a velocity of positive nine. We start here with positive nine. so we're gonna end with positive 11 meters per second. You might object. You might say, \"Wait a minute, hold on now. \"If we want delta v, \"right, and that's positive two, \"shouldn't delta v be the whole thing \"from like zero to six seconds? \"Shouldn't I say v at six seconds minus v at zero \"is positive two meters per second?\" I can't do that. The reason I can't do that is because look at what I did on the left hand side, my time interval goes from zero to six but on the right hand side, I only included the area from four to six. That's the area, there's a yellow triangle right here. If I wanted to put six and zero on this left hand side, I could do that but from my total area, I wouldn't use that. I have to use the total area. In other words, the total are from zero all the way to six"
    },
    {
        "Q": "what is meant by \"6 followed by roughly 230's of molecules\" at 1:50 ?\n",
        "A": "Well, a mole of any element is equal to 6.022*10^23 atoms. So, if we ignore .022, it means roughly 6*10^23 atoms. This means that there are roughly 6 followed by 23 zeroes(6000.....) atoms. That s what Sal meant at 1:50. Hope this helps :)",
        "video_name": "5B1i26dUwME",
        "timestamps": [
            110
        ],
        "3min_transcript": "Before I continue, I want to introduce you to what might be an unfamiliar concept, although if you've taken chemistry, you might know a little bit about it-- it's called a mole. This isn't the thing that grows on your face with the hair in it, or the animal that digs in your backyard, although those are also called moles. We're talking about the SI unit called a mole. A mole is just a number. It's like saying a mole of something means a certain number of something, just like a dozen-- it's like saying that a dozen eggs is 12 eggs. Just like that, a mole would be 6-- I always forget the exact number, but it's 6.023, or something of that nature. You could look it up-- I think it's 6.023 times 10 to the twentieth. Let me look up the exact number, just because I think The mole is 6.022 times 10 to the 23 of something, so it's a very number of something. Normally, we don't deal in moles of eggs-- I don't think there has been a mole of eggs ever produced in the history of the universe. 10 to the 23 is a very, very, very large number. Where is it useful? A mole is useful for counting atoms, and so what is a mole of atoms or molecules? What's that many molecules? It's 6 followed by roughly 23 0's of molecules-- a very, very big number. What's interesting about a mole is that when I have a mole of something, its mass-- let's say its mass in grams. A to-- so if I have this many carbon molecules, its mass in grams is x grams. It'll have a mass of x grams, where x is the atomic mass number of an atom of carbon, although if I was talking about a mole of a molecule, I would figure out the atomic mass of the entire molecule. What's an atomic mass number? Let me see if I can do a Web search on a periodic table. I'm showing you what I do here, and it's not fancy. Let me go to Google, and let's look up periodic table, and"
    },
    {
        "Q": "at 3:45 why didn't Sal just write C6H8O2 in stead of HC6H7O2 for sorbic acid?\n",
        "A": "Both formulas are correct for sorbic acid! Sal wrote it with one of the hydrogen atoms at the beginning to make it a little more clear that the underlying structure of sorbic acid has one proton (or hydrogen ion) that is reactive, since it can be donated to a base. By writing it as HC6H7O2 and knowing that is an acid, one can figure out that once sorbic acid donates a proton, you are left with H+ and C6H7O2-.",
        "video_name": "jzcB3faNdq0",
        "timestamps": [
            225
        ],
        "3min_transcript": "is we think about well, if these are dissolved in water, it's an aqueous solution, these are going to disassociate into their, into ions. And so we would write that out on both of the, on both the reactant and the product side. So the potassium sorbate, we can write that as, it's gonna be a potassium ion dissolved in an aqueous solution, plus the C six H seven O two, this is also going to be an ion dissolved in the aqueous solution, plus the hydrochloric acid will dissolve, so you have the hydrogen proton dissolved in the aqueous solution plus the chloride ion, or anion I guess we could say it. so that's going to be in our aqueous solution, What happens? Well, you're going to have the C six H seven O two react with the hydrogen proton to get to sorbic acid. So you're gonna have sorbic acid, H C six H seven O two, that's the sorbic acid. It's going to be in an aqueous solution. So I took care so far of that and that, and then you're going to have, and then you're going to have your potassium ions, your potassium ions and your chloride ions. It's going to be just like that. So this right over here is the ionic equation, not the net ionic equation. I have the ions on the reaction, on the reactant side, and then on the product side right over here. And did I, yep, I included everything. Now you do the net ionic reaction, you can imagine what's going to happen here. I have potassium ions on the right. I could net them out on both sides. So let's net them out on both sides. I have chloride ions on the left. I have chloride ions on the right. I could net them out on both sides. And then I can write the net ionic equation. So the net ionic equation, net ionic equation, is going to be well, I have my C six H seven O two ion dissolved in an aqueous solution. You combine that with the hydrogen proton dissolved in the aqueous solution, and it's going to, it's going to give us sorbic acid: H C six H seven O two in our aqueous solution. So there you have it. That is the net ionic equation."
    },
    {
        "Q": "\nat 8:18 sal says that helium is smaller than hydrogen because it has 2 electrons oin the valence shell but in the case of helium , how can the nucleus attract the valence electrons ,it is a noble gas\nplease explain ?",
        "A": "Valence electrons are the electron in the outer shell of an atom. Helium and hydrogen both have valence electrons: hydrogen has 1, helium has 2. Why does the nucleus attract electrons? Because the nucleus is made up from protons (positively charged) that are attracting electrons (negatively charged) and opposite charges attract. Whether or not it is a noble gas is irrelevant. Note that the valance electrons are simply those in the outer most layer and nothing more.",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            498
        ],
        "3min_transcript": "So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up."
    },
    {
        "Q": "At 8:10 Sal mentions Coulomb attraction; Exactly what is that?\n",
        "A": "It s the attraction between two opposite charges due to Coulombs law. Law of electrostatics: The force between two point charges (e.g. proton & electron) is proportional to the product of their charges, and inversely proportional to the square of the distance between them. So the more charged particles are there, the higher is the charge pulling the electrons to the nucleus, the smaller becomes the distance between them",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            490
        ],
        "3min_transcript": "So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up."
    },
    {
        "Q": "at 9:35 why does sal make a diagonal line? I understand the horizontal line but why is there anything vertical involved?\n",
        "A": "Each element in the table has one more proton than the last and that means one more electron if the element is not ionized. So as you go down the table, you get more and more electrons, and more importantly, more orbitals. The more orbitals, an element has, the larger it will be.",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            575
        ],
        "3min_transcript": "And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up. So, what are going to be, what's going to be the smallest ones? Well, we've already said helium is the smallest. So what are going to be some of the largest? What are going to be some of the largest atoms? Well that's going to be the atoms down here at the bottom left. So, these are going to be large, these are going to be small. So, large over here, small over here and the general trend, as you go from the bottom left to the top right you are getting, you are getting smaller."
    },
    {
        "Q": "\nAt 2:40 Sal draws this putting two atoms together WITHOUT bonding / sharing electrons. Then he draws the other example WITH bondings(electrons share the same atom) ... At the first example you can take half of it and the second you even though can take half of it. So it seems to be that there isn't a difference in Bonding or Not Bonding?!",
        "A": "No, these are different ways of estimating an atomic radius. They give different results because atoms do not have a fixed or definite boundary.",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            160
        ],
        "3min_transcript": "That would work except for the fact that this is not the right way to conceptualize how electrons or how they move or how they are distributed around a nucleus. Electrons are not in orbits the way that planets are in orbit around the sun and we've talked about this in previous videos. They are in orbitals which are really just probability distributions of where the electrons can be, but they're not that well defined. So, you might have an orbital, and I'm just showing you in 2 dimensions. It would actually be in 3 dimensions, where maybe there's a high probability that the electrons where I'm drawing it in kind of this more shaded in green. But there's some probability that the electrons are in this area right over here and some probability that the electrons are in this area over here, and let's say even a lower probability that the electrons are over this, like this over here. And so you might say, well at a moment the electron's there. You might say well that's the radius. But in the next moment, there's some probability it might be likely that it ends up here. But there's some probability that it's going to be over there. Then the radius could be there. So electrons, these orbitals, these diffuse probability distributions, they don't have a hard edge, so how can you say what the size of an atom actually is? There's several techniques for thinking about this. One technique for thinking about this is saying, okay, if you have 2 of the same atom, that are- 2 atoms of the same element that are not connected to each other, that are not bonded to each other, that are not part of the same molecule, and you were able to determine somehow the closest that you could get them to each other without them bonding. So, you would kind of see, what's the closest that they can, they can kind of get to each other? So let's say that's one of them and then this is the other one right over here. that closest, that minimum distance, without some type of, you know, really, I guess, strong influence happening here, but just the minimum distance that you might see between these 2 and then you could take half of that. So that's one notion. That's actually called the Van der Waals radius. Another way is well what about if you have 2 atoms, 2 atoms of the same element that are bonded to each other? They're bonded to each other through a covalent bond. So a covalent bond, we've already- we've seen this in the past. The most famous of covalent bonds is well, a covalent bond you essentially have 2 atoms. So that's the nucleus of one. That's the nucleus of the other. And they're sharing electrons. So their electron clouds actually, their electron clouds actually overlap with each other, actually overlap with each other"
    },
    {
        "Q": "\nDo we have a line diagram for methane as Sal explained for Propane at 6:42",
        "A": "There s no way to represent it using a line structure.As each line represents a carbon-carbon bond, and methane only has carbon-hydrogen bonds, there is no line structure.",
        "video_name": "pMoA65Dj-zk",
        "timestamps": [
            402
        ],
        "3min_transcript": "And the important thing is, no matter what the notation, as long as you can figure out the exact molecular structure, as long as you can-- so there's this last CH3. Whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others. Now, there's an even simpler way to write this. You could write it just like this. Let me do it in a different color. You literally could write it so we have three carbons. So one, two, three. Now, this seems ridiculously simple and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, and in organic chemistry in particular, any of these-- let me call it a line diagram or a line angle diagram. It's the simplest way and it's actually probably the most useful way to show chains of carbons or to show organic molecules. Once they start to get really, really complicated, because something like this, you assume that the end points of any lines have a carbon on it. So if you see something like that, you assume that there's a carbon at that end point, a carbon at that end point, and a carbon at that end point. And then you know that carbon makes four bonds. There are no charges here. All the carbons are going to make four bonds, and each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. So just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. So you're going to see a lot of this. This really simplifies things. And sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that. molecule where you write the CH3's for the end points, but then you implicitly have the CH2 on the inside. You assume that this end point right here is a C and it's bonded to two hydrogens. So these are all completely valid ways of drawing the molecular structures of these carbon chains or of these organic compounds."
    },
    {
        "Q": "At 4:30, do you need to show all electrons for propane.\n",
        "A": "Unless you are drawing a dot structure no it s a waste of time. That s why we almost always draw structures like he does at 5:35, it tells us exactly the same information as the other structures it just saves time.",
        "video_name": "pMoA65Dj-zk",
        "timestamps": [
            270
        ],
        "3min_transcript": "Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Well, let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly it might So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has-- let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon so we're going to have a three-carbon chain. It has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen, so let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a done, a hydrogen there, and then a hydrogen there. Now notice, in this molecular structure that I've drawn, I have three carbons. They were each able to form four bonds. This guy has bonds with three hydrogens and another carbon. This guy has a bond with two hydrogens and two carbons. This guy has a bond with three hydrogens and then this carbon And so this is a completely valid molecular structure, but it was kind of a pain to draw all of these valence electrons here. So what we typically would want to do is, at least in this structure, and we're going to see later in this video there's even simpler ways to write it, so if we want at least do it with these lines, we can draw it like this. So you have a carbon, carbon, carbon, and then they are bonded to the hydrogens. So you'll almost never see it written like this because this is just kind of crazy. Hyrdrogen, hydrogen-- at least crazy to write. It takes forever. And it might be messy, like it might not be clear where these I didn't write it as clearly as I could. So they have two electrons there. They share with these two guys. Hopefully, that was reasonably clear. But if we were to draw it with the lines, it looks just like that. So it's a little bit neater, faster to draw, same exact idea here and here. And in general, and we'll go in more detail on it, this three-carbon chain, where everything is a single bond, is propane. Let me write these words down because it's helpful to get. This is methane. And you're going to see the rhyme-- you're going to see the reason to this naming soon enough. This is methane; this is propane. And there's an even simpler way to write propane. You could write it like this. Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3, so you have a CH3, connected to a-- this is a CH2, that is CH2 which is then connected to"
    },
    {
        "Q": "\nAt 11:35, why does the horizontal velocity stay the same? Doesn't the object horizontally slow down as it goes further in space (by which the velocity would decrease)?",
        "A": "In the real world the horizontal velocity would decrease because of air resistance but usually in problems like this air resistance is ignored to illustrate the concept without complicating it with extra details.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            695
        ],
        "3min_transcript": "cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out. times 1.02. It gives us 8.83 meters, So this is going to be equal to, this is going to be equal to, this is going to be oh, sorry. this is going to be equal to 8.8, is that the number I got? 8.83, 8.83 meters. And we're done. And the next video, I'm gonna try to, I'll show you another way of solving for this delta t. To show you, really, that there's multiple ways to solve this. It's a little bit more complicated but it's also a little bit more powerful if we don't start and end at the same elevation."
    },
    {
        "Q": "\nAt 6:44, why is the delta velocity be -10?",
        "A": "What are the values that determine the change in velocity, and what numbers represent them in this video?",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            404
        ],
        "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air,"
    },
    {
        "Q": "\n6:45 why is velocity -5 m/s and not 0 m/s?",
        "A": "Why would it be 0? Because the projectile hits the ground and stops? What we are trying to find out is how fast the projectile is going when it hits the ground. We wouldn t need to do any work to know that after it hits the ground it just sits there.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            405
        ],
        "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air,"
    },
    {
        "Q": "\nAt 4:05, Sal takes only the MAGNITUDE of the vert component of the velocity vector. What happens to its Direction ?",
        "A": "The direction is up which is what has been chosen as the positive vertical direction",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            245
        ],
        "3min_transcript": "We want to break it down it with x- and y-components, or its horizontal and vertical components. so that's its horizontal, let me draw a little bit better, that's its horizontal component, and that its vertical component looks like this. This is its vertical component. So let's do the vertical component first. So how do we figure out the vertical component given that we know the hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side, this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude So this is the magnitude of velocity, I'll say the velocity in the y direction. That's the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator out if you want, It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation"
    },
    {
        "Q": "At around 6:48, why is the final vertical velocity -5 m/s? Wouldn't the final velocity be 0, since the ball momentarily comes to a complete stop when hitting the ground?\n",
        "A": "You re right, but at the very instant that the ball hits the ground, before the ground stops it, it will be moving at -5 m/s. Just like how the ball isn t moving before it is thrown, but its initial velocity is still 5 m/s.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            408
        ],
        "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air,"
    },
    {
        "Q": "\nAt 06:26, why does Sal say the final velocity is -5 m/s? Surely it would be 0 m/s when it hits the ground as it inevitably will?",
        "A": "We are interested in its velocity right before it touches the ground. It is pretty obvious that the body will be at rest when it touches the ground, so calculating the velocity at that time is pretty useless. Since we d rather take into account the velocity with which the body strikes the ground, we consider -5 m/s (which is the velocity RIGHT before the body touches the ground).",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            386
        ],
        "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information"
    },
    {
        "Q": "\ni am confused about the initial velocity and the final velocity when you talked about it in 5:50. how do you figure it out?",
        "A": "The initial velocity is the velocity you start with (in this case, in the vertical direction). The final velocity is the velocity you finish with, which is exactly the same one as you started with but in the opposite direction (going down, instead of going up).",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            350
        ],
        "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information"
    },
    {
        "Q": "At 6:28 Sal says that the final velocity is -5m/s. I thought it would be 0 since it hits the ground and at that moment in time, it will have no velocity.\n",
        "A": "We re studying projectile motion. It s not a projectile when it s lying on the ground,.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            388
        ],
        "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information"
    },
    {
        "Q": "At 7:34 I am still confused on how the acceleration in the vertical direction became -9.8 meter per second squared...?\n",
        "A": "Downward direction is always taken a negative acceleration, a deceleration and is thus represented by a -ve.!",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            454
        ],
        "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02."
    },
    {
        "Q": "\nAt about 8:14 in video, both sides are divided by -9.8m/s/s. The left side has a -10m/s unit, but when the division is done the units are ignored. Can you explain why it is ok to divide m/s by m/s/s?",
        "A": "It s just algebra. (m/s) / ((m/s)/s) = 1/ (1/s) = s",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            494
        ],
        "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02."
    },
    {
        "Q": "\nat 10:20 how is it that the velocity of the object in the horizontal direction remains the same. i can't seem to put my head around this one.",
        "A": "Why would the velocity in the horizontal direction change? Is there a force in the horizontal direction? When you are in an airplane and you drop your peanuts, they land right at your feet, don t they? Their horizontal velocity remained the same.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            620
        ],
        "3min_transcript": "negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02. delta t, I'm using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of its velocity the whole time, we just assume we can this multiply that times our change in time and we'll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, So this is the component of our velocity in the x direction, or the horizontal direction. Once again, we break out a little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse is the cosine of the angle. Cosine of an angle is adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times"
    },
    {
        "Q": "\n11:00\n\nSal simplified 10*cos30 to 10*((squareRoot of 3)/2) to 5*squareroot of 3... wouldn't it still be 5*((squareRoot of 3)/2)?\n\nSorry, it's late and maybe I need to take a break...",
        "A": "No it would not. 10/2 is 5 so 10(square root of 3)/2 = 5(square root of 3).",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            660
        ],
        "3min_transcript": "delta t, I'm using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of its velocity the whole time, we just assume we can this multiply that times our change in time and we'll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, So this is the component of our velocity in the x direction, or the horizontal direction. Once again, we break out a little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse is the cosine of the angle. Cosine of an angle is adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out."
    },
    {
        "Q": "\nAt 3:21 why does Sal put two lines in the front and back of the unknown magnitude?",
        "A": "That is a way to indicate take the magnitude of the vector in between these bars . It s sort of like the vector version of an absolute value sign.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            201
        ],
        "3min_transcript": "we can use that to figure out how long will this rock stay in the air. Because it doesn't matter what its horizontal component is. Its vertical component is gonna determine how quickly it decelerates due to gravity and then re-accelerated, and essentially how long it's going to be the air. And once we figure out how long it's in the air, we can multiply it by, we can multiply it by the horizontal component of the velocity, and that will tell us how far it travels. And, once again, the assumption that were making this videos is that air resistance is negligible. Obviously, if there was significant air resistance, this horizontal velocity would not stay constant while it's traveling through the air. But we're going to assume that it does, that this does not change, that it is negligible. We can assume that were doing this experiment on the moon if we wanted to have a, if we wanted to view it in purer terms. But let's solve the problem. So the first that we want to do is we wanna break down this velocity vector. We want to break down this velocity vector that has a magnitude of ten meters per second. We want to break it down it with x- and y-components, or its horizontal and vertical components. so that's its horizontal, let me draw a little bit better, that's its horizontal component, and that its vertical component looks like this. This is its vertical component. So let's do the vertical component first. So how do we figure out the vertical component given that we know the hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side, this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude So this is the magnitude of velocity, I'll say the velocity in the y direction. That's the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator out if you want,"
    },
    {
        "Q": "\nAt 6:25 why is the final velocity equal to -5m/s and not 0 m/s ?",
        "A": "He is talking about the vertical velocity at which it falls back down to the ground. You know the old saying, what goes up must come down . Well, in order to come back down, it must do so at some speed. Assuming no loss of energy due to air resistance, it will come back down at the same speed at which it went up, except in the opposite direction, hence negative rather than positive.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            385
        ],
        "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information"
    },
    {
        "Q": "\nAt about 7:42 Sal said that vertical velocity = acceleration x change in time. Then he said that 10m/s = -9.8m/s2 x t I do not understand why the acceleration equals -9.8m/s2. I know that -9.8m/s2 is the acceleration of an object falling, but couldn't a projectile be thrown up at ANY acceleration? (For example, if the projectile is thrown up at an acceleration of 7.3m/s2 then 10m/s2 = 7.3m/s2 x t and t = 1.37 seconds, not 1.02 seconds)",
        "A": "No, a projectile will accelerate at 9.8 m/s^2 downward as soon as it is released. We can give it whatever initial upward velocity we want, but the moment it is released, there s no upward force on the object, so gravity takes over.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            462
        ],
        "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02."
    },
    {
        "Q": "\nThroughout this lecture series, (at this video time 0:37) you state that a chiral carbon is \"usually\" a carbon bonded to four different groups. This begs the question, when is a carbon chiral, that is not bonded to four different groups? Thank you, Mike Johnston",
        "A": "A chiral carbon always has four different groups. But you can have a chiral molecule that contains no chiral atoms.",
        "video_name": "0XSSPow5oAc",
        "timestamps": [
            37
        ],
        "3min_transcript": "In the last video we learned a little bit about what a chiral molecule or what a chiral carbon or a chiral atom is. What I want to do in this video is go through a bunch of examples and see if we can identify if there are any chiral atoms and to also see if we're dealing with a chiral molecule. So let's look at our examples here. So here I have, what is this? This is chlorocyclopentane. So the first question is do we have any chiral atoms? And when we look at our definition that we thought of chiral atoms, it all comes from this notion of handedness and not being able to be superimposable on your mirror image, but we said that they're usually carbons bonded to four different groups. Let's see, do we have any carbons here bonded to four different groups? Well, all the CH2's, they're bonded to another CH2 and then two H'2. I could draw it like this: H and H. So they're bonded to two of the same group, so none of these CH2's are good candidates for being a chiral They're both bonded to-- or all of them are bonded to two hydrogens and two other very similar-looking CH2 groups, although you have to look at the entire group that it's bonded to. But they're all definitely bonded to two hydrogens, so it's not four different groups. If we look at this CH right here, we could separate it out like this. We could separate the H out like this, and so since it's bonded to a hydrogen. This carbon is bonded to a chlorine, and then it's bonded to-- well, it's not clear when you look at it right from the get-go whether this group is different than this group. But if you go around, if you were to split it half-way like this, or maybe another better way to think about is if you were to go around this molecule in that direction, the counterclockwise direction, you would encounter a CH2 group, and then you encounter a CH2 group, and then you would encounter a third, and then you would encounter a fourth CH2 group, then you would come back to So you would encounter four CH2's and then you'd come back to where you were before. If you go in this direction, what happens? You encounter one, two, three, four CH2's and you come back to where you were before. So all of this, this bottom group, depending on how far you want to extend it, and this top group, are really the same group. So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule, because it does not have a chiral center. And to see that it's not a chiral molecule-- let me see if I can backtrack this back to the way I wrote it right before. So you see that it's not a chiral molecule. There's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this."
    },
    {
        "Q": "\nI love the part when Sal is looking for a spot to draw the flipped molecule at 3:55",
        "A": "I like that part, too. :)",
        "video_name": "0XSSPow5oAc",
        "timestamps": [
            235
        ],
        "3min_transcript": "So you would encounter four CH2's and then you'd come back to where you were before. If you go in this direction, what happens? You encounter one, two, three, four CH2's and you come back to where you were before. So all of this, this bottom group, depending on how far you want to extend it, and this top group, are really the same group. So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule, because it does not have a chiral center. And to see that it's not a chiral molecule-- let me see if I can backtrack this back to the way I wrote it right before. So you see that it's not a chiral molecule. There's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this. Then you have a CH, CH2, CH2, and you have a CH2, CH2, and then you complete your cyclopentane. Now, in this situation, is there any way to rotate this to get this over there? Well, if you just took this molecule right here and you just rotated it 180 degrees, what would it look like? Well, maybe a little over-- yeah, well, not quite 180 degrees, but if you were to rotate it so that the chlorine goes about that far, you would get this exact molecule. You would get something. It would look a little bit different. It would look like this. Let me see if I can do it justice. It would look like this. You would have a CH2. little bit more space. If I were to rotate this about that far, I would get a CH. You get the chlorine and then you have your CH2, and then you have another CH2, CH2, and then you would have your CH2 up there. If you were to rotate this all the way around, or actually this is almost exactly 180 degrees, it would look like this. And the only difference between this and this is just how we drew this bond here. I could have easily, instead of drawing that bond like that, I could draw it facing up like that, and these are the exact same molecule. So this molecule is also not chiral. So let's go to this one over here. So what is this? This is a bromochlorofluoromethane, just"
    },
    {
        "Q": "\nat 5:14 why does it produce 2 halid groups and not a alcohol group?",
        "A": "Answered starting at 2:15.",
        "video_name": "1k6MUeM-pEo",
        "timestamps": [
            314
        ],
        "3min_transcript": "And our halogen now has an extra lone pair of electrons, giving it a negative charge, making it a halide anion. So in the next step of the mechanism, the halide anion will function as our nucleophile and attack the carbon bonded to the oxygen, kicks these electrons off onto the oxygen. And that's how we make our second alkyl halide. So we'd form R-X as our other product. And also water would be produced in this as well-- so H2O, like that. Now, I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol, and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance. However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether's going to go away. And we know that we're going to get two alkyl halides out So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups. And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide will be this methyl group over here. So I take a methyl group and I attach it to bromine. And we'd also produce water, as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make it a little bit tricky. So in three dimensions-- so we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen's going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether,"
    },
    {
        "Q": "At 7:16 what are the names of those molecule?\n",
        "A": "It can be named 3-aminopropanal or 3-aminopropionaldehyde",
        "video_name": "GuaozMpFS3w",
        "timestamps": [
            436
        ],
        "3min_transcript": "that there are only single-bonds around that carbon, only sigma bonds, and so, therefore we know that carbon is SP three hybridized, with tetrahedral geometry, so SP three hybridized, tetrahedral geometry. All right, let's look at this carbon right here; it's the exact same situation, right, only sigma, or single bonds around it, so this carbon is also SP three hybridized, and so, therefore tetrahedral geometry. Let's next look at the oxygen here, so if I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here, so let's go ahead and calculate the steric number of this oxygen. So that's number of sigma bonds, so here's a single-bond, so that's a sigma bond, and then here's another one; so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom, so here's a lone pair of electrons, and here's a lone pair of electrons. So, I have two lone pairs of electrons, so I need four hybridized orbitals for this oxygen, and we know that occurs when you have SP three hybridization, so therefore, this oxygen is SP three hybridized: There are four SP three hybrid orbitals around that oxygen. All right, let's do geometry of this oxygen. So, the electron groups, there are four electron groups around that oxygen, so each electron group is in an SP three hydbridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here, so the geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent, so even though that oxygen is SP three hybridized, it's geometry is not tetrahedral; the geometry of that oxygen there is bent or angual. All right, and because of symmetry, this carbon right here is the same as this carbon, so it's also is the same as this carbon, so it's also SP three hybridized, so symmetry made our lives easy on this one. All right, let's do one more example. So, once again, our goal is to find the hybridization states, and the geometries for all the atoms, except for hydrogen, and so, once again, let's start with carbon; let's start with this carbon, right here. All right, so once again, our goal is to find the hybridization state, so the fast way of doing it, is to notice there's one double-bond to that carbon, so it must be SP two hybridized, and therefore the geometry is trigonal planar, so trigonal planar geometry. Let's do the steric number way, so if I were to calculate the steric number: Steric number is equal to the number of sigma bonds. So here's a sigma bond, here's a sigma bond; I have a double-bond between the carbon and the oxygen, so one of those is a sigma bond,"
    },
    {
        "Q": "\nI also wanted to ask that though in the eg at 7:00, we figured out that the S.N. of oxygen and I have clearly understood the concept of finding the steric no. but I wanted to know that how can oxygen with its elec. confi. 1s2 2s2 2px2 2py1 2pz1 excite its electron(or by any other means) form 4 sp3 hybridised just as we have seen so clearly in the case of carbon.",
        "A": "You don t excite the electrons. You hybridize the orbitals first and then puts the electrons in. So you hybridize the s orbital and two 2pz orbitals. This gives 2sp\u00c2\u00b2, 2sp\u00c2\u00b2, 2sp\u00c2\u00b2 and an unhybridized 2pz. The 6 electrons of O give the configuration (2sp\u00c2\u00b2)\u00c2\u00b2, (2sp\u00c2\u00b2)\u00c2\u00b2, (2sp\u00c2\u00b2)\u00c2\u00b9, (2pz)\u00c2\u00b9. The filled sp\u00c2\u00b2 orbitals are the lone pairs, the half-filled sp\u00c2\u00b2 orbital forms the \u00cf\u0083 bond to C, and the half-filled pz orbital forms the \u00cf\u0080 bond to C.",
        "video_name": "GuaozMpFS3w",
        "timestamps": [
            420
        ],
        "3min_transcript": "that there are only single-bonds around that carbon, only sigma bonds, and so, therefore we know that carbon is SP three hybridized, with tetrahedral geometry, so SP three hybridized, tetrahedral geometry. All right, let's look at this carbon right here; it's the exact same situation, right, only sigma, or single bonds around it, so this carbon is also SP three hybridized, and so, therefore tetrahedral geometry. Let's next look at the oxygen here, so if I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here, so let's go ahead and calculate the steric number of this oxygen. So that's number of sigma bonds, so here's a single-bond, so that's a sigma bond, and then here's another one; so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom, so here's a lone pair of electrons, and here's a lone pair of electrons. So, I have two lone pairs of electrons, so I need four hybridized orbitals for this oxygen, and we know that occurs when you have SP three hybridization, so therefore, this oxygen is SP three hybridized: There are four SP three hybrid orbitals around that oxygen. All right, let's do geometry of this oxygen. So, the electron groups, there are four electron groups around that oxygen, so each electron group is in an SP three hydbridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here, so the geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent, so even though that oxygen is SP three hybridized, it's geometry is not tetrahedral; the geometry of that oxygen there is bent or angual. All right, and because of symmetry, this carbon right here is the same as this carbon, so it's also is the same as this carbon, so it's also SP three hybridized, so symmetry made our lives easy on this one. All right, let's do one more example. So, once again, our goal is to find the hybridization states, and the geometries for all the atoms, except for hydrogen, and so, once again, let's start with carbon; let's start with this carbon, right here. All right, so once again, our goal is to find the hybridization state, so the fast way of doing it, is to notice there's one double-bond to that carbon, so it must be SP two hybridized, and therefore the geometry is trigonal planar, so trigonal planar geometry. Let's do the steric number way, so if I were to calculate the steric number: Steric number is equal to the number of sigma bonds. So here's a sigma bond, here's a sigma bond; I have a double-bond between the carbon and the oxygen, so one of those is a sigma bond,"
    },
    {
        "Q": "\nThis might be a silly question, but at 8:45 when he draws in the H - does it matter that he draws it down? Could it drawn up?",
        "A": "The carbon bonding with the H you mentioned is sp3, so it tetrahedral geometric. It is a common way to draw the H like that",
        "video_name": "7p2qfyqiXHc",
        "timestamps": [
            525
        ],
        "3min_transcript": "we have carbon with three bonds to hydrogen and one loan pair of electrons on this carbon, a negative one formal charge, so we can represent that here with our negative sign next to that carbon. A carbon with a negative charge is called a carbanion, so this is a carbanion, and let's analyze the pattern that we have for our carbanion. We have one, two, three bonds, so let me write that down, we have three bonds, and this time we have one loan pair of electrons, so we have one loan pair, so three single bonds plus one loan pair of electrons for a carbon will give us a negative one formal charge on that carbon, we will have a carbanion. These also come up in mechanisms in organic chemistry, so let's analyze some carbanions, so down here, let's start with the carbanion on the left, and the negative one formal charge is on So we should have three bonds and one loan pair of electrons on that carbon, well, let's analyze it, the carbon in red is bonded to a CH3, a CH3, and a CH3, so that takes care of our three bonds, and of course, here's the one loan pair of electrons. Let's move on to the next example, so the carbon with the negative one formal charge is this carbon, that I just marked in red, the carbon in red is directly bonded to a carbon here, and directly bonded to a carbon here, so that's two bonds, I need a total of three bonds. 'Cause I already have a loan pair of electrons, right here's my loan pair of electrons on that carbon, so I need one more bond, and that bond, of course, must be to a hydrogen, so I can draw in a hydrogen here, again, that hydrogen is usually left off when you're drawing dot structures, but it's important to realize that hydrogen is actually there. Finally, one more example. The negative one formal charge is on this carbon, it already has a loan pair of electrons, so so far, we have one bond and one loan pair, we need a total of three bonds, so we need two more bonds on that carbon in red, and so those last two bonds, of course, must be to two hydrogens. So it's important to be able to assign formal charge and to do the math, it's important to be able to do these calculations, but eventually you won't need to do the math anymore, eventually you'll be able to look at a carbon and come with a formal charge after you've done enough problems."
    },
    {
        "Q": "At 2:53 it is mentioned that in this example carbon only has 6 electrons around it, but what about the single electron in P orbital that is not bonding?\n",
        "A": "With a carbocation there is no electron in that p orbital. Another atom has taken it when the carbocation was formed.",
        "video_name": "7p2qfyqiXHc",
        "timestamps": [
            173
        ],
        "3min_transcript": "so I could give that one valence electron back to hydrogen, and one valence electron to carbon, and so we're going to divide up all of our bonds that way, alright, give one valence electron to hydrogen, and the other valence electron to carbon, because that makes it easier for us to see that carbon has four valence electrons in our drawing, so let me highlight them here, one, two, three, four. So we're going to subtract four from four, so four minus four is equal to zero. So carbon has a formal charge of zero in methane. Let's do another example, this one down here. You can see it's different because this time we have three bonds. So let me draw in the electrons in those bonds, and let's find the formal charge on carbon. The formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that we subtract the number of valence electrons We divide up the electrons in our bonds, just like we did before, and we can see that carbon has only three electrons around it this time, so I'll highlight those, one, two, and three. So four minus three is equal to plus one, so carbon has a formal charge of plus one. So carbon's supposed to have four valence electrons, it has only three around it, so it lost one of its electrons, which gives it a formal charge of plus one. Let me go ahead and redraw that, so over here on the right, we have carbon with three bonds to hydrogen, and this carbon has a plus one formal charge, so we can represent that by putting a + charge here next to the carbon. Notice that carbon does not have an octet of electrons around it, it has only six electrons around it, and that's actually okay, carbon can never exceed an octet, but it's okay for carbon to have less than eight electrons. it's a positively charged carbon, we call those carbocations, so let me write down here, this is a carbocation. And carbocations come up a lot in organic chemistry mechanisms, so it's really important to understand them. Alright, let's think about the pattern that we see here, we have three single bonds around this carbon, let me go ahead and highlight them here, so here's one, two, and three, so we have three single bonds around our carbon, and we have zero lone pairs of electrons around that carbon, so three bonds plus zero loan pairs of electrons will give you a positively charged carbon, will give you a carbocation. What is the hybridization of this positively charged carbon? Well, there's one, two, three single bonds and zero loan pairs of electrons, and so from the videos on hybridization, you should know that"
    },
    {
        "Q": "When he was talking in the pic about 1:25 your mouse shifted over long waves what are they?\n",
        "A": "long waves are the lowest part of the electromagnetic radiation scale even less than your tv and radio produce",
        "video_name": "PX_XSnVWlNc",
        "timestamps": [
            85
        ],
        "3min_transcript": "I want to make a quick correction to the video on quasars. In that video I said, and I mistakenly said that the creation disk that's really releasing the energy of the quasar, that it's releasing energy predominately in the x-ray part of the electromagnetic spectrum. And that was incorrect. Most quasars are actually emitting electromagnetic radiation across the spectrum, all the way from x-rays, as high frequency as x-rays, all the way down to infrared. And some quasars even release super high frequency gamma rays, and they'll release low frequency electromagnetic waves all the way down to radio waves. So I just wanted to make that correction. It's not predominately in the x-ray part of the spectrum. It's across the spectrum right over here. It's this entire range of the spectrum, and sometimes even a wider range. Now, the other thing I want to clarify is this is the range of the spectrum that's being emitted. But we have to remember that most, or actually all of these quasars are quite far away. Many of them are many, many billions of light years away. And so they're moving away from us at a very fast speed, or they're getting redshifted because the universe is expanding so fast relative to us at that point, or that coordinate is moving so fast away from our coordinate. And so even though this is the spectrum that's being emitted, it's all going to be redshifted. It's all going to be redshifted down, and so we are going to observe things at a much lower frequency, maybe around the radio part of the frequency. So everything will be redshifted down. And that's why these were originally called quasi stellar radio sources. Anyway, hopefully that clears things up a little bit."
    },
    {
        "Q": "\n5:38 What does cohesion mean?",
        "A": "Cohesion is the intermolecular attraction between like-molecules.",
        "video_name": "6G1evL7ELwE",
        "timestamps": [
            338
        ],
        "3min_transcript": "This is going to form a partial negative charge at the, I guess you could say, the non-hydrogen end that is the end that has, that's well I guess this top end, the way I've drawn it right over here. And this Greek letter delta, this is to signify a partial charge, and it's a partial negative charge. Because electrons are negative. And then over here, since you have a slight deficiency of electrons, because they're spending so much time around the oxygen, it forms a partial positive charge right over there. So right when you just look at one water molecule, that doesn't seem so interesting. But it becomes really interesting when you look at many water molecules interacting together. So let me draw another water molecule right over here. So it's oxygen, you have two hydrogens, and then you have the bonds between them. You have a partially negative charge there. Partially positive charge on that end. And so you can imagine the partial, the side that has a partially negative charge is going to be And that attraction between these two, this is called a hydrogen bond. So that right over there is called a hydrogen bond. And this is key to the behavior of water. And we're going to see that in future videos. All the different ways that hydrogen bonds give water its unique characteristics. Hydrogen bonds are weaker than covalent bonds, but they're strong enough to give water that kind of nice fluid nature when we're thinking about kind of normal, or you could say, normal temperatures and pressures. This nice fluid nature, it allows these things to be attracted to each other, to have some cohesion, but also to break and reform and flow past each other. So you can imagine another hydrogen bond with another water molecule right over here. So put my hydrogens over there. Put my hydrogens, your bonds, partial negative, partial positive right over there. And so we'll see in future videos, hydrogen bonds, Key for its properties to its ability to take in heat. Key for its ability to regulate temperature. The key for its ability is why lakes don't freeze over. It's key for some of it's properties around evaporative cooling and surface tension and adhesion and cohesion, and we'll see that. And probably most important, and it's hard to rank of these things, if we're thinking about biological systems, this polarity that we have in water molecules and these hydrogen bonding, it's key for its ability to be a solvent, for it to be able to have polar molecules be dissolved inside of water. And we'll see that in future videos."
    },
    {
        "Q": "\nwhy is oxygen a electronegative element at 3:16- 3:18?? and why is it way more electronegative that hydrogen at 3:23-3:25??",
        "A": "See the electronegativity values for oxygen and hydrogen and you ll see that the difference is great that s why oxygen is more electronegative than hydrogen. Also, as you go through the periodic table, from right to left, electronegativity increases. See the periodic trends for electronegativity for better understanding.",
        "video_name": "6G1evL7ELwE",
        "timestamps": [
            196,
            198,
            203,
            205
        ],
        "3min_transcript": "And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability and they jump around. And what's interesting about water is oxygen is extremely electronegative. So oxygen, that's oxygen and that's oxygen, it is extremely electronegative, it's one of the more electronegative elements we know of. It's definitely way more electronegative than hydrogen. And you might be saying, \"Well, Sal, \"what does it mean to be electronegative?\" Well, electronegative is just a fancy way of saying that it hogs electrons. It likes to keep electrons for itself. Hogs electrons, so that's what's going on. Oxygen like to keep the electrons more around itself than the partners that it's bonding with. So even in these covalent bonds, you say, \"Hey, we're supposed to be sharing these electrons.\" Oxygen says, \"Well I still want them to \"spend a little bit more time with me.\" And so they actually do spend more time on the side without the hydrogens than they do around the hydrogens."
    },
    {
        "Q": "\n2:18 Why are only 4 electrons shown? There should be 8 electrons shown.",
        "A": "Each bubble represents either a lone pair of electrons or a covalent bond in both cases they represent 2 electrons therefore 8 in total",
        "video_name": "6G1evL7ELwE",
        "timestamps": [
            138
        ],
        "3min_transcript": "- [Voiceover] I don't think it's any secret to anyone that water is essential to life. Most of the biological, or actually in fact all of the significant biological processes in your body are dependent on water and are probably occurring inside of water. When you think of cells in your body, the cytoplasm inside of your cells, that is mainly water. In fact, me, who is talking to you right now, I am 60% to 70% water. You could think of me as kind of this big bag of water making a video right now. And it's not just human beings that need water. Life as we know it is dependent on water. That why when we have the search for signs of life on other planets we're always looking for signs of water. Maybe life can occur in other types of substances, but water is essential to life as we know it. And to understand why water is so special let's start to understand the structure of water and how it interacts with itself. And so water, as you probably already know, is made up of one oxygen atom and two hydrogen atoms. And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability"
    },
    {
        "Q": "\nat 5:31 sal mentions \"action potential\". what is it?",
        "A": "An action potential is a short-lasting event in which the electrical membrane potential of a cell rapidly rises and falls, following a consistent trajectory. You will see this often in physiology, particularly is neurons because this is how signals travel through our bodies",
        "video_name": "ob5U8zPbAX4",
        "timestamps": [
            331
        ],
        "3min_transcript": "actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath. So the general idea, as I mentioned, is that you get a signal here. We're going to talk more about what the signal means-- and then that signal gets-- actually, the signals can be summed, so you might have one little signal right there, another signal right there, and then you'll have maybe a larger signal there and there-- and that the combined effects of these signals get summed up and they travel to the hillock and if they're a large enough, they're going to trigger an action potential on the axon, which will cause a signal to travel down the balance of the axon and then over here it might be connected via synapses to other dendrites or muscles. And we'll talk more about synapses and those might help trigger other things. So you're saying, what's triggering these things here? Well, this could be the terminal end of other neurons' axons, like in the brain. This could be some type of sensory neuron. This could be on a taste bud someplace, so a salt molecule be some type of sensor. It could be a whole bunch of different things and we'll talk more about the different types of neurons."
    },
    {
        "Q": "at 4:22, sam says the axon is covered by schwann cells which form the myelin sheath. but as far as i know myelin sheath have cells called schwann cells. i am confused. please help.\n",
        "A": "The myelin sheath is made by Schwann cells. The Schwann cells wrap around the axon and produce myelin around it, creating the myelin sheath",
        "video_name": "ob5U8zPbAX4",
        "timestamps": [
            262
        ],
        "3min_transcript": "probably in the next few. So this is where it receives the signal. So this is the dendrite. This right here is the soma. Soma means body. This is the body of the neuron. And then we have kind of a-- you can almost view it as a tail of the neuron. It's called the axon. A neuron can be a reasonably normal sized cell, although there is a huge range, but the axons can be quite long. They could be short. Sometimes in the brain you might have very small axons, but you might have axons that go down the spinal column or that go along one of your limbs-- or if you're talking about one of a dinosaur's limbs. So the axon can actually stretch several feet. Not all neurons' axons are several feet, but they could be. And this is really where a lot of the distance of the signal gets traveled. Let me draw the axon. So the axon will look something like this. connect to other dendrites or maybe to other types of tissue or muscle if the point of this neuron is to tell a muscle to do something. So at the end of the axon, you have the axon terminal right there. I'll do my best to draw it like that. So this is the axon. This is the axon terminal. And you'll sometimes hear the word-- the point at which the soma or the body of the neuron connects to the axon is as often referred to as the axon hillock-- maybe you can kind of view it as kind of a lump. It starts to form the axon. And then we're going to talk about how the impulses travel. And a huge part in what allows them to travel efficiently are these insulating cells around the axon. actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath."
    },
    {
        "Q": "why does sal say temperature changes at 1:50, if PV is the only thing determining it? i mean, if PV is constant throughout the process, and there is no loss of kinetic energy, how can temperature become lower?\n",
        "A": "Temperature will change because PV is determining it. So you logic is right, and so is Sal s. I think you just misinterpreted the way Sal said it.",
        "video_name": "lKq-10ysDb4",
        "timestamps": [
            110
        ],
        "3min_transcript": "SAL: In the last video, where we talked about macrostates, we set up this situation where I had this canister, or the cylinder, and had this movable ceiling. I call that a piston. And the piston is being kept up by the pressure from the And it's being kept down by, in the last example I had, a rock or a weight on top. And above that I had a vacuum. So essentially there's some force per area, or pressure, being applied by the bumps of the particles into this piston. And if this weight wasn't here-- let's assume that the piston itself or this movable ceiling itself, it has no mass-- if that weight wasn't there it would just be pushed indefinitely far, because there'd be no pressure from But this weight is applying some force on that same area downwards. So we're at some equilibrium point, some stability. And we plotted that on this PV diagram right here. I'll do it in magenta. So that's our state 1 that we were in right there. And then what I did in the last video, I just blew away And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1."
    },
    {
        "Q": "At 0:26 can we determine the value of this pressure? If so, how?\n",
        "A": "i think it is P = t/v (temp/volume) as volume is inversely proportional to pressure i.e if p increases v decreases and vice-versa",
        "video_name": "lKq-10ysDb4",
        "timestamps": [
            26
        ],
        "3min_transcript": "SAL: In the last video, where we talked about macrostates, we set up this situation where I had this canister, or the cylinder, and had this movable ceiling. I call that a piston. And the piston is being kept up by the pressure from the And it's being kept down by, in the last example I had, a rock or a weight on top. And above that I had a vacuum. So essentially there's some force per area, or pressure, being applied by the bumps of the particles into this piston. And if this weight wasn't here-- let's assume that the piston itself or this movable ceiling itself, it has no mass-- if that weight wasn't there it would just be pushed indefinitely far, because there'd be no pressure from But this weight is applying some force on that same area downwards. So we're at some equilibrium point, some stability. And we plotted that on this PV diagram right here. I'll do it in magenta. So that's our state 1 that we were in right there. And then what I did in the last video, I just blew away And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1."
    },
    {
        "Q": "\nAt 03:06 Sal says\"the temperature also probably went down\" Why would that be?",
        "A": "When the volume increases, the pressure and temperature usually both go down. Both pressure and temperature are products of molecular motion. When the same amount of molecules occupy a larger space, then they bounce off the walls less frequently. Therefore, the pressure and temperature are lower.",
        "video_name": "lKq-10ysDb4",
        "timestamps": [
            186
        ],
        "3min_transcript": "And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1. The pressure went down, the volume went up. The temperature also probably went down. And so I ended up in this other state once I reached equilibrium. And that's all fair and good, but wouldn't it have been nice if there was some way? If we could have said, look, you know, there's some way that we got from this point to this point? If we could perform my little rock experiment in a slightly different manner, so that all this hell didn't break loose, so that maybe at every point in between my macro variables are actually defined? So how could I do that? Remember, I said that the macro variables, the macrostates, whether it's pressure, temperature, volume, and there are others, but I said these are only defined when we are in a thermodynamic equilibrium. And that just means that things have reached a stability point. That, for example, the temperature is consistent throughout the system. If it's not consistent throughout the system, I shouldn't be talking about it. If the temperature is different here than it is up here, I shouldn't say that the temperature of"
    },
    {
        "Q": "If calories, or kilocalories, (3:00) are heat or energy in your body, why is it unhealthy to have to many calories in your body?\n",
        "A": "The energy is stored in fat, and evidence suggests that excess fat is not good for your health.",
        "video_name": "h-31O7CaF2o",
        "timestamps": [
            180
        ],
        "3min_transcript": "I have to do a better job of drawing sand. So this is sand right over here. If I wanna raise that one degree celsius, I would need a different amount. It actually turns out that I need less heat, I need less heat to raise the sand one degree celsius than I need to raise the temperature of the water or one gram of the water one degree celsius. So let's say this is a gram of water and this is a gram of sand. I'm going to need more heat here to raise this one degree celsius than to raise that. That's because water has a higher specific heat. So higher, higher, relatively higher specific heat. Specific heat. And sand, or at least relative to water, has a lower specific heat. Lower specific heat. So two ways you could think about it. Let me write this. Lower, lower specific, specific heat. Two ways to think about it. If you wanna raise one gram of each of them into the water than you're going to have to put into the sand. Or the other way around, if you put the same amount of energy into both, you're gonna raise the temperature of the sand a lot more than you would raise the temperature of the water. And water, actually, its specific heat has a special name and this is a name that you have seen before. The specific heat of water is called the calorie. Specific, specific heat of water is called the calorie. And you have seen this word before. When you've wanted to cut calories, when you've looked at the back of nutritional labels on food. There's one clarification. The calorie that people talk about when they're talking about nutritional labels or how many calories are actually in food, that's actually kilocalories. So if you see, if someone hands you a, let's say a bowl of ice cream. Let me draw a bowl of ice cream here. So if someone hands you a bowl of ice cream right over here and they tell you that this is 500 calories. This is 500 calories. If we're thinking of it in terms of specific heat, it's actually 500 kilocalories. 500 kilocalories. So there's a couple of ways that you could think about 500 kilocalories. You could think about it as, this is, this ice cream has enough energy to raise, to raise 500 kilograms of water one degree celsius. You could also view it as the amount, well actually if you wanna think of it in more human terms, most humans are roughly grown people are between 50 kilograms or 60, 70 kilograms roughly over there. You could say 50 kilograms of water. Actually a grown male might be composed of about 50 kilograms of water and then there's obviously other things that make up their weight, I'm just approximating. 50 kilograms of water"
    },
    {
        "Q": "\nAt 6:00 you said that sand heats up faster than water due to less specific heat. could that also be because the sun heats up the top inch or so of sand but penetrates deeper into the water? So the sun is heating up a greater volume of water than sand per area? Just curious.",
        "A": "If you take the same amount of sand and water (and place it with the same shape), still sand is going to heat faster. You should see the last part of the video when Sal talks about the friction of the hydrogen bonds in water.",
        "video_name": "h-31O7CaF2o",
        "timestamps": [
            360
        ],
        "3min_transcript": "in the top case, but in this case you would raise it 10. You would raise it 10 degrees celsius. And that's actually happening in our body. Your body heat is actually caused, partially, the energy from food, some of it is to process your movement and the different functions of your muscles and the brain and all the things you body does, and some of it is just producing heat. Sometimes as a byproduct of that movement and sometimes, frankly, just for the sake of producing heat. So, 500 calories you see on a food label, that's really 500 kilocalories, and that's enough energy to raise 500,000 grams of water, remember 500 kilograms, 500,000 grams of water, one degree celsius, or 50,000 grams of water 10 degrees celsius. But anyway, water has a special name, it's calorie. It's neat to be able to connect it to, well, what we're eating and to think about what a calorie actually means, or what kilocalorie actually means. But this notion that water has a higher relative specific heat It's actually one of the reasons why it's often nice to live near the coast. Because, let me draw, let me draw a coastline. Right over here. This is, actually I'll draw it from, I'll draw it from above. So that's the coastline. This is land. Let's just say it's made up of sand, for the sake of argument, and other things. This is water. Let's think about it, first let's think about it in the summer. Let's think about in the summer, when it is hot. Think about just a sunny day. So in the summer, when things are hot, so you have the sun, you have the sun right over here and it's radiating energy and obviously the area also has, is also warming up the things that it comes in touch with. But you're having, at least at the coastline, the air is roughly the same temperature over both and, although there will be some variation, and they're getting the same amount of sunlight. But since water has a higher specific heat, So this is going to get less hot. Let me write it over here. This is going to get less hot. Less hot. And the land is going to get more hot. More hot. More hot. And that's why, when you're on the coastline, the air is also affected by what's in touch with it, it's gonna be in tough with this less hot water, and so as the air, especially if it's coming from the ocean, if the air is going in this direction, then if you're at the coast, you're gonna get probably a cooler breeze than if you're more inland over here. If you're more inland over here, the air is going to be hotter. So this is going to be hotter air if you're inland. And it's less hot air if you're at the coastline because the air is being cooled down to some degree, it's not being warmed up as much by the water. Now you have the opposite effect if you think about the coldest times of the year. If you think about the middle of the night in the winter. So let me draw that. If you think about nighttime in the winter,"
    },
    {
        "Q": "\nAt 7:40 Sal shifts the parallel component down and joins it up to the Fg component. My question is, how can you know that this line will meet up exactly with the Fg component and therefore become a right angled triangle? If you don't know the length(force component) of the perpendicular line, how can you assume that it will be at a 90deg angle to the Fg line?",
        "A": "The projected force is skewed by the angle theta of the ramp. Meaning if the ramp wasn t inclined then Fg1 would be a line perpendicular with the ground. Since the ramp is inclined we know that the force is displaced by the same angle, as Sal proved in the video. Thus, Fg1 forms a right angle with the ramp whether the ramp in inclined or not.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            460
        ],
        "3min_transcript": "This too will be 90 minus theta degrees. 90 minus theta degrees. Now, given that, can we figure out this angle? Well one thing, we're assuming that this yellow force vector right here is perpendicular to the surface of this plane or perpendicular to the surface of this ramp. So that's perpendicular. This right here is 90 minus theta. So what is this angle up here going to be equal to? This angle, let me do it in green. What is this angle up here going to be equal to? So this angle plus 90 minus theta plus 90 must be equal to 180, or this angle plus 90 minus theta must be equal to-- let me write this down. I don't want to do too much in your head. So let's call it x. So x plus 90 minus theta. Plus this 90 degrees right over here, plus this 90 degrees, needs to be equal to 180 degrees. So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember."
    },
    {
        "Q": "\nAt about 2:30, he draws the force due to gravity and the force of the gravity of the perpendicular. I'm confused on which one of those is the force of gravity normal.",
        "A": "I don t get it",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel."
    },
    {
        "Q": "\nIs the sign of parallel. || . and magnetude same?? Becuz in 7:56 sal tells the parallel lines as magnitude??",
        "A": "parallel: || magnitude: ||x||",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            476
        ],
        "3min_transcript": "This too will be 90 minus theta degrees. 90 minus theta degrees. Now, given that, can we figure out this angle? Well one thing, we're assuming that this yellow force vector right here is perpendicular to the surface of this plane or perpendicular to the surface of this ramp. So that's perpendicular. This right here is 90 minus theta. So what is this angle up here going to be equal to? This angle, let me do it in green. What is this angle up here going to be equal to? So this angle plus 90 minus theta plus 90 must be equal to 180, or this angle plus 90 minus theta must be equal to-- let me write this down. I don't want to do too much in your head. So let's call it x. So x plus 90 minus theta. Plus this 90 degrees right over here, plus this 90 degrees, needs to be equal to 180 degrees. So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember."
    },
    {
        "Q": "At 8:56 Sal refers to the perpendicular component Fg as the adjacent but I thought that was supposed to be the opposite component which kinda throws off all the calculations after this. Did I miss something because I am really confused.\n",
        "A": "A side is termed as Adjacent or Opposite in reference to the the Angle you are considering. Here, Sal refers \u00d1\u00b2 (Theta). So the Fg\u00e2\u008a\u00a5 is the Adjacent side here. Hypotenuse is constant in position. But Adjacent and Opposite are variable, Because they change with respect to your preferred Angle.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            536
        ],
        "3min_transcript": "So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember. We can do this in a new color. SOH CAH TOA. Cosine is adjacent over hypotenuse. So this is equal to cosine of the angle. So cosine of theta is equal to the adjacent over the hypotenuse. So if you multiply both sides by the magnitude of the hypotenuse, you get the component of our vector that is perpendicular to the surface of the plane is equal to the magnitude of the force due to gravity times the cosine of theta. Times the cosine of theta. We'll apply this in the next video just so you can make the numbers a lot more concrete. Sometimes just the notation makes it confusing. You'll see it's really actually pretty straightforward. And then this second thing, we can use the same logic. If we think about the parallel vector right over here, the magnitude of the component of the force"
    },
    {
        "Q": "At about 2:30, he draws the force due to gravity and the force of the gravity of the perpendicular. I'm confused on which one of those is the force of gravity normal.\n",
        "A": "The normal force is always perpendicular to the surface of contact.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel."
    },
    {
        "Q": "At 7:52, it is stated that v+ = v_in. But isn't v_in = v+ - v-? How are these two statements consistent with each other?\n",
        "A": "In this video, v_in is the name of the voltage created by the voltage source on the left side of the circuit. It is connected directly to the v+ input of the opamp. In other opamp videos, the same variable name might be assigned a different meaning. like v_in = v+ - v- as you suggest. There are no standard rules or conventions for which signal gets to be called v_in. It all depends on the person who draws the schematic.",
        "video_name": "_Ut-nQ535iE",
        "timestamps": [
            472
        ],
        "3min_transcript": "Kay, so what I'm going to do next is I'm going to take this expression and stuff it right in there. Let's do that. See if we got enough room, okay now let's go over here. Now I can say that V out equals A times V plus minus V out times R2 over R1 plus R2, alright so far so good. Let's keep going, let's keep working on this. V not equals A times V plus minus A V not, R2 over R1 plus R2. terms over on the left hand side. Let's try that. So that gives me, V not plus A V not, times R2 over R1 plus R2 and that equals A times V plus, and actually I can change that now V plus is what? V plus is V in. Okay let's keep going I can factor out the V not. V not is one plus A R2 over R1 plus R2 and that equals AV in. and our original goal, we want to find V out in terms of V in. So I'm going to take this whole expression here and divide it over to the other side, so then I have just V not on this side, and V in on the other side. Make some more room. I can do that, I can say V not equals A V in divided by this big old expression, one plus A R2 over R1 plus R2. Alright so that's our answer. That's the answer. That's V out equals some function of V in. Now I want to make a really important observation here."
    },
    {
        "Q": "At 3:05 he said \"the water isn't exerting force on anything.\" but isn't it exerting force on the molecules around it?\n",
        "A": "Everything has gravity, so yes. However, the gravity of a molecule of water is so minuscule it can largely be ignored.",
        "video_name": "NGpJPz44JYc",
        "timestamps": [
            185
        ],
        "3min_transcript": "You can get rid of the area 1 on both sides, and then you're saying that the velocity up here is equal to the rate at which the top of the surface moves down and is equal to 1/1,000 of the velocity of the liquid spurting out of this little hole. With that, we actually have the three variables for the left-hand side of Bernoulli's equation. What are the variables on the left-hand side? What is the pressure at this point where we have a hole? This is an important thin. When we talk about Bernoulli's-- let me rewrite Bernoulli's equation. It's P1 plus rho gh1 plus rho V1 squared over 2 is equal to We figured out all of these terms. Now let's figure out the things that we have to input here. What is the pressure at point two? You might want to say, and this was my initial reaction, too, and that why I made a mistake, is that what 's the pressure at this depth in the fluid? That's not what Bernoulli's equation is telling us. Bernoulli's equation is telling us actually what is the external pressure at that hole. When we did the derivation, we were saying how much work-- this was kind of the work term, although we played around with it a little bit. But if we look at the water that's spurting out of the exerting force against anything so it's not actually doing work. When we think about the pressure, the output pressure, it's not the pressure at that depth of the fluid. You should think of it as the external pressure at the hole. In this case, there is no external pressure at the hole. Let's say that if we closed the hole, then at that point, sure. The pressure would be the pressure that's being exerted by the outside of the canister to contain the water, in which case, we would end up with no velocity. The water wouldn't spurt anywhere. But now we're seeing the external pressure is zero. That's what the hole essentially creates. We're going to say that P2 is zero, so this pressure was zero, because we're in a vacuum. P2 is also zero, so both of these are zero. Remember, that's the external pressure."
    },
    {
        "Q": "At 10:50, when do you use the kb value or the ka value?\n",
        "A": "When you set up the equation for the equilibrium constant, the products are on the top line and the reactants are on the bottom line (for example, K = [NH4+][[OH-]/[NH3]). If the reactant is a weak base (ie, NH3 in this example) then K is Kb. But if the reactant is a weak acid (eg, CH3COOH), then K is Ka. (For example, K = [H3O+][CH3COO-]/[CH3COOH].)",
        "video_name": "223KLPnJCBI",
        "timestamps": [
            650
        ],
        "3min_transcript": "And we're saying that we have zero for our initial concentration of hydroxide. So when the reaction comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide we would have just x. So since ammonia is acting as a weak base here, let's go ahead and write our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8 times 10 to the negative five. So this is equal to concentration of our products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration of our reactants, leaving out water. So we just have ammonia here, So let's go ahead and plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times 10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is extremely small number, then we don't have to worry about it And so we just say this is equal to 0.35. So 0.35 plus x is pretty close to 0.35. So this is times x. And make sure you understand this x is this x. And then once again, 0.15 minus x, if x is a very small number, that's approximately equal to 0.15. So now, we would have this. And we need to solve for x. So let's go ahead and do that. Let's get out the calculator here. We need to solve for x. So 1.8 times 10 to the negative five times 0.15, and then we need to divide that by 0.35. And that gives us what x is equal to. And so x is equal to 7.7 times 10 to the negative six. So x is equal to 7.7 times 10 to the negative six. x represents the concentration"
    },
    {
        "Q": "\nAt 9:50 we have 0.35 M NH4NO3, but how is that the same as having 0.35 M NH4? How come we just completely ignore the NO3 part of the compound?",
        "A": "because nitrate ions are mainly found as free ions on both sides of the equation, i.e, they are not bonded to any other ion while they are in solution, and thus are called spectator ions and can be neglected while writing a net ionic equation.",
        "video_name": "223KLPnJCBI",
        "timestamps": [
            590
        ],
        "3min_transcript": "take an H+ away from H2O, we form OH- or the hydroxide ion. So once again we start with our initial concentration. And we're going to pretend like it's one of the problems that we've been doing in earlier videos. So if we have 0.15 molar concentration of ammonia, we go ahead and put 0.15 here. We think about the change. Whatever concentration we lose for ammonia is the same concentration that we gain for ammonium since ammonia turns into ammonium. And therefore that's also the same concentration we gain for hydroxide. So one source for the ammonium ion it would be the protonation of ammonia. So that's one source. But we have an additional source because we also have 0.35 molar ammonium nitrate. So there's another source for ammonium ions. So we have 0.35 molar, so we go ahead and put 0.35 molar in here And we're saying that we have zero for our initial concentration of hydroxide. So when the reaction comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide we would have just x. So since ammonia is acting as a weak base here, let's go ahead and write our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8 times 10 to the negative five. So this is equal to concentration of our products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration of our reactants, leaving out water. So we just have ammonia here, So let's go ahead and plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times 10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is extremely small number, then we don't have to worry about it"
    },
    {
        "Q": "\nAt 11:14, would I be correct if I said that increasing the resistance decreases the volume (I'm guessing that more resistance means less current to the speaker, and less current means less volume). Am I correct?",
        "A": "you are correct it like water, a small rock would have low resistance but a big rock would have high resistance and so would let less water through",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            674
        ],
        "3min_transcript": "When you press the button, the pin moves and the switch gets triggered. When you wanna snooze in the morning, you push this. It shifts the pin and it causes the sleep button to be triggered. That's kinda how that works. This is kind of ingenious in another way too because it holds a bunch of different things together. It's got the pins. It holds the speaker and it also holds the ferrite rod with the copper coil around it which functions as an antenna. That's an antenna for AM/FM radio. The signals come from here. We got a wire broken there. Signals come from here and they go to this thing which is a setup of four variable capacitors and they help to tune out frequencies we don't want. When we turn our dial, we can go right to 101.1 FM or 538 AMR, whichever station we want. This helps us to select those things. Those variable capacitors help us to filter out unwanted frequencies. and they can be used to oscillate at a particular frequency if they're coupled with a capacitor. That can be useful in performing radio functions as well. This guy right here is a, it's a radio chip. It's an IC chip that helps to demodulate or to separate the music or the signal that you want from the actual wave. AM is amplitude modulation so that means that the wave is changed in its height. FM is frequency modulation so that means that the wave is changed in how often it occurs in order to embed the signal that we get to listen to as radio sound. This chip basically decodes that and says, this is the original wave and then this is embedded signal. That's able to be then sent to our speaker Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that."
    },
    {
        "Q": "\nAt 13:10, what is the material they use to etch away the area that isn't protected?",
        "A": "It depends. There are many, and I mean VERY many materials on the market that can do this, but I personally use a non-corrosive drying paste that protects certain areas from the acid, although this is time-taking and is not favored by most industries -The Weekend tinkerer",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            790
        ],
        "3min_transcript": "Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that. That resist electric, current flow. That can be useful because it helps to prevent too much power from flowing to certain components on the board and things like that. This is a transistor. These things are transistors. They can function as switches. These guys right here are filters and they can help to reduce noise or electromagnetic interference, They're probably used in the radio circuit here to help to clean up the signal. On the back here, you can see again, these prongs connect to the battery. This is the printed circuit board here on the back. Basically, it's a thin layer of copper that's been applied to this fiber glass board. Then a chemical was used. They used basically a photomotion process which is like a similar to remove ... They shine a light on it and they use certain areas of the copper and keep other areas. They'll use an asset or a material to etch away the areas that aren't protected. Those result in copper traces. Those copper traces are basically very well ordered, little tiny wires that are very flat and they allow us to connect all these different components in a very small space very efficiently so we can just push the components. These are called through-hole components. Through-holes and then solder them on the back and then they're all wired up together. We don't have to worry about a lot of messy wires and things not being connected correctly. You can see there's different components, small components on the back. The little surface mount resistors and things like that. That's our alarm clock radio and those are the insides. Hope you've enjoyed it."
    },
    {
        "Q": "\nAt 12:56, he mentions this photo emulsion process, but I couldn't tell what he meant. What is it?",
        "A": "please do check the information and confirm: when a photograph (NON DIGITAL) is taken, photo-chemical compounds react and produce different colours and shades. i guess it is quite similar...",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            776
        ],
        "3min_transcript": "Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that. That resist electric, current flow. That can be useful because it helps to prevent too much power from flowing to certain components on the board and things like that. This is a transistor. These things are transistors. They can function as switches. These guys right here are filters and they can help to reduce noise or electromagnetic interference, They're probably used in the radio circuit here to help to clean up the signal. On the back here, you can see again, these prongs connect to the battery. This is the printed circuit board here on the back. Basically, it's a thin layer of copper that's been applied to this fiber glass board. Then a chemical was used. They used basically a photomotion process which is like a similar to remove ... They shine a light on it and they use certain areas of the copper and keep other areas. They'll use an asset or a material to etch away the areas that aren't protected. Those result in copper traces. Those copper traces are basically very well ordered, little tiny wires that are very flat and they allow us to connect all these different components in a very small space very efficiently so we can just push the components. These are called through-hole components. Through-holes and then solder them on the back and then they're all wired up together. We don't have to worry about a lot of messy wires and things not being connected correctly. You can see there's different components, small components on the back. The little surface mount resistors and things like that. That's our alarm clock radio and those are the insides. Hope you've enjoyed it."
    },
    {
        "Q": "\nIn the diagram at 10:13 , u say that the top ones are correct for IUPAC naming system .I would request u to please state that whether both are applicable or only one ? If one please state which one and also why ?",
        "A": "The top 2 would be named exactly the same way: 3,4-diethyl-2-methylheptane",
        "video_name": "F8RCR_1jIAk",
        "timestamps": [
            613
        ],
        "3min_transcript": "So let's see what we have. Let's see how many carbons we have if we said this is our longest carbon chain. So let's number them. Let's call this carbon one. Let's call this carbon two, three, four, five, six, So once again this would be called heptane. What sort of substituents do we have coming off this molecule? We have a methyl group coming off of carbon two. We have an ethyl group coming off of carbon three. And we have another ethyl group coming off of carbon four. So that's the exact same situation we had for the first example here. So these are the same thing. So it doesn't really matter which one of those you chose, you'd be naming it the exact same name. Let's compare those two to the molecule to the example down here. Again it's the same molecule, but let's say you chose a different path. Let's say you chose down here. So you said, oh, this looks like it's So you go like this, and you say, all right, that's my longest carbon chain. How many carbons are in that? Well, this would be one, two, three, four, five, six, and seven. So what sort of substituents do we have in this molecule? Well coming off of carbon four we can see there is an ethyl group. Coming off of carbon three we can see this looks kind of complicated. It's not really a straight chain. This is much more complex substituent, which we'll get to naming in a future video. So for this molecule we have a total of two substituents. For the top molecule we have an example of three substituents. So the question is which one of these will be the correct way to name my molecule according to IUPAC nomenclature? So I have two chains of equal length. Both of these chains are seven carbons. So how do I break that tie? IUPAC rules state you choose the parent chain So the top one has three subtituents, the bottom example has two substituents. So if you were to name this molecule using IUPAC packed nomenclature you would choose the top way of naming it, which again we will get to in more detail in the next few videos here. So let's look at cyclo alkanes now. So we've just done straight chain alkanes. We looked at branched chain alkanes. Let's look at cyclo alkanes. So this is a pretty funny dot structure here. Let's see how many carbons are in this triangle. Well, of course, there's one, two, and three carbons. So if I were to draw what this molecule looks like, if I were to draw all the atoms involved there'd be three carbons like that. And to complete the octet around carbon there'd have to be two hydrogens on each carbon like that. So that's a cyclo alkane."
    },
    {
        "Q": "At 0:26 why is the oxygen said to get the plus one charge instead of the hydrogen. If oxygen is more electronegative than the hydrogen, wouldn't it make more sense to say the hydrogen is positively charged. Is this because when you draw these molecules, you usually leave out the hydrogens?\n",
        "A": "The oxygen will have a +1 formal charge. To find formal charge, you take the number of valence electrons of a free atom, subtract 1/2 # of shared e-, and subtract #of lone e-. In this case for oxygen in H3O+: Oxygen has 6 valence e- , has three bonds, and has 2 electrons that fill its octet but aren t involved in bonding. Therefore the formal charge is 6-3-2=+1.",
        "video_name": "BeHOvYchtBg",
        "timestamps": [
            26
        ],
        "3min_transcript": "- [Voiceover] Let's look at this acid base reaction. So water is gonna function as a base that's gonna take a proton off of a generic acid HA. So lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the A. Oxygen, oxygen is now bonded to three hydrogens. So it picked up a proton. That's gonna give this oxygen a plus one formal charge and we can follow those electrons. So these two electrons in red here are gonna pick up this proton forming this bond. So we make hydronium H30 plus and these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that. Let's analyze what happened. HA donated a proton so this is our Bronsted-Lowry acid. Once HA donates a proton, we're left with the conjugate base Water, H2O accepted a proton, so this is our Bronsted-Lowry base and then once H2O accepts a proton, we turn into hydronium H3O plus. So this is the conjugate acid. So H3O plus, the conjugate acid and then A minus would be a base. If you think about the reverse reaction, H3O plus donating a proton to A minus then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression and we're gonna consider the stuff on the left to be the reactants. We're gonna think about the fourth reaction and the stuff on the right to be the products. Let's write our equilibrium expression. And so we write our equilibrium constant and now we're gonna write KA which we call the acid, the acid ionization constant. So this is the acid ionization constant so acid dissociation. So either one is fine. All right and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. Over here for our products we have H3O plus, so let's write the concentration of hydronium H3O plus times the concentration of A minus, so times the concentration of A minus. All over the concentration of our reactant, so we have HA over here, so we have HA. So we could write that in and then for water, we leave water out of our equilibrium expression. It's a pure liquid. Its concentration doesn't change and so we leave, we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant and let's apply this to a strong acid."
    },
    {
        "Q": "\nat 4:35, the equation had a value d. what is it?",
        "A": "The d in the capacitor and inductor equations is an expression used in calculus. You can read it as a tiny change in ... something . So dv means a tiny change in voltage , and dt means a tiny change in time . The quotient dv/dt represents the rate of change (slope) of voltage with respect to time.",
        "video_name": "l-h72j2-X0o",
        "timestamps": [
            275
        ],
        "3min_transcript": "and we're gonna be very consistent about this and that's gonna keep us from making mistakes. All right, so let's go back to our resistor and we're gonna do the equation for a resistor. What is the I-V equation for a resistor? I-V equation means what relates current to voltage and for a resistor, it's V equals i times R so the voltage across the resistor is equal to the current through the resistor times this constant of proportionality that we call the resistance. This has a very important name. This is called Ohm's law and you're gonna use this a lot so that's Ohm's law right there. This is Ohm's law. Now for the IV relationship for the capacitor, the capacitor has that property that the current of the voltage, not to the voltage but to the rate of change of the voltage and the way we write that is current equals, C is the proportionality constant, and we write dv, dt so this is the rate of change of voltage with respect to time. We multiply that by this property of this device called capacitance and that gives us the current. This doesn't have a special name but I'm gonna refer to it as the capacitor equation so now we have two equations. Let's do the third equation which is for the inductor. The inductor has the property very similar to the capacitor. It has the property that the voltage across is proportional to the time rate of change of the current flowing through the inductor so this is a similar but opposite of how a capacitor works. The voltage is proportional to the time rate of change is voltage equals L, di, dt. The voltage is proportional. The proportionality constant is the inductants. The inductance of the inductor and this is the time rate of change of voltage, OH sorry, the time rate of change of current flowing through the inductor so this gives us our three equations. Here they are. These are three element equations and we're gonna use these all the time, right there, those three equations. One final point I wanna make is for both these equations of components, these are ideal, ideal components."
    },
    {
        "Q": "\nWhen, at ~ 2:41 Sal is talking about the specific heats of water is it assuming atmospheric pressure? Just wondering because I've been taught that it'll have lower boiling and evaporating points at lower pressures and higher ones at higher pressures.\nOn a separate note, should I do the section on the gas laws before this section?\n\nI'm not trying to be a smart Alec.\nThanks.",
        "A": "Specific heat has nothing to do with boiling. It is the amount of heat you need to add to 1 kg of the material to get the temperature to go up by 1 K.",
        "video_name": "zz4KbvF_X-0",
        "timestamps": [
            161
        ],
        "3min_transcript": "that heat energy is being used to kind of break the lattice structure. To add potential energy to the ice. Or essentially melt it. So for it here, right here, we're ice. Right at this point, we're zero degree ice. And then as we add more and more heat we get to zero degree water. So at zero degrees, you can either have water or ice. And if you have water, to turn it into ice you have to take heat out of it. And if you have ice, and you want to turn it into water you have to put heat into it. And then the heat is used again to warm up the water at some rate. And then at 100 degrees, which is the boiling point of water, right here, a similar phase change happens. Where the increased heat is not used to increase the temperature of the water, it's used to put potential energy into the system. So the water molecules are forced away from each other. The same way that if I'm forced away from the planet Earth. I have potential energy because I can fall back to the earth. of falling back to each other. But this energy right here is the energy necessary to vaporize the water. Right here you have 100 degree water. 100 degree liquid. And here you have 100 degree vapor, water vapor And then as you add more and more heat once again it increases the temperature But you, Sal, I already learned this a few videos ago, I have the intuition. But I want to deal with real numbers. I want to know exactly how much heat is required for these different things to happen. And for that, we can get these numbers. And these are specific to the different states of water. If you looked up any other element or molecule, you would have different values for these numbers we're going to be dealing with right now. But this first number right here is the heat of fusion And this is the amount of heat that's required to fuse 100 degree water into 100 degree ice. Or the amount of energy you have to take out of the water. So this distance right here or along this axis, If you're going in the the leftward direction you have to take that much out of the system to turn into ice. If you're going in the rightward direction you have to add that much to turn into water. So heat of fusion. It's called the heat of fusion because when you fuse something together you make it solid. So it could also be considered the heat of melting. Just two different words for the same thing depending on what direction you going. The important thing is the number, 333. Similarly, you have the heat of vaporization 2257 joules per gram. That's this distance along this axis right here. So if you had one gram of 100 degree liquid water and you wanted to turn it into one gram of 100 degree liquid vapor. And in all of this we're assuming that nothing silly is happening to the pressure that we're under constant pressure. You would have to put 2257 joules into the system."
    },
    {
        "Q": "What is potential energy? I'm confused since you mentioned it at 1:48.\n",
        "A": "Any form of stored energy is potential energy. Most commonly in physics you have gravitational potential energy and electric potential energy. You can also have elastic potential energy (the energy stored in a spring) and chemical potential energy (stored in chemical bonds), A physicist might tell you those are ultimately forms of electric potential energy but don t worry about that for now.",
        "video_name": "zz4KbvF_X-0",
        "timestamps": [
            108
        ],
        "3min_transcript": "A couple of videos ago, we learned that --if we started with solid water or ice at a reasonably low temperature --maybe this temperature right here is minus 10 degrees Celsius. And we can deal with Celsius when we're dealing with these phase changes because we really just care about the difference in temperature and not necessarily the absolute temperature. So one degree in Celsius is the same thing as one degree in Kelvin. So the differences are the same whether you're dealing with Celsius or Kelvin. So we're starting with minus 10 degree Celsius ice or solid water. And we learned that as you heat it up as you add heat energy to the water, the temperature goes up. The molecules, at least while they're in that ice lattice network, they just start vibrating. And their average kinetic energy goes up until we get to zero degrees. Which is the melting point of water. And at zero degrees, we already learned something interesting happens. The added heat in the system does not increase the temperature of the ice anymore. At least over that little period right here. that heat energy is being used to kind of break the lattice structure. To add potential energy to the ice. Or essentially melt it. So for it here, right here, we're ice. Right at this point, we're zero degree ice. And then as we add more and more heat we get to zero degree water. So at zero degrees, you can either have water or ice. And if you have water, to turn it into ice you have to take heat out of it. And if you have ice, and you want to turn it into water you have to put heat into it. And then the heat is used again to warm up the water at some rate. And then at 100 degrees, which is the boiling point of water, right here, a similar phase change happens. Where the increased heat is not used to increase the temperature of the water, it's used to put potential energy into the system. So the water molecules are forced away from each other. The same way that if I'm forced away from the planet Earth. I have potential energy because I can fall back to the earth. of falling back to each other. But this energy right here is the energy necessary to vaporize the water. Right here you have 100 degree water. 100 degree liquid. And here you have 100 degree vapor, water vapor And then as you add more and more heat once again it increases the temperature But you, Sal, I already learned this a few videos ago, I have the intuition. But I want to deal with real numbers. I want to know exactly how much heat is required for these different things to happen. And for that, we can get these numbers. And these are specific to the different states of water. If you looked up any other element or molecule, you would have different values for these numbers we're going to be dealing with right now. But this first number right here is the heat of fusion And this is the amount of heat that's required to fuse 100 degree water into 100 degree ice. Or the amount of energy you have to take out of the water. So this distance right here or along this axis,"
    },
    {
        "Q": "\nShouldn't we always calculate with K in the equation not C? Especially @6:11 because that's how the K's cancel out.",
        "A": "When you re talking about change in temperature, C is actually the same as K. The difference between -10 C (263 K) and 0 C (273 K) is the same in both cases: 10 degrees. It s important to note the difference when you re talking in absolute values (0 C vs. 273 K) but when you re talking in relative values, the amounts will be the same. (Note that this does NOT apply to F. Each degree F is about half a degree C or K.)",
        "video_name": "zz4KbvF_X-0",
        "timestamps": [
            371
        ],
        "3min_transcript": "and you wanted to condense it, you would have to take that much energy out of the system. OK, fine, you know how much energy is required for the phase changes. But what about these parts right here? How much energy is required to warm up a gram of ice by one degree Celsius or Kelvin? And for that we look at the specific heat. It takes 2 joules of energy to warm up 1 gram 1 degree Kelvin. When water is in the solid state When it's in the liquid state, it takes about double that. It takes about 4 joules per gram to raise it 1 degree Kelvin. And when you're in the vapor state, it's actually more similar to the solid state. So given what we know now, we can actually figure out how much energy it would take to go from minus 10 degree ice to 110 degree vapor. So the first thing we're going to be doing is we're going to be going from minus 10 degree ice to zero degree ice. So we're going to go 10 degrees. We have to figure out how much heat does it take to warm up ice by 10 degrees. So the heat is going to be equal to the change in temperature. So actually let me write the specific heat first. So 2.05 joules per gram Kelvin. Oh, and I should tell you we can't different values for the amount of ice we're warming up to vapor, so let's say we're dealing with 200 grams. So it'll be the specific heat times the number of grams we're warming up of ice times the change in temperature So, times 10 degrees Kelvin Let's just say it's a 10 degrees Kelvin change, it doesn't matter if we're using Kelvin or Celsius. I could have written a Celsius here. Let's put Celsius right there. So what is that equal to? Get the calculator. Clear it out. 2.05 times 200 times 10 is equal to 4,100 joules. Let me do this in a different color This is 4,100 joules. Fair enough. Now, so what we've done is just this part right here. This distance right here is 4,100 joules. Now we have to turn that zero degree ice into zero degree water. So now we have to go from 0 degree ice to 0 degree water."
    },
    {
        "Q": "\nat 4:19 shouldnt x approach zero\nthen and only,'I think ' we might get closer to zero\nPLEASE correct if i am wrong\nI am talking about the slope at point 'a' on the second graph",
        "A": "Since x = a+h, when h approaches 0, x approaches a.",
        "video_name": "Df2escG-Vu0",
        "timestamps": [
            259
        ],
        "3min_transcript": "you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of-- every place you see an x, replace it with an a. f of-- I'll stay in this color for now-- blank plus h minus f of blank, all of that over h. And I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way-- and this is often used as the alternate form of the derivative-- would be to do it directly. So this is the point a comma f of a. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? Well, it would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. Actually, let me do that in that purple color. Over x minus a. Now, how could we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. We have an expression for the slope of a secant line. And then we're bringing those x values of those points closer and closer together. So the slopes of those secant lines better and better and better approximate that slope of the tangent line. And at the limit, it does become the slope of the tangent line. That is the definition of the derivative. So this is the more standard definition of a derivative. It would give you your derivative as a function of x. And then you can then input your particular value of x. Or you could use the alternate form of the derivative. If you know that, hey, look, I'm just looking to find the derivative exactly at a. I don't need a general function of f. Then you could do this. But they're doing the same thing."
    },
    {
        "Q": "So, for resistors in parallel, the circuit experiences no drop in voltage when the current reaches the resistors (11:19)? And for resistors in series, there is a voltage drop? Also, are the formulas for total resistance for resistors in parallel and for resistors in series different?\n",
        "A": "There is a voltage drop, but it is identical for each resistor. The equations are very different. Series: Rtotal = R1 + R2 +...+Rn Parallel: 1/Rtotal = (1/R1 +1/R2 + ... +1/R3)",
        "video_name": "ZrMw7P6P2Gw",
        "timestamps": [
            679
        ],
        "3min_transcript": "That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes. through here? I2? I'm going to do this in a different color. It's getting confusing. I'll do it in the vibrant yellow. So the current flowing through here, once again, the potential difference from here-- that's not different enough-- the potential difference from here to here is also 16 volts, right? So the current is going to be I2, is going to be equal to 16/5, which is equal to 3 1/5 amps. So most of the current is actually flowing through this, and that makes sense because the resistance is less, right? So that should hopefully give you a little bit of intuition And less current is flowing through here, so I2 through the 20-ohm resistor is 0.8 amps is I1, and I2 through the 5-ohm resistor is equal to 3.2 amps. And it makes sense that when you add these two currents together, the 3.2 amperes flowing through here and the"
    },
    {
        "Q": "9:37 - Could you not also use the formula R(total) = 1 / ((1 / R1) + (1 / R2) + ... (1 / Rn)) , where \"Rn\" is the last resistance that is compiled into the equation?\n",
        "A": "Hello Swaggy, Yes! Personally I prefer this the equation you mentioned as it is very fast to solve using a calculator and the 1/x key. Regards, APD",
        "video_name": "ZrMw7P6P2Gw",
        "timestamps": [
            577
        ],
        "3min_transcript": "drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes."
    },
    {
        "Q": "\nIf the voltage difference between two points in the circuit is always the same as long as the resistence is between these two points,is there no voltage difference between two points which are both in the blue region, for example (04:43)? So, in a circuit without a resistence, wouldn't there be a voltage difference? But then there would no current at all... I don't understand this at all. Also, isn't the voltage difference which provokes the movement of the electrons?\n\nThanks!",
        "A": "Virtually all conductors have resistance, albeit very small. If you short a battery with a wire, you will get a very large amount of current flowing (likely the largest that battery can output). There will be a voltage drop across that wire that is equal to the voltage increase across the battery. Also batteries do have some built in internal resistance which also limits the maximum output current and results in a small voltage drop within the battery when current is flowing.",
        "video_name": "ZrMw7P6P2Gw",
        "timestamps": [
            283
        ],
        "3min_transcript": "Or you could even think of it that the current entering-- when the currents I2 and I1 merge, that they combine and they become Current 1, right? I mean, think about it. In a given second, if this is 5 coulombs per second-- I'm just making up numbers-- and this is 6 coulombs per second, in a given second right here, you're going to have 5 coulombs coming from this branch and 6 coulombs coming from this branch, so you're going to have 11 coulombs per second coming out once they've merged, so this would be 11 So I think hopefully that makes sense to you that this current is equal to the combination of this current and that current. Now, what do we also know? We also know the voltage along this entire ideal wire is color, in blue. So, for example, the voltage anywhere along this blue that I'm filling in is going to be the same, because this wire is an ideal conductor, and you can almost view this blue part as an extension of the positive terminal of the battery. And very similarly-- I'll do it in yellow-- we could draw this wire as an extension of the negative terminal of the battery. This is an extension of the negative terminal of the battery. So the voltage difference between here and here-- so let's call that the total voltage, or let's just call it the voltage, right? The voltage difference between that point and that point is the exact same thing as the voltage difference between this point and this point, which is the exact same thing point and this point. So what can we say? What is the total current in the system? If we just viewed this as a black box, that this is some type of total resistance, well, the total current in the system would be the total voltage, the voltage divided by-- let's call this our total resistance, right? Let's say we couldn't see this and we just said, oh, that's just some total resistance, and that is equal to the current going through R1. This is I1. This is a 1 right here. This is current I1. What's current I1? Well, it's going to be the voltage across this resistor divided by the resistance, right? That's what Ohm's law tells us: V is equal to IR, or another way we could say it is V over R is equal to I, right? So I1 is equal to the voltage across this resistor, but we"
    },
    {
        "Q": "\nAt 9:25 in the video the resistance comes out to be 4 ohms. While the mathematics behind this makes perfect sense, the physics doesn't. The smallest resistor on the entire circuit is 5 ohms. I understand that most of the electrons will want to pass through this, but wouldn't that mean that the total resistance should never come out to be less than the smallest resistor?",
        "A": "No, the resistance of two resistors in parallel will always be less than the resistance of either of the two. Think about it this way. Take the bigger resistor out. Now the resistance is just 4 ohms, right. Now put it back. There s another path for current to go through that wasn t there before. Therefore more current is going to flow, therefore the total resistance is less than it was before.",
        "video_name": "ZrMw7P6P2Gw",
        "timestamps": [
            565
        ],
        "3min_transcript": "So let's see if we can use this information we have learned to actually solve a problem, and I actually find that it's always easier to solve a problem than to explain the theory behind a problem. You'll see that with most of these circuit problems, it's actually very basic mathematics. So let's say I have a 16 volt battery plus, minus, it's 16 volts. And just to hit the point home that you always don't have to draw circuits the same, although it is nice if you're actually drawing complicated circuits, I could draw it like this. I could draw the circuit like this, and let's say that there's a resistor here. And then let's say there's a wire and then there's another resistor here, and that this decides to do some random loopy thing here and that they connect here, and that they come back here. This strange thing that I have drawn, which you will never see in any textbook, because most people are more reasonable than me, is the exact same-- you can almost drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law."
    },
    {
        "Q": "\nAt 0:07, Sal mentioned that at higher temperatures you get a 4th state(plasma). But doesn't our blood contain plasma, and isn't our average temperature is 37C?",
        "A": "good question. blood plasma has nothing to do with the fourth state of matter. It s just has the same (inaccurate, of course) name.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            7
        ],
        "3min_transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules."
    },
    {
        "Q": "At 0:06, what is the fourth state of matter Sal refers to?\n",
        "A": "Plasma would be the fourth state of matter that Sal is referring to.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            6
        ],
        "3min_transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules."
    },
    {
        "Q": "\nAround 14:00 minutes, the temperature goes up but the state does not change (yet). So does that mean that there is some ice that can be colder or warmer that other ice?",
        "A": "Sure, ice can be -10 C, or -20C, or -100C, or 0C.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            840
        ],
        "3min_transcript": "So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up."
    },
    {
        "Q": "At about14:50, Sal said that the steam will get hotter. However, is there a limit to that, or does it just heat up forever?\n",
        "A": "No, you cannot heat steam forever . Like all molecules, when sufficiently hot it will decompose. There is not a set temperature at which it decomposes, it is more of a gradual process as the temperature increases. You start getting significant quantities of water decomposing at around 2500 K. At about 3000 K you get around 50% of water molecules decomposing. Of course, this is a rather hot temperature and it is rather impractical to generate, much less create a container to house the reaction.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            890
        ],
        "3min_transcript": "Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go"
    },
    {
        "Q": "\nat 10:07 he said that the months stay the same. Is that what causes leap years?",
        "A": "No AegonTargaryen is right, a year is not completly 365 days. If it were, there wouldn t be February 29th...which BTW is my birthday, but we usually celebrate it on February 28th",
        "video_name": "2o-Sef6wllg",
        "timestamps": [
            607
        ],
        "3min_transcript": "important to see what happens to our calendar-- if we wait 1,800 years this arrow, it will still have a tilt of 23.4 degrees, but instead of pointing in this direction, it might be pointing in this direction. Or in fact, it will be pointing in this direction. I'm obviously not drawing it that exact. And then the bottom of the arrow will come out over here. So if you think about that, if you wait 1,800 years, and once again, the tilt hasn't changed or it's changed a little bit, but what the precession has done, tracing out this circle has changed the direction of this arrow, it changed the direction of our axis of rotation. And if you wait 1,800 years when will the Northern Hemisphere be pointed most away from the sun. Well, now, it won't be pointed most away from the sun at this point in space relative to the sun anymore because now its axis of rotation looks something like this. it will be most pointed away, or the Northern Hemisphere will be most pointed away from the sun, about a month earlier. So about a month earlier. It'll be most pointed away from the sun about a month earlier. So this is when it will be most pointed away from the sun. To today's time, we would say, no, that's still not the most pointed away, but since we have this precession, since the direction of the tilt I guess we could say, or the direction of our rotational axis is changing, we are now at a different point in our orbit where we are most pointed away from the sun. So this is 1,800 years later, approximately. So now based on this, and I think this is what Vicksoma might have been hinting at, you say, look, OK, it's earlier in our orbit. Wouldn't this now be November? It will still be December 22nd. This will still be December 21st or 22nd Depending on the year. Still be the same date, and that's because our calendar is based on when we are most tilted away or when we are most tilted towards the sun. So by definition, this is when we are most tilted away so this will be the winter solstice. So what happens is every year-- so the way I drew it right over here and, actually, this perihelion actually changes over time as well. There's a precession of the perihelion as well, but I'm not going to go into that right now. So if you fast forward 1,800 years, all that's going to happen is that what we consider by our calendar to be December 22nd in an absolute point in our orbit we'll be earlier in our orbit, but we're still going to call it December 22nd. And so the perihelion is going to be further away from that December 22nd, it's actually going to be a month further"
    },
    {
        "Q": "\nAt 1:40, why must those electrons also attack? Why can't we just be left with a halogen ion and 1-halogen-ethane?",
        "A": "We could. If the molecule is ethane, the attack of the \u00cf\u0080 electrons would give a halide ion and a 1\u00c2\u00b0 2-haloethyl carbocation. But the 1\u00c2\u00b0 cation is less stable than the cyclic halonium ion, so the reaction does not use that pathway.",
        "video_name": "Yiy84xYQ3es",
        "timestamps": [
            100
        ],
        "3min_transcript": "Let's take a look at the halogenation of alkenes. So on the left, I have my alkene. And I'm going to add a halogen to it, so something like bromine or chlorine. And I can see that those 2 halogen atoms are going to add anti to each other-- so they'll add on opposite sides of where the double bond used Let's take a look at the mechanism so we can figure out why we get an anti addition. So I start with my alkene down here. And I'm going to show the halogen approaching that alkene. So it's going to approach this way. And I put in my lone pairs of electrons. Like that. If I think about that halogen molecule, I know that it's nonpolar. Because if I think about the electrons in the bond between my two halogens here, both halogen atoms, of course, have the exact same electronegativity. So neither one is pulling more strongly. And so overall, the molecule is nonpolar. However, if the pi electrons in my alkene-- so I'm going to say that these electrons right here are my pi electrons-- if those electrons get too close to the electrons in blue, they would, of course, since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen, like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge because it's losing a little bit of electron density. Now, I could think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile, like that. At the same time, these electrons over here-- this electron pair on the left side of the halogen-- is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen, like that. So let's go ahead and draw the result of all those electrons moving around. So now, I have carbon singly-bonded to another carbon, like that. on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step."
    },
    {
        "Q": "At 3:00 shouldn't both carbon atoms have a partial + charge?\n",
        "A": "Yes, they should. He just put the \u00ce\u00b4\u00e2\u0081\u00ba on one of the carbons to illustrate the mechanism.",
        "video_name": "Yiy84xYQ3es",
        "timestamps": [
            180
        ],
        "3min_transcript": "since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen, like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge because it's losing a little bit of electron density. Now, I could think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile, like that. At the same time, these electrons over here-- this electron pair on the left side of the halogen-- is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen, like that. So let's go ahead and draw the result of all those electrons moving around. So now, I have carbon singly-bonded to another carbon, like that. on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon"
    },
    {
        "Q": "At 3:42, does the halide ion attack the carbon that is most substituted or can it attack any carbon equally? I know in the halohydrin reaction the H2O attacks the most substituted carbon and the halogen goes on the least substituted.\n",
        "A": "The halide ion preferentially attacks the more substituted carbon, just like water does in the halohydrin reaction.",
        "video_name": "Yiy84xYQ3es",
        "timestamps": [
            222
        ],
        "3min_transcript": "on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon It picked up the electrons in magenta. So that's what the carbon on the left will look like. The carbon on the right is still bonded to 2 other things. And the halide anion had to add from below. So now we're going to have this halogen down here. Like that. And so now we understand why it's an anti addition of my 2 halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexane as our reactant And we're going to react cyclohexane with bromine-- so Br2. Now, if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this."
    },
    {
        "Q": "\nAt 3:08 Hank says Ground tissue does photosynthesis but how can it get light if there is Dermal Tissue on the outside layer? Is the dermal tissue transparent?",
        "A": "Very good question. I might not be the right person to answer this, and I could be wrong, but I think photosynthesis does not require all of the visible light, but only some energy from the light. There are wavelengths of the light that will continue through the dermal tissue. I suppose the energy from those light waves contain enough energy to keep the photosynthesis going.",
        "video_name": "VFtOcdXeP0Y",
        "timestamps": [
            188
        ],
        "3min_transcript": "But these things alone can't explain a vascular plants extraordinary evolutionary success. I mean, algae was photosynthesizing long before plants made it fashionable and as we learned last week, non-vascular plants have reproductive strategies that are tricked out six ways from Sunday. So, what gives? The secret to vascular plants success is their defining trait, conductive tissues that can take food and water from one part of the plant to another part of a plant. This may sound simple enough, but the ability to move stuff from one part of an organism to another was a huge evolutionary break through for vascular plants. It allowed them to grow exponentially larger, store food for lean times and develop some fancy features that allowed them to spread farther and faster. It was one of the biggest revolutions in the history of life on earth. The result? Plants dominated earth long before animals even showed up and even today, they hold most of the world records. The largest organism in the world is a redwood in northern California, 115 meters tall. Bigger than three blue whales laid end-to-end. of quaking aspen in Utah, all connected by the roots. Weighing a total of 13 million pounds and the oldest living thing, a patch of sea grass in the Mediterranean dating back 200,000 years. We spend a lot of time congratulating ourselves on how awesomely magnificent and complex the human animal is, but you guys, I got to hand it to you. (music) So you know by now, the more specialized tissues an organism has, the more complex they are and the better they typically do, but you also know that these changes don't take place overnight. The tissues that define vascular plants didn't evolve all at once, but today we recognize three types that make these plants what they are. Dermal tissues make up their outermost layers and help prevent damage and water loss. Vascular tissues do all of that conducting of materials I just mentioned and the most abundant tissue type, ground tissues, carry out some of the most important and the storage of left over food. Now, some plants never go beyond these basics. They sprout from a germinated seed, develop these tissues and then stop. This is called primary growth and plants that are limited to this stage are herbaceous. As the name says, they are like herbs; small, soft and flexible and typically they die down to the root or die completely after one growing season. Pretty much everything you see growing in a backyard garden; herbs and flowers and broccoli and that kind of stuff. Those are herbaceous. But a lot of vascular plants go on to secondary growth, which allows them to grow, not just taller, but wider. This is made possible by the development of additional tissues, particularly woody tissues. These are your woody plants, which include; shrubs and bark covered vines, called lianas and, of course, your trees, but no matter how big they may or may not grow all vascular plants are organized into three main organs. All of which you are intimately familiar with, not just because you knew what they were when you were in second grade, but also because you probably eat them everyday. First, the root. It absorbs water and nutrients"
    },
    {
        "Q": "At about 10:18, the sap Hank is talking about gives the Ponderosa (spelling??) its delicious smell. In a maple tree, is that sap we turn into maple syrup produced in the same way the Ponderosa-scent-sap is? Thanks! :D\n",
        "A": "Maple sap is very watery and needs to be boiled down into what you can buy in the store.",
        "video_name": "VFtOcdXeP0Y",
        "timestamps": [
            618
        ],
        "3min_transcript": "while in cold, dry years they're light and thin. These woody remains form tree rings, which scientists can use, not only to track the age of a tree, but also the history of the climate that it lived in. Now, at the top of the xylem, water arrives at it's final destination, the leaf. Here water travels through an increasingly minuscule network of vein-like structures until it's dumped into a new kind of tissue called the mesophyll. As you can tell from it's name, meso meaning middle and phyll meaning leaf, this layer sits between the top and the bottom epidermis of the leaf, forming the bacon in the BLT that is the leaf structure. This, my friends, marks our entry into the ground tissue. I'm sure you're as excited about that as I am. Despite it's name, ground tissue isn't just in the ground and it's actually just defined as any tissue that's either not dermal or vascular. Regardless of this low billing though, it's where the money is and by money, I mean food. And the mesophyll is chalk full of parenchyma cells of various shapes and sizes and many of them are arranged loosely to let CO2 and other materials These cells contain the photosynthetic organelles, chloroplasts, which as you know, host the process of photosynthesis, but where is this CO2 coming from? Well, some of the neatest features on the leaf are these tiny openings in the epidermis called stomata. Around each stoma are two guard cells connected at both ends that regulate it's size and shape. When conditions are dry and the guard cells are limp, they stick together, closing the stoma, but when the leaf is flush with water the guard cells plump up and bow out from each other opening the stoma to allow water to evaporate and let carbon dioxide in. This is what allows evapotranspiration to take place, as well as photosynthesis. And you remember photosynthesis through a series of brain-rackingly complicated reactions sparked by the energy from the sun, the CO2 combines with hydrogen from the water to create glucose. The left over oxygen is released through the stomata and the glucose is ready for shipping. Now, if you've been paying attention, you'll notice that earlier I said that there are two kinds of vascular tissue as the sugar exits the leaf through the phloem. The phloem is mostly made of cells stacked in tubes with perforated plates at either end. After the glucose is loaded into these cells called sieve cells or sieve-tube elements, they then absorb water from the nearby xylem to form a rich, sugary sap to transport the sugar. This sweet sap, by the way, is what gives the ponderosa it's delicious smell. By way of internal pressure and diffusion, the sap travels wherever it's needed to parts of the plant experiencing growth during the growing season or down to the root if it's dormant like during winter, where it's stored until spring. So, now that you understand everything that it takes for vascular plants to succeed, I hope you see why plants equals winning and I'm not just talking about them sweeping the contest for biggest, heaviest, oldest living things, though again, congrats on that guys. Plants are not only responsible for like making rain happen, they're also the first and most important link in our food chain and that's why the world's most plant rich habitats like rain forests and grasslands are so crucial to our survival."
    },
    {
        "Q": "At 0:44 sec in the video how is the slope coming from positive infinity at negative four?\n",
        "A": "At x = -4, the slope is vertical and has a positive infinity slope, thus, the graph is going from that positive infinity amount to 0 at y=4.",
        "video_name": "NFzma7NsHtI",
        "timestamps": [
            44
        ],
        "3min_transcript": "I have a function f of x here, and I want to think about which of these curves could represent f prime of x, could represent the derivative of f of x. Well, to think about that, we just have to think about, well, what is a slope of the tangent line doing at each point of f of x and see if this corresponds to that slope, if the value of these functions correspond to that slope. So we can see when x is equal to negative 4, the slope of the tangent line is essentially vertical. So you could say it's not really defined there. But as we go slightly to the right of x equals negative 4, we just have a very, very, very positive slope. So you could kind of view it as our slope is going from infinity to very, very positive to a little bit less positive to a little bit less positive, to a little bit less positive, to a little bit less positive. So which of these graphs here have that property? Remember, this is trying to graph the slope. So which of these functions down here, which of these graphs, have a value that is essentially kind of approaching infinity when x is equal to negative 4, and then it So this one, it looks like it's coming from negative infinity, and it's getting less and less and less negative. So that doesn't seem to meet our constraints. This one looks like it is coming from positive infinity, and it's getting less and less and less positive, so that seems to be OK. This has the same property. It's getting less and less and less positive. This one right over here starts very negative and gets less and less and less negative. So we can rule that out. Now let's think about what happens when x gets to 0. When x gets to 0, the tangent line is horizontal. We're at a maximum point of this curve right over here. The slope of a horizontal line is 0. Remember, we're trying to look for which one of these curves represent the value of that slope. So which one of these curves hit 0 when x is equal to 0? Well, this one doesn't. So the only candidate that we have left is this one, And let's see if it keeps satisfying what we need for f prime of x. So after that point, it should start getting more and more negative. The slope should get more and more and more negative, essentially approaching negative infinity as x approaches 4. And we see that here. The value of this function is getting more and more negative, and it's approaching negative infinity as x approaches 4. So we'll go with this one. This looks like a pretty good candidate for f prime of x."
    },
    {
        "Q": "Shouldn't Hydrogen be the one with the partially positive charge and Fluorine with partially negative charge? Then why in the video did Sal do oppositely at 10:09 mins?\n",
        "A": "You are correct, the H would be partially positive and the F should be partially negative due to the differences in electronegativities. Also, Fluorine has chemical symbol of F, not Fl. Fl is now the symbol for Flerovium, element number 114.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            609
        ],
        "3min_transcript": "And because this van der Waals force, this dipole-dipole interaction is stronger than a London dispersion force. And just to be clear, London dispersion forces occur in all molecular interactions. It's just that it's very weak when you compare it to pretty much anything else. It only becomes relevant when you talk about things with noble gases. Even here, they're also London dispersion forces when the electron distribution just happens to go one way or the other for a single instant of time. But this dipole-dipole interaction is much stronger. And because it's much stronger, hydrogen chloride is going to take more energy to, get into the liquid state, or even more, get into the gaseous state than, say, just a sample of helium gas. Now, when you get even more electronegative, when this guy's even more electronegative when you're dealing with nitrogen, oxygen or fluorine, you get into a special case of dipole-dipole interactions, and that's the hydrogen bond. So it's really the same thing if a bunch of hydrogen fluorides around the place. Maybe I could write fluoride, and I'll write hydrogen fluoride here. Fluoride its ultra-electronegative. It's one of the three most electronegative atoms on the Periodic Table, and so it pretty much hogs all of the electrons. So this is a super-strong case of the dipole-dipole interaction, where here, all of the electrons are going to be hogged around the fluorine side. So you're going to have a partial positive charge, positive, partial negative, partial positive, partial negative and so on. So you're going to have this, which is really a dipole interaction. But it's a very strong dipole interaction, so people call it a hydrogen bond because it's dealing with hydrogen and a very electronegative atom, is pretty much hogging all of hydrogen's one electron. So hydrogen is sitting out here with just a proton, so it's going to be pretty positive, and it's really attracted to the negative side of these molecules. But hydrogen, all of these are van der Waals. So van der Waals, the weakest is London dispersion. Then if you have a molecule with a more electronegative atom, then you start having a dipole where you have one side where molecule becomes polar and you have the interaction 404 00:10:31,330 --&gt; 00:10:32,470 between the positive and the negative side of the pole. It gets a dipole-dipole interaction. And then an even stronger type of bond is a hydrogen bond because the super-electronegative atom is essentially stripping off the electron of the hydrogen, or almost stripping it off. It's still shared, but it's all on that side of the molecule. Since this is even a stronger bond between molecules, it will have even a higher boiling point. So London dispersion, and you have dipole or polar bonds, and then you have hydrogen bonds."
    },
    {
        "Q": "\nAt 11:00, why is the partial positive at fluorine? If the fluorine is more electronegative, why does it have a partial positive?",
        "A": "I believe that is the video s mistake, because generally in hydrogen bonds it is the hydrogen side that is slightly positive and the more electronegative element is slightly negative. You are correct, probably just a small mistake on the video s end.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            660
        ],
        "3min_transcript": "a bunch of hydrogen fluorides around the place. Maybe I could write fluoride, and I'll write hydrogen fluoride here. Fluoride its ultra-electronegative. It's one of the three most electronegative atoms on the Periodic Table, and so it pretty much hogs all of the electrons. So this is a super-strong case of the dipole-dipole interaction, where here, all of the electrons are going to be hogged around the fluorine side. So you're going to have a partial positive charge, positive, partial negative, partial positive, partial negative and so on. So you're going to have this, which is really a dipole interaction. But it's a very strong dipole interaction, so people call it a hydrogen bond because it's dealing with hydrogen and a very electronegative atom, is pretty much hogging all of hydrogen's one electron. So hydrogen is sitting out here with just a proton, so it's going to be pretty positive, and it's really attracted to the negative side of these molecules. But hydrogen, all of these are van der Waals. So van der Waals, the weakest is London dispersion. Then if you have a molecule with a more electronegative atom, then you start having a dipole where you have one side where molecule becomes polar and you have the interaction 404 00:10:31,330 --&gt; 00:10:32,470 between the positive and the negative side of the pole. It gets a dipole-dipole interaction. And then an even stronger type of bond is a hydrogen bond because the super-electronegative atom is essentially stripping off the electron of the hydrogen, or almost stripping it off. It's still shared, but it's all on that side of the molecule. Since this is even a stronger bond between molecules, it will have even a higher boiling point. So London dispersion, and you have dipole or polar bonds, and then you have hydrogen bonds. but because the strength of the intermolecular bond gets stronger, boiling point goes up because it takes more and more energy to separate these from each other. In the next video-- i realize I'm out of time. So this is a good survey, I think, of just the different types of intermolecular interactions that aren't necessarily covalent or ionic. In the next video, I'll talk about some of the covalent and ionic types of structures that can be formed and how that might affect the different boiling points."
    },
    {
        "Q": "\nIf the sides of atoms become negative and positive (4:26), then that means at one moment in time, one side of the neon atom will be attracted to another neon atom. As the atoms change charges again, the neon atom will be attracted in the other direction. But that would mean, even at absolute zero, the particles would be moving. How can this be? This would mean the presence of kinetic energy even though the gas has no heat.",
        "A": "It is not correct to say that atoms are not moving at 0 K. If that were the case. we would then know both the position and the momentum of the atom. That would violate the Heisenberg Uncertainty Principle. Just as an electron in a hydrogen atom has a certain minimum allowed energy (a 1s orbital), an atom at 0 K must have a certain minimum energy. For example, liquid helium does not freeze under atmospheric pressure at any temperature because of its zero-point energy.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            266
        ],
        "3min_transcript": "-- it's a probability cloud and it's what neon's atomic configuration is. 1s2 and it's outer orbital is 2s2 2p6, right? So it's highest energy electron, so, you know, it'll look-- I don't know. It has the 2s shell. The 1s shell is inside of that and it has the p-orbitals. The p-orbitals look like that in different dimensions. That's not the point. And then you have another neon atom and these are-- and I'm just drawing the probability distribution. I'm not trying to draw a rabbit. But I think you get the point. Watch the electron configuration videos if you want more on this, but the idea behind these probability distributions is that the electrons could be anywhere. There could be a moment in time when all the electrons out over here. There could be a moment in time where all the electrons are over here. Same thing for this neon atom. If you think about it, out of all of the possible configurations, there's actually a very low likelihood that they're going to be completely evenly distributed. There's many more scenarios where the electron distribution is a little uneven in one neon atom or another. So if in this neon atom, temporarily its eight valence electrons just happen to be like, you know, one, two, three, four, five, six, seven,eight, then what does this neon atom look like? It temporarily has a slight charge in this direction, right? It'll feel like this side is more negative than this side or this side is more positive than that side. Similarly, if at that very same moment I had another neon that has 1 2 3 4 5 6 7 8... that had a similar-- actually, let me do that differently. Let's say that this neon atom is like this: one, two, three,four, five, six, seven, eight. So here, and I'll do it in a dark color So this would be a little negative. Temporarly, just for that single moment in time, this will be kind of negative. That'll be positive. This side will be negative. This side will be positive. So you're going to have a little bit of an attraction for that very small moment of time between this neon and this neon, and then it'll disappear, because the electrons will reconfigure. But the important thing to realize is that almost at no point is neon's electrons going to be completely distributed. So as long as there's always going to be this haphazar distribution, there's always going to be a little bit of a-- I don't want to say polar behavior, because that's almost too strong of a word. But there will always be a little bit of an extra charge on one side or the other side of an atom, which will allow it to attract it to the opposite side charges of other similarly imbalanced molecules. And this is a very, very, very weak force. It's called the London dispersion force. I think the guy who came up with this, Fritz London, who was neither--"
    },
    {
        "Q": "At 7:24, why isn't the hydrogen-chlorine bond a hydrogen bond?\n",
        "A": "Because a hydrogen bond is a weak intermolecular bond i.e between two or more molecules due to dipole-moment. The H-Cl bond is simply a bond between two atoms to form a molecule. Hope that helps.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            444
        ],
        "3min_transcript": "I think he was German-American. London dispersion force, and it's the weakest of the van der Waals forces. I'm sure I'm not pronouncing it correctly. And the van der Waals forces are the class of all of the intermolecular, and in this case, neon-- the molecule, is an atom . It's just a one-atom molecule, I guess you could say. The van der Waals forces are the class of all of the intermolecular forces that are not covalent bonds and that aren't ionic bonds like we have in salts, and we'll touch on those in a second. And the weakest of them are the London dispersion forces. So neon, these noble gases, actually, all of these noble gases right here, the only thing that they experience are London dispersion forces, which are the weakest of all of the intermolecular forces. And because of that, it takes very little energy to get them into a gaseous state. So at a very, very low temperature, That's why they're called noble gases, first of all. And they're the most likely to behave like ideal gases because they have very, very small attraction to each other. Fair enough. Now, what happens when we go to situations when we go to molecules that have better attractions or that are a little bit more polar? Let's say I had hydrogen chloride, right? Hydrogen, it's a little bit ambivalent about whether or not it keeps its electrons. Chloride wants to keep the electrons. Chloride's quite electronegative. It's less electronegative than these guys right here. These are kind of the super-duper electron hogs, nitrogen, oxygen, and fluorine, but chlorine is pretty electronegative. So if I have hydrogen chloride, so I have the chlorine atom right here, it has seven electrons and then it shares an electron with the hydrogen. It shares an electron with the hydrogen, Because this is a good bit more electronegative than hydrogen, the electrons spend a lot of time out here. So what you end up having is a partial negative charge on the side, where the electron hog is, and a partial positive side. And this is actually very analogous to the hydrogen bonds. Hydrogen bonds are actually a class of this type of bond, which is called a dipole bond, or dipole-dipole interaction. So if I have one chlorine atom like that and if I have another chlorine atom, the other chlorin eatoms looks like this. If I have the other chlorine atom-- let me copy and paste it-- right there, then you'll have this attraction between them. You'll have this attraction between these two chlorine atoms-- oh, sorry, between these two hydrogen chloride molecules. And the positive side, the positive pole of this dipole is the hydrogen side, because the electrons have kind of left it, will be attracted to the chlorine side"
    },
    {
        "Q": "\nAt 7:33, if the bond is between two atoms that are the same such as a C-C bond, which way would the dipole moment go towards?",
        "A": "when equally electronegative atoms form a bond the electrons do not prefer to go more towards either of the atoms i.e., they are shared equally between the atoms giving a zero dipole moment",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            453
        ],
        "3min_transcript": ""
    },
    {
        "Q": "What's the relation with center of mass, at 2:51?\n",
        "A": "The molecule has different centers of charge and mass, meaning that the position where charge is zero is not the same as the average mass position.",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            171
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 5:18, why do you care about the lone paris? Bc then wouldnt Sf2 be non-olar as well?",
        "A": "By Sf2 do you mean SF2 (sulfur difluoride)? If so it s polar. The lone pairs around S mean that the molecule is NOT linear, it looks sort of like a V instead. This means that the polar bonds between S and each F do not cancel one another out and the molecule is polar.",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            318
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At around 7:20, I don't understand how you can replace V with 4 m^3.\n\nmeters^3 is not a unit for volume so it is confusing.\n",
        "A": "meters^3 IS a unit for volume. cubic meters. any length cubed is a measure of volume.",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            440
        ],
        "3min_transcript": "So let's see, if we do 2100 grams, remember I want to do everything in grams, so I just want to do a unit conversion there. 2100 grams times 2.86% is equal to about 60 grams. So hydrogen, this 2% of that 2100 grams is 60 grams. And then what's the molar mass of one hydrogen. That's H2. So we know that the hydrogen atom by itself has a mass of 1, doesn't have a neutron in most cases. So the atomic mass of this is 2. Or the molar mass of this is 2 grams. So one mole of H2 is equal to two grams. We have 60 grams. So we clearly have 60 divided by 2, we have 30 moles. was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin."
    },
    {
        "Q": "\nat 9:50 sal said 101,325Pa=1atm , I thought it was 101.3Pa=1atm, thats what i remember learning in chemistry, or am I wrong.",
        "A": "1 atm = 101.3 kPa, that is kiloPascals, or 101 325 Pa",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            590
        ],
        "3min_transcript": "these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273. Which is nice, because that's a unit of pressure. So, let's do the math, 25 times 8.3145 times 273 is equal to 56,746 pascals. And that might seem like a crazy number. But the pascal is actually a very small amount of pressure. It actually turns out that 101,325 pascals is equal to one atmosphere. So if we want to figure out how many atmospheres this is, we could just divide that. Let me look it up on this table. Yes, 101,325. That's 56.746 kilopascals. Or if we wanted it in atmospheres we just take 56,746 divided by 101,325. It equals 0.56 atmospheres. So that's the total pressure being exerted from all of the gases. I deleted that picture. So this is the total pressure. So our question is, what's the partial pressure? We could use either of these numbers, they're just in different units. What's the partial pressure of just the oxygen by itself? Well, you look at the moles, because we don't care about the actual mass. Because we're assuming that they're ideal gases. We want to look at the number of particles. Because remember, we said pressure times volume is proportional to the number of particles times temperature. And they're all at the same temperature. So the number of particles is what matters. So oxygen represents 20% of the particles."
    },
    {
        "Q": "At 8:04 Sal mentions multiple values or R in PV=nRT, but my science book only has one. How do you know which to use?\n",
        "A": "You use the version of R that contains the same units that you have for volume and pressure.",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            484
        ],
        "3min_transcript": "was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273."
    },
    {
        "Q": "\nAt 4:33 he says no net work due to no change in kinetic energy. Is there not a change in potential energy that would account for work being done?",
        "A": "When PE increases, no NET work is done. For example, when you lift a 1 kg book 1 meter in the air, you did 1 Joule of work on it, but gravity did -1 Joule of work, so the net work is zero, and thats why the book is not moving. This is called the work-KE theorem.",
        "video_name": "udgMh3Y-dTk",
        "timestamps": [
            273
        ],
        "3min_transcript": "But the angle between the gravitational force and the direction of the displacement is 90 degrees in this case. And since cosine of 90 is 0, the gravitational force does no work on this trashcan. Similarly, if we were to find the work done by the normal force, the angle between the direction of the displacement and the normal force is 90 degrees. So the normal force also does no work on the trashcan. This makes sense because forces that are perpendicular to the motion can never do any work on that object. So that's how you can find the work done by individual forces. And if we wanted to know the net work done on this trashcan, we could just add up the work done by each individual force. So the net work is going to be 200 joules. Now that we know the net work done on the trashcan, we can use the work-energy principle to figure out the speed of the trashcan after it's slid the 10 meters. The work-energy principle says that the net work done in kinetic energy of that object. So 200 joules is going to equal the difference in kinetic energy. If we assume the trashcan started at rest, which seems reasonable, the initial velocity is 0. So we can solve for the final speed of the trashcan, which comes out to be 10 meters per second. This time, let's say you take the trashcan and lift it upwards with a constant velocity for a distance of 2 meters. In order to lift the trashcan up with a constant velocity, you need to push with a force equal to the weight of the trashcan, which means you have to push upwards with a force of 39.2 newtons. So to find the work done by the force that you exert, the force is going to be 39.2 newtons. The displacement is going to be 2 meters. And the angle between the force and the displacement is going to be 0 degrees because the direction of the force that you exert is in the same direction as the displacement of the trashcan. So the work that you've done in lifting up this 4 kilogram To find the work done by the force of gravity, we can use the force of gravity, which is again 39.2 newtons. The displacement is again 2 meters. But the angle between the direction of the displacement and the gravitational force is 180 degrees because the displacement points up and the gravitational force points down. So the work done by the gravitational force is negative 78.4 joules, which means the net work done on the trashcan is 0. And that makes sense. Because since the trashcan moved upwards with constant velocity, there was no change in the kinetic energy of this object."
    },
    {
        "Q": "\nAt 7:32 to 7:45 If an object gets to the center of some massive object will it be moving back and forth because gravity is pulling and pushing it or will it stay at center?",
        "A": "It will stay at the center as the gravity is 0 at that point.",
        "video_name": "RpOHZc6cDIw",
        "timestamps": [
            452,
            465
        ],
        "3min_transcript": "If this distance right here is r/1,000 wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? The entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here. Let me think of the best way that's doing it. So you can think of it all of this mass over here is pulling it in an outward direction. It's not telling. What that mass out there is doing, since that mass itself is being pulled inward, It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you could imagine if you were in the center of a really massive object-- so that's a really massive object. If you were in the center, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward. So at every point it will be pulling you outward. And so that's why if you were to enter the core of a star, if you were to get a lot closer to its center of mass, it's not going to be pulling on you with this type of force. And the only way you can get these types of forces is if the entire mass is contained in a very dense region, in a very small region. such strong gravity that not even light can escape. Hopefully that clarifies things a little bit."
    },
    {
        "Q": "\nAt around 3:20, Sal is talking about how, the faster the projectile is thrown, the further it goes before falling back down. How is this so? I thought that objects accelerated downward at the same rate regardless of initial horizontal velocity. Does this only apply to objects projected perfectly horizontally?",
        "A": "The downward acceleration is the same, so the fall time is the same, but the horizontal velocity is faster, so it goes further in the same amount time.",
        "video_name": "oIZV-ixRTcY",
        "timestamps": [
            200
        ],
        "3min_transcript": "this r isn't that different. It's a little bit further than if you were at the surface of the Earth. Remember that r is measured from wherever you are to the center, from the center of the Earth, or really the center of the object to the center of the Earth. The center of the Earth represents most of the distance here. So if I'm at the surface of the Earth or if I'm just a few hundred miles above the surface of the Earth, it's not going to change r that dramatically, especially in terms of percentage. So when you look at it this way, it seems pretty clear that the force of gravity for someone who is in space only a few hundred miles above the Earth should not be that different than the force of gravity for someone who is on the surface of the Earth. So my question to you is, what gives? If there should be gravity in space, how can we see all of these pictures of people floating around like this? And the answer is that there is gravity in space, They're just moving fast enough relative to the Earth that they keep missing it. And let me show you what I'm talking about there. Let's say I'm sitting here in Africa, and I were to shoot something, if maybe I have a really good sling shot, and I were to sling something super fast and maybe at a 45 degree angle, it might take off a little bit and eventually hit another point. And this would actually already be a super duper slingshot. I just made it travel a couple of thousand miles or at least over 1,000 miles. If I make it go a little bit faster, if I put a little bit more force on, if I just propelled the projectile a little bit faster, it might go a little bit further, but it will eventually fall back to the Earth. Let's try to propel it a little bit faster than that. Then it'll still fall to the Earth. Let's propel it even faster than that. Well then, it's still eventually going to fall to the Earth. I think you might see where this is going. Let's go even faster than that. eventually it'll fall to the Earth. Even faster than that, so if you were to throw an object even faster than that, it would then go really far and then fall to the Earth. I think you see what's happening. Every time you go faster and faster, you throw this projectile faster and faster, it gets further and further, up to some velocity that you release this projectile, and whenever it's trying to fall to the Earth, it's going so fast that it keeps missing the Earth. So it'll keep going around and around and around the Earth, and a projectile like that would essentially be in orbit. So what's happening is if there was no gravity for that projectile, if there was no gravity, the projectile would just go straight away into space. But because there's gravity, it's constantly pulling it towards the center of the Earth, or the center of that projectile and the center of the Earth are being pulled towards each other,"
    },
    {
        "Q": "At 1:56, Sal said, \"I'm making these numbers up on the fly, so bear with me.\" What did he mean? Did he mean that he's inventing these numbers temporarily and without preparing carefully before recording this video?\n",
        "A": "yes, that is what it means.",
        "video_name": "_k3aWF6_b4w",
        "timestamps": [
            116
        ],
        "3min_transcript": "Welcome to Level 4 multiplication. Let's do some problems. Let's see, we had 235 times-- I'm going to use two different colors here, so bear with me a second. Let's say times 47. So you start a Level 4 problem just like you would normally do a Level 3 problem. We'll take that 7, and we'll multiply it by 235. So 7 times 5 is 35. 7 times 3 is 21, plus the 3 we just carried is 24. 7 times 2 is 14, plus the 2 we just carried. This is 16. So we're done with the 7. Now we have to deal with this 4. Well, since that 4 is in the tens place, we add a 0 here. You could almost view it as we're multiplying 235, not by we put that 0 there. But once you put the 0 there, you can treat it just like a 4. So you say 4 times 5, well, that's 20. Let's ignore what we had from before. 4 times 3 is 12, plus the 2 we just carried, which is 14. 4 times 2 is 8, plus the 1 we just carried, so that's 9. And now we just add up everything. 5 plus 0 is 5, 4 plus 0 is 4, 6 plus 4 is 10, carry the 1, and 1 plus 9, well, that's 11. So the answer's 11,045. Let's do another problem. Let's say I had 873 times-- and I'm making these numbers up on the fly, so bear with me-- 873 times-- some high numbers-- hopefully get a better understanding of what I'm trying to explain. Let's say 97. No, I just used a 7. Let's make it 98. So just like we did before, we go to the ones place first, and that's where that 8 is, and we multiply that 8 times 873. So 8 times 3 is 24, carry the 2. 8 times 7 is 56, plus 2 is 58, carry the 5. 8 times 8 is 64, plus the 5 we just carried. That's 69. We're done with the 8. Now we have to multiply the 9, or we could just do it as we're multiplying 873 by 90. But multiplying something by 90 is just the same thing as multiplying something by 9 and then adding a 0 at the end, so that's why I put a 0 here."
    },
    {
        "Q": "\nAt 5:15, how did he get the excluded values?",
        "A": "he just found what values of x would make the denominator equal zero.",
        "video_name": "dstNU7It-Ro",
        "timestamps": [
            315
        ],
        "3min_transcript": "When I multiply them, I get 2 times 30, which is 60. And an a plus a b, when I add them, I get 17. Once again, 5 and 12 seem to work. So let's split this up. Let's split this up into 2x squared. We're going to split up the 17x into a 12x plus a 5x and that adds up to 17x. When you multiply 12 times 5, you get 60, and then plus 30. Then on this first group right here, we can factor out a 2x, so if you factor out a 2x, you get 2x times x plus 6. In that second group, we can factor out a 5, so you get Now, we can factor out an x plus 6, and we get we get x plus 6 times 2x plus 5. We've now factored the numerator and the denominator. Let's rewrite both of these expressions or write this entire rational expression with the numerator and the denominator factored. The numerator is going to be equal to x plus 4 times 2x plus 5. We figured that out right there. And then the denominator is x plus 6 times 2x plus 5. It might already jump out at you that you have 2x plus 5 in the numerator and the denominator, and we can cancel them out. We will cancel them out. But before we do that, let's work on the second part of this question. State the domain. A more interesting question is what are the x values that will make this rational expression undefined? It's the x values that will make the denominator equal to 0, and when will the denominator equal to 0? Well, either when x plus 6 is equal to 0, or when 2x plus 5 is equal to 0. We could just solve for x here. Subtract 6 from both sides, and you get x is equal to negative 6. If you subtract 5 from both sides, you get 2x is equal to negative 5. Divide both sides by 2. You get x is equal to negative 5/2. We could say the domain-- let me write this over here. The domain is all real numbers other than or except x is equal to negative 6 and x is equal to negative 5/2."
    },
    {
        "Q": "\nI think that I am on the wrong topic, but I really don't understand what Sal was doing at 1:16-1:28. Can someone please explain what he just did?",
        "A": "Adding on to what Jerry said, you would get the factors of 5 and -1. You could then plug it into the equation like this. x^2 + 5x - x - 5. Form here, you can solve it to this. x(x + 5) - 1(x + 5) Finally you would get (x + 5) and (x - 1) as your final answer",
        "video_name": "DRpdoZQtvOM",
        "timestamps": [
            76,
            88
        ],
        "3min_transcript": "We already have the tools in our tool kit to factor something like x squared plus 4x minus 5. And the way that we've already thought about it is we've said, hey, let's think of two numbers that if we were to take their product, we'd get negative 5. And if we were to add the two numbers, we'd get positive 4. And the fact that their product is negative tells you one of them is going to be positive and one of them is going to be negative. And so there's a couple of ways you could think about it. Well, you could say, well, maybe one of the numbers is negative 1 and then the other one is positive 5. Actually, this one seems to work. Negative 1 times 5 is negative 5. Negative 1 plus 5 is positive 4. So this one actually seems to work. The other option would have been-- since we're just going to deal with the factors of 5, and 5's a prime number, the other option would have been something like 1 and negative 5. There's only two factors for 5. So 1 and negative 5-- their product would have been negative 5. But if you were to add these two numbers, you would have gotten a negative 4 right over here. So we're going to go with this right over here. to factor this using the tools that we already know about, we will get-- And let me write these numbers in a different color so we can keep track of them. So negative 1 and 5. We know that this would factor out to be x minus 1 times x plus 5. And you can verify this for yourself that if you were to multiply this out, you will get x squared plus 4x minus 5. You can even see this here. x times x is x squared. Negative x plus 5x is going to be 4x. And then negative 1 times 5 is negative 5. This is all review for us at this point. Now I want to tackle something a little bit more interesting. Let's say we wanted to factor x squared plus 4xy minus 5y squared. All of a sudden, I've introduced a y and a y squared here. I have two variables. How would I tackle it? But the important thing is to just take a deep breath and realize that we're not fundamentally doing something different. Now, the one little tricky thing I've done when I've written it this way-- and I encourage you to pause this and try this on your own before I explain any further. But the one tricky thing I did right over here is I wrote the x before the y. And that tends to be the convention. You just write them kind of in alphabetical order. But if we wanted it in a form that's a little bit closer to this and something that would fit this mold a little bit more is if we swapped these two. Because then we could write it as x squared plus 4yx minus 5y squared. And now it becomes pretty clear that this 4y term right over here-- this right over here is the coefficient on the x term, the same way that 4 was the coefficient on x right here. And this negative 5y squared corresponds"
    },
    {
        "Q": "So we are supposed to guess at 2:16?\n",
        "A": "No, that side definitely goes there. Could wasn t the best choice of word.",
        "video_name": "m1ZTnl4CNQg",
        "timestamps": [
            136
        ],
        "3min_transcript": "Teddy knows that a figure has a surface area of 40 square centimeters. The net below has 5 centimeter and 2 centimeter edges. Could the net below represent the figure? So let's just make sure we understand what this here represents. So it tells us that it has 5 centimeter edges. So this is one of the 5 centimeter edges right over here. And we know that it has several other 5 centimeter edges because any edge that has this double hash mark right over here is also going to be 5 centimeters. So this edge is also 5 centimeters, this is also 5 centimeters, this is also 5 centimeters, and then these two over here are also 5 centimeters. So that's 5 centimeters, and that's 5 centimeters. And then we have several 2 centimeter edges. So this one has 2 centimeters. And any other edge that has the same number of hash marks, in this case one, is also going to be 2 centimeters. So all of these other edges, pretty much all the rest of the edges, are going to be 2 centimeters. Now, they don't ask us to do this in the problem, but it's always fun to start with a net like this It looks pretty clear this is going to be a rectangular prism. But let's actually draw it. So if we were to-- we're going to fold this in. We're going to fold this that way. You could view this as our base right over here. We're going to fold this in. We're going to fold that up. And then this is going to be our top. This is the top right over here. This polyhedron is going to look something like this. So you're going to have your base that has a length of 5 centimeters. So this is our base. Let me do that in a new color. So this is our base right over here. I'll do it in the same color. So that's our base, this dimension right over here. I could put the double hash marks if I want. 5 centimeters, and that's of course the same as that dimension up there. Now, when we fold up this side-- we'll do this in orange, could be this side right over here, along this 2 centimeter edge. So that's that side right over here. When you fold this side in right over here, that could be that. That's that side right over there. And then when of course we fold this side in-- that's the same color. Let me do a different color. When we fold this side in, that's the side that's kind of facing us a little bit. So that's that right over there. That's that right over there. Color that in a little bit better. And then we can fold this side in, and that would be that side. And then, of course, we have the top that's connected right over here. So the top would go-- this would be the top, and then the top would, of course, go on top of our rectangular prism. So that's the figure that we're talking about. It's 5 centimeters in this dimension. It is 2 centimeters tall, and it is 2 centimeters wide."
    },
    {
        "Q": "At 3:38, how does Sal get (1 - 4x^2)? I understand the -4x^2 part, but not how he gets the 1 preceding it when he factors out the e^-2x^2...\n",
        "A": "All he did is factor out the e^(-2x^2). e^(-2x^2) - 4x^2 * e^(-2x^2) = e^(-2x^2) * (1-4x^2) It is the same as factoring out the x term in this expression: x -2x^2 = x * (1-2x) When factoring it out, there is still the e^(-2x^2) * 1 term that you need to take into account.",
        "video_name": "MUQfl385Yug",
        "timestamps": [
            218
        ],
        "3min_transcript": "What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x, and this part is also going to be defined for any value of x. There's no point where this is undefined. Let's think about when this is going to be equal to zero. The product of these two expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say very negative number, you will approach zero but you will never get it to be zero. This part here can't be zero. If the product of two things are zero at least one of them has to be zero, so the only way we can get f prime of x to be equal to zero is when one minus four x squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get"
    },
    {
        "Q": "at 4:28 why the e^-2x^2 cant be zero ?\n",
        "A": "The exponential function (when considered a function of real numbers) is always positive. That is, for every real number u, we have exp(u) > 0. Now let u = -2x\u00c2\u00b2, which is a real number whenever x is a real number. It follows that exp(u) = exp(-2x\u00c2\u00b2) > 0. Hence never zero. The fact that exp(u) > 0 for every real number u requires a proof in itself, but I will not provide one here (unless you really want me to).",
        "video_name": "MUQfl385Yug",
        "timestamps": [
            268
        ],
        "3min_transcript": "What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x, and this part is also going to be defined for any value of x. There's no point where this is undefined. Let's think about when this is going to be equal to zero. The product of these two expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say very negative number, you will approach zero but you will never get it to be zero. This part here can't be zero. If the product of two things are zero at least one of them has to be zero, so the only way we can get f prime of x to be equal to zero is when one minus four x squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get"
    },
    {
        "Q": "\nAt 2:50, shouldn't d/dx e^-2x^2 = -2e^-2x^2*(-4x)?",
        "A": "f(x) = e^(-2x^2) f (x) = d/dx(e^(-2x^2)) Apply Chain Rule: f (x) = e^(-2x^2)*d/dx(-2x^2) Remove constant: f (x) = e^(-2x^2)*-2*d/dx(x^2) Apply Power Rule d/dx(x^n) = n*x^(n - 1): f (x) = e^(-2x^2)*-2*2*x^(2 - 1)*d/dx(x) f (x) = e^(-2x^2)*-2*2*x^(2 - 1)*dx/dx f (x) = e^(-2x^2)*-2*2*x^(2 - 1) f (x) = e^(-2x^2)*-2*2*x^(1) [ f (x) = -4x*e^(-2x^2) ]",
        "video_name": "MUQfl385Yug",
        "timestamps": [
            170
        ],
        "3min_transcript": "Let's think about how we can find the derivative of this. f prime of x is going to be, well let's see. We're going to have to apply some combination of the product rule and the chain rule. It's going to be the derivative with respect to x of x, so it's going to be that, times e to the negative two x squared plus the derivative with respect to x of e to the negative two x squared times x. This is just the product rule right over here. Derivative of the x times e to the negative of two x squared plus the derivative of e to the What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x,"
    },
    {
        "Q": "\nwhat is the associauive property again? Sal says it on 2:05 to 2:10",
        "A": "In the context of the video, Sal was talking about 3*a^1*a^7. The Associative Property refers to grouping numbers in a particular way to make solving the problem easier. It means that 3*(a^1*a^7) = a^7*(3*a^1) = a^1*(3*a^7). In the video at 2:10, Sal is choosing to make 3*a^1*a^7 into 3*(a^1*a^7) so he can simplify it into 3*(a^8) and into 3a^8.",
        "video_name": "-TpiL4J_yUA",
        "timestamps": [
            125,
            130
        ],
        "3min_transcript": "Simplify 3a times a to the fifth times a squared. So the exponent property we can use here is if we have the same base, in this case, it's a. If we have it raised to the x power, we're multiplying it by a to the y power, then this is just going to be equal to a to the x plus y power. And we'll think about why that works in a second. So let's just apply it here. Let's start with the a to the fifth times a squared. So if we just apply this property over here, this will result in a to the fifth plus two-th power. So that's what those guys reduce to, or simplify to. And of course, we still have the 3a out front. Now what I want to do is take a little bit of an aside and realize why this worked. Let's think about what a to the fifth times a squared means. A to the fifth literally means a times a times a times a times a. Now, a squared literally means a times a. So we're multiplying these five a's times these two a's. And what have we just done? We're multiplying a times itself five times, and then another two times. We are multiplying a times itself. So let me make it clear. This over here is a to the fifth. This over here is a squared. When you multiply the two, you're multiplying a by itself itself seven times. 5 plus 2. So this is a to the seventh power. a to the 5 plus 2 power. So this simplifies to 3a times a to the seventh power. Now you might say, how do I apply the property over here? What is the exponent on the a? And remember, if I just have an a over here, this is equivalent to a to the first power. So I can rewrite 3a is 3 times a to the first power. A to the first power-- and the association property of multiplication, I can do the multiplication of the a's before I worry about the 3's. So I can multiply these two guys first. So a to the first times a to the seventh-- I just have to add the exponents because I have the same base and I'm taking the product-- that's going to be a to the eighth power. And I still have this 3 out front. So 3a times a to the fifth times a squared simplifies to 3a to the eighth power."
    },
    {
        "Q": "\nAt 2:53 why is it that at most 1 Sals out of 100 may die in order for the insurance company not to loose money? What's the explanation? Sal didn't give any here.",
        "A": "Because each Sal only pays 1% of the insurance payout over the life of the policy. So in this case 100 Sals each pay $10,000 for a total of $1,000,000. For each Sal that dies, the insurance company needs to pay that Sal s family $1,000,000. So if 2 Sals die the Insurance pay out $2,000,000 but have only collected $1,000,000 so they are losing money.",
        "video_name": "NSSoMafbBqQ",
        "timestamps": [
            173
        ],
        "3min_transcript": "have a higher chance of dying at that point, then it's probably going to be more expensive for me to get insurance. But I really am just worried about the next 20 years. But what I want to do in this video is think about given these numbers that have been quoted to me by the insurance company, what do they think that my odds of dying are over the next 20 years? So what I want to think about is the probability of Sal's death in 20 years, based on what the people at the insurance company are telling me. Or at least, what's the maximum probability of my death in order for them to make money? And the way to think about it, or one way to think about it, kind of a back-of-the-envelope way, is to think about what's the total premiums they're getting over the life of this policy divided by how much they're insuring me for. So they're getting $500 times 20 years is equal to, that's $10,000 over the life of this policy. So they're getting-- let's see those 0s cancel out, this 0 cancels out-- they're getting, over the life of the policy, $1 in premiums for every $100 in insurance. Or another way to think about it. Let's say that there were 100 Sals, 100 34-year-olds looking to get 20-year term life insurance. And they insured all of them. So if you multiplied this times 100, they would get $100 in premiums. This is the case where you have 100 Sals, or 100 people who are pretty similar to me. 100 Sals. And the only way that they could make money is if, at most, one of those Sals-- or really just break even-- if, at most, 1 of those Sals were to die. So break even if only 1 Sal dies. I don't like talking about this. It's a little bit morbid. So one way to think about it, they're getting $1 premium for $100 insurance. Or if they had 100 Sals, they would get $100 in premium, and the only way they would break even, if only 1 of those Sal dies. So what they're really saying is that the only way they can break even is if the probability of Sal dying in the next 20 years is less than or equal to 1 in 100. And this is an insurance company. They're trying to make money. So they're probably giving these numbers because they think the probability of me dying is a good-- maybe it's 1 in 200 or it's 1 in 300."
    },
    {
        "Q": "what does denote mean at 0:08?\n",
        "A": "Denote means: to show, mark, or be a sign of (something). In his case, he is denoting or marking that the two lines are parallel.",
        "video_name": "Ld7Vxb5XV6A",
        "timestamps": [
            8
        ],
        "3min_transcript": "So I've got two parallel lines. So that's the first line right over there and then the second line right over here. Let me denote that these are parallel. These are parallel lines. Actually, I can do that a little bit neater. And let me draw a transversal, so a line that intersects both of these parallel lines, so something like that. And now let's say that we are told that this angle right over here is 9x plus 88. And this is in degrees. And we're also told that this angle right over here is 6x plus 182, once again, in degrees. So my goal here and my question for you is, can we figure out what these angles actually are, given that these are parallel lines and this is a transversal line? And I encourage you to pause this video to try this on your own. are related by the fact that they're formed from a transversal intersecting parallel lines. And we know, for example, that this angle corresponds to this angle right over here. They're going to be congruent angles. And so this is 6x plus 182. This is also going to be 6x plus 182. And then that helps us realize that this blue angle and this orange angle are actually going to be supplementary. They're going to add up to 180 degrees, because put together, when you make them adjacent, their outer rays form a line right over here. So we know that 6x plus 182 plus 9x plus 88 is going to be equal to 180 degrees. And now we just have to simplify this thing. So 6x plus 9x is going to give us 15x. Let's see, 182 plus 8, would get us to 190. And then we add another 80. It gets us to 270-- plus 270-- is equal to 180. If we subtract 270 from both sides, we get 15x is equal to negative 90. And now we can divide both sides by 15. And we get x is equal to-- what is this? Let's see, 6 times 15 is 60 plus 30 is 90. So x is going to be equal to negative 6. So far, we've made a lot of progress. We figured out what x is equal to. x is equal to negative 6, but we still haven't figured out what these angles are equal to. So this angle right over here, 9x plus 88, this is going to be equal to 9 times negative 6 plus 88."
    },
    {
        "Q": "Why would Sal not just subtract 34 degrees from one-hundred eighty and get the angle for the other side at 3:58?\n",
        "A": "Good eye. When I come up with an alternate correct answer to a problem, as I m learning, I consider it a good sign -- you re thinking.",
        "video_name": "Ld7Vxb5XV6A",
        "timestamps": [
            238
        ],
        "3min_transcript": "Let's see, 182 plus 8, would get us to 190. And then we add another 80. It gets us to 270-- plus 270-- is equal to 180. If we subtract 270 from both sides, we get 15x is equal to negative 90. And now we can divide both sides by 15. And we get x is equal to-- what is this? Let's see, 6 times 15 is 60 plus 30 is 90. So x is going to be equal to negative 6. So far, we've made a lot of progress. We figured out what x is equal to. x is equal to negative 6, but we still haven't figured out what these angles are equal to. So this angle right over here, 9x plus 88, this is going to be equal to 9 times negative 6 plus 88. Let me write this down before I make a mistake. Negative 54 plus 88 is going to be-- let's see, to go from 88 minus 54 will give us 34 degrees. So this is equal to 34, and it's in degrees. So this orange angle right here is 34 degrees. The blue angle is going to be 180 minus that. But we can verify that by actually evaluating 6x plus 182. So this is going to be equal to 6 times negative 6 is negative 36 plus 182. So this is going to be equal to-- let's see, if I subtract the 6 first, I get to 176. So this gets us to 146 degrees. And you can verify-- 146 plus 34 is equal to 180 degrees. as well. We know that if this is 34 degrees, then this must be 34 degrees as well. Those are opposite angles. This angle also corresponds to this angle so it must also be 34 degrees, which is opposite to this angle, which is going to be 34 degrees. Similarly, if this one right over here is 146 degrees, we already know that this one is going to be 146. This one's going to be 146 since it's opposite. And that's going to be 146 degrees as well."
    },
    {
        "Q": "\nAt 5:13, couldn't you have just found the total area of all the triangles and the square and have it equal to c^2?\n\n(1/2 ab)*4= area of 4 triangles\n(b-a)^2=area of square\n\n2ab+(b-a)^2=c^2\n2ab+b^2-2ab+a^2=c^2\nb^2+a^2=c^2\nWhat's wrong with this why do we have to shift all the triangles and stuff",
        "A": "I think its essentially the same thing. While you ve penned down an equation and played about with the variables algebraically, Bhaskara has played about geometrically and gotten the same answer. Also (correct me if I m wrong), you wrote down two equations before you arrived at the theorem, and Bhaskara shifted two triangles before he arrived at the theorem! Pretty neat I think.",
        "video_name": "1ul8g55dYA4",
        "timestamps": [
            313
        ],
        "3min_transcript": "the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle, so at minimum, they are similar, and their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length, b. So the longer side of these triangles I'm So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all-- this distance right over here, that these are of length, a. So if I were to say this height right over here, Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared. Now, what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's, and hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste. So this is our original diagram. And what I will now do-- and actually, let me clear that out. I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much-- well, the way I drew it, it's not that-- well, that might do the trick. I want to retain a little bit of the-- so let me copy, or let me actually cut it, and then let me paste it. So that triangle I'm going to stick right over there. And let me draw in the lines that I just erased. So just to be clear, we had a line over there, and we also had this right over here. And this was straight up and down, and these were straight side to side. Now, so I moved this part over down here. So I moved that over down there. And now I'm going to move this top right triangle down"
    },
    {
        "Q": "\nI REALLY don't understand why we rewrite (7x - 5) for a second time on the right hand side (2:59) sec in to the video?? why?",
        "A": "What sal is doing is the distribution property. He is distributing the value (7x-5) to g(x) s components. Recall: (a)(b+c) = ab+ac",
        "video_name": "JKvmAexeMgY",
        "timestamps": [
            179
        ],
        "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions."
    },
    {
        "Q": "At 0:25 He say (f*g)(x). then says f of x and g of x is the same thing, why is it like that? Also why doesn't he use y?\n",
        "A": "Because you are doing what he also says to do in the video, distributing. You are distributing the x in (f*g)(x) to f and g, thus getting f(x)+g(x)",
        "video_name": "JKvmAexeMgY",
        "timestamps": [
            25
        ],
        "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions."
    },
    {
        "Q": "At 2:28 why is it that 7x * x^3 equals to x^4 ?\n",
        "A": "Diego, 7x^3 means 7*x*x*x 7x^3 * x = 7*x*x*x * x 7x^4 means 7*x*x*x*x That is why 7x^3 * x = 7x^4 I hope that helps",
        "video_name": "JKvmAexeMgY",
        "timestamps": [
            148
        ],
        "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions."
    },
    {
        "Q": "at 1:17 how do you determine that x =2?\n",
        "A": "Because u need it to equal zero so u must replace x with what is necessary witch was 2 in that case.",
        "video_name": "7QMoNY6FzvM",
        "timestamps": [
            77
        ],
        "3min_transcript": "We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5,"
    },
    {
        "Q": "\n4:52 He labeled the postulate as AAS but once he made it clear BE was congruent to EC doesn't it become SAS after that is marked?",
        "A": "He used AAS to prove that the two triangles are congruent, which by definition means that all corresponding sides/angles are congruent, including \u00f0\u009d\u0090\u00b5\u00f0\u009d\u0090\u00b8 \u00e2\u0089\u008c \u00f0\u009d\u0090\u00b6\u00f0\u009d\u0090\u00b8 (which means that \u00f0\u009d\u0090\u00b5\u00f0\u009d\u0090\u00b8 and \u00f0\u009d\u0090\u00b6\u00f0\u009d\u0090\u00b8 are of equal length, which is what we wanted to prove).",
        "video_name": "RFesGHsuFZw",
        "timestamps": [
            292
        ],
        "3min_transcript": "I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with hash. This line segment right over here is congruent to this line segment right over here, because we know that those two triangles are congruent. And I've inadvertently, right here, done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason. And we're done."
    },
    {
        "Q": "\nat 4:21 Did you mean angle side angle?",
        "A": "NO because Angle Side Angle mean the side is in between the 2 angles. Since the side is not between them so it SSA. Hope this help.",
        "video_name": "RFesGHsuFZw",
        "timestamps": [
            261
        ],
        "3min_transcript": "So this line right over here, this is a transversal. And there's actually several ways that we can do this problem. But we know that this is a transversal. And there's a couple of ways to think about it right over here. So let me just continue the transversal, so we get to see all of the different angles. You could say that this angle right here, angle ABE-- so this is its measure right over here-- you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out at you, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles. And then this one is vertical. But either way, angle ABE-- let me be careful. Angle ABE is going to be congruent to angle DCE. I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE."
    },
    {
        "Q": "\nI'm a little confused on how he got 3 at 5:12... how did he get that?",
        "A": "Statement #3 - triangle-BEA is congruent to triangle-CED, comes from angle-angle-side. The first angle is from statement #2 (ABE = DCE via alternate interior angles), the second angle is from statement #1 (AEB = DEC via vertical angles), and the side is given in the original drawing (AE=DE).",
        "video_name": "RFesGHsuFZw",
        "timestamps": [
            312
        ],
        "3min_transcript": "I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with hash. This line segment right over here is congruent to this line segment right over here, because we know that those two triangles are congruent. And I've inadvertently, right here, done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason. And we're done."
    },
    {
        "Q": "\nAt 6:50, so for every eigenvector there is also a corresponding eigenvalue?",
        "A": "Actually, that s a good question. Some matrices transform vectors so that some of the vectors don t rotate (they re eigenvectors ). If a transformed vector x isn t rotated, what is x , the transformed vector? x is a scalar (called lambda - the eigenvalue ) times the original vector x - that is, Ax = x = lambda*x. So: Does every eigenvector have an eigenvalue?",
        "video_name": "PhfbEr2btGQ",
        "timestamps": [
            410
        ],
        "3min_transcript": "So in the example I just gave where the transformation is flipping around this line, v1, the vector 1, 2 is an eigenvector of our transformation. So 1, 2 is an eigenvector. And it's corresponding eigenvalue is 1. This guy is also an eigenvector-- the vector 2, minus 1. He's also an eigenvector. A very fancy word, but all it means is a vector that's just scaled up by a transformation. It doesn't get changed in any more meaningful way than just the scaling factor. And it's corresponding eigenvalue is minus 1. transformation matrix is. I forgot what it was. We actually figured it out a while ago. If this transformation matrix can be represented as a matrix vector product-- and it should be; it's a linear transformation-- then any v that satisfies the transformation of-- I'll say transformation of v is equal to lambda v, which also would be-- you know, the transformation of [? v ?] would just be A times v. These are also called eigenvectors of A, because A is just really the matrix representation of the transformation. So in this case, this would be an eigenvector of A, and this would be the eigenvalue associated with the eigenvector. So if you give me a matrix that represents some linear transformation. Now the next video we're actually going to figure out a way to figure these things out. But what I want you to appreciate in this video is that it's easy to say, oh, the vectors that don't get changed much. But I want you to understand what that means. It literally just gets scaled up or maybe they get reversed. Their direction or the lines they span fundamentally don't change. And the reason why they're interesting for us is, well, one of the reasons why they're interesting for us is that they make for interesting basis vectors-- basis vectors whose transformation matrices are maybe computationally more simpler, or ones that make for better coordinate systems."
    },
    {
        "Q": "At 4:11, Sal assumes he can draw a line through B, C, and D, making BCD a transversal of the two parallel lines, an assumption which allows him to solve the whole problem. I understand how that all works out. But unless I'm mistaken, it's never stated that points B, C, and D all lie on the same line. Sure, they look like they do, but without that being explicitly stated, isn't it possibly an erroneous assumption?\n",
        "A": "Let s say that BC and CD are subtly off kilter. The technique will still work because you can extend line segment BC along (infinite) line BC, and there will be a new point (call it Q) to use instead of D. From there, the technique will be the same. The important element is that AB and CE are parallel, which means that BC (not necessarily BC and CD) is the transversal.",
        "video_name": "0gzSreH8nUI",
        "timestamps": [
            251
        ],
        "3min_transcript": "Bring down the 0. 5 goes into 20 four times, and then you're not going to have a remainder. 4 times 5 is 20. No remainder. So it's 34 times. So x is equal to 34. So the second largest angle has a measure of 34 degrees. This angle up here is going to be 4 times that. So 4 times 34-- let's see, that's going to be 120 degrees plus 16 degrees. This is going to be 136 degrees. Is that right? 4 times 4 is 16, 4 times 3 is 120, 16 plus 120 is 136 degrees. So we're done. The three measures, or the sizes of the three angles, are 10 degrees, 34 degrees, and 136 degrees. Let's do another one. So let's see. We have a little bit of a drawing here. And what I want to do is-- and we could think about different things. We could say, let's solve for x. I'm assuming that 4x is the measure of this angle. 2x is the measure of that angle right over there. And then if we know x, we can figure out what the actual measures of these angles are, assuming that we can figure out x. And the other thing that they tell us is that this line over here is parallel to this line over here. And it was very craftily drawn. Because it's parallel, but one stops here, and then one starts up there. So the first thing I want to do-- if they're telling us that these two lines are parallel, there's probably going to be something involving transversals or something. It might be something involving-- the other option is something involving triangles. And at first, you might say, wait, is this angle and that angle vertical angles? But you have to be very careful. This is not the same line. This line is parallel to that line. This line, it's bending right over there, so we can't make any type of assumption like that. So the interesting thing-- and I'm not sure if this will lead in the right direction-- is to just make it clear that these two are part of parallel lines. So I could continue this line down like this. And then I can continue this line up like that. And then that starts to look a little bit more like we're used to when we're dealing with parallel lines. could even say line BC, if we were to continue it on. If we were to continue it on and on, even pass D, then this is clearly a transversal of those two parallel lines. This is clearly a transversal. And so if this angle right over here is 4x, it has a corresponding angle. Half of the-- or maybe most of the work on all of these is to try to see the parallel lines and see the transversal and see the things that might be useful for you. So that right there is the transversal. These are the parallel lines. That's one parallel line. That is the other parallel line. You can almost try to zone out all of the other stuff in the diagram. And so if this angle right over here is 4x, it has a corresponding angle where the transversal intersects the other parallel line. This right here is its corresponding angle. So let me draw it in that same yellow. This right over here is a corresponding angle. So this will also be 4x."
    },
    {
        "Q": "\nAt 5:04, instead of finding the corresponding angle for 4x, couldn't we find the corresponding angle for 2x and solve? I know you get the same answer, but I'm just clarifying if you can't.",
        "A": "there is more than one way to solve most problems. You can do it both ways in this problem.",
        "video_name": "0gzSreH8nUI",
        "timestamps": [
            304
        ],
        "3min_transcript": "And then if we know x, we can figure out what the actual measures of these angles are, assuming that we can figure out x. And the other thing that they tell us is that this line over here is parallel to this line over here. And it was very craftily drawn. Because it's parallel, but one stops here, and then one starts up there. So the first thing I want to do-- if they're telling us that these two lines are parallel, there's probably going to be something involving transversals or something. It might be something involving-- the other option is something involving triangles. And at first, you might say, wait, is this angle and that angle vertical angles? But you have to be very careful. This is not the same line. This line is parallel to that line. This line, it's bending right over there, so we can't make any type of assumption like that. So the interesting thing-- and I'm not sure if this will lead in the right direction-- is to just make it clear that these two are part of parallel lines. So I could continue this line down like this. And then I can continue this line up like that. And then that starts to look a little bit more like we're used to when we're dealing with parallel lines. could even say line BC, if we were to continue it on. If we were to continue it on and on, even pass D, then this is clearly a transversal of those two parallel lines. This is clearly a transversal. And so if this angle right over here is 4x, it has a corresponding angle. Half of the-- or maybe most of the work on all of these is to try to see the parallel lines and see the transversal and see the things that might be useful for you. So that right there is the transversal. These are the parallel lines. That's one parallel line. That is the other parallel line. You can almost try to zone out all of the other stuff in the diagram. And so if this angle right over here is 4x, it has a corresponding angle where the transversal intersects the other parallel line. This right here is its corresponding angle. So let me draw it in that same yellow. This right over here is a corresponding angle. So this will also be 4x. this angle that has measure 4x and this angle that measures 2x-- we see that they're supplementary. They're adjacent to each other. Their outer sides form a straight angle. So they're supplementary, which means that their measures add up to 180 degrees. They kind of form-- they go all the way around like that if you add the two adjacent angles together. So we know that 4x plus 2x needs to be equal to 180 degrees, or we get 6x is equal to 180 degrees. Divide both sides by 6. You get x is equal to 30, or x is equal to-- well, I shouldn't say-- well, x could be 30. And then this angle right over here is 2 times x. So it's going to be 60 degrees. So this angle right over here is going to be 60 degrees. And this angle right over here is 4 times x. So it is 120 degrees, and we're done."
    },
    {
        "Q": "Also, can you get a definite answer for part 2:30 where the area is going to be x?\n",
        "A": "Yes and No, it depends on what you mean. No if you do not know anything like in thew video. What Sal is showing is like a format. (you can fill out the sides and the inside) Yes if you know the side L*W. What he is trying to show you is that for a square each side, if it is squared will get you the area. Hope this helped!",
        "video_name": "87_qIofPwhg",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube,"
    },
    {
        "Q": "At 1:14, he mentions imaginary and complex numbers.\n\nWhat does he mean by \"imaginary\" numbers?\n",
        "A": "Imaginary numbers are complex numbers in the form a + bi, in which a = 0. So, you d simply have i and its coefficient.",
        "video_name": "87_qIofPwhg",
        "timestamps": [
            74
        ],
        "3min_transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube,"
    },
    {
        "Q": "at 5:08 where does the tan function come from?\n",
        "A": "tan theta = opp/adj opp = sin theta adj = cos theta A simple proof: I dont know how to do theta symbol, so just pretend the no. 0 is theta show that: tan0 = sin0/cos0 RHS = opp/hyp divided by adj/hyp = opp/hyp times hyp/adj (when dividing fractions, invert and multiply) = opp/adj = tan0 = LHS",
        "video_name": "8RasCV_Lggg",
        "timestamps": [
            308
        ],
        "3min_transcript": "Well, this is forming an angle of theta with a positive real axis and so the horizontal coordinate over here by definition is going to be cosine of theta. Cosine of theta. That's the unit circle definition of cosine of theta, and the vertical coordinate is going to be sine of theta. Sine of theta. And so what would the horizontal and vertical coordinates of this point be? We obviously know they're negative three and two, but what would they be in terms of cosine theta and sine theta? Well look, this point right over here is a radius of one away from the origin. So this distance right over here is one, but now we are r away from the origin. We're r times as far. So if we're r times as far in that direction, then we're going to be r times as far in the vertical direction and r times We're going to scale everything by r. So the horizontal coordinate of this point right here instead of being cosine of theta is going to be r times cosine of theta. So this point right over here, which we know is negative three, is going to be equal to r cosine of theta, and by the same logic, this point over here, the vertical coordinate, we're going to scale up sine theta by r, we're r times as far. So this point right over here is going to be r sine theta and we already know that that's equal to two. r sine theta. So given that, can we now figure out what r and theta are? So let's first think about figuring out what theta is. So to do that, let's think about some of our trig functions. So one trig function that involves sine theta and cosine theta is tangent theta. tangent of our angle, tangent of theta, is equal to sine of theta over cosine of theta. We could also multiply the numerator and denominator here by r. That won't change the value. So that's the same thing as r sine theta over r cosine theta and we know r sine theta is going to be equal to two and we know that r cosine theta is negative three. So this whole thing is going to be negative 2/3. Another way of thinking about it is the tangent of theta is going to be the same thing as the slope of this line right over here. And what is the slope of that line? Well if you start at z and you want to go to the origin you're going to go positive three in the x direction"
    },
    {
        "Q": "sal why are you dividing when you could be subtracting at 00:55\n",
        "A": "Because you wouldn t get the correct answer. Whenever dealing with decimals and percents, in situations like Sal had in 00:55, divide.",
        "video_name": "gKywkLHV6Ko",
        "timestamps": [
            55
        ],
        "3min_transcript": "A zoo has 15 emperor penguins who make up 30% of the total number of penguins at the zoo. How many penguins live at the zoo? So let's let x equal the total number of penguins who live at the zoo. So they're telling us that 30% of that are the emperor penguins, which are 15. So they're saying if we take 30%, so 30% of x is equal to 15. Or another way of saying this is we could say, Instead of writing 30%, we could write that as a decimal, as 0.30 of x. So it's 0.30 times x is equal to 15. And now to solve this, we just have to divide both sides by 0.30. So let's do that. 0.30, 0.30, we get x is equal to 15 divided by 0.30. this one. So if we take 0-- it's 15 divided by 0.30. So just as a refresher of dividing decimals, we could essentially multiply both of these by 100. And when you multiply both of them by 100, you can't just multiply one of them. Then you would get a different answer. But if you multiply both of them by 100, you move this decimal point over two spots, so it becomes a 30. And you move this decimal over two spots, so it becomes 1,500. So it really boils down to how many times-- the same number of times that 30 goes into 1,500 is the number of times that 0.30 will go into 15. So let's think about this. Let me rewrite it here, just so it's neat. 30 goes into 1,500, all we did is we moved both decimals over to the right twice. Or you could say we just multiplied here by 100, which wouldn't change the value of the fraction. So 30 goes into-- let's see, it doesn't go into 1. It does go into 150. It goes into 150 five times. 5 times 30 is 150. Subtract, you get 0. And then 30 goes into 0 zero times. So this 30 goes into 1,500 fifty times. So this right over here is equal to 50. And you can verify it. Multiply 0.3 times 50, and you will get 15. Now, there is another way of doing it. You could say the fraction of emperor penguins over the total number of penguins at the zoo. So you could say 15. You could say 15 over the total number of penguins at the zoo is equal to 30%. Percent literally means 30 hundreds. So it's equal to 30 for every 100."
    },
    {
        "Q": "Hi Sal, at 4:44 it says a mean of 7.5 is impossible for a sample of size n=2. But, wouldn't a mean of 7.5 be achieved by a random sample with values 6 and 9, which was a mean of (6+9)/2 = 7.5? Thanks, Chris Broski\n",
        "A": "Hi I just saw the answer to my questions below. I understand now, yes, it is possible to get 7.5 as the mean for a sample of size n =2. Thanks, Chris Broski",
        "video_name": "NYd6wzYkQIM",
        "timestamps": [
            284
        ],
        "3min_transcript": "When n is pretty small, it doesn't approach a normal distribution that well. So when n is small-- let's take the extreme case. What happens when n is equal to 1? And that literally just means I take one instance of this random variable and average it. Well, it's just going to be that thing. So if I just take a bunch of trials from this thing and plot it over time, what's it look like? Well, it's definitely not going to look like a normal distribution. You're going to have a couple of 1's. You're going to have a couple of 2's. You're going to have more 3's like that. You're going to have no 4's. You're going to have a bunch of 5's. You're going to have some 6's that look like that. And you're going to have a bunch of 9's. So there, your sampling distribution of the sample mean for an n of 1 is going to look-- I don't care how many trials you do, it's not going to look like a normal distribution. So the central limit theorem, although I said you do a bunch of trials, it'll look like a normal distribution, definitely doesn't work for n equals 1. As n gets larger, though, it starts to make sense. Let's see, if we've got n equals 2-- and I'm all just doing this in my head. I don't know what the actual distributions would look like. for it to become an exact normal distribution. But then you could get more instances, you could get more-- you might get things from all of the above. But in each of your baskets that you're averaging, you're only going to get two numbers. For example, you're never going to get a 7 and 1/2 in your sampling distribution of the sample mean for n is equal to 2, because it's impossible to get a 7, and it's impossible to get an 8. So you're never going to get 7 and 1/2 as-- so maybe when you plot it, maybe it looks like this. But there'll be a gap at 7 and 1/2 because that's impossible. And maybe it looks something like that. So it still won't be a normal distribution when n is equal to 2. So there's a couple of interesting things here. So one thing-- and I didn't mention this the first time, just because I really wanted you to get the gut sense of what the central limit theorem is. The central limit theorem says as n approaches, really as it approaches infinity, then is when you get the real normal distribution. you don't have to get that much beyond n equals 2. If you get to n equals 10 or n equals 15, you're getting very close to a normal distribution. So this converges to a normal distribution very quickly. Now, the other thing is you obviously want many, many trials. So this is your sample size. That is your sample size. That's the size of each of your baskets. In the very first video I did on this, I took a sample size of 4. In the simulation I did in the last video, we did sample sizes of 4 and 10 and whatever else. This is a sample size of 1. So that's our sample size. So as that approaches infinity, your actual sampling distribution of the sample mean will approach a normal distribution. Now, in order to actually see that normal distribution, and actually to prove it to yourself, you would have to do this many, many-- remember the normal distribution happens-- this is kind of the population, or this is the random variable. That tells you all of the possibilities. In real life, we seldom know all of the possibilities."
    },
    {
        "Q": "\nat around 4:25, Sal says we are never going to get 7.5 when n=2. But if 6 and 9 are randomly selected, then 7.5 would be the average, so I'm not quite understanding his reasoning here?",
        "A": "You re totally correct, Mike G! Your logic is solid; Sal made a mistake.",
        "video_name": "NYd6wzYkQIM",
        "timestamps": [
            265
        ],
        "3min_transcript": "So these boxes are really small. So we just do a bunch of these trials. At some point, it might look a lot like a normal distribution. Obviously, there are some average values. It won't be a perfect normal distribution, because you can never get anything less than 0, or anything less than 1, really, as an average. You can't get 0 as an average. And you can't get anything more than 9. So it's not going to have infinitely long tails but, at least for the middle part of it, a normal distribution might be good approximation. In this video, what I want to think about is what happens as we change n. So in this case, n was 4. n is our sample size. Every time we do a trial, we took four and we took their average, and we plotted it. We could have had n equal 10. We could've taken 10 samples from this population, you could say, or from this random variable, averaged them, and then plotted them here. And in the last video, we ran the simulation. I'm going to go back to that simulation in a second. We saw a couple of things. And I'll show it to you in a little bit more depth When n is pretty small, it doesn't approach a normal distribution that well. So when n is small-- let's take the extreme case. What happens when n is equal to 1? And that literally just means I take one instance of this random variable and average it. Well, it's just going to be that thing. So if I just take a bunch of trials from this thing and plot it over time, what's it look like? Well, it's definitely not going to look like a normal distribution. You're going to have a couple of 1's. You're going to have a couple of 2's. You're going to have more 3's like that. You're going to have no 4's. You're going to have a bunch of 5's. You're going to have some 6's that look like that. And you're going to have a bunch of 9's. So there, your sampling distribution of the sample mean for an n of 1 is going to look-- I don't care how many trials you do, it's not going to look like a normal distribution. So the central limit theorem, although I said you do a bunch of trials, it'll look like a normal distribution, definitely doesn't work for n equals 1. As n gets larger, though, it starts to make sense. Let's see, if we've got n equals 2-- and I'm all just doing this in my head. I don't know what the actual distributions would look like. for it to become an exact normal distribution. But then you could get more instances, you could get more-- you might get things from all of the above. But in each of your baskets that you're averaging, you're only going to get two numbers. For example, you're never going to get a 7 and 1/2 in your sampling distribution of the sample mean for n is equal to 2, because it's impossible to get a 7, and it's impossible to get an 8. So you're never going to get 7 and 1/2 as-- so maybe when you plot it, maybe it looks like this. But there'll be a gap at 7 and 1/2 because that's impossible. And maybe it looks something like that. So it still won't be a normal distribution when n is equal to 2. So there's a couple of interesting things here. So one thing-- and I didn't mention this the first time, just because I really wanted you to get the gut sense of what the central limit theorem is. The central limit theorem says as n approaches, really as it approaches infinity, then is when you get the real normal distribution."
    },
    {
        "Q": "\nAt 1:46, shouldn't Sal have noticed he wrote \"Armaan\" when it should be Arman?",
        "A": "This is a known problem with the video. A box pops up with the correction.",
        "video_name": "W-5liMGKgHA",
        "timestamps": [
            106
        ],
        "3min_transcript": "Let's say that Arman today is 18 years old. And let's say that Diya today is 2 years old. And what I am curious about in this video is how many years will it take-- and let me write this down-- how many years will it take for Arman to be three times as old as Diya? So that's the question right there, and I encourage you to try to take a shot at this yourself. So let's think about this a little bit. We're asking how many years will it take. That's what we don't know. That's what we're curious about. How many years will it take for Arman So let's set some variable-- let's say, y for years. Let's say y is equal to years it will take. So given that, can we now set up an equation, given this information, to figure out how many years it will take for Arman to be three times as old as Diya? Well, let's think about how old Arman will be in y years. How old will he be? Let me write here. In y years, Arman is going to be how old? Arman is going to be-- well, he's 18 right now-- and in y years, he's going to be y years older. So in y years, Arman is going to be 18 plus y. And what about Diya? How old will she be in y years? she will just be 2 plus y. So what we're curious about, now that we know this, is how many years will it take for this quantity, for this expression, to be three times this quantity? So we're really curious. We want to solve for y such that 18 plus y is going to be equal to 3 times 2 plus y. Notice, this is Arman in y years. This is Diya in y years. And we're saying that what Arman's going to be in y years is three times what Diya is going to be in y years. So we've set up our equation. Now we can just solve it. So let's take this step by step. So the left hand side-- and maybe I'll do this in a new color, just so I don't have to keep switching-- so on the left hand side,"
    },
    {
        "Q": "The third problem, The \"Bizarre looking shape\" at 5:09, would it be possible to find the area of that dodecagon?\n",
        "A": "Yes. Since the dodecagon is made up of all right angles, as Sal said, finding the area would be easy. All you need to do is cut it up into rectangles, find the rectangles areas, and add them up.",
        "video_name": "vWXMDIazHjA",
        "timestamps": [
            309
        ],
        "3min_transcript": "at 90 degrees and we could call this point E. And what is interesting here is we can split this up into something we recognize a rectangle and a right triangle. But you might say how do, how do we figure out what these you know we have this side and that side, so we can figure out the area of this rectangle pretty straight forwardly. But how would we, how would we figure out the area of this triangle? Well if this side is 6 then that means that this that EC is also going to be 6. If AB is 6, notice we have a rectangle right over her, opposite side of a rectangle are equal. So if AB equals 6, implies that EC is equal to 6, EC is equal to 6, so EC is equal to 6 and if EC is equal to 6 then that tells us that DE is going to be 3. DE is going to be 3, this distance right over here is going to be 3. 9 was the length of this entire, of the entire base of this figure right over here. 9 was this entire distance so 9 minus 6 gives us 3, and now we have all the information that we need to figure out the area. The area of this part right over here of this rectangle is just going to be 6 times 7, so is going to be equal to 42 plus the area of this triangle right over here. Plus the area of this triangle right over here, and that is one half base times height one half. The base over her is 3, one half times 3 and the height over here is once again going to be 7 this is a rectangle, opposite sides are equal, so if this is 7, this is also going to be 7 one half times 3 times 7, so it is going be 42, lets see. 3 times 7 is 21, 21 divided by 2 is 10.5, 10.5 so this is going to be equal to 52.5 Lets do one more. So here I have a bizarre looking, a bizarre looking shape, and we need to figure out its perimeter. And it it first seems very daunting because they have only given us this side and this side and they have only given us this side right over here. And one thing that we are allowed to assume in this and you don't always have to make you can't always make that assumption and I just didn't draw it here I had time because it would had really crowded out this this diagram. Is it all of the angles in this diagrams are right angles,so i could have drawn a right angle here a right angle here, a right angle there, right angle there, but as you can see it kind of makes things a little bit, it makes things a little bit messy. But how do we figure out the perimeter if we don't know these little distances, if we don't know these little distances here. And the secret here is to kind of shift the sides because all we want to care about is the sum of the sides of the sides. So what I will do is a little exercise in shifting the sides. So this side over here I am going to shift"
    },
    {
        "Q": "\nI'm really confused about why the top equation was multiplied by -2 at 17:20. Surely it's not an arbitrary number, right?",
        "A": "Sal was setting up the elimination step. The next thing he does is add the two equations and the C_1 variable is eliminated allowing us to solve for C_2. Multiplying by -2 was the easiest way to get the C_1 term to cancel. Another question is why he chooses to use elimination. The first equation is already solved for C_1 so it would be very easy to use substitution. He may have chosen elimination because that is how we work with matrices.",
        "video_name": "Qm_OS-8COwU",
        "timestamps": [
            1040
        ],
        "3min_transcript": "It was 1, 2, and b was 0, 3. Let me remember that. So my vector a is 1, 2, and my vector b was 0, 3. Now my claim was that I can represent any point. Let's say I want to represent some arbitrary point x in R2, so its coordinates are x1 and x2. I need to be able to prove to you that I can get to any x1 and any x2 with some combination of these guys. So let's say that my combination, I say c1 times a plus c2 times b has to be equal to my vector x. Let me show you that I can always find a c1 or c2 given that you give me some x's. So let's just write this right here with the actual vectors So we have c1 times this vector plus c2 times the b vector 0, 3 should be able to be equal to my x vector, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2-- sorry. c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now, if I can show you that I can always find c1's and c2's any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns. This is just 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here. So we get minus 2, c1-- I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two. You get 3c2, right? These cancel out. You get 3-- let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3, you get c2 is equal to 1/3 x2 minus x1. Now we'd have to go substitute back in for c1."
    },
    {
        "Q": "\nAround 13:50 when Sal gives a generalized mathematical definition of \"span\" he defines \"i\" as having to be greater than one and less than \"n\". Is this because \"i\" is indicating the instances of the variable \"c\" or is there something in the definition I'm missing?",
        "A": "Its because we are looking at the Span of the Vectors v1 to vn, so for every vi there is a ci. Thats why i has to be between 1 and n.",
        "video_name": "Qm_OS-8COwU",
        "timestamps": [
            830
        ],
        "3min_transcript": "And in our notation, i, the unit vector i that you learned in physics class, would be the vector 1, 0. So this is i, that's the vector i, and then the vector j is the unit vector 0, 1. This is what you learned in physics class. Let me do it in a different color. This is j. j is that. And you learned that they're orthogonal, and we're going to talk a lot more about what orthogonality means, but in our traditional sense that we learned in high school, it means that they're 90 degrees. But you can clearly represent any angle, or any vector, in R2, by these two vectors. And the fact that they're orthogonal makes them extra nice, and that's why these form-- and I'm going to throw out a word here that I haven't defined yet. These form the basis. These form a basis for R2. In fact, you can represent anything in R2 by these two I'm not going to even define what basis is. But let me just write the formal math-y definition of span, just so you're satisfied. So if I were to write the span of a set of vectors, v1, v2, all the way to vn, that just means the set of all of the vectors, where I have c1 times v1 plus c2 times v2 all the way to cn-- let me scroll over-- all the way to cn vn. So this is a set of vectors because I can pick my ci's to be any member of the real numbers, and that's true for i-- so I should write for i to be anywhere between 1 and n. All I'm saying is that look, I can multiply each of these vectors by any value, any arbitrary value, real value, And now the set of all of the combinations, scaled-up combinations I can get, that's the span of these vectors. You can kind of view it as the space of all of the vectors that can be represented by a combination of these vectors right there. And so the word span, I think it does have an intuitive sense. I mean, if I say that, you know, in my first example, I showed you those two vectors span, or a and b spans R2. I wrote it right here. That tells me that any vector in R2 can be represented by a linear combination of a and b. And actually, just in case that visual kind of pseudo-proof doesn't do you justice, let me prove it to you algebraically. I'm telling you that I can take-- let's say I want to represent, you know, I have some-- let me rewrite my a's and b's again."
    },
    {
        "Q": "At 1:10 why does he ask that question?\n",
        "A": "he was adding the number thats it",
        "video_name": "Oe1PKI_6-38",
        "timestamps": [
            70
        ],
        "3min_transcript": "Add, and they give us 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So we can really view this as just adding two polynomials. And actually, the second polynomial here can be simplified right off the bat. We have two like terms here. It's not the 2x squared. There's no other 2x squared term in this polynomial. But you have a negative 7x term. And then you have a plus 13x. So we could actually add these two terms. What is the negative 7 of something plus 13 of that something? Or another way to view it, what's 13x minus 7x? 13 of something minus 7 of that something? Well, that's just going to be 6 of that something. So that is just going to be 6x. And then you have your 2x squared. And you have your 2x squared right over here. Let me write it. So 2x squared. So this polynomial right over here simplifies to 2x squared plus 6x. And then we want to add it to this polynomial right And what we can do is, what I like to do is just rewrite this polynomial under this polynomial. or the like terms, the terms that have the same variable raised to the same power. So we have x raised to the second power. So let's put the 5x squared over here. We have an x raised to the first power. We have an x raised to the first power. So let's put the 8x over here. And then we don't have a constant term in this yellow polynomial. We don't have anything raised to the zeroth power, any constant terms. We do here. So let's just put a minus 3. And then we could add these two things up. We have 2x squared plus 5x squared. That is 2 of something plus 5 of something is 7 of that something. So it's going to be 7x squared. And then to that, if I 6 of something and I add 8 of that something to it, well, I'm going to have 14 of that something. If I have 6x's and I give you 8 more x's, you're going to 14x's. And then we have nothing here. And so if we just add nothing to negative 3, you're just going to get a negative 3. we get 7x squared plus 14x minus 3."
    },
    {
        "Q": "at time 9:10 when he puts the boundary conditions, he replaces each of the y variables with -1. i do not get this point as boundary condition means the whole function should evaluate to -1 as x=0. some one please help me with that.\n",
        "A": "thanks a lot i got it now....!!!....",
        "video_name": "C5-lz0hcqsE",
        "timestamps": [
            550
        ],
        "3min_transcript": "can use this initial condition, when x is 0, y is 1, to figure out the constant. So let's first separate this equation. So let's multiply both sides by 2 times y minus 1. And you get 2 times y minus 1 times dy dx is equal to 3x squared plus 4x plus 2. Multiply both sides times dx. This is really just an exercise in algebra. And I can multiply this one out, too, you get 2y minus 2, that's just this, dy. I multiplied both sides times dx, so that equals 3x squared plus 4x plus 2 dx. I have separated the equations. I've separated the independent from the dependent variable, and their relative differentials, and so now I can integrate. What's the antiderivative of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus c, I'll just do it on the right hand side. That is equal to 3x squared. Well, the antiderivative is x to the third, plus 2x squared, plus 2x plus c. And that c kind of takes care of the constant for both sides of the equation, and hopefully you understand why from the last example. But we can solve for c using the initial condition y of 0 is equal to negative 1. So let's see. When x is 0, y is negative 1. So let's put y as negative 1, so we get negative 1 squared minus 2 times negative 1, that's the value of y, is equal to when x is equal to 0. So when x is equal to 0, that's 0 to the third plus 2 times 0 squared plus 2 times 0 plus c. All of these, this is all 0. This is, let's see, negative 1 squared, that's 1. Minus 2 times minus 1, that's plus 2, is equal to c. And we get c is equal to 3. So, the implicit exact solution, the solution of our differential equation-- remember now, it's not a class, because they gave us an initial condition-- is y squared minus 2y is equal to x to the third plus 2x squared plus 2x plus 3. We figured out that's what c was. And actually, if you want, you could write this in an explicit form by completing the square. This is just algebra this point. This is an implicit form. If you wanted to make it explicit, you could add 1 to I'm just completing the square here. So y squared minus 2y plus 1."
    },
    {
        "Q": "\n1:10 I don't understand Sal's explanation of the vertical tangent as having infinite dy. I know that a vertical line has an undefined slope because there's no change in x so dy is being divided by zero, but what does infinity have to do with that?\nAlso, how does a small change in x give an undefined slope when it's still a defined number that returns another real number when divided by a defined change in y?",
        "A": "But we don t have a defined change in y for a vertical line. The change could be anything. If we have a vertical line, then y is not a function of x. Would you be happier if we said dy was undefined for a vertical line, rather than infinite ? To be honest I m not sure if you think a vertical line is differentiable with respect to x or something else is troubling you.",
        "video_name": "pwh1dK3vTkM",
        "timestamps": [
            70
        ],
        "3min_transcript": "- [Voiceover] The graph of function f is given below. It has a vertical tangent at the point three comma zero. Three comma zero has a vertical tangent, let me draw that. It has a vertical tangent right over there, and a horizontal tangent at the point zero comma negative three. Zero comma negative three, so it has a horizontal tangent right over there, and also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent, just like that. Select all the x-values for which f is not differentiable. Select all that apply. F prime, f prime, I'll write it in short hand. We say no f prime under it's going to happen under three conditions. The first condition you could say well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well, remember, our derivative is with respect to x, but when you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. That's one situation where you have no derivative. They tell us where we have a vertical tangent in here, where x is equal to three. We have no ... F is not differentiable at x equals three because of the vertical tangent. You might say what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. F prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not gonna have a defined derivative is where the graph is not continuous. Not continuous. We see right over here at x equals negative three, our graph is not continuous. Those are thee only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These there I guess would be interesting cases. They haven't given us those choices here. We already said, at x equals 0, the derivative is zero. It's defined. At x equals six, the derivative is zero. We have a flat tangent. Once again it's defined there as well. Let's do another one of these. Actually, I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. This isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that, or like, well no, that doesn't look too sharp, or like this. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that."
    },
    {
        "Q": "\nat 0:53 it is said that a point having a vertical tangent is not differentiable because for a change in x there is an infinite change in y at that particular point . how is it so?",
        "A": "Sal is a tad off with his wording there. The change in y is finite but it doesn t get smaller as x gets smaller so you have a limit of dy/dx where dy decreases much slower than dx as dx --> 0 so the limit is unbounded.",
        "video_name": "pwh1dK3vTkM",
        "timestamps": [
            53
        ],
        "3min_transcript": "- [Voiceover] The graph of function f is given below. It has a vertical tangent at the point three comma zero. Three comma zero has a vertical tangent, let me draw that. It has a vertical tangent right over there, and a horizontal tangent at the point zero comma negative three. Zero comma negative three, so it has a horizontal tangent right over there, and also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent, just like that. Select all the x-values for which f is not differentiable. Select all that apply. F prime, f prime, I'll write it in short hand. We say no f prime under it's going to happen under three conditions. The first condition you could say well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well, remember, our derivative is with respect to x, but when you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. That's one situation where you have no derivative. They tell us where we have a vertical tangent in here, where x is equal to three. We have no ... F is not differentiable at x equals three because of the vertical tangent. You might say what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. F prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not gonna have a defined derivative is where the graph is not continuous. Not continuous. We see right over here at x equals negative three, our graph is not continuous. Those are thee only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These there I guess would be interesting cases. They haven't given us those choices here. We already said, at x equals 0, the derivative is zero. It's defined. At x equals six, the derivative is zero. We have a flat tangent. Once again it's defined there as well. Let's do another one of these. Actually, I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. This isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that, or like, well no, that doesn't look too sharp, or like this. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that."
    },
    {
        "Q": "At 2:52 how can we be sure that the three angles equal 180 degrees? It may appear to be a straight line, but without any notation ensuring that it is, how can we be sure?\n",
        "A": "Hi J, Right at the very beginning of the video, at 0:01, Sal uses the words in this larger triangle here as he outlines triangle ABE. If this was a straight word question however, the question would start out with something like, Given triangle ABE, prove that ... Hope that helps!",
        "video_name": "aDCXPdzyS0s",
        "timestamps": [
            172
        ],
        "3min_transcript": "corresponds to the E vertex in ECD. So all-- everything that I've done in magenta, all of these angles are congruent, and then we also know that the C angle. So in BCA-- sorry, BCD, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's keep looking at everything else that they're telling us. Finally, we have vertex D over here. So angle-- so this is the last one in where we listed-- so corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow. And so we have all these congruences, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent, and when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle, if you look at their outer sides. So you have, if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. Well, we have these two characters up here. They are both equal and they add up to 180 degrees. They are supplementary, the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees. Or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees, and then we're left with these magenta parts of the angle. And here, we could just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150. So this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing-- we've actually done what we said we would do, we found out all of the angles. We could also think about these outer angles. So this-- or not the outer angles, or these combined angles. So angle say AC-- or say, angle ABE, so this whole angle we see is 60 degrees."
    },
    {
        "Q": "At \"3:12\" I do not exactly understand how Sal pinpoints point C in the designated location? Please explain how he knew where to find all of the points of the hexagon given the special set of values for y, thank you.\n",
        "A": "because the hexagon had to be convex and with all sides equal.",
        "video_name": "Ec-BKdC8vOo",
        "timestamps": [
            192
        ],
        "3min_transcript": "So let me draw my x-axis. That is my x-axis. Right over there. And then my y-axis. My y-axis would look like that. Y-axis. We know that the vertex A sits at the point 0,0. That is the vertex A. Now, we know that all of the vertices have y-coordinates that are either 0, 2, 4, 6, 8, 10. And they are distinct members of the set. Which means no 2 of the vertices share the same y-coordinate. So they're not going to be on the same horizontal line. So let me draw these horizontal lines-- the x-axis is 0. Then you have y is equal to 2. Then you have 4, then you have 6, and then you have 8, and then you have 10 up here. Now, B we already know. So first of all, we've already used up to the 0 for A. A Is already using up the 0. is 2. So we use that as well. Let me see if I can draw B over here-- It sits on this horizontal someplace. And the hexagon has side length S, where we don't know what that length is, but they're all the same. So let's just call this S it's going to help me think about it. Now that I know that this is equilateral equaliateral hexagon, all the sides are going to be the same length. We're going to go out here the coordinate B comma 2. We don't know what B is, but that is our vertex B. Now, F is the other vertex that is connected to A. F cannot sit on this horizontal-- cannot sit on y is equal to 2. It can't sit on y is equal to 6, because then this distance would be super far. Clearly much further than this distance over here. Or, actually, you could have that. But then you would you wouldn't be able to draw really a convex hexagon. So it's going to be S away. Maybe it will be something like that. So let me draw it-- so that is the next vertex. That is vertex F. Because we're going A, B,C,D, E, F, and then back to A. Fair enough. Now what about vertex C? Well, vertex C can't be on the 4 horizontal. So it's going to have to be on the 6 horizontal. So vertex C is going to have to be someplace like that. That's vertex C. And once again, that length is S this length is S. Now what about vertex E? Can't be on the 6 horizontal, already taken up by vertex C. So the 4 the 6 are taken up. So it has to be at the 8 horizontal. And so this is length S. And we also know that we're going back to the origin now. So this is the vertex E right here."
    },
    {
        "Q": "At 3:07 why does he add 16?\n",
        "A": "I ve just looked at this video and cannot find any 16 mentioned. Did you reference the correct video?",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            187
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 7:11 why do you all of a sudden flip 81 into 1/81",
        "A": "Sal was substituting into A / C. A was 1 / 9 and C was 81. So, A / C = ( 1 / 9 ) / 81 which is the same as ( 1 / 9 ) * ( 1 / 81 ). In words, dividing by a number is the same as multiplying by its reciprocal.",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            431
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In 2:14, we have the same bases on one side of the equation but what if there was a value on the other side of the equation without the same base. For example, logbase3 of 5 +logbase3of 2= 8 How would you approach the problem?\n",
        "A": "First you would combine the bases. Then you will exponentiate both sides of the equation. What this means is you make all of those numbers into a power. For example, x = y. When you exponentiate it, you will get something like 3^x = 3^y. You can use this to your advantage to eliminate logs.",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            134
        ],
        "3min_transcript": ""
    },
    {
        "Q": "is any number in the sequence of numbers 1 2 4 8 16 32 64 128 256 512 etc... able to some how convert into a higher number in the sequence? this question came from 3:33 in the video, it might have been before, when you converted the number 8 in the sequence into 25, because i have noticed that many numbers are multiples of those higher numbers.\n",
        "A": "I m not exactly sure what you are asking but the sequence you wrote down is the sequence you get by multiplying by 2 to get the next number. Another way to write that sequences is 2^0,2^1,2^2,2^3,2^4,2^5,2^6,.... I m not sure what you mean by converting 8 into 25.",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            213
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Why does Sal change colors, like at 4:37?\n",
        "A": "It lets the viewer understand things more easily. He also does it so that it isn t boring and he color codes some things. Try to learn things the KHAN way!!",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            277
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 1:58 why did u divide the number ab=-3 by a , why not multiplied it by a?\n",
        "A": "Sal is solving the b . To move the a , we look at the relationship between the a and b in ab . It is multiplication. Then we always use the opposite operation to move the a . The opposite operation to multiplication is division. If you multiplied both sides by a , you would get a*ab = -3a or a^2b = -3a . The b is not by itself. Hope this helps.",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            118
        ],
        "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m."
    },
    {
        "Q": "at around 4:25 he says that you can manipulate it to make one variable seem more DEPENDENT then another. What does he mean? If anyone could help me that would be nice, thanks.\n",
        "A": "m=n/7, m/n = 1/7, n = 7m, n/m = 7. The only thing I can come up with is that whichever variable has the number 7 on the same side with it might seem to be the independent variable, as in y = f(x) , y = 3x .",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            265
        ],
        "3min_transcript": "isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2, So m and n. So when-- and the way I've written it here, although you could algebraically manipulate it so that one looks more dependent than the other. But in this situation where n is 1, m is 1/7. And when n is 7, m is going to be 1. So you have the situation that if n is scaled up by 7, then m is also scaled up by 7, or vice versus. So it's more of a relationship. I could have expressed n in terms of m, but when you scale one variable up by 7, you also have to scale up the other variable by 7. When you scale it up by some amount, you have to scale the other variable by the same amount. So this is direct variation. Let's take the inverse, or when two variables vary inversely, this situation right over here. Let's take a and b."
    },
    {
        "Q": "\ncan anyone hrlp me. at 2:15 he says 1/10/x = 1/10x. I can't see how he got there..",
        "A": "(1/10)/x is the same as multiplying 1/10 by the reciprocal of x i.e. 1/x. So, (1/10)/x = 1/10 * 1/x = 1/10x. HOPE THAT HELPED YOU!",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            135
        ],
        "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m."
    },
    {
        "Q": "At 3:41 I didn't understand how it is neither\n",
        "A": "Cause the funcion has a sign minus (-) and is not clear if a is negative or is a sustraction. If you see the direct/inverse variation has no addtions or sustractions, just multiplication and divisions. I hope being helpfully",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            221
        ],
        "3min_transcript": "isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2, So m and n. So when-- and the way I've written it here, although you could algebraically manipulate it so that one looks more dependent than the other. But in this situation where n is 1, m is 1/7. And when n is 7, m is going to be 1. So you have the situation that if n is scaled up by 7, then m is also scaled up by 7, or vice versus. So it's more of a relationship. I could have expressed n in terms of m, but when you scale one variable up by 7, you also have to scale up the other variable by 7. When you scale it up by some amount, you have to scale the other variable by the same amount. So this is direct variation. Let's take the inverse, or when two variables vary inversely, this situation right over here. Let's take a and b."
    },
    {
        "Q": "\nwhat does he mean at 1:11 when he says constant?",
        "A": "Well, what he means by constant is that when one side of a function goes up or down, multiplying or dividing, the other side will go up by the same amount, since it is a direct variation. If the function was a inverse variation, it wouldn t be a constant, because when one side of the function goes up, the other side of the function goes down, so therefore the variable isn t constant, because it doesn t do the same thing for both sides, it does the opposite. HOPE THIS HELPS!!",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            71
        ],
        "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m."
    },
    {
        "Q": "At 2:53, why doesn't y = -2x count as inverse variation? The values move in opposite directions; as x increases y decreases. Thanks for any help.\n",
        "A": "Inverse variation is when the letter is under, so divides the coefficient and doesn t multiply it.",
        "video_name": "3_DxJwDTbyQ",
        "timestamps": [
            173
        ],
        "3min_transcript": "So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2,"
    },
    {
        "Q": "\nDid you know that the reason things float to the edge of a overfilled cup if you put something in it (say, a penny) is because that the middle of the water is actually just a teeny bit higher then the sides, so the penny automatically goes to the lowest point in the cup.\nVice Versa, if you were to underfill a cup, and put a penny on the side, it would automatically go towards the middle because that is the lowest point (3:02)",
        "A": "Yep. Surface tension is an amazing thing, isn t it? In science, when you re measuring liquid in a beaker or graduated cylinder, you would call the lowest point the meniscus",
        "video_name": "lOIP_Z_-0Hs",
        "timestamps": [
            182
        ],
        "3min_transcript": "would be directly above it. So that's no good, because it blocks all the light or something. You can go 180 degrees, to have the next leaf directly opposite, which seems ideal. Only once you go 180 again, the third leaf is right over the first. In fact, any fraction of a circle with a whole number as a base is going to have complete overlap after that number of turns. And unlike when you're doodling, as a plant you're not smart enough to see you've gone all the way around and now should switch to adding things in between. If you try and postpone the overlap by making the fraction really small, you just get a ton of overlap in the beginning, and waste all this space, which is completely disastrous. Or maybe other fractions are good. The kind that position leaves in a star like pattern. It will be a while before it overlaps, and the leaves will be more evenly spaced in the meantime. But what if there were a fraction that never completely overlapped? For any rational fraction, eventually the star will close. But what if you used an irrational number? The kind of number that can't be expressed as a whole number What if you used the most irrational number? If you think it sounds weird to say one irrational number is might want to become a number theorist. If you are a number theorist, you might tell us that phi is the most irrational number. Or you might say, that's like saying, of all the integers, 1 is the integer-iest. Or you might disagree completely. But anyway, phi. It's more than 1, but less than 2, more than 3/2, less than 5/3. Greater than 8/5, but 13/8 is too big. 21/13 is just a little too small, and 34/21 is even closer, but too big, and so on. Each pair of adjacent Fibonacci numbers creates a ratio that gets closer and closer to Phi as the numbers increase. Those are the same numbers on the sides of these squares. Now stop being a number theorist, and start being a plant again. You put your first leaf somewhere, and the second leaf at an angle, which is one Phi-th of a circle. Which depending on whether you're going one way or the other, could be about 222.5 degrees, or about 137.5. Great, your second leaf is pretty far from the first, gets lots of space in the sun. And now let's add the next one a Phi-th of the circle away. You can see how new leaves tend to pop up in the spaces left between old leaves, but it never quite fills things evenly. So there's always room for one more leaf, without having to do a whole new layer. It's very practical and as a plant you probably like this. It would also be a good way to give lots of room to seed pods and petals and stuff. As a plant that follows this scheme, you'd be at an advantage. Where do spirals come in? Let's doodle a pinecone using this same method. By the way, you can make your own phi angle-a-tron by dog-earing a corner of your notebook. If you folded it so the edges of the line-- You have 45 degrees plus 90, which is 135. Pretty close to 137.5. If you're careful, you can slip in a couple more degrees. Detach your angle-a-tron and you're good to go. Add each new pine cone-y thing a phi angle around, and make them a little farther out each time. Which you can keep track of by marking the distance on your angle-a-tron. The spirals form by themselves. And if we count the number of arms, look it's five and eight."
    },
    {
        "Q": "\nCan you explain the sentence written at 2:12 to me: \"Of all the integers, 1 is the integeriest.\"?",
        "A": "She is saying that some people say that saying that phi is the most irrational number is like saying that of all the integers 1 is the most integer-y.",
        "video_name": "lOIP_Z_-0Hs",
        "timestamps": [
            132
        ],
        "3min_transcript": "Say you're me and you're in math class, and you're doodling flowery petally things. If you want something with lots of overlapping petals, you're probably following a loose sort of rule that goes something like this. Add new petals where there's gaps between old petals. You can try doing this precisely. Start with some number of petals, say five, then add another layer in between. But the next layer, you have to add 10, then the next has 20. The inconvenient part of this is that you have to finish a layer before everything is even. Ideally, you'd have a rule that just lets you add petals until you get bored. Now imagine you're a plant, and you want to grow in a way that spreads out your leaves to catch the most possible sunlight. Unfortunately, and I hope I'm not presuming too much in thinking that, as a plant, you're not very smart. You don't know how to add number to create a series, you don't know geometry and proportions, and can't draw spirals, or rectangles, or slug cats. But maybe you could follow one simple rule. Botanists have noticed that plants seem to be fairly consistent when it comes to the angle between one leaf and the next. So let's see what you could do with that. So you grow your first leaf, and if you didn't change angle would be directly above it. So that's no good, because it blocks all the light or something. You can go 180 degrees, to have the next leaf directly opposite, which seems ideal. Only once you go 180 again, the third leaf is right over the first. In fact, any fraction of a circle with a whole number as a base is going to have complete overlap after that number of turns. And unlike when you're doodling, as a plant you're not smart enough to see you've gone all the way around and now should switch to adding things in between. If you try and postpone the overlap by making the fraction really small, you just get a ton of overlap in the beginning, and waste all this space, which is completely disastrous. Or maybe other fractions are good. The kind that position leaves in a star like pattern. It will be a while before it overlaps, and the leaves will be more evenly spaced in the meantime. But what if there were a fraction that never completely overlapped? For any rational fraction, eventually the star will close. But what if you used an irrational number? The kind of number that can't be expressed as a whole number What if you used the most irrational number? If you think it sounds weird to say one irrational number is might want to become a number theorist. If you are a number theorist, you might tell us that phi is the most irrational number. Or you might say, that's like saying, of all the integers, 1 is the integer-iest. Or you might disagree completely. But anyway, phi. It's more than 1, but less than 2, more than 3/2, less than 5/3. Greater than 8/5, but 13/8 is too big. 21/13 is just a little too small, and 34/21 is even closer, but too big, and so on. Each pair of adjacent Fibonacci numbers creates a ratio that gets closer and closer to Phi as the numbers increase. Those are the same numbers on the sides of these squares. Now stop being a number theorist, and start being a plant again. You put your first leaf somewhere, and the second leaf at an angle, which is one Phi-th of a circle. Which depending on whether you're going one way or the other, could be about 222.5 degrees, or about 137.5. Great, your second leaf is pretty far from the first, gets lots of space in the sun. And now let's add the next one a Phi-th of the circle away."
    },
    {
        "Q": "\nAt 9:10 to 9:33 Sal says that a rational can be a repeating decimal but it can't go backwards. But his answer was backwards. Is he wrong or am I not getting something.",
        "A": "Sal is completely right! What he meant was that you can t say (by that statement) that a number is rational because it can be written as a repeating decimal. To conclude that a number that can t be written as a repeating decimal is not rational, he didn t goes backwards. He did use a strait foward reasoning to deduce. If all rationals can be expressed that way, a number that can t be written as a repeting decimal is not rational.",
        "video_name": "GluohfOedQE",
        "timestamps": [
            550,
            573
        ],
        "3min_transcript": "All right. And then, the last one, 2 is congruent to 3. 2 is congruent to 3. Well, by the same logic, if 1 is congruent to 4, and since 1 and 2 are opposite, it's also the same as 2. And 4, because it's opposite of 3, it's congruent to that, all these angles have to be the same thing. So 2 and 3 would also be congruent angles. So all of the other ones must be true, B, C and D. So A is definitely our choice. Next problem. Let me copy and paste it. OK. Consider the arguments below. Every multiple of 4 is even, 376 is a multiple of 4. Fair enough. A number can be written as a repeating decimal if it is rational. Pi cannot be written as a repeating decimal. Therefore, pi is not rational. Which ones, if any, use deductive reasoning? OK, so statement one, every multiple of 4 is even. 376 is a multiple of 4. So this is deductive reasoning. Because you know that every multiple of 4 is even. So you pick any multiple of 4, it's going to be even. 376 is a multiple of 4. Therefore, it has to be even. So this is correct logic. So statement one is definitely deductive reasoning. Let's see, statement number two. A number can be written as a repeating decimal if it is rational. So if you're rational, that means that you can write it as a repeating decimal. That's like 0.33333. That's 1/3. That's all they mean by a repeating decimal. written as a repeating decimal if it is rational. That doesn't say that a repeating decimal means that It just means that a rational number can be written as a repeating decimal. This statement doesn't let us go the other way. It doesn't say that a repeating decimal can definitely be written as a rational number. It just says that if it is rational, a number can be written as a repeating decimal. Fair enough. And then it says pi cannot be written as a repeating decimal. Pi cannot be written as a repeating decimal. So if pi cannot be written as a repeating decimal, pi be rational? Well if pi was rational, if pi was in this set, if pi were rational, then you could write it as a repeating decimal. But it said that you cannot write it as a repeating decimal. So therefore, pi cannot be rational."
    },
    {
        "Q": "\nSal converted the fractions into improper fractions in 1:38. Is that the easiest way?",
        "A": "converting mixed fractions into improper fractions helps in comparing mixed fractions",
        "video_name": "QS1LMomm0Gk",
        "timestamps": [
            98
        ],
        "3min_transcript": "Find the sum 3 and 1/8 plus 3/4 plus negative 2 and 1/6. Let's just do the first part first. It's pretty straightforward. We have two positive numbers. Let me draw a number line. So let me draw a number line. And I'll try to focus in. So we're going to start at 3 and 1/8. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 and 1/8 is going to be right about there. So let me just draw its absolute value. So this 3 and 1/8 is going to be 3 and 1/8 to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 and 1/8. Now whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction, and, actually, the multiplication and the division plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 and 1/8. Another way to think about it, 3 is 24 over 8. And then you add 1/8 to that, so you get 25 over 8. So this is our starting point. Now to that, we are going to add 3/4. We are going to add 3/4. So we're going to move another 3/4. We are going to move another 3/4. It's hard drawing these arrows. We're going to move another 3/4 to the right. So this right here, the length of this that we're moving to the right is 3/4. So plus 3/4. Now where does this put us? Well, both of these are positive integers. So we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3/4. That's the same thing as we need to find a common denominator of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3/4 is the same thing as 6/8. If we have 25/8 and we're adding 6/8 to that, that gives us 25 plus 6 is 31/8. So this number right over here, this number right over here, is 31/8. And it makes sense because 32/8 would be 4. So it should be a little bit less than four. So this number right over here is 31/8. Or the length of this arrow, the absolute value of that number, is 31/8, a little bit less than 4. If you wanted to write that as a mixed number, it would be what? It would be 3 and 7/8. So that's that right over here. This is 31/8."
    },
    {
        "Q": "At 1:57, Why do we need to find a common denominator? Can't we just add 25/8 + 3/4 directly?\n",
        "A": "Cause a common denominators solves all of life s problems",
        "video_name": "QS1LMomm0Gk",
        "timestamps": [
            117
        ],
        "3min_transcript": "Find the sum 3 and 1/8 plus 3/4 plus negative 2 and 1/6. Let's just do the first part first. It's pretty straightforward. We have two positive numbers. Let me draw a number line. So let me draw a number line. And I'll try to focus in. So we're going to start at 3 and 1/8. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 and 1/8 is going to be right about there. So let me just draw its absolute value. So this 3 and 1/8 is going to be 3 and 1/8 to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 and 1/8. Now whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction, and, actually, the multiplication and the division plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 and 1/8. Another way to think about it, 3 is 24 over 8. And then you add 1/8 to that, so you get 25 over 8. So this is our starting point. Now to that, we are going to add 3/4. We are going to add 3/4. So we're going to move another 3/4. We are going to move another 3/4. It's hard drawing these arrows. We're going to move another 3/4 to the right. So this right here, the length of this that we're moving to the right is 3/4. So plus 3/4. Now where does this put us? Well, both of these are positive integers. So we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3/4. That's the same thing as we need to find a common denominator of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3/4 is the same thing as 6/8. If we have 25/8 and we're adding 6/8 to that, that gives us 25 plus 6 is 31/8. So this number right over here, this number right over here, is 31/8. And it makes sense because 32/8 would be 4. So it should be a little bit less than four. So this number right over here is 31/8. Or the length of this arrow, the absolute value of that number, is 31/8, a little bit less than 4. If you wanted to write that as a mixed number, it would be what? It would be 3 and 7/8. So that's that right over here. This is 31/8."
    },
    {
        "Q": "\nReferencing the Banach-Tarski Paradox made my day. (2:24)\nIf only physics wouldn't get in the way of my math!",
        "A": "I know, right? Physics rocks UNTIL it gets in the way of doing things like the Banach-Tarski paradox!",
        "video_name": "F5RyVWI4Onk",
        "timestamps": [
            144
        ],
        "3min_transcript": "At my house, no Thanksgiving dinner is complete without mathed potatoes. To make mathed potatoes, start by boiling the potatoes until they're soft, which will take about 15 to 20 minutes. After you drain them and let them cool slightly, you're ready for the math. Take one potato and divide evenly to get half a potato, plus half a potato. Then divide the halves into fourths and the fourths into eights and so on. Eventually, you will have a completely mathed potato that looks like this. Once you have proven this result for one potato, you can apply it to other potatoes without going through the entire process. That's how math works. While I prefer refined and precise methods for mathing a potato, many people just apply brute force algorithms. You can also add other variables like butter, cream, garlic, salt, and pepper. Place in a hemisphere and garnish with an organic hyperbolic plane, and your math potatoes are ready for the table. Together with a cranberry cylinder and a nice basket of bread spheres with butter prism, you'll be well on your way to creating a delicious and engaging Thanksgiving meal. Here's a serving tip. When arranging mathed potatoes on your plate, If you just make a mound, the gravy will fall off. It's best to create some kind of trough or pool. But what shape will maximize the amount of gravy it can hold? Due to the structural properties of mathed potatoes, this can essentially be reduced to a two-dimensional gravy pool problem, where you want the most gravy area given a certain potato perimeter. When I think of this question, I like to think about inflating shapes. Say you inflated a triangle. It would add more area and round out into a circle. And then, if the perimeter can't change, it would pop. In fact, all 2D shapes inflate into circles. And in 3D, it's spheres, which is my bubbles like to be round. And turkeys are spheroid, because that optimizes for maximum stuffing. The limited three-dimensional capacity of mathed potatoes may confuse things a little. But since a mathed potato sphere can't support itself, you're really stuck with extrusions of 2D shapes. So what's better? A deep mathed potato cylinder or a shallow but wider one? Well, think of it like this. If you slice the deep version in half, you'll see it has equivalent gravy-holding capacity to two And the perimeter of two circles would be more efficient if combined into one bigger circle. So the solution is to create the biggest, roundest, shallowest gravy pool you can. In fact, maybe you should just skip the mathed potatoes and get a bowl. Anyway, I hope this simple recipe helps you have an optimal Thanksgiving experience. Advanced chef-amaticians may wish to try Banach-Tarski potatoes, wherein after you cut a potato in a particular way, you put the pieces back together and get two potatoes. Stay tuned for more delicious and extremely practical Thanksgiving recipes this week."
    },
    {
        "Q": "\n2:27 If I had this sort of problem on a test, what are other ways of stating that there's no solution? I'm assuming the symbol for the empty set would be one of them? Or is that basically it?",
        "A": "Yes, the symbol for the empty set would be an acceptable way of saying there is no solution.",
        "video_name": "ZF_cZ-GX9PI",
        "timestamps": [
            147
        ],
        "3min_transcript": "Solve for x, 5x - 3 is less than 12 \"and\" 4x plus 1 is greater than 25. So let's just solve for X in each of these constraints and keep in mind that any x has to satisfy both of them because it's an \"and\" over here so first we have this 5 x minus 3 is less than 12 so if we want to isolate the x we can get rid of this negative 3 here by adding 3 to both sides so let's add 3 to both sides of this inequality. The left-hand side, we're just left with a 5x, the minus 3 and the plus 3 cancel out. 5x is less than 12 plus 3 is 15. Now we can divide both sides by positive 5, that won't swap the inequality since 5 is positive. So we divide both sides by positive 5 and we are left with just from this constraint that x is less than 15 over 5, which is 3. So that constraint over here. But we have the second constraint as well. is greater than 25. So very similarly we can subtract one from both sides to get rid of that one on the left-hand side. And we get 4x, the ones cancel out. is greater than 25 minus one is 24. Divide both sides by positive 4 Don't have to do anything to the inequality since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember there was that \"and\" over here. We have this \"and\". So x has to be less than 3 \"and\" x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3 so this is 3 on the number line. We're saying x has to be less than 3 so it has to be in this shaded area right over there. greater than 6. So if this is 6 over here, it says that x has to greater than 6. It can't even include 6. And since we have this \"and\" here. The only x-es that are a solution for this compound inequality are the ones that satisfy both. The ones that are in the overlap of their solution set. But when you look at it right over here it's clear that there is no overlap. There is no x that is both greater than 6 \"and\" less than 3. So in this situation we have no solution."
    },
    {
        "Q": "At 8:54, why is tan theta described as sine theta/ Cosine theta?\n",
        "A": "You will need to know these definitions: sin\u00ce\u00b8 = y/r cos\u00ce\u00b8 = x/r tan\u00ce\u00b8 = y/x Since we re in a unit circle, r = 1, so sin\u00ce\u00b8 = y and cos\u00ce\u00b8 = x. Since tan\u00ce\u00b8 = y/x, y = sin\u00ce\u00b8, and x = cos\u00ce\u00b8, we can make the substitutions. y = sin\u00ce\u00b8 x = cos\u00ce\u00b8 tan\u00ce\u00b8 = y/x = sin\u00ce\u00b8/cos\u00ce\u00b8",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            534
        ],
        "3min_transcript": "is the same thing as cosine of theta. And b is the same thing as sine of theta. Well, that's interesting. We just used our soh cah toa definition. Now, can we in some way use this to extend soh cah toa? Because soh cah toa has a problem. It works out fine if our angle is greater than 0 degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But soh cah toa starts to break down as our angle is either 0 or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90-degree angles in it. It starts to break down. Let me make this clear. So sure, this is a right triangle, so the angle is pretty large. I can make the angle even larger and still have a right triangle. Even larger-- but I can never get quite to 90 degrees. have a right triangle any more. It all seems to break down. And especially the case, what happens when I go beyond 90 degrees. So let's see if we can use what we said up here. Let's set up a new definition of our trig functions which is really an extension of soh cah toa and is consistent with soh cah toa. Instead of defining cosine as if I have a right triangle, and saying, OK, it's the adjacent over the hypotenuse. Sine is the opposite over the hypotenuse. Tangent is opposite over adjacent. Why don't I just say, for any angle, I can draw it in the unit circle using this convention that I just set up? And let's just say that the cosine of our angle is equal to the x-coordinate where we intersect, where the terminal side of our angle intersects the unit circle. to be equal to the y-coordinate where the terminal side of the angle intersects the unit circle? So essentially, for any angle, this point is going to define cosine of theta and sine of theta. And so what would be a reasonable definition for tangent of theta? Well, tangent of theta-- even with soh cah toa-- could be defined as sine of theta over cosine of theta, which in this case is just going to be the y-coordinate where we intersect the unit circle over the x-coordinate. In the next few videos, I'll show some examples where we use the unit circle definition to start evaluating some trig ratios."
    },
    {
        "Q": "At 2:34, shouldn't the point on the circle be (x,y) and not (a,b)? [Since horizontal goes across 'x' units and vertical goes up 'y' units--- A full explanation will be greatly appreciated]\n",
        "A": "It would be x and y, but he uses the letters a and b in the example because a and b are the letters we use in the Pythagorean Theorem a\u00c2\u00b2+b\u00c2\u00b2 = c\u00c2\u00b2 and they re the letters we commonly use for the sides of triangles in general. It doesn t matter which letters you use so long as the equation of the circle is still in the form a\u00c2\u00b2+b\u00c2\u00b2 = 1.",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            154
        ],
        "3min_transcript": "And the way I'm going to draw this angle-- I'm going to define a convention for positive angles. I'm going to say a positive angle-- well, the initial side of the angle we're always going to do along the positive x-axis. So you can kind of view it as the starting side, the initial side of an angle. And then to draw a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise direction. So let me draw a positive angle. So a positive angle might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate"
    },
    {
        "Q": "\nat 4:29, why do we find cosine?",
        "A": "Sal is just demonstrating how all of the trig functions work as they are defined on a unit circle.",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            269
        ],
        "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's"
    },
    {
        "Q": "At 3:48, how do we know that side b of the triangle is the same as the chord \"b\", like Sal says?\n",
        "A": "He never says that the length of the chord you are creating is equivalent to the length of side b. He s simply saying that the length of side b, or the height as he says, is the y-value, b, which is what he marks on the y-axis. He s not creating a chord b, it s just a dashed line to show that the y-value corresponds to the height of side b of the triangle. Hope this helps!",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            228
        ],
        "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's"
    },
    {
        "Q": "At 7:05,Sal said \"bottom boundary\" as z=2-2/3*x-y/3 but from graph it seems\nz=0 to z=z=2-2/3*x-y/3. I think the bottom limit for dz is zero? i am confused?\n",
        "A": "Here is the clarification: Review from 1:13 where Sal defines the plane (via x,y,z intercepts). Then at 2:11 he says he cares about the volume above the plane (to make the problem more complicated). So he is asking for the volume between this plane and the plane z=2.",
        "video_name": "ZN2PfqZ4ihM",
        "timestamps": [
            425
        ],
        "3min_transcript": "And that is equal to dx -- no, sorry, this is dy. Let me do this in yellow, or green even better. So dy, which is this. dy times dx, dx times dz. That's the volume of that little cube. And if we wanted to know the mass of that cube, we would multiply the density function at that point times this dv. So the mass, you could call it d -- I don't know, dm. The mass differential is going to be equal to that times that. So x squared y z times this. dy, dx, and dz. And we normally switch this order around, depending on what we're going to integrate with respect to first so we don't get confused. So let's try to do this. Let's try to set up this integral. So let's do it traditionally. The last couple of triple integrals we did we integrated So let's do that. So we're going to integrate with respect to z first. so we're going to take this cube and we're going to sum up all of the cubes in the z-axis. So going up and down first, right? So if we do that, what is the bottom boundary? So when you sum up up and down, these cubes are going to turn to columns, right? So what is the bottom of the column, the bottom bound? What's the surface? It's the surface defined right here. So, if we want that bottom bound defined in terms of z, we just have to solve this in terms of z. So let's subtract. So what do we get. If we want this defined in terms of z, we get 3z is equal to 6 minus 2x minus y. Or z is equal 2 minus 2/3x minus y over 3. This is the same thing as that. But when we're talking about z, explicitly defining a So the bottom boundary -- and you can visualize it, right? The bottom of these columns are going to go up and down. We're going to add up all the columns in up and down direction, right? You can imagine summing them. The bottom boundary is going to be this surface. z is equal to 2 minus 2/3x minus y over 3. And then what's the upper bound? Well, the top of the column is going to be this green plane, and what did we say the green plane was? It was z is equal to 2. That's this plane, this surface right here. Z is equal to 2. And, of course, what is the volume of that column? Well, it's going to be the density function, x squared yz times the volume differential, but we're integrating with respect to z first. Let me write dz there. I don't know, let's say we want to integrate with respect to -- I don't know, we want to integrate with respect"
    },
    {
        "Q": "\nAt 10:10, I don't get it",
        "A": "There is an exact same question at the back, with the same point of time reference which has some good answers. You should have a look at them.",
        "video_name": "ZN2PfqZ4ihM",
        "timestamps": [
            610
        ],
        "3min_transcript": "In the last couple of videos, I integrated So let's do x just to show you it really doesn't matter. So we're going to integrate with respect to x. So, now we have these columns, right? When we integrate with respect to z, we get the volume of each of these columns wher the top boundary is that plane. Let's see if I can draw it decently. The top boundary is that plane. The bottom boundary is this surface. Now we want to integrate with respect to x. So we're going to add up all of the dx's. So what is the bottom boundary for the x's? Well, this surface is defined all the way to -- the volume under question is defined all the way until x is equal to 0. And if you get confused, and it's not that difficult to get confused when you're imagining these three-dimensional things, say you know what, we already integrated with respect to z. The two variables I have left are x and y. Let me draw the projection of our volume onto the xy plane, and what does that look like? Because that actually does help simplify things. So if we twist it, if we take this y and flip it out like that, and x like that we'll get in kind of the traditional way that we learned when we first learned algebra. The xy-axis. So this is x, this is y. And this point is what? Or this point? That's x is equal to 3. So it's 1, 2, 3. That's x is equal to 3. And this point right here is y is equal to 6. So 1, 2, 3, 4, 5, 6. So on the xy-axis, kind of the domain -- you can view it that -- looks something like that. So one way to think about it is we've figured out if these columns -- we've integrated up/down or along the z-axis. But when you view it looking straight down onto it, you're looking on the xy plane, each of our columns are going to of your screen in the z direction. But the base of each column is going to dx like that, and then dy up and down, right? So we decided to integrate with respect to x next. So we're going to add up each of those columns in the x direction, in the horizontal direction. So the question was what is the bottom boundary? What is the lower bound in the x direction? Well, it's x is equal to 0. If there was a line here, then it would be that line probably as a function of y, or definitely as a function of y. So our bottom bound here is x is equal to 0. What's our top bound? I realize I'm already pushing. Well, our top bound is this relation, but it has to be in terms of x, right? So, you could view it as kind of saying well, if z is equal to 0, what is this line? What is this line right here? So z is equal to 0. We have 2x plus y is equal to 6. We want the relationship in terms of x. So we get 2x is equal to 6 minus y where x is equal"
    },
    {
        "Q": "at 4:35 cant the x and y axi, axises, axis(multiple axis) be a line and go on forever in any direction and seems how forever is a relative term go on in both directions in so far it is impossible to reach the end what so ever.\n",
        "A": "The axes do go on forever, so to speak, in both directions, and it is indeed impossible to reach the end (if we progress by finite distances). Did it seem to you that Sal was implying something different? Is your question related to the restriction in the domain of arctangent?",
        "video_name": "QGfdhqbilY8",
        "timestamps": [
            275
        ],
        "3min_transcript": "so inverse tangent of x minus six plus three pi over two so let me write that down. Let me type this. G inverse of x is going to be the inverse tangent so I can write it like this, the inverse tangent of x minus six and yes it interpreted it correctly. Inverse tangent you can do that as arctangent of x minus six plus three pi over two and it did interpret it correctly but then we have to think about what is the domain of g inverse? What is the domain of g inverse of x? Let's think about this a little bit more. The domain of g inverse of x, so let's just think about what tangent is doing. The tangent function if we imagine a unit circle, so that's a unit circle right over there. Guess we can imagine to be a unit circle. My pen tool is acting up a little bit it's putting this little gaps and things Let's just say for the sake of argument that that's a unit circle, that's the x axis and that's the y axis. If you form an angle theta. If you form some angle theta right over here, the tangent of theta is essentially the slope of this terminal ray of the angle or the ... Or I guess we can call it the terminal ray of the angle. The angles form by that ray and this ray along the positive x axis. The tangent of theta is the slope right over there and you can get a tangent of any theta except for a few. You can find the tangent of that, you could find the slope there, you could find the slope there, you could also find the slope there, you could find the slope there but the place where you can't find the slope is when this ray goes straight up, or this ray goes straight down. Those were the cases where you can't find the slope. They are the slope you could say is approaching The domain of tangent, so tangent domain so the domain is essentially all real numbers, all reals except multiples of pi over \u2026 I guess you can say pi over two plus multiples of pi, except pi over two plus multiples of pi where k could be any integer so you could also be subtracting pi because if you have pi over two, if you add pi, you go straight down here. You add another pi you go up there, if you subtract pi you go down here, add, subtract another pi you go over there. This is the domain but given this domain you can get any real number. The range here is all reals because you can get any slope here,"
    },
    {
        "Q": "\nAt 0:43 what did mean by 90 degrees. Do you mean that all sides are the same length for a square?",
        "A": "It is true that all sides of a square have equal length but when he said the square had 90 degrees he meant that all the angles of the square were equal to 90 degrees. Have an awesome day!\u00f0\u009f\u0098\u008a",
        "video_name": "1pHhMX0_4Bw",
        "timestamps": [
            43
        ],
        "3min_transcript": "A parallelogram is a blank with two sets of parallel lines. So let's see what the options are. So one option is a quadrilateral. And a parallelogram is definitely a quadrilateral. A quadrilateral is a four-sided figure, and it is definitely a four-sided figure. A parallelogram is not always a rhombus. A rhombus is a special case of a parallelogram where not only do you have to sets of parallel lines as your sides, two sets of parallel sides, but all of the sides are the same length in a rhombus. And a square is a special case of a rhombus where all of the angles are 90 degrees. So here, all we can say is that a parallelogram is a quadrilateral. And so let's check our answer. And it's always a good idea to look at hints. And so it'll kind of say the same thing that we just said, but it would say it for the particular problem that you're actually looking at. Let's do a few more of these. Suzanne is on an expedition to save the universe. Sounds like a reasonable expedition to go on. a game called Find the Rhombuses. A wizard tells her that she has a square, a quadrilateral, and a parallelogram, and she must identify which of the shapes are also rhombuses. Which of these shapes should she pick to save the universe? So a square is a special case of a rhombus. Just to remind ourselves, a rhombus, the opposite sides are parallel to each other. You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides. So all rhombuses are quadrilaterals. But not all quadrilaterals are rhombuses. You could have a quadrilateral where none of the sides are parallel to each other. So we won't click this one. Once again, a parallelogram. So all rhombuses are parallelograms. two sets of parallel line segments representing their sides. But all parallelograms are not rhombuses. So I would say that if someone gives you square, you can say, look, a square is always going to be a rhombus. A quadrilateral isn't always going to be a rhombus, nor is a parallelogram always going to be a rhombus. We got it right."
    },
    {
        "Q": "At 0:34-0:39, did he also mean to say that all sides are of equal length? A rectangle is also a quadrilateral with only 90-degree angles, but all the sides aren't always the same length.\n",
        "A": "Yes, rhombus is a quadrilateral with 4 sides of equal length. So rectangles are not always rhombus.",
        "video_name": "1pHhMX0_4Bw",
        "timestamps": [
            34,
            39
        ],
        "3min_transcript": "A parallelogram is a blank with two sets of parallel lines. So let's see what the options are. So one option is a quadrilateral. And a parallelogram is definitely a quadrilateral. A quadrilateral is a four-sided figure, and it is definitely a four-sided figure. A parallelogram is not always a rhombus. A rhombus is a special case of a parallelogram where not only do you have to sets of parallel lines as your sides, two sets of parallel sides, but all of the sides are the same length in a rhombus. And a square is a special case of a rhombus where all of the angles are 90 degrees. So here, all we can say is that a parallelogram is a quadrilateral. And so let's check our answer. And it's always a good idea to look at hints. And so it'll kind of say the same thing that we just said, but it would say it for the particular problem that you're actually looking at. Let's do a few more of these. Suzanne is on an expedition to save the universe. Sounds like a reasonable expedition to go on. a game called Find the Rhombuses. A wizard tells her that she has a square, a quadrilateral, and a parallelogram, and she must identify which of the shapes are also rhombuses. Which of these shapes should she pick to save the universe? So a square is a special case of a rhombus. Just to remind ourselves, a rhombus, the opposite sides are parallel to each other. You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides. So all rhombuses are quadrilaterals. But not all quadrilaterals are rhombuses. You could have a quadrilateral where none of the sides are parallel to each other. So we won't click this one. Once again, a parallelogram. So all rhombuses are parallelograms. two sets of parallel line segments representing their sides. But all parallelograms are not rhombuses. So I would say that if someone gives you square, you can say, look, a square is always going to be a rhombus. A quadrilateral isn't always going to be a rhombus, nor is a parallelogram always going to be a rhombus. We got it right."
    },
    {
        "Q": "\nShouldn't it be (2x-2y) - (2x-2y)(dy/dx) instead of (2x-2y)+(2x-2y)(dy/dx) at 2:15",
        "A": "He has switched the order of the variables, so that it is (2x-2y)+(2 y-2 x)(dy/dx). -(2x-2y) = 2y-2x",
        "video_name": "9uxvm-USEYE",
        "timestamps": [
            135
        ],
        "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides."
    },
    {
        "Q": "At 2:12, why does Sal simply multiply (2x-2y) *1 and then by the (dy/dx), instead of FOIL'ing it out and getting something like 2x-2x(dy/dx)-2y+2y(dy/dx) ?\n",
        "A": "He could have done that. Then he would have had to put the dy/dx s back together: 2x - 2y + 2y(dy/dx) - 2x(dy/dx) 2x - 2y + (2y - 2x)(dy/dx) He wants to get all the dy/dx s together such that he can manipulate them into one instance of dy/dx multiplied by a factor (in this case that factor ends up being (2y - 2x - 1)), such that he can isolated dy/dx on the left and have his answer.",
        "video_name": "9uxvm-USEYE",
        "timestamps": [
            132
        ],
        "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides."
    },
    {
        "Q": "at 3:05 shouldnt it be (2y-2x) time negative dy/dx but why is there no negative\n",
        "A": "at 2:15 - 2:25 he explains that instead of doing -(2x - 2y) he chose to do (2y - 2x) (they are equal).",
        "video_name": "9uxvm-USEYE",
        "timestamps": [
            185
        ],
        "3min_transcript": "of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. 2y from that side. And then we could also subtract a dy dx from both sides, so that all of our dy dx's are on the left hand side, and all of our non dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So this is we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going"
    },
    {
        "Q": "At 0:52, why does he say she wants to divide the blueberries into 6 groups? That number doesn't take Kali into account, and we don't know how many berries she may want. She could want 1 blueberry, or 0, or 12. Why then, does he not at least say \"we can assume that Kali doesn't want any berries, if she is giving them all too her friends.\"?\n",
        "A": "That s right. I agree",
        "video_name": "QXNg_u5Tv8Q",
        "timestamps": [
            52
        ],
        "3min_transcript": "- Kali is having a picnic for her six friends. The oldest friend, Vikram, is 10 years old. The youngest, Diya, is six years old. She has a total of 48 blueberries and wants to split them evenly between her friends. How many blueberries does each friend get? And I encourage you to now, pause this video and try to figure it out on your own. How many blueberries does each friend get? Let's think about this a little bit. So she has six friends. She has six friends and she wants them all to be able to get the same amount. So, she wants to split the 48 blueberries evenly amongst her six friends. The ages of her friends don't matter. So, she's going to take the 48 blueberries. She's going to take the 48 blueberries and divide it, and divide it by six. She wants to divide it into six groups. So she wants to divide it into six groups. 48 divided by 6. And so this, is going to be equal to the number of blueberries that each friend gets. So 48 divided by six is equal to question mark, is the same thing as saying that 48, 48 is equal to, is equal to question mark times six, times six. So if we could figure out what number we can multiply by six to get 48, then we know what 48 divided by six actually is. For example, this question mark, this is the number of blueberries per friend. The number of blueberries per friend times six friends, well that should tell us the total number of blueberries, which is 48. So what is this number? Well, let's think about, let's just think about all of the multiples of six. So, six times one is equal to six. Six times two is equal to 12. And really we're just increasing by six each time. Six times three is equal to 18. Six times four is equal to 24. Six times five is equal to 30. Six times six is equal to 36. Six times seven is equal to 42. Notice we're just adding six every time. Six times eight is equal to 48. Is equal to 48. So we now know that question mark, we now know that the question mark must be eight. Six times eight and eight times six is the same thing. So this is going to be equal to six times question mark. Six times question mark and now we learned that question mark is equal to 48. Sorry, question mark is equal to eight. So, each of her friends are going to get eight blueberries. So this is, right over here, 48 divided by six is equal to eight blueberries."
    },
    {
        "Q": "so it ends at 4:42 ?\n",
        "A": "Yes. Why wouldn t it its the end of the video.",
        "video_name": "vAlazPPFlyY",
        "timestamps": [
            282
        ],
        "3min_transcript": "it's a vertical angle, it's the one on the opposite side of the intersection. It's one of these angles that it is not adjacent to. So it would be this angle right over here. So going back to the question, a vertical angle to angle EGA, well if you imagine the intersection of line EB and line DA, then the non-adjacent angle formed to angle EGA is angle DGB. Actually, what we already highlighted in magenta right over here. So this is angle DGB. Which could also be called angle BGD. These are obviously both referring to this angle up here. Name an angle that forms a linear pair with the angle DFG. So we'll put this in a new color. Angle DFG. Sorry, DGF, all of these should have G in the middle. So linear pair with angle DGF, so that's this angle right over here. So an angle that forms a linear pair will be an angle that is adjacent, where the two outer rays combined will form a line. So for example, if you combine angle DGF, which is this angle, and angle DGC, then their two outer rays form this entire line right over here. So we could say angle DGC. Or, if you look at angle DFG, you could form a line this way. If you take angle AGF, so if you take this one, then the outer rays will form this line. So angle AGF would also work. Angle AGF. Let's do one more. Name a vertical angle to angle FGB. So this is FGB right over here. when CF-- let me highlight this, that's hard to see. This is the last one, so I can make a mess out of this. That angle is formed when CF and EB intersect with each other. And four angles are formed. The one question, FGB, these two angles that are adjacent to it, it shares a common ray. And then the vertical angle, the one that sits on the opposite side. So this angle, this angle right over here, which is angle EGC. Or you could also call it angle CGE. So angle CGE."
    },
    {
        "Q": "at 0:22, what is a better under standing of what adjacent is?\n",
        "A": "It just means next to",
        "video_name": "vAlazPPFlyY",
        "timestamps": [
            22
        ],
        "3min_transcript": "We're asked to name an angle adjacent to angle BGD. So angle BGD, let's see if we can pick it out. So here is B, here is G, and here is D, right over here. So angle BGD is this entire angle right over here. So when we talk about adjacent angles, we're talking about an angle that has one of its rays in common. So for example, angle AGB has one of the rays in common, it has GB in common with angle BGD. So we could say angle AGB, which could obviously also be called angle BGA, BGA and AGB are both this angle right over here. You could also go with angle FGB, because that also has GB in common. So you go angle FGB, which could also be written as angle BGF. So you could do this angle right over here, angle EGD. Or you could go all the way out here, angle FGD. These last two sharing ray GD in common. So any one of these responses would satisfy the question of just naming an angle, just naming one. Let's do this next one. Name an angle vertical to angle EGA. So this is this angle right over here. And the way you think about vertical angles is, imagine two lines crossing. So imagine two lines crossing, just like this. And they could literally be lines, and they're intersecting at a point. This is forming four angles, or you could imagine it's forming two sets of vertical angles. it's a vertical angle, it's the one on the opposite side of the intersection. It's one of these angles that it is not adjacent to. So it would be this angle right over here. So going back to the question, a vertical angle to angle EGA, well if you imagine the intersection of line EB and line DA, then the non-adjacent angle formed to angle EGA is angle DGB. Actually, what we already highlighted in magenta right over here. So this is angle DGB. Which could also be called angle BGD. These are obviously both referring to this angle up here. Name an angle that forms a linear pair with the angle DFG. So we'll put this in a new color. Angle DFG. Sorry, DGF, all of these should have G in the middle."
    },
    {
        "Q": "\nAt 4:00 Sal cancels out the terms that approach zero as if they dont exist and states that the limit is 3/6 or 1/2.\n\nI dont understand how he can be sure the function isn't defined for 1/2.\n\nAs it approaches infinity it is (3 minus a tiny fraction) over (6 minus a tiny fraction)\n\nAs I see it, unless those tiny fractions are exactly the same, the limit will be a tiny amount greater or less than 1/2, making the function defined for the point Sal claims it never reaches.\n\nIs there something I'm misunderstanding?",
        "A": "I think you just answered your own question without realizing it. The value will always be a tiny amount greater or less than 1/2. It will never be exactly equal to 1/2.",
        "video_name": "P0ZgqB44Do4",
        "timestamps": [
            240
        ],
        "3min_transcript": "These other terms are going to matter less obviously minus 54 isn't going to grow at all and minus 18X is going to grow much slower than the three X squared, the highest degree terms are going to be what dominates. If we look at just those terms then you could think of simplifying it in this way. F of X is going to get closer and closer to 3/6 or 1/2. You could say that there's a horizontal asymptote at Y is equal to 1/2. Another way we could have thought about this if you don't like this whole little bit of hand wavy argument that these two terms dominate is that we can divide the numerator and the denominator by the highest degree or X raised to the highest power in the numerator and the denominator. The highest degree term is X squared in the numerator. Let's divide the numerator and the denominator or I should say the highest degree term in the numerator and the denominator is X squared. and denominator by that. If you multiply the numerator times one over X squared and the denominator times one over X squared. Notice we're not changing the value of the entire expression, we're just multiplying it times one if we assume X is not equal zero. We get two. In our numerator, let's see three X squared divided by X squared is going to be three minus 18 over X minus 81 over X squared and then all of that over six X squared times one over X squared, this is going to be six and then minus 54 over X squared. What's going to happen? If you want to think in terms of if you want to think of limits as something approaches infinity. If you want to say the limit as X approaches infinity here. What's going to happen? Well this, this and that are going to approach zero Now, if you say this X approaches negative infinity, it would be the same thing. This, this and this approach zero and once again you approach 1/2. That's the horizontal asymptote. Y is equal to 1/2. Let's think about the vertical asymptotes. Let me write that down right over here. Let me scroll over a little bit. Vertical asymptote or possibly asymptotes. Vertical maybe there is more than one. Now it might be very tempting to say, \"Okay, you hit a vertical asymptote\" \"whenever the denominator equals to zero\" \"which would make this rational expression undefined\" and as we'll see for this case that is not exactly right. Just making the denominator equal to zero by itself will not make a vertical asymptote. It will definitely be a place"
    },
    {
        "Q": "\nAt 7:55, there was a reference to the function g:R^2-->R. How did that function morph into g(x1,x2)=2.",
        "A": "so you have a vector space that is R^2, and a vector space that is R. the R^2 vector space simply has more data than the vector space R. the transformation G fits vectors from the R^2 vector space into vectors in R, or at least attempts to.",
        "video_name": "BQMyeQOLvpg",
        "timestamps": [
            475
        ],
        "3min_transcript": "It could be equal to the codomain. It's some subset. A set is a subset of itself, every member of a set is also a member of itself, so it's a subset of itself. So range is a subset of the codomain which the function actually maps to. So let me give you an example. Let's say I define the function g, and it is a mapping from the set of real numbers. Let me say it's a mapping from R2 to R. mapping it to R. And I will define g, I'll write it a couple of different ways. So now I'm going to take g of two values, I could say xy or I could say x1, x2. Let me do it that way. g of x1, x2 is always equal to 2. It's a mapping from R2 to R, but this always equals 2. And let me actually write the other notation just because you probably haven't seen this much. g maps any points x1 and x2 to the point 2. This makes the mapping a little bit clearer. But just to get the notation right, what is our domain? That was part of my function definition, I said we're mapping from R2, so my domain is R2. Now what is my codomain? My codomain is the set that I'm potentially mapping to, and is part of the function definition. This by definition is the codomain. So my codomain is R. Now what is the range of my function? The range is the set of values that the function actually maps to. In this case, we always map to the value 2, so the range is actually just the value 2. And if we were to visualize this-- R2 is actually-- I wouldn't draw it as a blurb, I would draw it as the entire Cartesian space, but I'm just giving you an abstract notion. If I really have to draw R, I'd draw it as some type of a number line."
    },
    {
        "Q": "\n@ 4:00 why is the exponent -10?",
        "A": "At around 2:10 you will see that it is counting the number of places from the standard decimal place, and it is negative because you re moving it to the right, towards the negative side of the number line.",
        "video_name": "6phoVfGKKec",
        "timestamps": [
            240
        ],
        "3min_transcript": "to 0-- let me actually-- I skipped a step right there. Let me add 1 times 10 to the 0, so we have something natural. So this is one times 10 to the first. One times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1/10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared or 1/100. So this is going to be 1/100, which is 0.01. What's happening here? When I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I've moved it from there to there. When I raise it to the negative 2, I moved it two over to the left. So how many times are we going to have to move it over to the left to get this number right over here? So we have to move it one time just to get in front of the 3. And then we have to move it that many more times to get all of the zeroes in there so that we have to move it one time to get the 3. So if we started here, we're going to move 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 times. So this is going to be 3.457 times 10 to the negative 10 power. Let me just rewrite it. So 3.457 times 10 to the negative 10 power. So in general, what you want to do is you want to find the first non-zero number here. Remember, you want a number here that's between 1 and 10. And it can be equal to 1, but it has to be less than 10. 3.457 definitely fits that bill. It's between 1 and 10. And then you just want to count the leading zeroes between the decimal and that number and include the number because that tells you to actually get this number up here. And so we have to shift this decimal 10 times to the left to get this thing up here."
    },
    {
        "Q": "At 9:45, how does \u00e2\u0088\u009a10 . \u00e2\u0088\u009a10 = 10 in the denominator?\n",
        "A": "That equals 10 because at 9:04, he says that 1/radical 10 is irrational. If you wanted to get rid of the radical, you would have to multiply square root 10 over square root 10. 1x square root 10= 10. Square root 10x square root of 10=10 because 10 times 10 is 100. 100 is a perfect square so it simplifys back to 10 again. If my explaination dosen t help much, watch other videos on this topic.",
        "video_name": "BpBh8gvMifs",
        "timestamps": [
            585
        ],
        "3min_transcript": "recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1. It's being written as a fraction. This one up here was also rational. It can be written expressed as a fraction."
    },
    {
        "Q": "\nWhy in example D @ 4:37 the square of 2*2=2 and in example E @ 6:24 the square of 2*2=4? What makes the two identically appearing values different?",
        "A": "Because in D There was only one 2*2 and in E there are two: (2*2)*(2*2)*5*5*5 And we can simplify that as 4.",
        "video_name": "BpBh8gvMifs",
        "timestamps": [
            277,
            384
        ],
        "3min_transcript": "Square root of 20. Once again, 20 is 2 times 10, which is 2 times 5. So this is the same thing as the square root of 2 times 2, right, times 5. Now, the square root of 2 times 2, that's clearly just going to be 2. It's going to be the square root of this times square root of that. 2 times the square root of 5. And once again, you could probably do that in your head The square root of the 20 is 4 times 5. The square root of 4 is 2. You leave the 5 in the radical. So let's do part D. We have to do the square root of 200. Same process. Let's take the prime factors of it. So it's 2 times 100, which is 2 times 50, which is 2 times 25, which is 5 times 5. Let me scroll to the right a little bit. This is equal to the square root of 2 times 2 times 2 times 5 times 5. Well we have one perfect square there, and we have another perfect square there. So if I just want to write out all the steps, this would be the square root of 2 times 2 times the square root of 2 times the square root of 5 times 5. The square root of 2 times 2 is 2. The square root of 2 is just the square root of 2. The square root of 5 times 5, that's the square root of 25, that's just going to be 5. So you can rearrange these. 2 times 5 is 10. 10 square roots of 2. And once again, this it is irrational. You can't express it as a fraction with an integer and a numerator and the denominator. And if you were to actually try to express this number, it Well let's do part E. The square root of 2000. I'll do it down here. Part E, the square root of 2000. Same exact process that we've been doing so far. Let's do the prime factorization. That is 2 times 1000, which is 2 times 500, which is 2 times 250, which is 2 times 125, which is 5 times 25, which is 5 times 5. And we're done. So this is going to be equal to the square root of 2 times 2-- I'll put it in parentheses-- 2 times 2, times 2 times 2, times 2 times 2, times 5 times 5,"
    },
    {
        "Q": "At 8:36, how come the fraction isn't reduced?\n",
        "A": "Zoe.daniele, you can either reduce the fraction, or you can put it into decimal form. Either way, your answer will be correct (make sure to check with your teacher whether he or she wants you to use fractions or decimals in your answer). Hope that helps!",
        "video_name": "BpBh8gvMifs",
        "timestamps": [
            516
        ],
        "3min_transcript": "which is equal to 1/2. Which is clearly rational. It can be expressed as a fraction. So that's clearly rational. Part G is the square root of 9/4. Same logic. This is equal to the square root of 9 over the square root of 4, which is equal to 3/2. Let's do part H. The square root of 0.16. recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1."
    },
    {
        "Q": "\nat 6:30 why we can't write \"36 35 34 33 32 1 1 1 1\" instead of \"1 1 1 1 32 31 30 29 28\"",
        "A": "Because knowing you have all the 1s, you can t have the whole 36 options. The 1s are part of the 36 unique cards, so you can t pick one of them and then add them again, that would make it five 1s. So once you locked the 1s to be in your hand, you only have 32 cards left as the first choice for the non-1s card.",
        "video_name": "ccrYD6iX_SY",
        "timestamps": [
            390
        ],
        "3min_transcript": "So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards. One, two, three, four. But the other five cards are going to be different. So one, two, three, four, five. So for the other five cards-- if you imagine this slot-- considering that of the 36 we would have to pick four of them already in order for us to have four 1's. Well, we've used up four of them, so there's 32 possible cards over in that position of the hand. And then there'd be 31 in that position of the hand. And then there'd be 30 because every time we're picking a card, were using it up. And now we only have 30 to pick from. Then we only have 29 to pick from. And then we have 28 to pick from. And just like we did before, we don't care about order. We don't care if we pick the 5 of clubs first or whether we pick the 5 of clubs last. So we shouldn't double count it. five cards can be arranged. So we have to divide this by the different ways that five cards can be arranged. The first card or the first position can be any one of five cards, then four cards, then three cards, then two cards, then one cards. So the number of hands with four 1's is actually just this number. You're actually looking at all of the different ways you can fill up the remaining cards. These four 1's are just going to be four 1's. There's only one way to get that if the remaining cards that's going to give all of the different combinations of having four 1's. So this will be a count of all of the different combinations because all of the different extra stuff that you have will be all of the different hands. Now we know the total number of hands with four 1's is this number. And now we can divide it by the total number of possible hands. And I didn't multiply them out on purpose so that we can cancel things out. So let's do that. Let's take this and divide by that. So let me just copy and paste it."
    },
    {
        "Q": "In 0 4:04 Sal tells about a factorial. What is the definition of a factorial\n",
        "A": "In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5! is 5*4*3*2*1. 76! would be 76*75*74*73*72*71*...*3*2*1. They are essential in probability.",
        "video_name": "ccrYD6iX_SY",
        "timestamps": [
            244
        ],
        "3min_transcript": "number of hands. Now let's figure out the total number of hands first, because on some level this might be more intuitive and we've actually done this before. Now, the total number of hands, we're picking nine cards. And we're picking them from a set of 36 unique cards. And we've done this many, many times. Let me write this, total number of hands, or total number of possible hands. That's equal to-- you can imagine, you have nine cards to pick from. The first card you pick, it's going to be 1 of 36 cards. Then the next one is going to be 1 of 35. Then the next one is going to be 1 of 34, 33, 32, 31. We're going to do this nine times, one, two, three, four, five, six, seven, eight, and nine. So that would be the total number of hands if order mattered. But we know-- and we've gone over this before-- that we don't care about the order. So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards."
    },
    {
        "Q": "\nat 03:55 , why does it equal to 36!/(36-9)!9! ??",
        "A": "Before that in the numerator we have 36*35*...*28, and 9! in the denominator. 36! is 36*35*....28*27...*2*1. So in writing 36! as the new numerator, we have to multiply by 27*26...*2*1 both the numerator and denominator. (36-9)! = 27!, which is exactly that factor.",
        "video_name": "ccrYD6iX_SY",
        "timestamps": [
            235
        ],
        "3min_transcript": "number of hands. Now let's figure out the total number of hands first, because on some level this might be more intuitive and we've actually done this before. Now, the total number of hands, we're picking nine cards. And we're picking them from a set of 36 unique cards. And we've done this many, many times. Let me write this, total number of hands, or total number of possible hands. That's equal to-- you can imagine, you have nine cards to pick from. The first card you pick, it's going to be 1 of 36 cards. Then the next one is going to be 1 of 35. Then the next one is going to be 1 of 34, 33, 32, 31. We're going to do this nine times, one, two, three, four, five, six, seven, eight, and nine. So that would be the total number of hands if order mattered. But we know-- and we've gone over this before-- that we don't care about the order. So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards."
    },
    {
        "Q": "From minute 5:07 to 5:53, why does it have to be 1x1x1x1x32x31x30x29x28 instead of 36x35x34x33x32x1x1x1x1? Obviously the end result will be different but I'm asking why or what is the rule, idea, concept or whatever that says to go with the former rather than the latter.\n",
        "A": "Even if you used your logic, you are still saying that the 4 ones are drawn at the end, so the first 5 cards could not be ones, so it is still 32*31*30*29*28*1*1*1*1.",
        "video_name": "ccrYD6iX_SY",
        "timestamps": [
            307,
            353
        ],
        "3min_transcript": "So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards. One, two, three, four. But the other five cards are going to be different. So one, two, three, four, five. So for the other five cards-- if you imagine this slot-- considering that of the 36 we would have to pick four of them already in order for us to have four 1's. Well, we've used up four of them, so there's 32 possible cards over in that position of the hand. And then there'd be 31 in that position of the hand. And then there'd be 30 because every time we're picking a card, were using it up. And now we only have 30 to pick from. Then we only have 29 to pick from. And then we have 28 to pick from. And just like we did before, we don't care about order. We don't care if we pick the 5 of clubs first or whether we pick the 5 of clubs last. So we shouldn't double count it."
    },
    {
        "Q": "\nat 1:26, i think you meant to write DCE = BAE. (You wrote DEC = BAE)",
        "A": "Indeed, it should be DCE = BAE.",
        "video_name": "TErJ-Yr67BI",
        "timestamps": [
            86
        ],
        "3min_transcript": "So we have a parallelogram right over here. And what I want to prove is that its diagonals bisect each other. So the first thing that we can think about-- these aren't just diagonals. These are lines that are intersecting, parallel lines. So you can also view them as transversals. And if we focus on DB right over here, we see that it intersects DC and AB. And since we know that they're parallel-- this is a parallelogram-- we know the alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Now, if we look at diagonal AC-- or we should call it transversal AC-- we can make the same argument. These two lines are parallel. So alternate interior angles must be congruent. So angle DEC must be-- so let me write this down-- angle DEC must be congruent to angle BAE, for the exact same reason. Now we have something interesting, if we look at this top triangle over here and this bottom triangle. We have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know-- and we proved this to ourselves in the previous video-- that parallelograms-- not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent, we have a side in between of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle-side-angle. So we know that triangle-- I'm going to go from the blue to the orange to the last one-- triangle ABE is congruent to triangle-- blue, orange, then the last one-- CDE, by angle-side-angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. They're corresponding sides of congruent triangles,"
    },
    {
        "Q": "I really don't get it. At 1:42, why do you have to do f(x) times g(x)^-1?\n",
        "A": "Dividing by a number is the same as multiplying by the inverse of that number, which you can get by setting the exponent equal to -1 : Thus, the inverse of B is B^-1 since B^-1 = 1/B^1 = 1/B. Therefore, A/B = A*B^-1. That s why f(x)/g(x) = f(x)*g(x)^-1.",
        "video_name": "ho87DN9wO70",
        "timestamps": [
            102
        ],
        "3min_transcript": "We already know that the product rule tells us that if we have the product of two functions-- so let's say f of x and g of x-- and we want to take the derivative of this business, that this is just going to be equal to the derivative of the first function, f prime of x, times the second function, times g of x, plus the first function, so not even taking its derivative, so plus f of x times the derivative of the second function. So two terms, in each term we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit rule. And I frankly always forget the quotient rule, and I just rederive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business, the derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of-- instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here-- so it's going to be f prime of x-- times just the second function, which is just which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of view as something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your math book. So let's see if we can simplify this a little bit. All of this is going to be equal to-- we can write this term right over here as f prime of x over g of x."
    },
    {
        "Q": "\nAt 3:54 Sal says f(x)=1/(2+2cos(theta)), but it should be f(theta)=\u00e2\u0080\u00a6.\nIt's not a function of x, but it is a function of theta.",
        "A": "Yup You got it right Sal Made A Mistake, it should be f(theta).",
        "video_name": "d8qtbGMB2gI",
        "timestamps": [
            234
        ],
        "3min_transcript": "zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta and we know from the Pythagorean, Pythagorean Identity in Trigonometry, it comes straight out of the unit circle definition of sine and cosine. We know that, we know that sine squared theta plus cosine squared theta is equal to one or, we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is one minus cosine theta. This is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, as a difference of squares. If you view this as, if you view this as A squared minus B squared, we know that this can be factored as A plus B times A minus B. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have one minus cosine theta in the numerator and I have a one minus cosine theta in the denominator. Now we might be tempted to say, \"Oh, let's just cross that out with that \"and we would get, we would simplify it \"and get F of X is equal to one over \"and we could distribute this two now.\" We could say, \"Two plus two cosine theta.\" \"Well, aren't these the same thing?\" And we would be almost right, because F of X, this one right over here, this, this is defined this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this F of X or in order to be, for this to be the same thing, we have to say, theta cannot be equal to zero. But now let's think about the limit again. Essentially, what we want to do is we want to find the limit as theta approaches zero of F of X. And we can't just do direct substitution into, if we do, if we really take this seriously, 'cause we're gonna like, \"Oh well, if I try to put zero here, \"it says theta cannot be equal to zero \"F of X is not defined at zero.\" This expression is defined at zero but this tells me, \"Well, I really shouldn't apply zero to this function.\" But we know that if we can find another function"
    },
    {
        "Q": "2:12 to 2:23 is so unclear, why do they take out the 2s and 3s? How did they get the numbers? I took the test and did it another way...\n",
        "A": "that is how it goes",
        "video_name": "-UagBvxCReA",
        "timestamps": [
            132,
            143
        ],
        "3min_transcript": "All right. What we've got here are 12 pirates. They're going to divide out a treasure chest of gold. And here's how they're going to do it. First pirate's going to come along, take 1/12 of the gold that's in the chest. Second pirate's going to come along, take 2/12 of whatever's left after the first pirate is Third pirate's going to take 3/12 of whatever's left after the second pirate finished, and on, and on, and on. Let's see what happens here. Each pirate gets a positive whole number of coins. And the number of coins that was in the chest is the smallest number of coins for which it's possible for each pirate to get, a positive number of coins, a positive whole number of coins using this process. We're going to start with x because x marks the spot. x is the number of coins that was in the chest at the beginning. And the first pirate comes along, takes 1/12. That leaves 11/12 remaining for the next pirate who And the next pirate, second pirate takes 2/12 and leaves 10/12 of what was there So she gets there and there's this much. She's going to leave 10/12 of this amount for the next pirate. The next pirate comes along, takes 3/12, leaves 9/12 of this for the following pirate, and on, and on, and on we go until we get to the last few pirates. The 11th pirate takes 11/12, leaves 1/12 of what was there for the last pirate, who comes along and takes everything that's left. Well, that's what we want to figure out. How much does the last pirate receive? So we want to figure out what the value of this expression We can write this a lot shorter as x times 11 factorial over 12 to the 11th. And we want to figure out what this is. Now, x is the smallest value that makes this an integer. Actually, x is the smallest value that makes sure each pirate gets an integer number of coins. I'm not going to worry about that right now. I'm just going to worry about the last pirate and figure what the last pirate gets. well, this will just come out to be an integer. But 11 factorial is not any of these choices. So we can simplify this fraction. We can take out all the factors of 2 and 3 from this 11 factorial and see what's left. We're going to be left with a factor of 11. And then we're going to have two 5's from the 5 and the 10. And we're going to have a 7 sitting in there. And then we need to figure out well, we're going to simplify this fraction, take out factors of 2. We could stop right here, just compute this and call that the answer. But I'm a little bothered by that whole every pirate has to get an integer number of coins thing. But let's go ahead and simplify this fraction. The number of 2's in 11 factorial, the number of factors of 2, you get 2, 4, 6, 8, 10. That's 5. You get an extra one from the 4, two extra ones from the 8. 8 factors of 2 up here are 22 down there. That leaves us 2 to the 14th. And then the factors of 3, you have 3, 6, and 9 up there. You get an extra factor of 3 in the 9."
    },
    {
        "Q": "\nAt 1:30, how did he get x((4x)2",
        "A": "You mean, how he got to x( (4x)^2 + 2*4*3 + 3^2) ?",
        "video_name": "BFW2lHobO4E",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] So let's say that we've got the polynomial 16x to the third plus 24x squared plus nine x. Now what I'd like you to do is pause the video and see if you could factor this polynomial completely. Now let's work through it together. So the first thing that you might notice is that all of the terms are divisible by x so we can actually factor out an x. So let's do that, and actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is gonna be x times. When you factor out an x from 16x to the third you're gonna be left with 16x squared, and then plus 24x and then plus nine. Now this is starting to look interesting so let me just rewrite it. It's gonna be x times. This part over here looks interesting this looks like a perfect square. Let me write it out. 16x squared. That's the same thing as four x squared, and then we have a nine over there, which is clearly a perfect square. That is three squared. Three squared. And when we look at this 24x, we see that it is four times three times two, and so we can write it as, let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times three x. Now why did I take the trouble, why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that Ax plus B, and you were to square it, you're going to get Ax squared plus two ABx plus B squared, and we have that form right over here. This is the Ax squared. Let me do the same color. The Ax squared. Ax squared. We have the B squared. You have the B squared. And then you have the two ABx. Two ABx right over there. So this section, this entire section, we can rewrite as being we know it, what A and B are. A is four and B is three so this is going to be Ax, so four x plus four x plus b, which we know to be three. That whole thing, that whole thing squared, and now we"
    },
    {
        "Q": "\nWould using a different trig identity work @4:00. For instance, we could use the \"Product-to-Sum\" trig I.D of Sin(t-Tau)Cos(Tau). Then we could just split the integrals and it would make life a bit nicer right?\nTrig I.D.: sin(u)cos(v) = sin(u + v) + sin(u \u00e2\u0088\u0092 v)",
        "A": "Yes, you can use that identity, and the integration does indeed is simpler. But that identity is not so easy to memorise, Sal decided to use only the most basic of identities.",
        "video_name": "IW4Reburjpc",
        "timestamps": [
            240
        ],
        "3min_transcript": "one of these things? And to kind of give you that comfort, let's actually compute a convolution. Actually, it was hard to find some functions that are very easy to analytically compute, and you're going to find that we're going to go into a lot of trig identities to actually compute this. But if I say that f of t, if I define f of t to be equal to the sine of t, and I define cosine of t-- let me do it in orange-- or I define g of t to be equal to the cosine of t. Now let's convolute the two functions. So the convolution of f with g, and this is going to be a function of t, it equals this. I'm just going to show you how to apply this integral. integral from 0 to t of f of t minus tau. This is my f of t. So it's is going to be sine of t minus tau times g of tau. Well, this is my g of t, so g of tau is cosine of tau, cosine of tau d tau. So that's the integral, and now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So let's evaluate this. But I wanted to evaluate this in this video because I want to show you that this isn't some abstract thing, that you can actually evaluate these functions. So the first thing I want to do-- I mean, I don't know what the antiderivative of this is. It's tempting, you see a sine and a cosine, maybe they're the derivatives of each other, but this is the sine of t minus tau. So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is the sine of tau times the cosine of t. And actually, I just made a video where I go through all of these trig identities really just to review them for myself and actually to make a video in better quality on them as well. So if we make this subsitution, this you'll find on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to-- I'll just write that f-star g; I'll just write it with that-- is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this thing right there. So I'm going to write the sine of t times the cosine of tau minus the sine of tau times the cosine of t, and then all"
    },
    {
        "Q": "in the first example at 0:42, what where the m's and the b's?\n",
        "A": "Here, m and b are constants; in a linear equation, m is traditionally used to represent slope, and b is used to represent the y-intercept. In other words, they re just the good old slope-intercept form pieces from algebra.",
        "video_name": "zid7J4EhZN8",
        "timestamps": [
            42
        ],
        "3min_transcript": "- So let's get a little bit more comfort in our understanding of what a differential equation even is. So here we have a differential equation. We haven't started exploring how we find the solutions for a differential equations yet. But let's just say you saw this, and someone just walked up to you on the street and says, \"Hey, I will give you a clue, \"that there's a solution to this differential equation \"that is essentially a linear function, \"where y is equal to mx plus b, \"and you just need to figure out \"the m's and the b's, or maybe the m and the b \"that makes this linear function \"satisfy this differential equation.\" What I now encourage you to do, is pause the video and see if you can do it. So I'm assuming you have paused it, and had a go at it. So let's think this through together. If we know that this kind of a solution can be described in this way, we need to figure out some m's and b's here. This is telling us that if we were to take the derivative of this with respect to x, if we take the derivative of mx plus b with respect to x, that that should be equal to negative 2 times x well, we know y is this thing, minus 5. And that should be true for all x's, in order for this to be a solution to this differential equation. Remember, the solution to a differential equation is not a value or a set of values. It is a function or a set of functions. So in order for this to satisfy this differential equation, it needs to be true for all of these x's here. So let's work through it. Let's figure out first what our dy dx is. So dy dx. We'll just take the derivative here with respect to x dy dx is derivative of mx with respect to x, is just going to be m. And of course derivative of b with respect to x, just a constant, so it's just going to be zero. So dy dx is m. So we could write m is equal to negative 2x, is equal to negative 2x, plus 3 times, and instead of putting y there I could write mx plus b. Remember y is equal to mx plus b. this has to be true for all x's. mx plus b, and then of course we have the minus 5. So if you weren't able to solve it the first time, I encourage you to start from here, and now figure out what m and b needs to be in order for this equation right over here, in order for this to be true for all x's. In order for this to be true for all x's. So assuming you have paused again and had a go at it, let's just keep algebraically manipulating this. I'll just switch to one color here. So we have m. m is equal to negative 2x plus, if we distribute this 3 we're going to have 3mx plus 3b, and then of course we're going to have minus 5. And now we can group the x terms. So if we were to group, if we were to group ... Let me find a new color here, maybe this blue. So if we were to take these two and add them together"
    },
    {
        "Q": "\nat 3:14, can you explain the use of the chain rule that results in 2y * y'? thank you.",
        "A": "The derivative of y^2 is 2y * (dy/dx). (dy/dx) = y so then 2y * (dy/dx) = 2y * y",
        "video_name": "ZtI94pI4Uzc",
        "timestamps": [
            194
        ],
        "3min_transcript": "with different notation. So let's take the derivative of this thing right over here. Well we're going to apply the chain rule. Actually, we're going to apply the chain rule multiple times here. The derivative of e to the something with respect to that something is going to be e to the something times the derivative of that something with respect to x. So times the derivative of xy squared. So that's our left-hand side. We aren't done taking the derivative yet. And on our right-hand side, the derivative of x is just 1. And the derivative with respect to x of y is just going to be minus-- or I could write-- negative dy dx. But instead of writing dy dx, I'm going to write y prime. As you can tell, I like this notation and this notation more because it makes it explicit Here, we just have to assume that we're taking the derivative with respect to x. Here, we have to assume that's the derivative of y with respect to x. But anyway let's stick with this notation right over here. Actually, let me make all of my y primes, all my derivatives of y with respect to x, let me make them pink so I keep track of them. So once again, this is going to be equal to e to the xy squared times the derivative of this. Well the derivative of this, we can just use the product and actually a little bit of the chain rule So the derivative of x is just 1 times the second function. So it's going to be times y squared. And then to that, we're going to add the product of the first function which is this x times the derivative of y squared with respect to x. Well that's going to be the derivative of y squared times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect"
    },
    {
        "Q": "At 6:09, Sal describes taking the integral at a specific value c. What would happen if you took the derivate of the Dirac delta function combined with a certain function at value c and then applied the Laplace transform?\n",
        "A": "The derivative of the Dirac delta function is undefined at c, so it would not work very well.",
        "video_name": "vhfjEpQWWeE",
        "timestamps": [
            369
        ],
        "3min_transcript": "Let me draw this, what we're trying to do. So let me draw what we're trying to take the integral of. And we only care from zero to infinity, so I'll only do it from zero to infinity. And I'll assume that c is greater than zero, that the delta function pops up someplace in the positive t-axis. So what is this first part going to look like? What is that going to look like? e to the minus st times f of t? I don't know. It's going to be some function. e to the minus st starts at 1 and drops down, but we're multiplying it times some arbitrary function, so I'll just draw it like this. Maybe it looks something like this. This right here is e to the minus st times f of t. And the f of t is what kind of gives it its arbitrary shape. Fair enough. Now, let's graph our Dirac delta function. With zero everywhere except right at c, right at c right there, it pops up infinitely high, but we only draw an I mean, normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care about the area under this whole thing. When we multiply these two functions, when we multiply this times this times the delta function, this is-- let This is the delta function shifted to c. If I multiply that times that, what do I get? This is kind of the key intuition here. Let me redraw my axes. Let me see if I can do it a little bit straighter. Don't judge me by the straightness of my axes. So that's t. So what happens when I multiply these two? Everywhere, when t equals anything other than c, the Dirac delta function is zero. So it's zero times anything. I don't care what this function is going to do, it's going to be zero. So it's going to be zero everywhere, except something At t equals c, what's the value of the function? Well, it's going to be the value of the Dirac delta function. It's going to be the Dirac delta function times whatever height this is. This is going to be this point right here or this right there, that point. This is going to be this function evaluated at c. I'll mark it right here on the y-axis, or on the f of t, whatever you want to call it. This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c, so that's the point right there. So if you take this point, which is just some number, it could be 5, 5 times this, you're just getting 5 times the Dirac delta function. Or in this case, it's not 5. It's this little more abstract thing. I could just draw it like this. When I multiply this thing times my little delta function"
    },
    {
        "Q": "In the first equation, Sal simplifies the equation to: (7x - 2) / (15 - 5/x). He states at 2:32 that 7x will approach negative infinity, however he simplified this value from 7x^2 which by definition will always turn negative numbers positive. Shouldn't he have simplified the equation to state: 7|x|?\n",
        "A": "Sal simplified the expression down to it s dominate terms: 7x/15. If you are going to use the 7x^2, then you also need to use the 15x in the denominator. Yes, 7x^2 would be positive, but 15x would be negative. A positive / a negative = a negative. Hope this helps.",
        "video_name": "Vtcmyr5IGYY",
        "timestamps": [
            152
        ],
        "3min_transcript": "by one over x, or another way of thinking about it is we're dividing both the numerator and the denominator by x. And if we're doing the same thing to the numerator and the denominator, if we're multiplying or dividing them by the same value, I should say, well then, I'm just really just multiplying it by one. So, I'm not changing its value. This will make it a little bit more interesting, and a little bit easier for us to think about what happens when x becomes very, very, very negative. So, 7x-squared divided by x, or being multiplied by one over x, is going to be 7x. 2x times one over x, or 2x divided by x, is just two. And then all of that over 15x divided by x, or 15x over x, is just going to be 15. And then you have five over x. Five times one over x is equal to five over x. Minus five over x. Now, this is equivalent, for our purposes, to think about what happens when x gets very, very, very, very negative. Well, when x gets very, very, very, very, very, very, very negative, this is going to become a very large negative number. You subtract two from it, it really won't matter much. You divide that by 15, well, that's not gonna matter much. And this is just going to become very, very, very small. You're taking five and you're dividing it by ever-larger negative numbers, or more and more negative numbers. So, this right over here is gonna go to zero. This thing over here is gonna go towards infinity. Or, I should say, it's gonna go towards negative infinity. Seven times a negative trillion, seven times a negative googol, seven times a negative googolplex, we're getting more and more negative numbers, this is gonna get, this is going to approach negative infinity. Doesn't matter that you're subtracting two from that. In fact, that'll get even more negative. And it doesn't matter if you then divide that by 15, you're still approaching negative infinity. If you had a arbitrarily negative number, an arbitrarily negative number. And, so, you could say that this is going to go to negative infinity. Now, another way that you could've thought about it. This is actually how I do think about it when I'm trying to, when I see these types of problems. I say, well which terms in the numerator and the denominator are going to dominate? And what do I mean by \"dominate\"? Well, as x gets very positive or x gets very negative, another way to think about it is the magnitude of x gets large, the absolute value of x gets large. The higher degree terms are going to grow much faster than the lesser degree terms. And so, we could say that for large x, for large x, and when I say \"large\" I mean high absolute value. High absolute value. And if we're going to negative infinity, that's high absolute value. So, f of x is going to be approximately equal to"
    },
    {
        "Q": "\nAt 2:00, why would I need to see if the sides are parallel to each other?",
        "A": "To know if any of the shapes are the right answer. And to know for sure that what you think is right",
        "video_name": "vsgrWDLEzcQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "Classify quadrilateral ABCD. Choose the option that best suits the quadrilateral. We're going to pick whether it's a square, rhombus, rectangle, parallelogram, trapezoid, none of the above. And I'm assuming we're going to pick the most specific one possible, because obviously all squares are rhombuses, or rhombi, I guess you'd say. Not all rhombi are squares. All squares are also rectangles. All squares, rhombi, and rectangles are parallelograms. So we want to be as specific as possible in picking this. So let's see, point A is at 1 comma 6. I encourage you to pause this video and actually try this on your own before seeing how I do it, but I'll just proceed. So that's point A right over there. Point B is at negative 5 comma 2. That's point B. Point C is at-- just had some carbonated water, so some air is coming up. Point C is at negative 7 comma 8. And then finally, point D is at 2 comma 11. And actually that kind of goes off the screen. This is 10, 11 would be right like this. So that would be 2 comma 11. If we were to extend this, this is 10 and this is 11 right up here. 2 comma 11. So let's see what this quadrilateral looks like. You have this line right over here, this line right over there, that line right over there, and then you have this line like this, and then you have this like this. So right off the bat, well it's definitely a quadrilateral. I have four sides. But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. Now it also looks like CD is not parallel to BA, but maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope of AB, or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5 comma 2, and we're ending at the point 1 comma 6. So what's our change in y? Our change in y, we go from 2-- we're going from 2 all the way to 6."
    },
    {
        "Q": "\nAt 7:17, how does Sal choose the starting point? Does it matter? Could it be the other point?",
        "A": "The starting/ending points could certainly be any as long as they are on the line c. I think these were chosen because they both had integer values for x and y and thus made the computation a little easier.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            437
        ],
        "3min_transcript": "So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to-- how did I know to do the 0 first and the negative 6 second and then the 1 first and then the negative 2 second. And the answer is you could've done it in either order as long as you keep them straight. So you could have also have done change in y over change in x. We could have said, it's equal to negative 2 minus 1. So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is a negative of that. That is the negative of that. But since we have a negative over negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out. This is also equal to 1/2. So the important thing is if you use this y-coordinate x-coordinate first as well. If you use this y-coordinate first, as we did here, then you have to use this x-coordinate first, as you did there. You just have to make sure that your change in x and change in y are-- you're using the same final and starting points. Just to interpret this, this is saying that for every minus 6 we go in x. So if we go minus 6 in x, so that's going backwards, we're going to go minus 3 in y. But they're essentially saying the same thing. The slope of this line is 1/2. Which tells us for every 2 we travel in x, we go up 1 in y. Or if we go back 2 in x, we go down 1 in y. That's what 1/2 slope tells us. Notice, the line with the 1/2 slope, it is less steep than the line with a slope of 3. Let's do line c right here. I'll do it in pink. Let's say that the starting point-- I'm just picking this arbitrarily. Well, I'm using these points that they've drawn here. The starting point is at the coordinate negative 1, 6 and that my finishing point is at the point 5, negative 6. Our slope is going to be-- let me write this-- slope is going to be equal to change in x-- sorry, change in y. I'll never forget that. Change in y over change in x. Sometimes it's said rise over run. Run is how much you're moving in the horizontal direction. Rise is how much you're moving in the vertical direction. Then we could say our change in y is our finishing y-point minus our starting y-point. This is our finishing y-point."
    },
    {
        "Q": "At 2:27,Can it still be -12 over -4? (-12/-4)\n",
        "A": "yes it can be (-12/-4)",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            147
        ],
        "3min_transcript": "In this video I'm going to do a bunch of example slope problems. Just as a bit of review, slope is just a way of measuring the inclination of a line. And the definition-- we're going to hopefully get a good working knowledge of it in this video-- the definition of it is a change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here. Line a. Let's figure out its slope. They've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3. Its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1 and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. We could say change in y-- so slope is change in y over change in x. We can figure it out numerically. I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6 and it went all the way up to positive 6. So what's this distance right here? It's going to be your end point y value. It's going to be 6 minus your starting point y value. Minus negative 6 or 6 plus 6, which is equal to 12. You say one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So when we changed our y value by 12, we had to change our x same change in y? Well we went from x is equal to negative 1 to x is equal to 3. Right? x went from negative 1 to 3. So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12/4 or if we want to write this in simplest form, this is the same thing as 3. Now the interpretation of this means that for every 1 we move over-- we could view this, let me write it this way. Change in y over change in x is equal to-- we could say it's 3 or we could say it's 3/1. Which tells us that for every 1 we move in the positive x-direction, we're going to move up 3 because this is a positive 3 in the y-direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. When we moved 1 in the x, we moved up 3 in the y."
    },
    {
        "Q": "At 2:29 isn't it suppose to be y2-y1 over x2-x1?\n",
        "A": "Yes . . . and that s what Sal did. y2 is 6, and y1 is -6, so we get y2 - y1 = 6 - (-6) = 12 For the change in x Sal basically just counted the number of squares horizontally between the two points, but that s the same as finding x2 - x1.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            149
        ],
        "3min_transcript": "In this video I'm going to do a bunch of example slope problems. Just as a bit of review, slope is just a way of measuring the inclination of a line. And the definition-- we're going to hopefully get a good working knowledge of it in this video-- the definition of it is a change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here. Line a. Let's figure out its slope. They've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3. Its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1 and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. We could say change in y-- so slope is change in y over change in x. We can figure it out numerically. I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6 and it went all the way up to positive 6. So what's this distance right here? It's going to be your end point y value. It's going to be 6 minus your starting point y value. Minus negative 6 or 6 plus 6, which is equal to 12. You say one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So when we changed our y value by 12, we had to change our x same change in y? Well we went from x is equal to negative 1 to x is equal to 3. Right? x went from negative 1 to 3. So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12/4 or if we want to write this in simplest form, this is the same thing as 3. Now the interpretation of this means that for every 1 we move over-- we could view this, let me write it this way. Change in y over change in x is equal to-- we could say it's 3 or we could say it's 3/1. Which tells us that for every 1 we move in the positive x-direction, we're going to move up 3 because this is a positive 3 in the y-direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. When we moved 1 in the x, we moved up 3 in the y."
    },
    {
        "Q": "At 4:39, Sal didn't label the x values as x1 and x2, and the y-values as y1 and y2. Why is that?\n",
        "A": "You don t need to label the points as x1, x2, y1 and y2. You just need to know which is which.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            279
        ],
        "3min_transcript": "move 6 in the y. 6/2 is the same thing as 3. So this 3 tells us how quickly do we go up as we increase x. Let's do the same thing for the second line on this graph. Graph b. Same idea. I'm going to use the points that they gave us. But really you could use any points on that line. So let's see, we have one point here, which is the point 0, 1. You have 0, 1. And then the starting point-- we could call this the finish point-- the starting point right here, we could view it as x is negative 6 and y is negative 2. So same idea. What is the change in y given some change in x? So let's do the change in x first. So what is our change in x? So in this situation, what is our change in x? delta x. It's one, two, three, four, five, six. It's going to be 6. But if you didn't have a graph to count from, you could literally take your finishing x-position, so it's 0, and subtract from that your starting x-position. 0 minus negative 6. So when your change in x is equal to-- so this will be 6-- what is our change in y? Remember we're taking this as our finishing position. This is our starting position. So we took 0 minus negative 6. So then on the y, we have to do 1 minus negative 2. What's 1 minus negative 2? That's the same thing as 1 plus 2. That is equal to 3. So it is 3/6 or 1/2. So notice, when we moved in the x-direction by 6, we moved So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to-- how did I know to do the 0 first and the negative 6 second and then the 1 first and then the negative 2 second. And the answer is you could've done it in either order as long as you keep them straight. So you could have also have done change in y over change in x. We could have said, it's equal to negative 2 minus 1. So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is a negative of that. That is the negative of that. But since we have a negative over negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out. This is also equal to 1/2. So the important thing is if you use this y-coordinate"
    },
    {
        "Q": "In the end of 10:00 , we can see, slope of one line is -2 (negative two) and another is +1. So which one's rate of change will be higher ?\n",
        "A": "The higher rate of change should be -2 as the rate of change deals with the magnitude of the slope.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            600
        ],
        "3min_transcript": "minus our starting x-point. If that confuses you, all I'm saying is, it's going to be equal to our finishing y-point is negative 6 minus our starting y-point, which is 6, over our finishing x-point, which is 5, minus our starting x-point, which is negative 1. So this is equal to negative 6 minus 6 is negative 12. 5 minus negative 1. That is 6. So negative 12/6. That's the same thing as negative 2. Notice we have a negative slope here. That's because every time we increase x by 1, we go down in the y-direction. So this is a downward sloping line. It's going from the top left to the bottom right. As x increases, the y decreases. And that's why we got a negative slope. This line over here should have a positive slope. So I'll use the same points that they use right over there. So this is line d. Slope is equal to rise over run. How much do we rise when we go from that point to that point? We could do it this way. We are rising-- I could just count it out. We are rising one, two, three, four, five, six. We are rising 6. How much are we running? We are running-- I'll do it in a different color. We're running one, two, three, four, five, six. We're running 6. So our slope is 6/6, which is 1. Which tells us that every time we move 1 in the x-direction-- positive 1 in the x-direction-- we go positive 1 in the y-direction. For every x, if we go negative 2 in the x-direction, we're going to go negative 2 in the y-direction. y in this slope. Notice, that was pretty easy. If we wanted to do it mathematically, we could figure out this coordinate right there. That we could view as our starting position. Our starting position is negative 2, negative 4. Our finishing position is 4, 2. So our slope, change in y over change in x. I'll take this point 2 minus negative 4 over 4 minus negative 2. 2 minus negative 4 is 6. Remember that was just this distance right there. Then 4 minus negative 2, that's also 6. That's that distance right there. We get a slope of 1. Let's do another one."
    },
    {
        "Q": "\nAt 2:11, isn't it a change of -4 because y=-4?",
        "A": "y=-4 is like y= 0x -4. In order for the x to be gone you would need a zero at its coefficient. since the coefficient of the x is the slope then the slope is zero.",
        "video_name": "J43CIbKpdWc",
        "timestamps": [
            131
        ],
        "3min_transcript": "- [Instructor] What is the equation of the horizontal line through the point negative four comma six? So let's just visualize this. Once you get the hang of it, you might not have to draw a graph, but for explanatory purposes, it might be useful. So negative four comma six, that's going to be in the second quadrant. So if this is my That is my y-axis. I'm going to go negative four in the x direction. So one, two, three, four. Negative four. And then one, three, four, five, six, in the y direction. So the point that we care about is going to be right over there. Negative four comma six. And what is the equation of the horizontal line? It is a horizontal line. So it's just going to go straight left, right like this. That is what the line would actually look like. So what is that equation? Well, for any x, y is going to be equal to six. y is equal to six. Doesn't matter what x you input here, you're gonna get y equals six. It just stays constant right over there. So the equation is y is equal to six. Let's do another one of these. So here we are asked what is the slope of the line y is equal to negative four? So let's visualize it and then in the future, you might not have to draw it like this. But let's just draw our axis again. X-axis y-axis and the slope of line y equals negative four. So for whatever x you have, y is going to be negative four. Let's say that's negative four right over there. And so, the line is y the line is y equals negative four. So I can draw it like this. So what's the slope of that? Well, slope is change in y for given change in x. y doesn't change. It stays at negative four. My change in y over change in x. Doesn't matter what my change in x is. My change in y is always going to be zero. It's constant. So the slope here is going to be equal to zero. Y doesn't change, no matter how much you change x. Let's do another one of these. This is fun. So now they are asking us, what is the slope of the line x equals negative three? Let me graph that one. So, I'm just going to draw my axis real fast. X-axis y-axis X is equal to negative three. So negative one, negative two, negative three. And so, this line is going to look let me, it's going to look like this. No matter what y or you can say no matter what y is. X is going to be equal to negative three. So,"
    },
    {
        "Q": "\nAt 3:45 he points out that there's a new slowest time, and uses that to determine that the 4th bullet point is wrong. Does that new slowest time even matter though? Even if that new slowest dot wasn't there (ie, it had been at the 52 mark in the final round), does that change anything regarding the 4th bullet point question?",
        "A": "It does matter because, as pointed out, one swimmer did swim slower so the statement all swam faster cannot be true.",
        "video_name": "KXDOOmquZag",
        "timestamps": [
            225
        ],
        "3min_transcript": "So his time definitely got worse. And this is at 53.8 seconds. So let's look at the statements and see which of these apply. The swimmers had faster times, on average, in the finals. Is this true? Faster times on average in the finals? So if we look at the finals right over here, we could take each of these times, add them up, and then divide by 8, the number of times we have. But let's see if we can get an intuition for where this is, because we're really just comparing these two plots, or these two distributions, we could say. And so let's see, if all the data was these three points and these three points, we could intuit that the mean would be right around there. It would be around 53.2 or 53.3 seconds, right around there. And then we have this point and this point, if you just found the mean of that point and that point, would get you right around there. So the mean of those two points would bring down the mean a little bit. And once again, I'm not figuring out the exact number. But maybe it would be around 53.2, 53.1, or 53.2 seconds. So that's my intuition for the mean of the final round. And now let's think about the mean of the semifinal round. Let's just look at these bottom five dots. If you find their mean, you could intuit it would be maybe someplace around here, pretty close to around 53.3 seconds. And then you have all these other ones that are at 53.5 and 53.3, which will bring the mean even higher. So I think it's fair to say that the mean in the final around and the time is less than the mean up here. And you could calculate it yourself, but I'm just trying to look at the distributions and get an intuition here. And at least in this case, it looks pretty clear that the swimmers had faster times, on average, It took them less time. One of the swimmers was disqualified from the finals. Well, that's not true. We have 8 swimmers in the semifinal round. And we have 8 swimmers in the final round. So that one's not true. The times in the finals vary noticeably more than the times in the semifinals. That does look to be true. We see in the semifinals, a lot of the times were clumped up right around here at 53.3 seconds and 53.5 seconds. The high time isn't as high as this time. The low time isn't as low there. So the final round is definitely-- they vary noticeably more. Individually, the swimmers all swam faster in the finals than they did in the semifinals. Well, that's not true. Whoever this was, clearly they were one of these data points up here. This data point took more time than all of these data points. So this represents someone who took more time in the finals than they did in the semifinals. And we got it right."
    },
    {
        "Q": "At 4:35, I still don't get why -(-e)/f is equal to e/f.\n",
        "A": "When there are 2 negatives, it equals to a positive. When someone says Jump , it s a positive. When someone says Don t eat , it is negative. Meanwhile, if someone says Don t not eat , that s back to saying Eat which is a positive.",
        "video_name": "9eSPhvhuInw",
        "timestamps": [
            275
        ],
        "3min_transcript": "So this is going to be right. Now this one, negative divided by a negative, well that's just going to be positive. So that's the same thing as five over b. One way to think about it is that well the negatives kind of cancel each other out. So five over b, that looks good too. And of course I won't select none of the above because I found two choices that worked. All right, let's do one more. Which of the following expressions are equal to negative e over negative f? And remember we just have to take this step by step here. Actually let's try to just simplify this directly. So negative e over negative f. Well we just need to remind ourselves that this part right over here. Negative e over negative f. Let me write an equal sign. Negative e over, and I'm gonna put this negative. Let me do this in a different color. Let me do this in purple. So we have this purple. So we have that purple negative right over there. And negative e over negative f. That's the same thing as negative's divided by a negative is a positive. That's the same thing as e over f, as positive e over f. So this whole thing will simplify to negative e over f. So let's see which of these choices are that. Well this right here is positive e over f. So that's not the choice. This one over here. This one we could write it several ways actually. We could write it negative negative e over f. Which of course is equal to positive e over f. We could also write this. We could put the negative in the denominator. We could say that this thing. Actually let me write it over here as negative e over negative f. This is also a legitimate thing to do. You could take this negative and multiply it times the denominator. Right over here. But either way it's going to be equal to positive e over f. These two are actually evaluate to the same expression. So here, I would select. Finally, I would select. All right, hopefully that helps."
    },
    {
        "Q": "at 1:08 he says that the sin of 7pi/12 is the length of the magenta line however isn't that the sin of pi/2 and shouldn't that length be one considering it is a unit circle?\n",
        "A": "Sal has drawn a right triangle. The hypotenuse (the diagonal line, or the green line furthest to the left) is = 1 because it goes from the center of the circle out to the circle. The angle 7pi/12 goes up to the hypotenuse. You are trying to make the angle into 90 degrees (pi/2)rather than the 7pi/12. Next, it is hard to see, but the magenta line is slightly shorter than 1. This is why it doesn t = 1. Hope this helps.",
        "video_name": "2RbKfRfzD-M",
        "timestamps": [
            68
        ],
        "3min_transcript": "Voiceover:What I want to attempt to do in this video is figure out what the sine of seven pi over 12 is without using a calculator. And so let's just visualize seven pi over 12 in the unit circle. One side of the angle is going along the positive x-axis if we go straight up, that's pi over two, which is the same thing as six pi over 12, so then we essentially just have another pi over 12 to get right over there. This is the angle that we're talking about that is seven pi over 12 radians, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. This is the unit circle, has radius one where it intersects the unit. The y-coordinate is the sine. Another way to think about it, it's the length of this line right over here. I encourage you to pause the video right now and try to think about it on your own. See if you can use your powers of trigonometry to figure out what sine of seven pi over 12 is I'm assuming you've given a go at it, and if you're like me, your first temptation might have been just to focus on this triangle right over here that I drew for you. The triangle looks like this. It looks like this, where that's what you're trying to figure out, this length right over here, sine of seven pi over 12. We know the length of the hypotenuse is one. It's a radius of the unit circle. It's a right triangle right over there. We also know this angle right over here, which is this angle right over here, this gets us six pi over 12, and then we have another pi over 12, so we know that that is pi over 12, not pi over 16. We know that this angle right over here is pi over 12. Given this information, we can figure out this, or we can at least relate this side to this other side using a trig function relative to this angle. This is the adjacent side. this magenta side over one, or you could just say it's equal to this magenta side. You could say that this is cosine of pi over 12. We just figured out that sine of seven pi over 12 is the same thing as cosine of pi over 12, but that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. Instead of thinking about it this way, let's see if we can compose this angle or if we can decompose it into some angles for which we do know the sine and cosine. What angles are those? Those are the angles in special right triangles. For example, we are very familiar with 30-60-90 triangles. 30-60-90 triangles look something like this. This is my best attempt at hand drawing it. Instead of writing 30-degree side, since we're thinking in radians, I'll write that as pi over six radians. The 60-degree side, I'm going to write that as pi over three radians, and of course, this is the right angle."
    },
    {
        "Q": "At 6:16, why do you need to collect multiple data points to determine if a question is a statistical question?\n",
        "A": "Statistics is the comparison of data at different points. If you have a single data point then there is nothing to compare it to. So you need multiple points so you can compare the data, an thus be using statistics.",
        "video_name": "OjzfQDFf7Uk",
        "timestamps": [
            376
        ],
        "3min_transcript": "variability. So you're saying, okay that kinda makes sense, but I need to see some tangible questions or tangible examples of things that are statistical questions and things that are not statistical questions. I would say fair enough. Let's look at some examples. So here I have six questions, and I encourage you to pause this video right now. Before I work through it, think about it. Based on this definition of a statistical question, which of these questions are statistical, would require your statistical toolkit, and which of these are not statistical? So assuming you had a go at it, let's go through these one by one. So the first question, how much does my pet grapefruit weigh? You know, it's bizarre to begin with to have a pet grapefruit, but is this a statistical question? What do I need to do to answer it? I have to take my pet grapefruit out. I have to weigh it. Then I have to just write that down. Just doing that I am collecting data, to mess with statistics a little bit, but I'm just getting one data point. So I might weigh it and I might see my grapefruit weighs one pound, but that's not data with variability. That's just one data point. In order to have variability you have to have multiple data points and should be at least possible that they could vary. So, for example, all of these folks ate zero bricks but maybe it was possible that someone actually ate a brick. But here I have just one data point. With one data point, you can't have variability, so this is not a statistical question. I just collect a data point. Next question, what is the average number of cars in a parking lot on Monday mornings? To think about whether it is a statistical question, we just have to think about what do I have to do to answer that question? I would have to go out to the parking lot on multiple Monday mornings, and measure the number of cars. So on the first Monday morning I might see there are 50 cars. The next Monday morning I might go out there and count there's 49 cars. The next Monday morning I might see 63 cars. So I'm collecting multiple data points to answer this question. Then I'm going to take the average of all these, but I'm collecting multiple data points to answer this question. It's definitely possible that there could be variation here, that there could be variability, so this is a statistical question. Next question, am I hungry? It's an important question. We ask it to ourselves multiple times. In fact, sometimes our bodies just tell it to us. But I am definitely not collecting ... I guess you could say I'm collecting some type of feelings from my stomach or how weak I feel or not, but it's definitely not data with variability. I'm either hungry or not hungry on a given day. I mean if you said broader, how does my hunger change from day to day and you came up some type of a scale for rating your hunger, all right maybe that's more statistical. But just am I hungry, a yes-no question. This is not ..."
    },
    {
        "Q": "\nAt 3:31, Why does Sal multiply 6x5x4x3?",
        "A": "nPr is used because nPr counts all answer sets (including duplicate unordered sets, which is what we want since the duplicates have different meaning than the ordered version) theres 6 total items so n=6 we re picking 4 items so r=4 the formula is n!/(n-r)!=6!/2!=6*5*4*3",
        "video_name": "oQpKtm5TtxU",
        "timestamps": [
            211
        ],
        "3min_transcript": "different than green, red, yellow and blue. We're going to assume that these are not the same code. Even though we've picked the same 4 colors, we're going to assume that these are 2 different codes, and that makes sense because we're dealing with codes. So these are different codes. So this would count as 2 different codes right here, even though we've picked the same actual colors. The same 4 colors, we've picked them in different orders. Now, with that out of the way, let's think about how many different ways we can pick 4 colors. So let's say we have 4 slots here. 1 slot, 2 slot, 3 slot and 4 slots. And at first, we care only about, how many ways can we We haven't picked any colors yet. Well, we have 6 possible colors, 1, 2, 3, 4, 5, 6. So there's going to be 6 different possibilities for this slot right there. So let's put a 6 right there. Now, they told us that the colors cannot be repeated, so whatever color is in this slot, we're going to take it out of the possible colors. So now that we've taken that color out, how many possibilities are when we go to this slot, when we go to the next slot? How many possibilities when we go to the next slot right here? Well, we took 1 of the 6 out for the first slot, so there's only 5 possibilities here. And by the same logic when we go to the third slot, we've used up 2 of the slots-- 2 of the colors already, so there would only 4 possible colors left. And then for the last slot, we would've used up 3 of the colors, so there's only 3 possibilities left. So if we think about all of the possibilities, all of the permutations-- and permutations are when you think about all the possibilities and you do care about order; where you say that this is different than So all of the different permutations here, when you pick 4 colors out of a possible of 6 colors, it's going to be 6 possibilities for the first 1, times 5 for the second bucket, times 4 for the third or the third bucket of the third position, times 3. So 6 times 5 is 30, times 4 is times 3. So 30 times 12. So this is 30 times 12, which is equal to their 360 possible 4-color codes."
    },
    {
        "Q": "\nhow is 1 m = to 100 cm 02:57",
        "A": "Centi refers to 100. The root is the same as in century (100 years). A centimeter is one one-hundredth of a meter (1/100). So 100 centimeters are equal to 1 meter. Note that a decimeter is 1/10 of a meter. That word looks a little like decade, doesn t it?",
        "video_name": "byjmR7JBXKc",
        "timestamps": [
            177
        ],
        "3min_transcript": "that would be represented by 1 centimeter in the floor plan. And it goes the other way around. 1 centimeter on the floor plan would represent 80 centimeters in the house. And it's always important to do-- if this confuses you, just always do a reality check that the house should be bigger than the floor plan. So if the floor plan for this dimension of our living room is 4 centimeters, the actual house will be 80 times that. And 80 times 4 is 320-- let me do that in a blue color-- is equal to 320 centimeters. And we can do the same thing for the length of the living room. So 80 times 5 centimeters is going to get us to-- is going to be-- 80 times 5 is 400 centimeters. So we could figure out the area of this room in centimeters, if we like, and I guess, why not? It might be easier to convert it to meters later. Let me write this down. 400 times 320. Let's think about it. 4 times 32 is going to be 120, plus 8, 128. And I have 1, 2, 3 zeroes. 1, 2, 3. So it's going to be 128,000 centimeters squared. Now that's a lot of square centimeters. What would we do if we wanted to convert it into meters? Well, we just have to figure out how many square centimeters are there in a square meter. So let's think about it this way. A meter is equal to-- 1 meter is equal to 100 centimeters. So a square meter, so that's right over there. is the same thing as 100 centimeters by 100 centimeters. And so if you were to calculate this area in centimeters, 100 times 100 is 10,000, is equal to 10,000 centimeters squared. So you have 10,000 square centimeters for every square meter. And so, if you want to convert 128,000 centimeters squared to meters squared, you would divide by 10,000. So dividing that by 10,000 would give us 12.8 square meters. Now, another way you could've done it, and maybe this would have been easier, is to convert it up here. Instead of saying 400 centimeters times 320 centimeters, you would say, well,"
    },
    {
        "Q": "\nwhy does it say the problem reads 80:1 scale but should read 1:80 scale",
        "A": "They corrected it",
        "video_name": "byjmR7JBXKc",
        "timestamps": [
            4801,
            140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "i am confused, is the 80 in 80:1 centimeters as well?\n",
        "A": "It is saying that 80 centimeters in the real world is 1 centimeter on the blueprint",
        "video_name": "byjmR7JBXKc",
        "timestamps": [
            4801
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 1:21 i never understand how he got y/x 3/1 6/2=27/9 im soooooooooooooooooooooooooooooooo lost right now",
        "A": "A proportional relationship just means that the 2 fractions are equal. Your 3 fractions are all equal fractions. Start with 3/1 If you multiply 3/1 * 2/2 = 6/2. If you reduce it, you get back 3/1 If you multiply 3/1 * 9/9 = 27/9. If you reduce it, you get back 3/1. Hope this helps.",
        "video_name": "qYjiVWwefto",
        "timestamps": [
            81
        ],
        "3min_transcript": "What I want to introduce you to in this video is the notion of a proportional relationship. And a proportional relationship between two variables is just a relationship where the ratio between the two variables is always going to be the same thing. So let's look at an example of that. So let's just say that we want to think about the relationship between x and y. And let's say that when x is one, y is three, and then when x is two, y is six. And when x is nine, y is 27. Now this is a proportional relationship. Why is that? Because the ratio between y and x is always the same thing. And actually the ratio between y and x or, you could say the ratio between x and y, is always the same thing. So, for example-- if we say the ratio y over x-- this is always equal to-- it could be three over one, which is just three. It could be six over two, It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value,"
    },
    {
        "Q": "\nIn the charts that Sal makest at 3:36, is the y side always going to be numerator in fraction form?",
        "A": "It doesn t matter which side you use as the numerator, as long as the proportions are consistent. So if you have x/y = 2x/2y = 3x/3y, that will work for sure. But if you have x/y = 2y/2x = 3x/3y, that won t work. Sure you can place the y on the top but make sure it stays on the top.",
        "video_name": "qYjiVWwefto",
        "timestamps": [
            216
        ],
        "3min_transcript": "It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value, And then take the ratio between them. Here we took the ratio y to x, and you see y to x, or y divided by x-- the ratio of y to x is always going to be the same here so this is proportional. And you could actually gone the other way. You could have said, well what's the ratio of x to y? Well over here it would be one to three, which is the same thing as two to six, which is the same thing as nine to 27. When you take this ratio-- if you say the ratio of x to y instead of y to x, you see that it is always one third. But any way you look at it-- the ratio between these two variables-- if you say y to x, it's always going to be three. Or x to y is always going to be one third. So this is proportional while this one is not."
    },
    {
        "Q": "At 7:34, why are all the sides equal to the square root of a^2 + b^2?\n",
        "A": "If one of the sides is equal to the square root of a^2 + b^2 (using Pythagoras theorem), then all of the sides are equal to it (since all the sides of a rhombus are equal).",
        "video_name": "jpKjXtywTlQ",
        "timestamps": [
            454
        ],
        "3min_transcript": "so all four of these triangles are congruent. So you could take the area of this triangle, multiply it by 4-- you have the area of the rhombus. Let me write this down. The rhombus area is equal to 4 times 1/2ab. 1/2ab gives us just this triangle right over here. 4 times that, so 4 times 1/2ab is 2ab, is going to be the area of the rhombus. If we can somehow figure out a and b, we can figure out the rhombus' area. So let's focus on this first piece of information. Let's focus on triangle AB and D. Now they tell us that its circumradius is 12.5. So let's just use this formula right over here. We get 12.5. It's circumradius is 12.5 is equal to the product So what's the length of the sides here? So we have this side right over here, side bd. That's just going to be 2a, right? That's an a plus another a, so it's going to be 2a, times this side right over here. What's this side, which is just one of the sides of the rhombus. Well, this is the hypotenuse of this right triangle right over here, right? This is a right angle. So it's going to be the square root of a squared plus b squared. But all of the sides are going to be that. It's a rhombus. All the sides are the same-- a squared plus b squared. They're all going to have that exact same length. So the product of the sides-- you have 2a-- that's the length of bd-- times the length of ba, which is going to be the square root of a squared plus b squared, times the length of ad, which is the square root of a squared plus b squared. 4 times the area of ABD. Now, what's the area of a, b, and d? Well, ABD is just two of these triangles right over here. This guy right over here is 1/2ab. This guy over here is also 1/2ab. So the entire area is going to be two of these guys, so it's just going to be a times b. It gives you the area of both of these triangles. Each of them are 1/2ab. So instead of writing area right here, I could write ab. Now let's see. This simplifies to 12.5, is equal to. Divide the numerator and the denominator by 2. So that becomes a 1. That becomes a 2, divided by a. That becomes a 1. That becomes a 1. Square root of a squared plus b squared times square root of a squared plus b squared is just a squared plus b squared."
    },
    {
        "Q": "\nAt 12:19, Sal simplified (2a)^2 to 4a^2. Wouldn't it just be 4a?",
        "A": "(2a)^2 = (2a)*(2a) = 2*2*a*a = (2*2)*(a*a) = 2^2*a^2.",
        "video_name": "jpKjXtywTlQ",
        "timestamps": [
            739
        ],
        "3min_transcript": "You add them together, you just get ab. You just get ab, 2 divided by 2, you get a 1 there. You get a 2 here. Divide by b, get a 1. That just becomes an a. And so you get 25 is equal to the numerator, square root of a squared b squared times itself is just going to be a squared plus b squared over 2a. So that second triangle, its circumradius being 25 gives us this equation right over here. Now we can use both of this. We have two equations with two unknowns. Let's solve for a and b. If we know a and b, we can then go back here and figure out the rhombus's area. So over here, we get-- let's multiply both sides by 2b. We get 25b is equal to a squared plus b squared. Over here, if we multiply both sides by 2a, we get 50a is equal to a squared plus b squared. 25b is equal to a squared plus b squared. So 25b must be the same thing as 50a. They're both a squared plus b squared. So we get 25b must be equal to 50a. They're both equal to a squared plus b squared. Now, divide both sides by 25, you get b is equal to 2a. b is equal to-- actually I wanted to do that in the magenta. b is equal to 2a. So we can take this information and then now substitute back into either one of these equations to solve for b, and then we can solve for a. So let's go back into this one. So we get 50a-- actually we'll solve for a first. 50a is equal to a squared plus b squared. so let's write 2a squared. So we get 50a is equal to a squared plus 4a squared. Or we get 50a is equal to 5a squared. We could divide both sides by 5a. If we divide this side by 5a, we get 10. And if we divide this side by 5a, we get a. So a is equal to 10. And then we could just substitute back here to figure out b. 2 times a is equal to b. b is equal to 2 times 10, which is equal to 20. So we know a is 10. b is 20. We just have to go right back here to figure out the area of the rhombus. The area of the rhombus is equal to 2 times-- a was 10-- 2 times 10 times 20. This is 20 times 20."
    },
    {
        "Q": "at 12:23 , he said 50=5a^2 , then he divided by 5 and got 10=a. what about that squared symbol?\n",
        "A": "The equation was 50a=5a^2. He then took each side divided by 5a, which came out to be 10=a, or a=10.",
        "video_name": "jpKjXtywTlQ",
        "timestamps": [
            743
        ],
        "3min_transcript": "You add them together, you just get ab. You just get ab, 2 divided by 2, you get a 1 there. You get a 2 here. Divide by b, get a 1. That just becomes an a. And so you get 25 is equal to the numerator, square root of a squared b squared times itself is just going to be a squared plus b squared over 2a. So that second triangle, its circumradius being 25 gives us this equation right over here. Now we can use both of this. We have two equations with two unknowns. Let's solve for a and b. If we know a and b, we can then go back here and figure out the rhombus's area. So over here, we get-- let's multiply both sides by 2b. We get 25b is equal to a squared plus b squared. Over here, if we multiply both sides by 2a, we get 50a is equal to a squared plus b squared. 25b is equal to a squared plus b squared. So 25b must be the same thing as 50a. They're both a squared plus b squared. So we get 25b must be equal to 50a. They're both equal to a squared plus b squared. Now, divide both sides by 25, you get b is equal to 2a. b is equal to-- actually I wanted to do that in the magenta. b is equal to 2a. So we can take this information and then now substitute back into either one of these equations to solve for b, and then we can solve for a. So let's go back into this one. So we get 50a-- actually we'll solve for a first. 50a is equal to a squared plus b squared. so let's write 2a squared. So we get 50a is equal to a squared plus 4a squared. Or we get 50a is equal to 5a squared. We could divide both sides by 5a. If we divide this side by 5a, we get 10. And if we divide this side by 5a, we get a. So a is equal to 10. And then we could just substitute back here to figure out b. 2 times a is equal to b. b is equal to 2 times 10, which is equal to 20. So we know a is 10. b is 20. We just have to go right back here to figure out the area of the rhombus. The area of the rhombus is equal to 2 times-- a was 10-- 2 times 10 times 20. This is 20 times 20."
    },
    {
        "Q": "at 0:42 , how do we get to know if an equation is homogeneous equation. i mean what is the working rule for homogeneous equation?\n",
        "A": "For a first order equation you can determine it s not separable because an x term is added to a y term. Homogeneous is probably the next thing to try. It works because the exponent on the x and y terms are the same. I don t know of any perfect test to determine if it s homogeneous. I would say it s trial and error and, of course, practice.",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            42
        ],
        "3min_transcript": "Let's do one more homogeneous differential equation, or first order homogeneous differential equation, to differentiate it from the homogeneous linear differential equations we'll do later. But anyway, the problem we have here. It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. I'm writing it a little bit small, so that I don't run out of space. Divided by 2xy. So in all of these homogeneous equations-- and obviously, we don't know it's homogeneous yet. So we have to try to write it as a function of y divided by x. So it looks like we could do that, if we divide the top and the bottom by x squared. So if we just multiply-- let me do it in a different color-- 1 over x squared, or x to the negative 2, over 1 over x squared. We're essentially just multiplying by 1. And then, that is equal to what? 1 plus 3y squared over x squared divided by 2-- if you times y over x. Or we could just rewrite the whole thing, and this is just equal to 1 plus 3y over x squared divided by 2 times y over x. So yes, this is a homogeneous equation. Because we were able to write it as a function of y divided by x. So now, we can do the substitution with v. And hopefully, this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then, of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v, this is just the product rule, plus x times the And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep,"
    },
    {
        "Q": "\nI have an dummy question, but at minute 1:03, i dont get why the multiplication (1/x\u00c2\u00b2)/1/x\u00c2\u00b2 makes the after result.\nThanks",
        "A": "Distribute the 1/x^2 on top to both terms, then multiply the bottom term by 1/x^2. You ll get the same result as the video.",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            63
        ],
        "3min_transcript": "Let's do one more homogeneous differential equation, or first order homogeneous differential equation, to differentiate it from the homogeneous linear differential equations we'll do later. But anyway, the problem we have here. It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. I'm writing it a little bit small, so that I don't run out of space. Divided by 2xy. So in all of these homogeneous equations-- and obviously, we don't know it's homogeneous yet. So we have to try to write it as a function of y divided by x. So it looks like we could do that, if we divide the top and the bottom by x squared. So if we just multiply-- let me do it in a different color-- 1 over x squared, or x to the negative 2, over 1 over x squared. We're essentially just multiplying by 1. And then, that is equal to what? 1 plus 3y squared over x squared divided by 2-- if you times y over x. Or we could just rewrite the whole thing, and this is just equal to 1 plus 3y over x squared divided by 2 times y over x. So yes, this is a homogeneous equation. Because we were able to write it as a function of y divided by x. So now, we can do the substitution with v. And hopefully, this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then, of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v, this is just the product rule, plus x times the And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep,"
    },
    {
        "Q": "I believe at 2:23 and other instances during previous videos he makes the mistake of putting the 'approx. equal to' sign. It is exactly equal to e^x.\n",
        "A": "Well, technically, yes and no, since we are summing something to infinity, it can never be exact exact, because it has to go on forever(e is transcendental anyway), but he probably could have used equals, or approx. equals, the definitions get quite similiar while summing things to infinity.",
        "video_name": "JYQqml4-4q4",
        "timestamps": [
            143
        ],
        "3min_transcript": "Now let's do something pretty interesting. And this will, to some degree, be one of the easiest functions to find the Maclaurin series representation of. But let's try to approximate e to the x. f of x is equal to e to the x. And what makes this really simple is, when you take the derivative-- and this is, frankly, one of the amazing things about the number e-- is that when you take the derivative of e to the x, you get e to the x. So this is equal to f prime of x. This is equal to f, the second derivative of x. This is equal to the third derivative of x. This is equal to the n-th derivative of x. It's always equal to e to the x. That's kind of the first mind blowing thing about the number e. It's just, you could keep taking its derivative. The slope at any point on that curve is the same as the value of that point on that curve. That's kind of crazy. Anyway, with that said, let's take its Maclaurin representation. So we have to find f of 0, f prime of 0, the second derivative at 0. Well, when you take e to the 0, e to the 0 is just equal to 1. This is going to be equal to f prime of 0. It's going to be equal to any of the derivatives evaluated at 0. The n-th derivative evaluated at 0. And that's why it makes applying the Maclaurin series formula fairly straightforward. If I wanted to approximate e to the x using a Maclaurin series-- so e to the x-- and I'll put a little approximately over here. And we'll get closer and closer to the real e to the x as we keep adding more and more terms. And especially if we had an infinite number of terms, it would look like this. f of 0-- let me do it in-- what colors did I use for cosine and sine? So I used pink and I used green. So let me use a non-pink, non-green. I'll use the yellow here. So f of 0 is 1 plus f prime of 0 times x. f prime of 0 is also 1. So plus x plus, this is also 1, so it's going to be x squared over 2 factorial. All of these things are going to be 1. This is 1, this is 1, when we're talking about e to the x. So you go to the third term. This is 1. You just have x to the third over 3 factorial. Plus x to the third over 3 factorial. And I think you see the pattern here. We just keep adding terms. x to the fourth over 4 factorial plus x to the fifth over 5 factorial plus x to the sixth over 6 factorial. And something pretty neat is starting to emerge. Is that e to x, 1-- this is just really cool-- that e to the x can be approximated by 1 plus x plus x squared over 2 factorial plus x to the third over 3 factorial. Once again, e to the x is starting to look like a pretty cool thing. This also leads to other interesting results. That if you wanted to approximate e, you just evaluate this at x is equal to 1. So if you wanted to approximate e, you'd say"
    },
    {
        "Q": "at 0:24 cant you just do -31+(1-1)-7 ?\n",
        "A": "yes, you can but the way Sal did it is simpler.",
        "video_name": "GA_yxxeFYBU",
        "timestamps": [
            24
        ],
        "3min_transcript": "- So I have a function here, h of n, and let's say that it explicitly defines the terms of a sequence. Let me make a little... Let me make a quick table here. We have n, and then we have h of n. When n is equal to one, h of n is negative 31, minus seven times one minus one, which is going to be... This is just going to be zero, so it's going to be negative 31. When n is equal to two, it's going to be negative 31 minus seven times two minus one, so two minus one. This is just going to be one, so it's negative 31 minus seven, which is equal to negative 38. When n is equal to three, it's gonna be negative 31 minus seven times three minus one, which is just two, so we're gonna subtract seven twice. It's gonna be negative 31 minus 14, which is equal negative 45. We're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven. We keep subtracting negative seven from that. In fact we subtract negative seven one less than the term... We subtract negative seven one less times than the term we're dealing with. If we're dealing with the third term we subtract negative seven twice. If we're dealing with the second term we subtract negative seven once. This is all nice, but what I want you to do now is pause the video and see if you can define this exact same sequence. The sequence here is you start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45. The next one is gonna be negative 52, and you go on and on and on. You keep subtracting negative seven. Can we define this sequence in terms of a recursive function? Why don't you have a go at that. Let's try to define it in terms of a recursive function. Let's just call that g of n, In some ways a recursive function is easier, because you can say okay look. The first term when n is equal to one, if n is equal to one, let me just write it, If n is equal to one, if n is equal to one, what's g of n gonna be? It's gonna be negative 31, negative 31. And if n, if n is greater than one and a whole number, so this is gonna be defined for all positive integers, and whole, and whole number, it's just going to be the previous term, so g of n minus one minus seven, minus seven. We're saying hey if we're just picking an arbitrary term we just have to look at the previous term and then subtract, and then subtract seven. It all works out nice and easy, because you keep looking at previous, previous, previous terms all the way until you get to the base case, which is when n is equal to one, and you can build up back from that. You get this exact same sequence."
    },
    {
        "Q": "During 0:21, he screen became blurred and it was tough to concentrate. Still, love your videos( Your ScratchPad makes me jealous)!!\u00f0\u009f\u0098\u0083\n",
        "A": "HaCk(ERROR<1>) X>0 Y<0{heading 0 Ipx} =hAck",
        "video_name": "qW-Ce44ll0Q",
        "timestamps": [
            21
        ],
        "3min_transcript": "- We have three different number lines here, and on each number line they mark off a couple of these marks, this is negative two this is negative 10, negative five, negative 11, and then we need to figure out what the blue dot represents. And as a little bit of a hint, each mark here is not necessarily incrementing by one, it could be by more than one. So I encourage you to pause this video, and to try it on your own. Try to figure out what number does this blue dot represent on these different number lines? So let's tackle this first one. So we're gonna go where they gave us this mark is negative two, this is negative 10, and we need to figure out this blue one that's further to the left of negative 10. Well just going from negative two to negative 10, what has to happen to negative two to get to negative 10? Well we'll have to subtract eight. And so if we move two to the left, that's the equivalent of subtracting eight. So if we move two to the left again, that's subtracting eight again. So this must be negative 18 and so if we move two to the left again, that also must be subtracting eight. that gets us to negative 26. Now another way we could have thought about it is, if we jump two to the left and that's negative eight, then one jump to the left is gonna be negative four. So you could say \"Well this is negative four\", that would get you to negative six, then negative four again gets you to negative 10, then this would be negative 14, negative 18, negative 22, negative 26. Now let's tackle this one. So here we're gonna, we have to figure out what this blue dot here is on the right. So if we started at 11, you make two jumps to go to negative five. So what do I have to add to negative 11 to get to negative five? Well I have to add six. So if I add six over two jumps, that means that each of these jumps, that must be plus three, and this one must be plus three. So negative five plus three... So plus three is gonna get us to negative two. Plus three gets us to positive one, negative two plus three is positive one, plus three gets us to four. So here we got to negative 26, here we get to four. Now let's tackle this one. So we're at negative seven, or this is negative seven this is one and then we have to figure out this mark right over here. So if you take two jumps to the right, from negative seven to one, how much did we have to add? Well to get from negative seven to one, you have to add eight. You have to add eight. So then if we make another two jumps to the mystery number that means we added eight again. So one plus eight is nine. Another way we could have thought about it, if we added eight over two jumps that means we added four on each jump. So this one must be negative three, add four you get to one,"
    },
    {
        "Q": "\nAt 2:09 he basically tells us we can't assume and we don't know the angels.\nWell.. We know the angle bisector is equal (no duh...) But doesn't that mean that the 2 angels at D are 90 degrees. And if we know that we also know that the Angle A and Angle C are equal am I right?",
        "A": "We actually don t know that the 2 angles at D are 90\u00c2\u00b0. They look like they could be, but that information wasn t given in the question. If they were, then you re correct that we could find the other angles and prove that the triangles are congruent by ASA (or AAAS).",
        "video_name": "TpIBLnRAslI",
        "timestamps": [
            129
        ],
        "3min_transcript": "What I want to do first is just show you what the angle bisector theorem is and then we'll actually prove it for ourselves. So I just have an arbitrary triangle right over here, triangle ABC. And what I'm going to do is I'm going to draw an angle bisector for this angle up here. And we could have done it with any of the three angles, but I'll just do this one. I'll make our proof a little bit easier. So I'm just going to bisect this angle, angle ABC. So let's just say that's the angle bisector of angle ABC, and so this angle right over here is equal to this angle right over here. And let me call this point down here-- let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector-- so when I put this angle bisector here, it created two smaller triangles out of that larger one. The angle bisector theorem tells us the ratios between the other sides of these two triangles So it tells us that the ratio of AB to AD is going to be equal to the ratio of BC to, you could say, CD. So the ratio of-- I'll color code it. The ratio of that, which is this, to this is going to be equal to the ratio of this, which is that, to this right over here-- to CD, which is that over here. So once you see the ratio of that to that, it's going to be the same as the ratio of that to that. So that's kind of a cool result, but you can't just accept it on faith because it's a cool result. You want to prove it to ourselves. And so you can imagine right over here, we have some ratios set up. So we're going to prove it using similar triangles. And unfortunate for us, these two triangles right here We know that these two angles are congruent to each other, but we don't know whether this angle is equal to that angle or that angle. We don't know. We can't make any statements like that. So in order to actually set up this type of a statement, we'll have to construct maybe another triangle that will be similar to one of these right over here. And one way to do it would be to draw another line. And this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you. What happens is if we can continue this bisector-- this angle bisector right over here, so let's just continue it. It just keeps going on and on and on. And let's also-- maybe we can construct a similar triangle to this triangle over here if we draw a line that's parallel to AB down here. So let's try to do that. So I'm just going to say, well, if C is not on AB, you could always find a point or a line that goes through C that is parallel to AB."
    },
    {
        "Q": "How did Sal know to divide by two at 2:04. Thats part of taking the average... right?\n",
        "A": "Sal was trying to find the midpoint of thouse two points. He averaged the coordinates to find a new set of coordinates exactly in the middle. The midpoint formula is... (x+x/2, y+y/2)",
        "video_name": "63mWxNXQQAk",
        "timestamps": [
            124
        ],
        "3min_transcript": "- [Voiceover] We're asked to use the \"Reflection\" tool to define a reflection that will map line segment ME, line segment ME, onto the other line segment below. So we want to map ME to this segment over here and we want to use a Reflection. Let's see what they expect from us if we want to add a Reflection. So if I click on this it says Reflection over the line from, and then we have two coordinate pairs. So they want us to define the line that we're going to reflect over with two points on that line. So let's see if we can do that. To do that I think I need to write something down so let me get my scratch pad out and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E, we want to map point E, to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the This and this should be E, and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see that is x equals negative four, y is equal to negative four, and the coordinates for the corresponding point to E in the image. This is x is equal to two, x is equal to two, and y is equal to negative six. So what's the midpoint between negative four, negative four, and two, comma, negative six? Well you just have to take the average of the x's and take the average of the y's. Let me do that, actually I'll do it over here. So if I take the average of the x's it's going to be negative four, negative four, plus two, plus two, And then the average of the y's, it's going to be negative four plus negative six over two. Negative four plus negative six, over two and then close the parentheses. Let's see, negative four plus two is negative two, divided by two is negative one. So it's going to be negative one, comma. Negative four plus negative six, that's the same thing as negative four minus six which is going to be negative 10. Divided by two is negative five. Let me do that in a blue color so you see where it came from. Is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one, comma, negative five. So x is negative one, y is negative five. So it's this point right over here and it does indeed look like the midpoint."
    },
    {
        "Q": "\nI still don't understand. e.g. Sharing $20 in the ratio of 2:3. How would you do it? As well as both numbers wouldn't be equal if you separated. Any advice?",
        "A": "Well ratios don t have to be equal that is the thing.2+3=5 so 20\u00c3\u00b75=4 so the first person who got 2 would get 4x2=8 so that would be 8 for the first person then 4x3=12 and so to check do 8+12=20 tada",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            123
        ],
        "3min_transcript": "Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things. So there's a couple of ways that we can specify this. We can literally count the number of apples. 1, 2, 3, 4, 5, 6. So we have 6 apples. And we can say the ratio is going to be 6 to, 6 to ... And then how many oranges do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. It is 6 to 9. The ratio of apples to oranges is 6 to 9. And you could use a different notation. You could also write it this way. 6 to ... You would still read the ratio as being 6 to 9. because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges"
    },
    {
        "Q": "\nInstead of using \"to\" for the ratio, we could also tell it as \"is to\"....For example 6:9 could be told as 6 is to 9 too...",
        "A": "Yes, I think using is to is the same as using to",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            369
        ],
        "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples."
    },
    {
        "Q": "\nIn ratio when you write something like 6:10, why do you have to write a colon??",
        "A": "The colons are used to seperate two different amounts of something/item.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            370
        ],
        "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples."
    },
    {
        "Q": "If the problem is like this one where it says find the ratio to apples and oranges 6:9 does it matter which one goes first could it be 9:6\n",
        "A": "It matters which goes first and it is very important. In your example, we have the ratio of apples to oranges so the number of apples will be first, 6:9 (I assume that 6 is the number of apples). If you write it as 9:6, you also need to switch your sentence into the ratio of oranges to apples or it won t be the same.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            369,
            546
        ],
        "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples."
    },
    {
        "Q": "\nat 3:13 what does he say",
        "A": "This is what he said 3:1= Here we 9 oranges for every 6 apples. This can also be shown as 6 apples for every 9 oranges Hope this helps.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            193
        ],
        "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples."
    },
    {
        "Q": "\nAt 3:54,is 9:6 the same as 6:9,is it only switched around?",
        "A": "Uh, no. For example, you d have 9 red cars for every 6 blue cars, while if it was 6:9, it would be 6 red cars for every 9 blue cars.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            234,
            546,
            369
        ],
        "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples."
    },
    {
        "Q": "At 3:09, Sal said that 0^0th power = 1. But I used my calculator and saw that 0^0 = undefined.\nIn the previous videos, he showed that decreasing the index/exponent would mean dividing by the number whose power is taken. In other words, 0^1 to 0^0= 1*0/0=0/0=undefined.\nBut then there is the case of other powers. what is actually going on?\n",
        "A": "Well, what Sal is saying is that 0^0th power could equal one, depending on what theory you think is correct. It comes out undefined, because the true answer to 0^0 power could be 0 or 1: we just don t know. Because theorists can t agree, they settled with undefined, which pretty much means, we don t know for sure .",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            189
        ],
        "3min_transcript": "once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1. should be equal to 1. So you see a conundrum here, and there's actually really good cases, and you can get actually fairly sophisticated with your mathematics. And there's really good cases for both of these, for 0 to 0-th being 0, and 0 to the 0-th power being 1. And so when mathematicians get into this situation, where they say, well, there's good cases for either. There's not a completely natural one. Either of these definitions would lead to difficulties in mathematics. And so what mathematicians have decided to do is, for the most part-- and you'll find people who will dispute this; people will say, no, I like one more than the other-- but for the most part, this is left undefined. 0 to the 0-th is not defined by at least just kind of more conventional mathematics. In some use cases, it might be defined to be one of these two things. So 0 to any non-zero number, you're going to get 0."
    },
    {
        "Q": "At 0:12, what is the answer\n",
        "A": "Zero, because multiplying ANYTHING by zero is going to be zero.",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            12
        ],
        "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1."
    },
    {
        "Q": "At about the time frame 5:20 to 5:40 you told us that the dot plot would probably be the easiest to use. did you mean for everything as it was stressed that way in \" and this is once again where maybe the dot plot is the most (pause), the most, it jumps out at you.\" So do mean that it is the easiest for all the questions or just those few? -thanks\n",
        "A": "No, he was talking about those that were shown there. So, the dot plot was the best for that in comparison with the list and the table. Hope this helps!",
        "video_name": "gdE46YSedvE",
        "timestamps": [
            320,
            340
        ],
        "3min_transcript": "A dot plot. And a dot plot, we essentially just take the same information, and even think about it the same way. But we just show it visually. In a dot plot, what we would have ... In a dot plot, what we would have ... Actually, let me just not draw an even arrow there. We have the different age groups, so five, six, seven, eight, nine, 10, 11, and 12, and we have a dot to represent, or we use a dot for each student at that age. So there's two five-year-olds, so I'll do two dots. One, and two. There's one six-year-old, so that's gonna be one dot, right over here. There's four seven-year-olds, so one, two, three, four dots. There's four nine-year-olds, so one, two, three, and four. There's one 10-year-old. So let's put a dot, one dot, right over there for that one 10-year-old. There's no 11-year-olds. I'm not gonna put any dots there. And then there's two 12-year-olds. So one 12-year-old, and another 12-year-old. So there you go, we have a frequency table, dot plot, list of numbers. These are all showing the same data, just in different ways. Once you have it represented in any of these ways, we can start to ask questions about it. So we could say, \"What is the most frequent age?\" Well, the most frequent age, when you look at it visually, or the easiest thing might be just to look at the dot plot because you see visually, the most frequent age are the two highest stacks. There's actually seven and nine are tied for the most frequent age. You'd have also seen it here, where seven and nine are tied at four. you would actually, you'd have to count all of them to kind of come up with this again and say, \"Okay, there's four sevens, four nines. \"That's the largest number.\" So this is, if you're looking for, what's the most frequent age? When you just visually inspect here, probably pops out at you the fastest. But there's other questions we can ask ourselves. We can ask ourselves, \"What is the range? \"What is the range of ages in the classroom?\" And this is once again where maybe the dot plot jumps out at you the most, because the range is just the maximum age ... or, the maximum data point minus the minimum data point. So what's the maximum age here? Well, the maximum age here, we see it from the dot plot, is 12. And the minimum age here, you see, is five. So there's a range of seven. The difference between the maximum and the minimum is seven. But you could have also done that over here. You could say, \"The maximum age here is 12. \"Minimum age here is five. \"And so let's subtract ...\" You find the difference between 12 and five, which is seven."
    },
    {
        "Q": "I thought I understood this but at 6:18 I stopped understanding it.:(\n",
        "A": "What exactly did you not understand? The range is a measure that helps you figure how large the difference between the smallest and largest values of your data set is, and to calculate it you take the smallest value and subtract it from the largest one. It s really just a way to figure out how spread out your data is. Hope this helped, if not don t hesitate to ask away :)",
        "video_name": "gdE46YSedvE",
        "timestamps": [
            378
        ],
        "3min_transcript": "There's four nine-year-olds, so one, two, three, and four. There's one 10-year-old. So let's put a dot, one dot, right over there for that one 10-year-old. There's no 11-year-olds. I'm not gonna put any dots there. And then there's two 12-year-olds. So one 12-year-old, and another 12-year-old. So there you go, we have a frequency table, dot plot, list of numbers. These are all showing the same data, just in different ways. Once you have it represented in any of these ways, we can start to ask questions about it. So we could say, \"What is the most frequent age?\" Well, the most frequent age, when you look at it visually, or the easiest thing might be just to look at the dot plot because you see visually, the most frequent age are the two highest stacks. There's actually seven and nine are tied for the most frequent age. You'd have also seen it here, where seven and nine are tied at four. you would actually, you'd have to count all of them to kind of come up with this again and say, \"Okay, there's four sevens, four nines. \"That's the largest number.\" So this is, if you're looking for, what's the most frequent age? When you just visually inspect here, probably pops out at you the fastest. But there's other questions we can ask ourselves. We can ask ourselves, \"What is the range? \"What is the range of ages in the classroom?\" And this is once again where maybe the dot plot jumps out at you the most, because the range is just the maximum age ... or, the maximum data point minus the minimum data point. So what's the maximum age here? Well, the maximum age here, we see it from the dot plot, is 12. And the minimum age here, you see, is five. So there's a range of seven. The difference between the maximum and the minimum is seven. But you could have also done that over here. You could say, \"The maximum age here is 12. \"Minimum age here is five. \"And so let's subtract ...\" You find the difference between 12 and five, which is seven. You still could have done it. You'd say, \"Okay, what's the lowest? \"Let's see, five. Are there any fours here? \"Nope, there's no fours. \"So five's the minimum age. \"And what's the largest? \"Is it seven? No. \"Is it nine? Nine, maybe 10. \"Oh, 12. 12. \"Are there any 13s? No. \"12 is the maximum.\" So you say, \"12 minus five is seven\" to get the range. But then we could ask ourselves other questions. We could say, \"How many ... \"How many older ... \"older than nine?\" is a question we could ask ourselves. And then, if we were to look at the dot plot, we'd say, \"Okay, this is nine.\" And we'd care about how many are older than nine. So that would be this one, two and three. Or you could look over here. How many are older than nine? Well, it's the one person who's 10 and then the two who are 12. So there are three. And over here, if you said, \"How many are older than nine?\" Well, then you'd just have to go through the list and say, \"Okay, no, no, no, no, no, no, no,"
    },
    {
        "Q": "I am wondering if it's possible to raise a number to a power to result in 0. x^x = 0 I've been graphing functions with the variable in the exponent (i.e. y = 4^x) and was wondering if it ever actually reaches zero. @ 4:13 Sal says it never quite reaches zero, so is it impossible? The line keeps going on forever in the X direction?\n",
        "A": "It s my firm belief that infinity doesn t exist. As you can see by the graphs of y = a^x, y is always positive, and can only get close to zero where x is negative. No matter how large a negative number -x you get, you can always find another so large that the first one is negligible in comparison. And a^(-x) = 1/(a^x) is never zero. (Kind of crazy, insn t it?)",
        "video_name": "9SOSfRNCQZQ",
        "timestamps": [
            253
        ],
        "3min_transcript": "That could be my x-axis. And then let's make this my y-axis. I'll draw it as neatly as I can. So let's make that my y-axis. And my x values, this could be negative 2. Actually, make my y-axis keep going. So that's y. This is x. That's a negative 2. That's negative 1. That's 0. That is 1. And that is positive 2. And let's plot the points. x is negative 2. y is 1/25. Actually, let me make the scale on the y-axis. So let's make this. So we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that, too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them. Negative 2, 1/25. So 1/25 is going to be really, really close to the x-axis. That's about 1/25. So that is negative 2, 1/25. It's not going to be on the x-axis. 1/25 is obviously greater than 0. It's going to be really, really, really, really, close. Now let's do this point here in orange, negative 1, 1/5. Negative 1/5-- 1/5 on this scale is still pretty close. It's pretty close. So that right over there is negative 1, 1/5. And now in blue, we have 0 comma 1. 0 comma 1 is going to be right about there. If this is 2 and 1/2, that looks about right for 1. And then we have 1 comma 5. 1 comma 5 puts us right over there. And then finally, we have 2 comma 25. When x is 2, y is 25. 2 comma 25 puts us right about there. And so I think you see what happens with this function, The further in the negative direction we go, 5 to ever-increasing negative powers gets closer and closer to 0, but never quite. So we're leaving 0, getting slightly further, further, further from 0. Right at the y-axis, we have y equal 1. Right at x is equal to 0, we have y is equal to 1. And then once x starts increasing beyond 0, then we start seeing what the exponential is good at, which is just this very rapid increase. Some people would call it an exponential increase, which is obviously the case right over here. So then if I just keep this curve going, you see it's just going on this sometimes called a hockey stick. It just keeps on going up like this at a super fast rate, ever-increasing rate. So you could keep going forever to the left, and you'd get closer and closer and closer to 0 without quite getting to 0. So 5 to the negative billionth power"
    },
    {
        "Q": "\n1:30 how can 5^0=1?",
        "A": "You have to think of it in groups a bit, but also considering the increase/decrease. Ex: 2^3, or 2x2x2=8 2^2, or 2x2=4, with a decrease/increase of 4. 2^1 or 2 = 2, with a decrease/increase of 2. So the number is basically halved each time, specifically only for 2s. For 3s it would be divided in to thirds, 4s would be fourths and so forth. So what s one half of 2? 2 x 1/2 cross multiply: 1 x 1/1 = 1. for 3s: 3 x 1/3 cross multiply: 1 x 1/1= 1. and so forth. Posted this above, hope it helps :D",
        "video_name": "9SOSfRNCQZQ",
        "timestamps": [
            90
        ],
        "3min_transcript": "We're asked to graph y is equal to 5 to the x-th power. And we'll just do this the most basic way. We'll just try out some values for x and see what we get for y. And then we'll plot those coordinates. So let's try some negative and some positive values. And I'll try to center them around 0. So this will be my x values. This will be my y values. Let's start first with something reasonably negative but not too negative. So let's say we start with x is equal to negative 2. Then y is equal to 5 to the x power, or 5 to the negative 2 power, which we know is the same thing as 1 over 5 to the positive 2 power, which is just 1/25. Now let's try another value. What happens when x is equal to negative 1? Then y is 5 to the negative 1 power, which is the same thing as 1 over 5 to the first power, or just 1/5. Now let's think about when x is equal to 0. Then y is going to be equal to 5 to the 0-th power, which is going to be equal to 1. So this is going to be equal to 1. And then finally, we have-- well, actually, let's try a couple of more points here. Let me extend this table a little bit further. Let's try out x is equal to 1. Then y is 5 to the first power, which is just equal to 5. And let's do one last value over here. Let's see what happens when x is equal to 2. Then y is 5 squared, 5 to the second power, which is just equal to 25. And now we can plot it to see how this actually looks. So let me get some graph paper going here. My x's go as low as negative 2, as high as positive 2. And then my y's go all the way from 1/25 all the way to 25. So I have positive values over here. So let me draw it like this. That could be my x-axis. And then let's make this my y-axis. I'll draw it as neatly as I can. So let's make that my y-axis. And my x values, this could be negative 2. Actually, make my y-axis keep going. So that's y. This is x. That's a negative 2. That's negative 1. That's 0. That is 1. And that is positive 2. And let's plot the points. x is negative 2. y is 1/25. Actually, let me make the scale on the y-axis. So let's make this. So we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that, too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them. Negative 2, 1/25."
    },
    {
        "Q": "on 4:11 it says to the 3rd power but how because in the formula it says to the 2nd power so I don't know how he got 3rd power\n",
        "A": "The units of volume is to the third power because it is related to multiplying length \u00e2\u0080\u00a2 width \u00e2\u0080\u00a2 height. Radius squared provides two, height is the third direction.",
        "video_name": "hC6zx9WAiC4",
        "timestamps": [
            251
        ],
        "3min_transcript": "of this entire can, this entire cylinder. But if you just want the cone, it's 1/3 of that. It is 1/3 of that. And that's what I mean when I say it's surprisingly clean that this cone right over here is 1/3 the volume of this cylinder that is essentially-- you could think of this cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1/3 times pi or pi/3 times hr squared. However you want to view it. The easy way I remember it? For me, the volume of a cylinder is very intuitive. You take the area of the base. And then you multiply that times the height. And so the volume of a cone is just 1/3 of that. It's just 1/3 the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers, just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, And let's say that we're told that it holds 131 cubic centimeters of water. And let's say that we're told that its height right over here-- I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest 10th of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1/3 times pi times the height, which is 5 centimeters, times the radius squared. If we wanted to solve for the radius squared, we could just divide both sides by all of this business. equal to 131 centimeters to the third-- or 131 cubic centimeters, I should say. You divide by 1/3. That's the same thing as multiplying by 3. And then, of course, you're going to divide by pi. And you're going to divide by 5 centimeters. Now let's see if we can clean this up. Centimeters will cancel out with one of these centimeters. So you'll just be left with square centimeters only in the numerator. And then to solve for r, we could take the square root of both sides. So we could say that r is going to be equal to the square root of-- 3 times 131 is 393 over 5 pi. So that's this part right over here. Once again, remember we can treat units just"
    },
    {
        "Q": "\nI know at 3:04 he gave the formula for finding the radius, but how do you change it to find the height?",
        "A": "If you already know the volume and the radius, to find the height, the formula wound be the following: h = 3 time v divided by pi times r squared.",
        "video_name": "hC6zx9WAiC4",
        "timestamps": [
            184
        ],
        "3min_transcript": "You could call this distance right over here h. And the formula for the volume of a cone-- and it's interesting, because it's close to the formula for the volume of a cylinder in a very clean way, which is somewhat surprising. And that's what's neat about a lot of this three-dimensional geometry is that it's not as messy as you would think it would be. It is the area of the base. Well, what's the area of the base? The area of the base is going to be pi r squared. It's going to be pi r squared times the height. And if you just multiplied the height times pi r squared, that would give you the volume of an entire cylinder that looks something like that. So this would give you this entire volume of the figure that looks like this, where its center of the top is the tip right over here. So if I just left it as pi r squared of this entire can, this entire cylinder. But if you just want the cone, it's 1/3 of that. It is 1/3 of that. And that's what I mean when I say it's surprisingly clean that this cone right over here is 1/3 the volume of this cylinder that is essentially-- you could think of this cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1/3 times pi or pi/3 times hr squared. However you want to view it. The easy way I remember it? For me, the volume of a cylinder is very intuitive. You take the area of the base. And then you multiply that times the height. And so the volume of a cone is just 1/3 of that. It's just 1/3 the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers, just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, And let's say that we're told that it holds 131 cubic centimeters of water. And let's say that we're told that its height right over here-- I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest 10th of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1/3 times pi times the height, which is 5 centimeters, times the radius squared. If we wanted to solve for the radius squared, we could just divide both sides by all of this business."
    },
    {
        "Q": "\nAt 15:20 Would we be wrong if we write \u00e2\u0088\u00ab f cos\u00ce\u00b8 dt?",
        "A": "You would want \u00e2\u0088\u00ab ||f||.||r|| cos\u00ce\u00b8 dt, with the magnitude of f and r, and \u00ce\u00b8 all defined as functions of t, which given the way Sal has set up the problem (with f and r defined as rectangular functions of t) would be difficult.",
        "video_name": "t3cJYNdQLYg",
        "timestamps": [
            920
        ],
        "3min_transcript": "How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is f dot r? Or what is f dot dr? Well, actually, to answer that, let's remember what dr looked like. If you remember, dr/dt is equal to x prime of t, I'm writing it like, I could have written dx dt if I wanted to do, times the i-unit vector, plus y prime of t, times the j-unit vector. And if we just wanted to dr, we could multiply both sides, if we're being a little bit more hand-wavy with the differentials, not too rigorous. We'll get dr is equal to x prime of t dt times the unit vector i plus y prime of t times the differential dt So this is our dr right here. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And we'll see that the dot product is actually not so crazy. So copy, and let me paste it down here. So what's this integral going to look like? This integral right here, that gives the total work done by the field, on the particle, as it moves along that path. Just super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee. It's going to be the integral, let's just say from t is equal to a, to t is equal to b. Right? a is where we started off on the path, t is equal to a to t is equal to b. You can imagine it as being timed, as a particle And then what is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your of vector, and add them up. So this is going to be the integral from t equals a to t equals b, of p of p of x, really, instead of writing x, y, it's x of t, right? x as a function of t, y as a function of t. So that's that. Times this thing right here, times this component, right? We're multiplying the i-components. So times x prime of t d t, and then that plus, we're going to do the same thing with the q function. So this is q plus, I'll go to another line. Hopefully you realize I could have just kept writing, but I'm running out of space. Plus q of x of t, y of t, times the component of our dr. Times"
    },
    {
        "Q": "\nHolla. At 06:00, \"add one to itself and times.\" I know, sounds silly, but cannot get it, since adding one to itself is two. What am I missing here?",
        "A": "He actually said (or at least he meant) add one to itself n times . If n = 4 then it would be 1 + 1 + 1 + 1 = 4 = n.",
        "video_name": "sRVGcYGjUk8",
        "timestamps": [
            360
        ],
        "3min_transcript": "this is just a constant term. This is just a constant term. So you can take it out. Times mu squared times the sum from i equals 1 to n. And what's going to be here? It's going to be a 1. We just divided a 1. We just divided this by 1. And took it out of the sigma sign, out of the sum. And you're just left with a 1 there. And actually, we could have just left the mu squared there. But either way, let's just keep simplifying it. So this we can't really do-- well, actually we could. Well, no, we don't know what the x sub i's are. So we just have to leave that the same. So that's the sum. Oh sorry, and this is just the numerator. This whole simplification, we're just simplifying the numerator. And later, we're just going to divide by n. So that is equal to that divided by n, which is equal to this thing divided by n. Because it's the numerator that's the confusing part. We just want to simplify this term up here. So let's keep doing this. So this equals the sum from i equals 1 to n of x sub i squared. And let's see, minus 2 times mu-- sorry, that mu doesn't look good. Edit, Undo, minus 2 times mu times the sum from i is equal to 1 to n of xi. And then, what is this? What is another way to write this? Essentially, we're going to add 1 to itself n times. This is saying, just look, whatever you have here, just iterate through it n times. If you had an x sub i here, you would use the first x term, then the second x term. When you have a 1 here, this is just essentially saying, add one to itself n times, which is the same thing as n. And then see if there's anything else we can do here. Remember, this was just the numerator. So this looks fine. We add up each of those terms. So we just have minus 2 mu from i equals 1 to-- oh well, think about this. What is this? What is this thing right here? Well actually, let's bring back that n. So this simplified to that divided by n, which simplifies to that whole thing, which is simplified to this whole thing, divided by n, which simplifies to this whole thing divided by n, which is the same thing as each of the terms divided by n, which"
    },
    {
        "Q": "\nIsn't the \"proof\" for the equivalence of the angles formed by a transversal across two parallel lines (4:09) simply that if the angles were not equivalent the lines would meet and not be parallel? Is that not considered a proof? If not, why not?",
        "A": "Yes it is. People think it isn t but Khan knows better\u00f0\u009f\u0098\u009c",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            249
        ],
        "3min_transcript": "This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle"
    },
    {
        "Q": "At 6:49, what does \"deduce\" mean?\n",
        "A": "Deduce means to make a conclusion given the evidence that you have,",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            409
        ],
        "3min_transcript": "If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e."
    },
    {
        "Q": "\nAt 1:46, do the lines have to be parallel for the line to be a transversal?",
        "A": "No, not necessarily. A transversal is a line/line segment/ ray intersecting two other lines/line segments/ rays.",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            106
        ],
        "3min_transcript": "Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write"
    },
    {
        "Q": "at 3:30 when Sal is discussing lim x--> c for f(x)/g(x), you can only calculate the limit if the bottom is not zero!\n",
        "A": "That is true until you learn L Hospital s rule and then it will be possible to calculate the limit of a function when g(x) is zero or when f(x) and g(x) are both infinity",
        "video_name": "lSwsAFgWqR8",
        "timestamps": [
            210
        ],
        "3min_transcript": "plus the limit of g of x as x approaches c. Which is equal to, well this right over here is-- let me do that in that same color-- this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule, or the sum property, of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x, is just going to be L minus M. It's just the limit of f of x as x approaches c, minus the limit of g of x as x approaches c. So it's just going to be L minus M. And we also often call it the difference rule, or the difference property, of limits. intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well lucky for us, this is going to be equal to the limit of f of x as x approaches c, times the limit of g of x, as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to, this is L times M. This is just going to be L times M. Same thing, if instead of having a function here, we had a constant. So if we just had the limit-- let me do it in that same color-- the limit of k times f of x, as x approaches c, where k is just some constant. This is going to be the same thing as k times the limit And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x. This is the exact same thing as the limit of f of x as x approaches c, divided by the limit of g of x as x approaches c. Which is going to be equal to-- I think you get it now-- this is going to be equal to L over M. And finally-- this is sometimes called the quotient property-- finally we'll look at the exponent property."
    },
    {
        "Q": "At 2:52, Sal said, \"anything divided by anything is just one.\" I think what he meant to say was \"anything divided by itself is just one.\" :)\n",
        "A": "Yeah, Sal was referring to the first anything when he said anything the second time : )",
        "video_name": "UquFdMg6Z_U",
        "timestamps": [
            172
        ],
        "3min_transcript": "The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, Either way, you knew how to do this before you even learned that exponent property. Because x divided by x is 1, and then assuming x does not equal to 0. And we kind of have to assume x doesn't equal 0 in this whole thing. Otherwise, we would be dividing by 0. And then finally, we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is negative 4 over 6 is the same thing as negative 2/3. Just simplified that fraction. And we're multiplying that times 1/x. So we can view this 4 times 1/x. Another way to think about it is you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the first power. And then when you tried to simplify it using your exponent properties, you would have-- well, that would be x to the 0 minus 1 power, which is x to the negative 1 power."
    },
    {
        "Q": "\nAt 0:30, why couldn't long division work?",
        "A": "That is another way to do basically the same thing. Whichever way is easiest for you!",
        "video_name": "UquFdMg6Z_U",
        "timestamps": [
            30
        ],
        "3min_transcript": "Simplify the expression 18x to the fourth minus 3x squared plus 6x minus 4, all of that over 6x. So there's a couple ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x, minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decomposed this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a/d plus b/d plus c/d. Or maybe not so clearly, but hopefully that helps clarify it. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1,"
    },
    {
        "Q": "At 2:14 to 2:32, could the -\u00c2\u00bdx also be written as -x/2?\n",
        "A": "Yes, they are the same.",
        "video_name": "UquFdMg6Z_U",
        "timestamps": [
            134,
            152
        ],
        "3min_transcript": "Simplify the expression 18x to the fourth minus 3x squared plus 6x minus 4, all of that over 6x. So there's a couple ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x, minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decomposed this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a/d plus b/d plus c/d. Or maybe not so clearly, but hopefully that helps clarify it. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1,"
    },
    {
        "Q": "at around 1:15 you say that you subtract 4x from each side so wouldn't it be 2y = 8-4x not 2y=4x-8\n",
        "A": "The equation was 4x + 2y = - 8 You simply missed out the negative signs. Subtracting 4x from the equation above will give us: 2y = -8 -4x which is the same as 2y = -4x -8 If you cannot understand why changing orders will still be the same, try taking -2 -4 and -4 -2. Either way, you will get -6! Same analogy here :) Hope this helped",
        "video_name": "V6Xynlqc_tc",
        "timestamps": [
            75
        ],
        "3min_transcript": "We're asked to convert these linear equations into slope-intercept form and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of a review, slope-intercept form is a form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A, so start with a line A. So line A, it's in standard form right now, it's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to. or negative 8 minus 4, however you want to do it. Now we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2 and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, let me graph it right now. So line A, its y-intercept is negative 4. So the point 0, negative 4 on this graph. If x is equal to 0, y is going to be equal to negative 4, you So 0, 1, 2, 3, 4. That's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1 that y goes down by negative 2. So let's do that. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive two, because two divided by negative one is still negative two, so I go over here. If I go back 2, I'm going to go up 4. Let me just do that. Back 2 and then up 4. So this line is going to look like this."
    },
    {
        "Q": "10:40. How can you define a function S such that S(y) is the unique solution in X to f(x) = y? How can you know such a function even exists?\n",
        "A": "Sal assumed that for any point in Y there is a unique solution to f(x) = y. That means that there must be a function. Remember, a function is anything that maps each value of y to a unique x.",
        "video_name": "7GEUgRcnfVE",
        "timestamps": [
            640
        ],
        "3min_transcript": "you have to be a little bit precise about it in order to get to the point you want. Let's see if the opposite is true. Let's see if we assume-- let's see if we start from the assumption, that for all y that is a member of our set Y, that the solution, that the equation f of x, is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. So let's think about the first way. So we're saying that for any y-- let me draw my sets again. So this is my set X and this is my set Y right there. Now we're working for the assumption that you can pick here has a unique solution. Let's call that unique solution. Well, we could call it whatever. But a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that, look, you pick a point in Y, I can find you some point in X such that f of x is equal to y. And not only can I find that for you, that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x and s of, let's say s of y, where, of course, y is a member of our set capital-Y. s of y is equal to the unique solution in x to f of x is Now, you're saying, hey, Sal, it looks a little convoluted. But think about it. This is a completely valid function definition. Right? We're starting with the idea that you give me any y here. You give me any member of this set, and I can always find you a unique solution to this equation. Well OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So, why don't I just define a function that says, look I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. And, of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value because this, any value y, any lower-case value y, in this set has a unique solution to f of x is equal to y."
    },
    {
        "Q": "\n@ 3:16 I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?",
        "A": "e^x and lnx are opposite functions. It would be akin to multiplying by some number and then dividing by that same number. The operations cancel one another out. e to the natural log of elephant = elephant.",
        "video_name": "C5Lbjbyr1t4",
        "timestamps": [
            196
        ],
        "3min_transcript": "log of x is equal to 1/x. So we have some expression, and we have its derivative, which tells us that we can use substitution. Sometimes you can do in your head, but this problem, it's still not trivial to do in your head. So let's make the substitution. Let's substitute this right here with a u. So let's do that. So if you define u, and it doesn't have to be u, it's just, that's the convention, it's called u-substitution, it could have been s-substitution for all we care. Let's say u is equal to the natural log of x, and then du dx, the derivative of u with respect to x, of course is equal to 1/x. Or, just the differential du, if we just multiply both sides by dx is equal to 1 over x dx. So let's make our substitution. So this will be equal to the indefinite integral, or the antiderivative, of 2 to the now u, so 2 to the u, Now what is 1 over x dx? That's just du. So this term times that term is just our du. Let me do it in a different color. 1 over x times dx is just equal to du. That's just equal to that thing, right there. Now, this still doesn't look like an easy integral, although it's gotten simplified a good bit. And to solve, you know, whenever I see the variable that I'm integrating against in the exponent, you know, we don't have any easy exponent rules here. The only thing that I'm familiar with, where I have my x or my variable that I'm integrating against in my exponent, is the case of e to the x. We know that the integral of e to the x, dx, is equal to e to the x plus c. So if I could somehow turn this into some variation of e to the x, maybe, or e to the u, maybe I can make this integral a So let's see. How can we redefine this right here? Well, 2, 2 is equal to what? 2 is the same thing as e to the natural log of 2, right? The natural log of 2 is the power you have to raise e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? Well, if I take something to an exponent, and then to another"
    },
    {
        "Q": "At 2:42 Sal says he wants to multiply the length of vector v by the vector itself (||v|| v), but then immediately after he defines the unit vector as \"1/||v|| v\". Where did he get the \"1/\" from?\n",
        "A": "If a vector v has a length of e.g. 5, then v/5 has length 1.",
        "video_name": "lQn7fksaDq0",
        "timestamps": [
            162
        ],
        "3min_transcript": "u, it's just equal to the square root of the squared sums of all of its components. And if you think about it, this is just an extension of the Pythagorean theorem, to some degree. So it's u1 squared plus u2 squared all the way to un squared. And it's the square root of that. If this is a unit vector, if this is a unit vector, so this is a unit vector, that implies that the length of u will be equal to 1. And that doesn't matter in what dimension space we are. This could be R100 this could be R2. For it to have a unit vector in any of those spaces, their length is 1. The next obvious question is, how do you construct a unit vector? Let's say that I have some vector, v. And let's say it's not a unit vector. And I want to turn it into some vector u that is a unit vector, that just goes in the same direction. So u will go in the same direction, as v, but the length of u is going to be equal to 1. How do I construct this vector u here? What I could do is, I could take the length of v. I could find out what the length of v is, and we know how to do that. We just apply this definition of vector length. And what happens if I figure out the length of v, and then I multiply the vector v times that? What if I make my u, what if I say u is equal to, 1 over the length of v times v itself? What happens here? here, what do I get? The length of u is equal to the length of this scalar. Remember this is just some number, right? It's equal to this scalar, and I'm assuming v is a non-zero vector. The length of whatever this scalar number is times v. And we know that we can take this scalar out of the formula, we can show that -- I think I've shown it in a previous video -- that the length of c times v is equal to c times the length of v. Let me write that down. And that's essentially what I'm assuming right here. That if I take the length of c times some vector, v, that is equal to c times the length of v. I think we showed this when we first were introduced to the idea of length. So we know that this is going to be equal to 1 over the"
    },
    {
        "Q": "At 0:50,why is it that dividing is the same as multiplying the fractions?\n",
        "A": "Dividing by a fraction is the same as multiplying by that fraction with it s denominator and numerator swapped.",
        "video_name": "yb7lVnY_VCY",
        "timestamps": [
            50
        ],
        "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject."
    },
    {
        "Q": "\nAt 2:32 you say that 5 + 5 cannot represent a vector. (5 represents a possible starting point on the number line and the additional 5 can represent a distance in the positive direction on the number line.)",
        "A": "A vector has to have BOTH a magnitude and a direction. As is, it is just a magnitude. Had there been a direction attached then it would be a vector.",
        "video_name": "n8Ic2Oj-zvA",
        "timestamps": [
            152
        ],
        "3min_transcript": "Voiceover:Which of the following can represent a vector? A vector is something that has both a magnitude and a direction. Let's see which one of these could represent something that has a magnitude and a direction. The first choice right over here is the number 5. 5 could represent a magnitude, you could say, essentially, how large something is. That's all the information has. It doesn't say 5 in a certain direction. This one by itself would not be a or it couldn't represent a vector. You need to specify a direction as well. The angle measure 5 degrees. Well the angle measure 5 degrees could represent a direction. If you say it's 5 degrees. Let's say that's the positive X axis that's the positive Y axis. If you said it's 5 degrees relative to the positive X axis. If you said something like this, so if you had 5 degrees over there, that would be specifying a direction, but it's not giving us a magnitude. It's not saying how far in that direction or how large in that direction. you're giving us a direction, but no magnitude. Let's just cross these out. The point (5,5). Now this is interesting. If we say that the point is the end point of a vector that starts at the origin, so lets draw that out. That's the positive X axis, positive Y axis. This is 1,2,3,4,5 in the horizontal direction. 1,2,3,4,5 in the positive vertical direction. The .5,5 is going to be right over there. If you said that this represents the head of a vector who's tail is at the origin. The vector would look something like this. Now, you're giving both a magnitude and a direction. What's the magnitude? What's the distance between the origin and this point, right over here. and the direction, it's this general direction. This could specify a magnitude and a direction. This could represent a vector. The outcome of 5+5, obviously that's just 10. That's just a number. For the same reason, this first choice, if I just have a number, that could be a magnitude, but we don't have a direction as well. This could not represent and that one could not represent a vector."
    },
    {
        "Q": "I haven't had the time to find the video to explain this type of equation so I'll just ask it here. If I had an equation such as 6t3rd + 9t2nd \u00e2\u0080\u0093 15t(that's 6t to the third plus 9t to the second minus 15t)how would I factor this to a form similar to the one shown at 3:35?\n",
        "A": "first factor out 3t so you get 3t(2t2nd+3t-5). Then factor (2t2nd+3t-5). to factor it, you need to find 2 numbers that multiply to make -10 (2*-5) and add together to make 3 (the middle number). those two numbers are 5 and -2. Split the middle term into those 2 numbers which gives you 3t(2t2nd-2t+5t-5). Then factor by grouping. that gives you 3t(2t(t-1)+5(t-1)). Simplify to 3t(2t+5)(t-1). The next video is very helpful in understanding this. Hope that helps!",
        "video_name": "u1SAo2GiX8A",
        "timestamps": [
            215
        ],
        "3min_transcript": "So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y. And so we have 2y times 2y minus 3, plus 5 times 2y minus 3. So now you have two terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3, so this is equal to 2y minus 3, times 2y, times that 2y, plus that 5. There's no magic happening here, all I did is undistribute the 2y minus 3. I factored it out of both of these guys. I took it out of the parentheses. If I distribute it in, you'd get back to this expression. But we're done, we factored it. We factored it into two binomial expressions. 4y squared plus 4y, minus 15 is 2y minus 3, times 2y plus 5."
    },
    {
        "Q": "\nAt 3:25, I'm confused about where the 5 came from. Can someone shed some light on this?",
        "A": "They divided 10y-15 by two and got 5(2y-3).",
        "video_name": "u1SAo2GiX8A",
        "timestamps": [
            205
        ],
        "3min_transcript": "So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y. And so we have 2y times 2y minus 3, plus 5 times 2y minus 3. So now you have two terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3, so this is equal to 2y minus 3, times 2y, times that 2y, plus that 5. There's no magic happening here, all I did is undistribute the 2y minus 3. I factored it out of both of these guys. I took it out of the parentheses. If I distribute it in, you'd get back to this expression. But we're done, we factored it. We factored it into two binomial expressions. 4y squared plus 4y, minus 15 is 2y minus 3, times 2y plus 5."
    },
    {
        "Q": "In 3:49, how does he know that the triangle is equilateral?\n",
        "A": "EC = EB as he has proven in 3:17 , and EB and CB are radiuses of the same circle, therefore EB = CB. And, if you think about it you will find out that you can say EC=EB=CB by combining the two relationships above. That s why it is equilateral.",
        "video_name": "3n0LvI99-KM",
        "timestamps": [
            229
        ],
        "3min_transcript": "So let me draw segment EC. I'll draw that as straight as possible. I can draw a better job of that, so segment EC. Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG. And then BG is equal to GC, and they both have 90-degree angles. You have a 90-degree angle here. You have a 90-degree angle there. So you see by side-angle-side that these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG-- I should emphasize the C, not the E-- And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC. So I could draw the three things right here. I'm referring to the entire thing, not just one of the segments, all of BC. So what kind of a triangle is this right over here, triangle BEC? And we know that because all three sides are equal, so that tells us that all of its angles are equal. So that tells us that the measure of angle BEC-- we're not done yet, but it gets us close-- is 60 degrees. So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees, and we're done. We've figured out the entire BED. Now, let's think about how we can do this right over here. So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here-- this is a square-- we know that this length down here is the same as this length up here, that these"
    },
    {
        "Q": "\nat 5:18 were did he get 9/3?????i understand the 3 but not the 9 plz help me",
        "A": "i am the helper",
        "video_name": "Zn-GbH2S0Dk",
        "timestamps": [
            318
        ],
        "3min_transcript": "And in general if you wanna work this out before I give you how I do it that now's a good time to actually pause the video. And you could, you could try to work it out and then, play it again and, and see what I have to say about it. But assuming you wanna hear it, let me go and do it. So let's do the same thing. We, first of all, we can merge these two Xs on the left-hand side. Remember, you can't add the 5 and the 3 because the 3 is just a constant term while the 5 is 5 times x. But the 5 times x and the negative 7 times actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is equal to x plus 8. Now, if we wanna take this x that's on the right-hand side and put it over the left-hand side, we can just subtract x from both sides. to, these two Xs cancel out, is equal to 8. Now, we can just add 3 to both sides to get rid of that constant term 3 on left hand-side. These two 3's will cancel out. And you get minus 3x is equal to 11. Now, you just multiply both sides by negative one-third. And once again, this is just the same thing as dividing both sides by negative 3. And you get x equals negative 11 over 3. Actually let's, let's, just for fun, let's check this just to see. And the cool thing about algebra is if you have enough time, you can always make sure you got the right answer. So we have 5x, so we have 5 times negative 11 over 3. So that's, I'm just, I'm just gonna take this and substitute it back into the original equation. So you have minus 55 over 3, that's just 5 times negative 11 over 3, that's a 3, minus 3. And what's 3? Three could also be written as, minus 9 over 3. I'm skipping some steps, but I think you, you know your fractions pretty good by this point. So that's minus 9 over 3. And then, minus 7x is the same thing, as plus 77 over 3. Because we have the minus 7 times minus 11, so it's plus 77. And, and the equation is saying that should equal minus 11 over 3, that's what x is, plus an 8 is nothing more than 24 over 3. Let's add this up. Minus 55 minus 9, that's minus 64, if I'm right, yeah, that's minus 64."
    },
    {
        "Q": "\nAt 3:00 i didnt get how he got 8",
        "A": "Sal doesn t seem to arriving at an 8 at 3:00. Maybe you mean how he gets -8 at 2:00? At that point he is adding -3 to both sides of the equation so he can isolate the term with x (i.e. the variable) on the left side of the = sign and the contants on the right side. When adding -3 to -5 you get -8.",
        "video_name": "Zn-GbH2S0Dk",
        "timestamps": [
            180
        ],
        "3min_transcript": "And we could do that by adding 7x to both sides, 7x. This is a review. We, we're adding the opposite. So, it's negative 7x, so we add 7x so that's why. And we do that, become the right side, these two will cancel. And the left side, we get 10x plus 3 equals, and on the right side, all we have left is the negative 5. Almost there, now we're at a level, what is this, a level two problem. And now we just have to take this 3 and move it to the other side. And we can do that by subtracting 3 from both sides. That's a 3 minus 3. The left-hand side, the 3s cancel out, that's why we subtract it in the first place. And you have 10x equals and then minus 5 minus 3, well that equals minus 8. the reciprocal of 10, which is the coefficient on x, times 1 over 10. You could also, some people would say, well, we're just dividing both side by 10 which is essential what we're doing. If you divide by 10, that's the same thing as multiplying by 1 over 10. Well, anyway, the left-hand side, 1 over 10 times 10. Well, that equals 1, so we're just left with x equals negative 8 over 10. And that can be reduced further. They both share the common factor 2. So you divide by 2. So it's minus 4 over 5. I think that's right, assuming that I didn't make any careless mistakes. Let's do another problem. Let's say I had 5, that's a 5x minus 3 minus And in general if you wanna work this out before I give you how I do it that now's a good time to actually pause the video. And you could, you could try to work it out and then, play it again and, and see what I have to say about it. But assuming you wanna hear it, let me go and do it. So let's do the same thing. We, first of all, we can merge these two Xs on the left-hand side. Remember, you can't add the 5 and the 3 because the 3 is just a constant term while the 5 is 5 times x. But the 5 times x and the negative 7 times actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is equal to x plus 8. Now, if we wanna take this x that's on the right-hand side and put it over the left-hand side, we can just subtract x from both sides."
    },
    {
        "Q": "At 5:41, Sal says that he could define the f(x) as -x^2-3x+28 or (x+7)(x-4), as they are the same thing.\nWhen you graph -x^2-3x+28 you get a parabola opening up\nWhen you graph (x+7)(x-4) you get a graph opening down\n\nHow can they be the same thing if their graphs are not the same?\n",
        "A": "Ok, so I think he misspoke when he said he could use either one, yes? Or is the sign for some reason not important in this case?",
        "video_name": "xdiBjypYFRQ",
        "timestamps": [
            341
        ],
        "3min_transcript": "try out the number 0 and make sure that 0 doesn't work, right, because 0 is between the two roots. It actually turns out that when x is equal to 0, f of x is minus 6, which is definitely less than 0. So I think this will give you a visual intuition of what this quadratic inequality means. Now with that visual intuition in the back of your mind, let's do some more problems and maybe we won't have to go through the exercise of drawing it, but maybe I will draw it just to make sure that the point hits home. Let me give you a slightly trickier problem. Let's say I had minus x squared minus 3x plus 28, let me say, is greater than 0. Well I want to get rid of this negative sign in front of the x squared. I just don't like it there because it makes it look more confusing to factor. I'm going to multiply everything by negative 1. I get x squared plus 3x minus 28, and when you multiply or to swap the sign. So this is now going to be less than 0. And if we were to factor this, we get x plus 7 times x minus 4 is less than 0. So if this was equal to 0, we would know that the two roots of this function -- let's define the function f of x -- let's define the function as f of x is equal to -- well we can define it as this or this because they're the same thing. But for simplicity let's define it as x plus 7 times x minus 4. That's f of x, right? Well, after factoring it, we know that the roots of this, the roots are x is equal to minus 7, and x is equal to 4. this inequality true? If this was any equality we'd be done. But we want to know what makes this inequality true. I'll give you a little bit of a trick, it's always going to be the numbers in between the two roots or outside of the two roots. So what I do whenever I'm doing this on a test or something, I just test numbers that are either between the roots or outside of the two roots. So let's pick a number that's between x equals minus 7 and x equals 4. Well let's try x equals 0. Well, f of 0 is equal to -- we could do it right here -- f of 0 is 0 plus 7 times 0 minus 4 is just 7 times minus 4, which is minus 28. So f of 0 is minus 28."
    },
    {
        "Q": "\n7:47 Is there a video where Sal proves this? Thanks",
        "A": "using the same diagram as in the video, imagine the following: draw segment AD and segment BC, angle AED is congruent to angle BEC by vertical angles theorem, angle DAE is congruent to angle BCE because angles that subtends the same arc are congruent (arc DB) so, triangle ADE is similar to triangle CBE by aa similarity postulate. corresponding sides of similar triangles are proportional, so AE/CE = DE/BE, cross multiply and you get the CHORD-CHORD PRODUCT THEOREM, AE*BE = CE*DE.",
        "video_name": "FJIZPvE3O1A",
        "timestamps": [
            467
        ],
        "3min_transcript": "This is its radius. Let's call that r. Well what's the area of the square going to be? If that's the radius, this is also the radius. So one side of this square up here, is going to be 2r. So this side is also going to be 2r. It's a square, all the sides are the same. So they want to know the ratio of the area of the circle to the area of the square. The area of the square is just 2r times 2r. Which is 4r squared. Area of the circle is just pi r squared. You hopefully learned the formula for area of a circle. Divide the numerator and the denominator by r squared. You're left with pi/4. That's choice D. In the circle below, AB and CD are chords intersecting at E. Fair enough. If AE is equal to 5, BE is equal to 12, what is the value of DE? CE is equal to 6. What is the value of DE. Let's call that x. Now, I'm not going to prove it here, just for saving time. But there's a neat property of chords within a circle. That if I have two chords intersecting a circle, it turns out that the two segments when you multiply them times each other, are always going to be equal to the same thing. So in this case, 5 times 12. That's going to be equal to these two segments multiplied by each other. It's going to be equal to x times 6. So you get 60 is equal to 6x. Divide both sides by 6, you get x is equal to 10. And that is choice C. That might be a fun thing for you to think about after this video of why that is. And maybe you want to play around with chords and prove to yourself that that's always the case. At least that it makes intuition for you, makes sense. RB is tangent to a circle. Tangent means that it just touches the outside of the circle right there at only one point. And it's actually perpendicular to the radius at that point. So this is the radius of that point. The center is at A. This is a radius. And it's tangent at point B, so it's perpendicular to the"
    },
    {
        "Q": "at 4:18, couldnt you go over 6 THEN up 4? if not why? please help! i just joined the pre-algebra class, and im way behind.\n",
        "A": "no because 6 would be a negative number (moving to the left) and 4 would be a positive number. If you did that,then you would have a negative slope. The answer, is a positive slope. Hope that helps :)",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            258
        ],
        "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope."
    },
    {
        "Q": "\nOn 3:36 he goes to the right but it still comes out negative. Why? If anyone knows the answer to this, please reply.",
        "A": "At 3:36 he goes 6 units to the left (-6) - he doesn t go to the right.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            216
        ],
        "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope."
    },
    {
        "Q": "can you go up first? So like at 3:24 he is going down first then over six. But from the point at -3 could you go up 4 and then over 6 to get to your other point?\n",
        "A": "Yes you can. Sal likely was more comfortable going down then over, but it will work either way.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            204
        ],
        "3min_transcript": "that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6?"
    },
    {
        "Q": "\nAt 2:57 Sal said that the change was -4. Why can't that number be 4? Does that change the outcome of the answer? Thanks!",
        "A": "You basically switched one of the values in y2-y1/x2-x1 so you made it y1-y2 as long as you switch the x value as well to x1-x2, then nothing changes but if you only switch the y, then it changes the slope.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            177
        ],
        "3min_transcript": "that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6?"
    },
    {
        "Q": "at 1:21, Sal draws a line going across, but I thought it was rise (which is going up), then run, (go across). Am I misunderstanding something?\n",
        "A": "Slope is rise/run, but it doesn t matter what order you calculate them in. If you find the run (x) first, just make sure you are putting that value in the denominator of your slope.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            81
        ],
        "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is"
    },
    {
        "Q": "At 0:25, why is it y over x and not x over y?\n",
        "A": "Slope is always rise over run.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            25
        ],
        "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is"
    },
    {
        "Q": "\nAt 1:04 how can a numerator be equal to 0 when there is only a value such 4/x-2",
        "A": "Well... When does 4 = 0? Let s keep an eye out for pigs on the wing.",
        "video_name": "fvC0dm2wzIo",
        "timestamps": [
            64
        ],
        "3min_transcript": "Voiceover:Right over here, I have the graph of f of x, and what I want to think about in this video is whether we could have sketched this graph just by looking at the definition of our function, which is defined as a rational expression. We have 2x plus 10 over 5x minus 15. There is a couple of ways to do this. First, you might just want to pick out any numbers that are really easy to calculate. For example, what happens when X is equal to 0? We could say f of 0 is going to be equal to, well, all the x term is going to be 0, so you're going to be left with 10 over negative 15, which is negative 10/15, which is negative 2/3. You can plot that one. x equals 0. f of x or y equals f of x is negative 2/3, and we see that point, let me do that in a darker color, you see that point right over there, so we could have plotted that point. We could also say, \"When does this function equal 0?\" Well, the function is equal to 0 when ... is if you get this numerator equal to 0, so you could try to solve 2x plus 10 is equal to 0. That's going to happen when 2x is equal to negative 10. I just subtracted 10 from both sides. If I divide both sides by 2, that's going to happen when x is equal to negative 5. You see that, you see this right over here. When x is equal to negative 5, the function intersects the x-axis. but that still doesn't give us enough to really form this interesting shape over here. You could think about what other functions have this type of shape. Now what I want to think about is the behavior of the function at different points. First, I want to think about when this function is undefined and what type of behavior we might expect for that function when it's undefined. This function is going to be undefined. The only way I can think of to make this undefined is if I make the denominator equal to 0. We don't know what it means to divide by 0. That is undefined. The function is going to be undefined let me do this in blue, when 5x minus 15 is equal to 0, or adding 15 to both sides, when 5x is equal to 15, or dividing both sides by 5, when x is equal to 3, f is undefined. Now, there is a couple of ways for a function to be undefined at a point. You could have something like this. Let me draw some axes right over here. Let's say that this is 3. You could have your function, it could look like this. It could be defined. It might approach something but just not be defined right at 3 and then just keep on going like that, or the other possibility is it might have a vertical asymptote there. If it has a vertical asymptote, it's going to look something like this. It might be approach, it might just pop up to infinity, and it might pop down from infinity on this side, or it might go from negative infinity right over here."
    },
    {
        "Q": "3:19 Shouldn't it be commutative instead of associative, since commutative is about switching orders? Thanks.\n",
        "A": "Idk! I m just a fifth grader, so... -Brianna-",
        "video_name": "vr0sTKbV7lI",
        "timestamps": [
            199
        ],
        "3min_transcript": "This is x is equal to-- let's see, 1, 2, 3. y is between 0 and 4. 1, 2, 3, 4. So the x-y plane will look something like this. The kind of base of our cube will look something like this. And then z is between 0 and 2. So 0 is the x-y plane, and then 1, 2. So this would be the top part. And maybe I'll do that in a slightly different color. So this is along the x-z axis. You'd have a boundary here, and then it would come in like this. You have a boundary here, come in like that. A boundary there. So we want to figure out the volume of this cube. And you could do it. You could say, well, the depth is 3, the base, the width is 4, so this area is 12 times the height. 12 times 2 is 24. You could say it's 24 cubic units, whatever units we're doing. But let's do it as a triple integral. Well, what we could do is we could take the volume of a very small-- I don't want to say area-- of a very small volume. So let's say I wanted to take the volume of a small cube. Some place in this-- in the volume under question. And it'll start to make more sense, or it starts to become a lot more useful, when we have variable boundaries and surfaces and curves as boundaries. But let's say we want to figure out the volume of this little, small cube here. That's my cube. It's some place in this larger cube, this larger rectangle, cubic rectangle, whatever you want to call it. So what's the volume of that cube? Let's say that its width is dy. So that length right there is dy. It's height is dx. Sorry, no, it's height is dz, right? The way I drew it, z is up and down. And it's depth is dx. This is dx. This is dz. This is dy. So you can say that a small volume within this larger volume differential. And that would be equal to, you could say, it's just the width times the length times the height. dx times dy times dz. And you could switch the orders of these, right? Because multiplication is associative, and order doesn't matter and all that. But anyway, what can you do with it in here? Well, we can take the integral. All integrals help us do is help us take infinite sums of infinitely small distances, like a dz or a dx or a dy, et cetera. So, what we could do is we could take this cube and first, add it up in, let's say, the z direction. So we could take that cube and then add it along the up and down axis-- the z-axis-- so that we get the volume of a column. So what would that look like? Well, since we're going up and down, we're adding-- we're taking the sum in the z direction. We'd have an integral."
    },
    {
        "Q": "At 5:54 is the trapezoid the same thing as a trapezium?\n",
        "A": "This is a very good question. If you are american then the accepted answer would be no. By american definition a trapezium has no parralel sides while a trapezoid has 1 set of sides that are parallel to each other. The British definition states that a trapezium has 2 sides that are parallel which would be the same thing as a trapezoid.",
        "video_name": "10dTx1Zy_4w",
        "timestamps": [
            354
        ],
        "3min_transcript": "now you might recognize this purple one on the left is a special type a not only are all the corners square not all do they have these right angles but all the sides are the same length at least it looks like it from the picture this side is the same length as this side is the same length as that side same length as that side. So if the sides are of same length and all the corners are square where they have right angles, then we call it a square so this right over here is a, this right over here is a square special type a rectangle this one not all the side to the same this one all the sides are the same out if we go down here looks like all the sides are the same one two three four but it doesn't have square corners. Notice if I tried to put a little square here it wouldn't fit perfectly, if I tried square here doesn't fit perfectly these So if all four sides are the same and that's all we know about it we call it a rhombus. We call it a rhombus so you might be saying, the square all four sides are the same, is that a rhombus? yes it is. So the square is a rectangle it's a rectangle and is a rhombus so this last shape here this last quadrilateral what's interesting about it is you have two of the side's going in the same direction so this one right over here knows what I mean by the same direction it sometimes is called parallel if I draw two lines here these two lines are never going to cross each other so these two lines are parallel if you have opposite one pair of opposite sides that are parallel we call them trapezoids we call them trapezoids. Notice the other blinds the other opposite sides these will cross, these will cross"
    },
    {
        "Q": "\nI dont understand 3:17?",
        "A": "You can t have a domain that maps to more than 1 number in the range.",
        "video_name": "Uz0MtFlLD-k",
        "timestamps": [
            197
        ],
        "3min_transcript": "maybe that is associated with 2 as well. So 2 is also associated with the number 2. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number 3. 3 is in our domain. Our relation is defined for number 3, and 3 is associated with, let's say, negative 7. So this is 3 and negative 7. Now this type of relation right over here, where if you give me any member of the domain, and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me 1, I say, hey, it definitely maps it to 2. You give me 2, it definitely maps to 2 as well. You give me 3, it's definitely associated with negative 7 as well. So this relation is both a-- it's obviously a relation-- but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here, and I do this big, fuzzy cloud-looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big, fuzzy cloud-looking thing is the range. And let's say in this relation-- and I'll build it the same way that we built it over here-- let's say in this relation, 1 is associated with 2. So let's build the set of ordered pairs. So 1 is associated with 2. Let's say that 2 is associated with, let's say that 2 is associated with negative 3. And let's say on top of that, we also associate, we also associate 1 with the number 4. So we also created an association with 1 with the number 4. So we have the ordered pair 1 comma 4. Now this is a relationship. We have, it's defined for a certain-- if this was a whole relationship, then the entire domain is just the numbers 1, 2-- actually just the numbers 1 and 2. It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is, if you tell me, OK I'm giving you 1 in the domain, what member of the range is 1 associated with? Over here, you say, well I don't know, is 1 associated with 2, or is it associated with 4? It could be either one. So you don't have a clear association. If I give you 1 here, you're like, I don't know, do I hand you a 2 or 4? That's not what a function does. A function says, oh, if you give me a 1, I know I'm giving you a 2. If you give me 2, I know I'm giving you 2. Now with that out of the way, let's actually"
    },
    {
        "Q": "At 1:05, I was wondering if there was an actual name for the type of function which maps 2 numbers in the domain to just 1 in the range, e.g. f(x)=x^2\n",
        "A": "Good question! We say that functions are not one-to-one. :) Therefore, what we actually say is that one-to-one functions are functions for which each element in the domain is paired with its own unique element of the range (or, that each element of the range is paired with exactly one element from the domain). Hope that helps!",
        "video_name": "Uz0MtFlLD-k",
        "timestamps": [
            65
        ],
        "3min_transcript": "Is the relation given by the set of ordered pairs shown below a function? So before we even attempt to do this problem, right here, let's just remind ourselves what a relation is and what type of relations can be functions. So in a relation, you have a set of numbers that you can kind of view as the input into the relation. We call that the domain. You can view them as the set of numbers over which that relation is defined. And then you have a set of numbers that you can view as the output of the relation, or what the numbers that can be associated with anything in domain, and we call that the range. And it's a fairly straightforward idea. So for example, let's say that the number 1 is in the domain, and that we associate the number 1 with the number 2 in the range. So in this type of notation, you would say that the relation has 1 comma 2 in its set of ordered pairs. These are two ways of saying the same thing. maybe that is associated with 2 as well. So 2 is also associated with the number 2. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number 3. 3 is in our domain. Our relation is defined for number 3, and 3 is associated with, let's say, negative 7. So this is 3 and negative 7. Now this type of relation right over here, where if you give me any member of the domain, and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me 1, I say, hey, it definitely maps it to 2. You give me 2, it definitely maps to 2 as well. You give me 3, it's definitely associated with negative 7 as well. So this relation is both a-- it's obviously a relation-- but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here, and I do this big, fuzzy cloud-looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big, fuzzy cloud-looking thing is the range. And let's say in this relation-- and I'll build it the same way that we built it over here-- let's say in this relation, 1 is associated with 2. So let's build the set of ordered pairs. So 1 is associated with 2. Let's say that 2 is associated with, let's say that 2 is associated with negative 3."
    },
    {
        "Q": "\nat 1:37 why did he multiply 5x5 ?",
        "A": "He evaluated what was inside the parentheses first: (9-4) = 5 So, 5(9-4) = 5(5) = 25",
        "video_name": "Badvask-UDU",
        "timestamps": [
            97
        ],
        "3min_transcript": "Rewrite the expression five times 9 minus 4-- that's in parentheses-- using the distributive law of multiplication over subtraction. Then simplify. So let me just rewrite it. This is going to be 5 times 9 minus 4, just like that. Now, if we want to use the distributive property, well, You could just evaluate 9 minus 4 and then multiply that times 5. But if you want to use the distributive property, you distribute the 5. You multiply the 5 times the 9 and the 4, so you end up with 5 times 9 minus 5 times 4. Notice, we distributed the 5. We multiplied it times both the 9 and the 4. In the first distributive property video, we gave you an idea of why you have to distribute the 5, why it makes And we're going to verify that it gives us the same answer as if we just evaluated the 9 minus 4 first. But anyway, what are these things? So 5 times 9, that is 45. So we have 45 minus-- what's 5 times 4? Well, that's 20. 45 minus 20, and that is equal to 25, so this is using the distributive property right here. If we didn't want to use the distributive property, if we just wanted to evaluate what's in the parentheses first, we would have gotten-- let's go in this direction-- 5 times-- what's 9 minus 4? 9 minus 4 is 5. Let me do that in a different color. 5 times 9 minus 4. So it's 5 times 5. 5 times 5 is just 25, so we get the same answer either way. This is using the distributive law of multiplication over subtraction, usually just referred to as the This is evaluating the inside of the parentheses first and then multiplying by 5."
    },
    {
        "Q": "I dont get what he says at 2:05\n",
        "A": "That 74.7 is one sigma away from the mean. What he should have said maybe would be like this. Where does that get us?Well, 81-6.3 is 74.7 which is one standard deviation from the mid",
        "video_name": "Wp2nVIzBsE8",
        "timestamps": [
            125
        ],
        "3min_transcript": "Here's the second problem from CK12.org's AP statistics FlexBook. It's an open source textbook, essentially. I'm using it essentially to get some practice on some statistics problems. So here, number 2. The grades on a statistics midterm for a high school are normally distributed with a mean of 81 and a standard deviation of 6.3. All right. Calculate the z-scores for each of the following exam grades. Draw and label a sketch for each example. We can probably do it all on the same example. But the first thing we'd have to do is just remember what is a z-score. What is a z-score? A z-score is literally just measuring how many standard deviations away from the mean? Just like that. So we literally just have to calculate how many standard deviations each of these guys are from the mean, and So let me do part a. So we have 65. So first we can just figure out how far is 65 from the mean. Let me just draw one chart here that we can use the entire time. So it's just our distribution. Let's see. We have a mean of 81. That's our mean. And then a standard deviation of 6.3. So our distribution, they're telling us that it's normally distributed. So I can draw a nice bell curve here. They're saying it's normally distributed, so that's as good of a bell curve as I'm capable of drawing. This is the mean right there at 81. And the standard deviation is 6.3. So one standard deviation above and below is going to be 6.3 away from that mean. So if we go 6.3 in the positive direction, that value right there is going to be 87.3. where does that get us? What, 74.7? Right, if we add 6, it'll get us to 80.7, and then 0.3 will get us to 81. So that's one standard deviation below and above the mean, and then you'd add another 6.3 to go 2 standard deviations, so on and so forth. So that's a drawing of the distribution itself. So let's figure out the z-scores for each of these grades. 65 is how far? 65 is maybe going to be here someplace. So we first want to say, well how far is it just from our mean? So the distance is, you just want to positive number here. Well actually, you want a negative number. Because you want your z-score to be positive or negative. Negative would mean to the left of the mean and positive would mean to the right of the mean. So we say 65 minus 81. So that's literally how far away we are."
    },
    {
        "Q": "second q is 1/4??? Paused @ o2:42\n",
        "A": "there are 13 hearts in a 52 card standard ddeck of playing cards...1/4.....13/52",
        "video_name": "obZzOq_wSCg",
        "timestamps": [
            162
        ],
        "3min_transcript": "a Jack of spades, or a Jack of hearts. So if you just multiply these two things-- you could take a deck of playing cards, take out the jokers and count them-- but if you just multiply this you have four suits, each of those suits have 13 types. So you're going to have 4 times 13 cards, or you're going to have 52 cards in a standard playing deck. Another way you could have said, look, there's 13 of these ranks, or types, and each of those come in four different suits-- 13 times 4. Once again, you would have gotten 52 cards. Now, with that of the way, let's think about the probabilities of different events. So let's say I shuffle that deck. I shuffle it really, really well and then I randomly pick a card from that deck. And I want to think about what is the probability that I pick a Jack. Well, how many equally likely events are there? Well, I could pick any one of those 52 cards. So there's 52 possibilities for when I pick that card. And how many of those 52 possibilities are Jacks? the Jack of clubs, and the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these are both divisible by 4-- 4 divided by 4 is 1, 52 divided by 4 is 13. Now, let's think about the probability. So I'll start over. I'm going to put that Jack back and I'm going to reshuffle the deck. So once again, I still have 52 cards. So what's the probability that I get a hearts? What's the probability that I just randomly pick a card from a shuffled deck and it is a heart? Well, once again, there's 52 possible cards I could pick from. 52 possible, equally likely events that we're dealing with. And how many of those have our hearts? Well, essentially 13 of them are hearts. For each of those suits you have 13 types. So there are 13 hearts in that deck. There are 13 spades in that deck. There are 13 clubs in that deck. So 13 of the 52 would result in hearts, and both of these are divisible by 13. This is the same thing as 1/4. One in four times I will pick it out, or I have a one in four probability of getting a hearts when I randomly pick a card from that shuffled deck. Now, let's do something that's a little bit more interesting, or maybe it's a little obvious. What's the probability that I pick something that is a Jack-- I'll just write J-- and it is a hearts? Well, if you are reasonably familiar with cards you'll know that there's actually only one card that is both a Jack and a heart. It is literally the Jack of hearts. So we're saying, what is the probability that we pick the exact card, the Jack of hearts? Well, there's only one event, one card, that meets this criteria right over here, and there's 52 possible cards."
    },
    {
        "Q": "\nin 00:20, what did Sal mean by Suits??",
        "A": "thanks a lot!",
        "video_name": "obZzOq_wSCg",
        "timestamps": [
            20
        ],
        "3min_transcript": "Let's do a little bit of probability with playing cards. And for the sake of this video, we're going to assume that our deck has no jokers in it. You could do the same problems with the joker, you'll just get slightly different numbers. So with that out of the way, let's first just think about how many cards we have in a standard playing deck. So you have four suits, and the suits are the spades, the diamonds, the clubs, and the hearts. You have four suits and then in each of those suits you have 13 different types of cards-- and sometimes it's called the rank. You have the ace, then you have the two, the three, the four, the five, the six, seven, eight, nine, ten, and then you have the Jack, the King, and the Queen. And that is 13 cards. So for each suit you can have any of these-- you can have any of the suits. a Jack of spades, or a Jack of hearts. So if you just multiply these two things-- you could take a deck of playing cards, take out the jokers and count them-- but if you just multiply this you have four suits, each of those suits have 13 types. So you're going to have 4 times 13 cards, or you're going to have 52 cards in a standard playing deck. Another way you could have said, look, there's 13 of these ranks, or types, and each of those come in four different suits-- 13 times 4. Once again, you would have gotten 52 cards. Now, with that of the way, let's think about the probabilities of different events. So let's say I shuffle that deck. I shuffle it really, really well and then I randomly pick a card from that deck. And I want to think about what is the probability that I pick a Jack. Well, how many equally likely events are there? Well, I could pick any one of those 52 cards. So there's 52 possibilities for when I pick that card. And how many of those 52 possibilities are Jacks? the Jack of clubs, and the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these are both divisible by 4-- 4 divided by 4 is 1, 52 divided by 4 is 13. Now, let's think about the probability. So I'll start over. I'm going to put that Jack back and I'm going to reshuffle the deck. So once again, I still have 52 cards. So what's the probability that I get a hearts? What's the probability that I just randomly pick a card from a shuffled deck and it is a heart? Well, once again, there's 52 possible cards I could pick from. 52 possible, equally likely events that we're dealing with. And how many of those have our hearts? Well, essentially 13 of them are hearts. For each of those suits you have 13 types. So there are 13 hearts in that deck."
    },
    {
        "Q": "\nGreat video! but just a word not clear... at 5:48 what is being said..four fifty what?",
        "A": "4/52=1/13 is the probability of a jack being drawn.",
        "video_name": "obZzOq_wSCg",
        "timestamps": [
            348
        ],
        "3min_transcript": "pick the Jack of hearts-- something that is both a Jack and it's a heart. Now, let's do something a little bit more interesting. What is the probability-- you might want to pause this and think about this a little bit before I give you the answer. What is the probability of-- so I once again, I have a deck of 52 cards, I shuffled it, randomly pick a card from that deck-- what is the probability that that card that I pick from that deck is a Jack or a heart? So it could be the Jack of hearts, or it could be the Jack of diamonds, or it could be the Jack of spades, or it could be the Queen of hearts, or it could be the two of hearts. So what is the probability of this? And this is a little bit more of an interesting thing, because we know, first of all, that there are 52 possibilities. But how many of those possibilities meet these conditions that it is a Jack or a heart. And to understand that, I'll draw a Venn diagram. Sounds kind of fancy, but nothing fancy here. represents all of the outcomes. So if you want, you could imagine it has an area of 52. So this is 52 possible outcomes. Now, how many of those outcomes result in a Jack? So we already learned, one out of 13 of those outcomes result in a Jack. So I could draw a little circle here, where that area-- and I'm approximating-- represents the probability of a Jack. So it should be roughly 1/13, or 4/52, of this area right over here. So I'll just draw it like this. So this right over here is the probability of a Jack. There's four possible cards out of the 52. So that is 4/52, or one out of 13. Now, what's the probability of getting a hearts? Well, I'll draw another little circle here that represents that. 13 out of 52 cards represent a heart. and a Jack. So I'm actually going to overlap them, and hopefully this will make sense in a second. So there's actually 13 cards that are a heart. So this is the number of hearts. And actually, let me write this top thing that way as well. It makes it a little bit clearer that we're actually looking at the number of Jacks. And of course, this overlap right here is the number of Jacks and hearts-- the number of items out of this 52 that are both a Jack and a heart-- it is in both sets here. It is in this green circle and it is in this orange circle. So this right over here-- let me do that in yellow since I did that problem in yellow-- this right over here is a number of Jacks and hearts. So let me draw a little arrow there. It's getting a little cluttered, maybe"
    },
    {
        "Q": "This is where the ray terminates.\n1:40\nIt's an endpoint.\nLOOK at the above statement In regarding to the ray, he messed up as there is not the end point but rather the commencement point or the source of the ray ,that is not where it terminates but starts were it a line transfoming into a ray then we would say thats where it terminates and an end point but since its genesis is where there is a point it is wrong to refer it as an end or use terminate what do you think?\n",
        "A": "It starts there, endlessly going in whatever direction. But it ends there, because when the ray reaches that part, it ends.",
        "video_name": "DkZnevdbf0A",
        "timestamps": [
            100
        ],
        "3min_transcript": "Any pair of points can be connected by a line segment. That's right. Connect two pairs of black points in a way that creates two parallel line segments. So let's see if we can do that. So I could create one segment that connects this point to this point and then another one that connects this point to this point. And they look pretty parallel. In fact, I think this is the right answer. If we did it another way, if we had connected that point to that point and this point to this point, then it wouldn't look so parallel. These clearly, if they were to keep going, they would intersect at some point. So let me set it back up the way I did it the first time. Let me make these two points parallel. And these are line segments because they have two end points. They each have two end points. And they continue forever. Well they don't continue forever. They continue forever in no directions, in zero directions. If it was a ray, it would continue forever If it's a line, it continues forever In fact, it wouldn't have end points because it would just continue forever in both of these directions. Let's do one more. Drag the ray so it has an endpoint at A, so we want to make its endpoint at A where the ray terminates and goes through one of the other black points. The ray should also be parallel to the pink line. So I have two options. I could make it go through this black point, but it's clearly not parallel. In fact, it looks perpendicular here. So let's try to make it go through this point. Well, yes, when I do that, it does indeed look like my ray is parallel to the pink line. And this is a ray because it has one endpoint. This is where the ray terminates. It's an endpoint. It literally ends there. And it continues forever in one direction. In this case, the direction is to the right. It continues forever to the right. So it continues forever in one-- continues forever in one direction."
    },
    {
        "Q": "At 0:40, he puts a zero as a place holder. my teachers at school say to just drop the number. Wouldn't that change the answer? Could you tell me which one is correct and why?\n",
        "A": "i think you have gotten confused. you can only do that while you are adding decimals. but i guess you can also do that if the number is at the top and it is the larger one over zero hope this helps!",
        "video_name": "MufbvU4tGh8",
        "timestamps": [
            40
        ],
        "3min_transcript": "Let's try to calculate 10.1 minus 3.93. And I encourage you to pause this video and try it on your own first, and then we can think about whether we did it the same way. So let's just rewrite it, aligning the decimal and the different place values. So 10.1 minus-- the 3 is in the ones place, so I'll put it right under the 0-- 3.93. Now, let's just try to calculate this. Now, before we subtract, we want all the numbers on top to be larger than the numbers on the bottom. And we don't even have a number here. We could stick a 0 here. Let me do that in a different color here. We could stick a 0 here. 10.1 is the same thing as 10.10, but we still face an issue 0 is less than 3. 1 is less than 9. 0 is less than 3. So we're going to do a little bit of regrouping. So let's do that regrouping. So we could take a 10 away, one 10 away, and then one 10 is the same thing as 10 ones. So I could write a 10 in the ones place. away so I'm going to have nine ones, and give that one to the tenths place. Well, one is 10 tenths. 10 tenths plus 1 tenth is going to be 11 tenths. Now, I could take one of those tenths away and give it to the hundredths place. 1 tenth is 10 hundredths. And now I have a higher digit in the numerator, or at least as equal in the numerator as I have in the denominator. So 10 minus 3 is 7. 10 minus 9 is 1. I have the decimal. 9 minus 3 is 6. And then I have nothing over here. So 10.1 minus 3.93, 6.17."
    },
    {
        "Q": "\nWait I do not get it do you have to do what he said at 0:55",
        "A": "You sort of do. Since 3 is larger than 0, you need to borrow. The 1 ten is the same as 10 ones. So if you add them together, you d get ten. Now you can subtract with 10 and 3.",
        "video_name": "MufbvU4tGh8",
        "timestamps": [
            55
        ],
        "3min_transcript": "Let's try to calculate 10.1 minus 3.93. And I encourage you to pause this video and try it on your own first, and then we can think about whether we did it the same way. So let's just rewrite it, aligning the decimal and the different place values. So 10.1 minus-- the 3 is in the ones place, so I'll put it right under the 0-- 3.93. Now, let's just try to calculate this. Now, before we subtract, we want all the numbers on top to be larger than the numbers on the bottom. And we don't even have a number here. We could stick a 0 here. Let me do that in a different color here. We could stick a 0 here. 10.1 is the same thing as 10.10, but we still face an issue 0 is less than 3. 1 is less than 9. 0 is less than 3. So we're going to do a little bit of regrouping. So let's do that regrouping. So we could take a 10 away, one 10 away, and then one 10 is the same thing as 10 ones. So I could write a 10 in the ones place. away so I'm going to have nine ones, and give that one to the tenths place. Well, one is 10 tenths. 10 tenths plus 1 tenth is going to be 11 tenths. Now, I could take one of those tenths away and give it to the hundredths place. 1 tenth is 10 hundredths. And now I have a higher digit in the numerator, or at least as equal in the numerator as I have in the denominator. So 10 minus 3 is 7. 10 minus 9 is 1. I have the decimal. 9 minus 3 is 6. And then I have nothing over here. So 10.1 minus 3.93, 6.17."
    },
    {
        "Q": "\nAt 5:00, all I have to do is graph is x=5 and x=-1 correct?",
        "A": "I don t believe you would graph x=5 and x=-1 because the solutions are points, not lines. What you have are vertical lines, and we don t really want that, because the solutions to the quadratic are one x with one y only. When asked for the answers, you are correct to say that x=5 and -1. But do not actually graph that.",
        "video_name": "VTlvg4wJ1X0",
        "timestamps": [
            300
        ],
        "3min_transcript": "that looks something like that. Let's say that the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x-values where you intersect, where you intersect the x-axis. Well, in order to intersect the x-axis, y must be equal to zero. So, y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So, figuring out the x-values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying, \"For what x-values does f of x equal zero?\" So we could just say, \"For what x-values does this thing right over here \"equal zero?\" So, let me just write that down. So we could rewrite this as x, x minus two squared minus nine equals zero. x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So, we could say x minus two is equal to positive three or x minus two is equal to negative three. Well, you add two to both sides of this, you get x is equal to five, or x is equal to, if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three, squared is nine, minus nine is zero. So, the point five comma zero is going to be on this graph. And also, if x is equal to negative one, negative one minus two, negative three. Squared is positive nine, minus nine is zero. So, also the point negative one comma zero So those are the points where, those are the x-values where the function intersects the x-axis."
    },
    {
        "Q": "At about 7:56 Sal simplifies 2x^2 - 3x^2 into -1x^2... How come the outcome isn't -1x^4?\n",
        "A": "If we let x\u00c2\u00b2 = y Then our problem becomes: 2y - 3y Which is obviously equal to: -1y But since y = x\u00c2\u00b2 We have: 2x\u00c2\u00b2 - 3x\u00c2\u00b2 = -1x\u00c2\u00b2 = -x\u00c2\u00b2",
        "video_name": "aoXUWSwiDzE",
        "timestamps": [
            476
        ],
        "3min_transcript": "negative 4 is negative 8, or 15 plus negative 8 is 15 minus 8. And now, the numerator becomes negative 3 minus 8, which is negative 11. And the denominator is 15 minus 8, which is 7. So problem 10, we simplified it to negative 11 over 7. Right there. Let's do a couple of these over here. OK, we see some exponents. I'll pick one of the harder ones. Let's do this one over here, problem 18. So 2x squared minus 3x squared, plus 5x minus 4. OK, well, this wasn't that hard. But what we could do here-- let me write this down. 2x squared minus 3x squared, plus 5x minus 4. And they tell us that x is equal to negative 1. One thing we could do is simplify this before we even substitute for negative 1. So what's 2 of something minus 3 of something? This is 2x squareds minus 3x squareds. So 2 of something minus 3 of something, that's going to be minus 1 of that something. So that right there-- or negative 1 of that something-- that would be negative 1x squared plus 5x minus 4. And they tell us x is equal to negative 1. So this is negative 1 times x squared, negative 1 squared, plus 5 times x, which is negative 1, minus 4. So what is this? Negative 1 squared is just 1. So this whole expression simplifies to negative 1 plus 5 times negative 1-- we do the multiplication first, of course. So that's minus 5, or negative 5 minus 4. So negative 1 minus 5 is negative 6, minus 4 is equal to negative 10. And I'll do these last two just to get a sample of all of the types of problems in this variable expression section. The weekly cost, c, of manufacturing x remote controls-- so the cost is c, x is the remote controls-- is given by this formula. The cost is equal to 2,000 plus 3 times the number of remote controls, where cost is given in dollars. Question a, what is the cost of producing 1,000 remote controls?"
    },
    {
        "Q": "\n@ 4:43 #6 Why is \"\" + -4 \"' The same as \"\" - 4 \"\" ?",
        "A": "The plus and the negative sign before that four immediately cancel each other out. Same as -(-4) : the minus negative x become a plus x.",
        "video_name": "aoXUWSwiDzE",
        "timestamps": [
            283
        ],
        "3min_transcript": "And every time we see a d, we'll put a minus 4 there. And I'll do a couple of these. I won't do all of them, just for the sake of time. So let's say problem number 5. They gave us 2 times a plus 3 times b. Well, this is the same thing as 2 times-- instead of an a, we know that a is going to be equal to negative 3. So 2 times minus 3, plus 3 times b-- what's b? They're telling us that b is equal to 2-- so 3 times 2. And what is this equal to? 2 times minus 3-- let me do it in a different color-- 2 times negative 3 is negative 6, plus 3 times 2. 3 times 2 is 6. That's positive 6. So that is equal to 0. And notice the order of operations. before we added the two numbers. Multiplication and division takes precedence over addition and subtraction. Let's do problem 6. I'll do that right here. So you have 4 times c. 4 times-- now what's c equal to? They tell us c is equal to 5. So 4 times 5, that's our c, plus d. d is minus or negative 4. So we have 4 times 5 is 20, plus negative 4-- that's the same thing as minus 4, so that is equal to 16. Problem 6. Now, let's do one of the harder ones down here. This problem 10 looks a little bit more daunting. Problem 10 right there. So we have a minus 4b in the numerator, if you can read it, it's kind of small. a is minus 3. b is 2. So 4 times 2. Remember, this right here is a, that right there is b. They're telling me up here. And then all of that over-- all of that is over 3c plus 2d. So 3 times-- what was c? c is 5 plus 2 times d. What is d? d is negative 4. So let's figure this out. So we have to do order of operations. Multiplication comes first before addition and subtraction. So this is going to be equal to minus 3 minus 4 times 2,"
    },
    {
        "Q": "\nat 00:41, where did you get the x and it shows 2 x11r?",
        "A": "it is actually an x it looks like an r",
        "video_name": "aoXUWSwiDzE",
        "timestamps": [
            41
        ],
        "3min_transcript": "Let's do some practice problems dealing with variable expressions. So these first problems say write the following in a more condensed form by leaving out the multiplication symbol or leaving out a multiplication symbol. So here we have 2 times 11x, so if we have 11 x's and then we're going to have 2 times those 11 x's, we're going to have 22 x's. So another way you could view this, 2 times 11x, you could view this as being equal to 2 times 11, and all of that times x, and that's going to be equal to 22 x's. You had 11 x's, you're going to have 2 times as many x's, so you're going to have 22 x's. Let's see, you have 1.35 times y. Now here we're just going to do a straight simplifying how we write it. So 1.35 times y-- I'll do it in a different color-- 1.35 times y-- that's a little dot there. If we have a variable following a number, we know that means 1.35 times that variable. So that, we could rewrite as just being equal to 1.35y. We've condensed it by getting rid of the multiplication sign. Let's see, here we have 3 times 1/4. Well, this is just straight up multiplying a fraction. So in problem 3-- this was problem 1, this is problem 2, problem 3-- 3 times 1/4, that's the same thing as 3 over 1 times 1/4. Multiply the numerators, you get 3. Multiply the denominators, 1 times 4, you get 4. So number 3, I got 3/4. And then finally, you have 1/4 times z. We could do the exact same thing we did up here in problem number 2. This was the same thing as 1.35y. That's the same thing as 1.35 times y. 1/4z, or we could view this as being equal to 1 over 4 times z over 1, which is the same thing as z times 1, over 4 times 1, or the same thing as z over 4. So all of these are equivalent. Now, what do they want us to do down here? Evaluate the following expressions for a is equal to 3, b is equal to 2, c is equal to 5, and d is equal to minus 4-- or, actually, I should say negative 4 is the correct terminology. Negative 4. So we just substitute. Every time we see an a, we're going to put a minus 3 there, or a negative 3 there. Every time we see a b, we'll put a positive 2 there."
    },
    {
        "Q": "\nAt 18:30 al finishes a problem as 40x10^-7 could that be simplified?",
        "A": "Yes it can, the number is really 0.000040, he simply wrote it in scientific notation (which should be explained in the video). You could think of scientific notation like a kind of shorthand. 0.000040 equals 40 \u00c3\u0097 10^-7, so you can write is either way, it s simply easier, and less cumbersome, to write it in scientific notation.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            1110
        ],
        "3min_transcript": "at this small a dose. Or maybe you would, I don't want to get into that. But how would I write this in scientific notation? So I start off with the first non-zero number, if I'm starting from the left. So it's going to be 8.192. I just put a decimal and write 0.192 times-- times 10 to what? Well, I just count. Times 10 to the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I have to include that number, 10 to the minus 10. And I think you'll find it reasonably satisfactory that this number is easier to write than that number over there. Now, and this is another powerful thing about scientific notation. Let's say I have these two numbers and I want to multiply them. Let's say I want to multiply the number 0.005 times the number 0.0008. This is actually a fairly straightforward one to do, And especially if you're dealing with twenty or thirty 0's on either sides of the decimal point. Put a 0 here to make my wife happy. But when you do it in scientific notation, it will actually simplify it. This guy can be rewritten as 5 times 10 to the what? I have 1, 2, 3 spaces behind the decimal. 10 to the third. And then this is 8, so this is times 8 times 10 to the-- sorry, this is 5 times 10 to the minus 3. That's very important. 5 times 10 to the 3 would have been 5,000. Be very careful about that. Now, what is this guy equal to? This is 1, 2, 3, 4 places behind the decimal. So it's 8 times 10 to the minus 4. If we're multiplying these two things, this is the same thing as 5 times 10 to the minus 3 times 8 times 10 to the minus 4. There's nothing special about the scientific notation. It literally means what it's saying. So for multiplying, you could write it out like this. And multiplication, order doesn't matter. 10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have"
    },
    {
        "Q": "\n20:26 wouldnt 6.022x7.23 be att least 42 because 6x7=42",
        "A": "Yes, you are correct. Well done for noticing Sal s mistake :)",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            1226
        ],
        "3min_transcript": "10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have So this is a really small number. But the multiplication, when you do it in scientific notation, is actually fairly straightforward. This is going to be equal to 6.0-- let me write it properly. 6.022 times 10 to the 23rd times 7.23 times 10 to the minus 22. We can change the order, so it's equal to 6.022 times 7.23. That's that part. So you can view it as these first parts of our scientific notation times 10 to the 23rd times 10 to the minus 22. And now, this is-- you're going to do some little decimal multiplication right here. It's going to be-- some number-- 40 something, I think. I can't do this one in my head. But this part is pretty easy to calculate. I'll just leave this the way it is. But this part right here, this will be times. 10 to the 23rd times 10 to the minus 22. You get times 10 to the first power. And then this number, whatever it's going to equal, I'll just leave it out here since I don't have a calculator. 0.23. Let's see, it will be 7.2. Let's see, 0.2 times-- it's like a fifth. It'll be like 41-something. So this is approximately 41 times 10 to the 1. Or, another way is approximately-- it's going to be 410-something. And to get it right, you just have to actually perform this multiplication. So hopefully you see that scientific notation is, one, really useful for super large and super small numbers. And not only is it more useful to kind of understand the numbers and to write the numbers, but it also simplifies operating on the numbers."
    },
    {
        "Q": "\nAt 6:01, what is avogadro's number?",
        "A": "it s the number of atoms in one mole of a substance, and it s equal to 6.02214179 x 10^23",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            361
        ],
        "3min_transcript": "And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you a very popular search engine-- Google. Google is essentially just a misspelling of the word \"googol\" with the O-L. And I don't know why they called it Google. Maybe they got the domain name. Maybe they want to hold this much information. Maybe that many bytes of information. Or, it's just a cool word. Whatever it is-- maybe it was the founder's favorite number. But it's a cool thing to know. But anyway, I'm digressing. This is a googol. It's just 1 with a hundred 0's. But I could equivalently have just written that as 10 to 100, which is clearly an easier way. This is an easier way to write this. This is easier. In fact, this is so hard to write that I didn't even take the trouble to write it. It would have taken me forever. This was just twenty 0's right here. A hundred 0's I would have filled up this screen and you have found it boring. So I didn't even write it. So clearly, this is easier to write. But how can we write something that isn't a direct power of 10? How can we use the power of this simplicity? How can we use the power of the simplicity somehow? And to do that, you just need to make the realization. This number, we can write it as-- so this has how many digits in it? It has 1, 2, 3, and then twenty 0's. So it has 23 digits after the 6. 23 digits after the 6. So what happens if I use this-- if I try to get close to it with a power of 10? So what if I were to say 10 to the 23? Do it in this magenta. 10 to the 23rd power. That's equal to what? That equals 1 with 23 0's. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23. You get the idea, that's 10 to the 23rd."
    },
    {
        "Q": "at 12:44 Sal says with negative exponents with a base of 10 (unlike positive exponents) it cannot be thought of as the number of zeroes. (i.e. 10^5 = 10 000, but 10^-5 = .00001) That only has 4 zeroes. :(\nBut couldn`t you write the answer to 10^-5 as 0.00001? That has 5 zeroes. (Sorry this has been really bugging me for a while. :) )\n",
        "A": "No, the decimal is supposed to move 5 times. Think about it. 10^-5 can also be written as 1/10^5. 10^5=100000, and you do 1/100000. When you divide, you find the quotient to be 0.00001. Hope this helps!",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            764
        ],
        "3min_transcript": "Well, it's just 6. So 6 is equal to 6 times 10 to the 0. You wouldn't actually have to write it this way. This is much simpler, but it shows you that you really can express any number in scientific notation. Now, what if we wanted to represent something like this? I had started off the video saying in science you deal with very large and very small numbers. So let's say you had the number-- do it in this color. And you had 1, 2, 3, 4. And then, let's say five 0's. And then you have followed by a 7. Well, once again, this is not an easy number to deal with. But how can we deal with it as a power of 10? As a power of 10? So what's the largest power of 10 that fits into this number, that this number is divisible by? So let's think about it. All the powers of 10 we did before were going to positive or going to-- well, yeah, positive powers of 10. We could also do negative powers of 10. We know that 10 to the 0 is 1. 10 to the minus 1 is equal to 1/10, which is equal to 0.1. Let me switch colors. I'll do pink. 10 to the minus 2 is equal to 1 over 10 squared, which is equal to 1/100, which is equal to 0.01. And you I think you get the idea that the--, well, let me just do one more so that you can get the idea. 10 to the minus 3. 10 to the minus 3 is equal to 1 over 10 to the third, which is equal to 1/1,000, which is equal to the 0.001. So the general pattern here is 10 to the whatever negative power is however many places you're going to have behind the decimal point. So here, it's not the number of 0's. In here, 10 to the minus 3, you only have two 0's but you have three places behind the decimal point. So what is the largest power of 10 that goes into this? Well, how many places behind the decimal point do I have? I have 1, 2, 3, 4, 5, 6. and we're going to have six places behind the decimal point. And the last place is going to be a 1. So you're going to have Five 0's and a 1. That's 10 to the minus 6. Now, this number right here is 7 times this number. If we multiply this times 7, we get 7 times 1. And then we have 1, 2, 3, 4, 5, 6 numbers behind the decimal point. So 1, 2, 3, 4, 5, 6. So this number times 7 is clearly equal to the number that we started off with. So we can rewrite this number. Instead of writing this number every time, we can write it as being equal to this number. Or, we could write it as 7. This is equal to 7 times this number. But this number is no better than that number. But this number is the same thing as 10 to the minus 6. 7 times 10 to the minus 6."
    },
    {
        "Q": "\nWhat is the difference between estimating and guestimating? 4:57",
        "A": "An estimate is more certain than a guestimate",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            297
        ],
        "3min_transcript": "That's equal to 10. What's 10 squared? That's 10 times 10. That's 100. What is 10 to the third? 10 to the third is 10 times 10 times 10, which is equal to 1,000. I think you see a general pattern here. 10 to the 0 has no 0's. No 0's in it. 10 to the 1 has one 0. 10 to the second power-- I was going to say the two-th power. 10 to the second power has two 0's. Finally, 10 to the third has three 0's. Don't want to beat a dead horse here, but I think you get the idea. Three 0's. If I were to do 10 to the 100th power, what would that look like? I don't feel like writing it all out here, but it would be 1 followed by-- you could guess it-- a hundred 0's. So it would just be a bunch of 0's. And if we were to count up all of those 0's, you And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you a very popular search engine-- Google. Google is essentially just a misspelling of the word \"googol\" with the O-L. And I don't know why they called it Google. Maybe they got the domain name. Maybe they want to hold this much information. Maybe that many bytes of information. Or, it's just a cool word. Whatever it is-- maybe it was the founder's favorite number. But it's a cool thing to know. But anyway, I'm digressing. This is a googol. It's just 1 with a hundred 0's. But I could equivalently have just written that as 10 to 100, which is clearly an easier way. This is an easier way to write this. This is easier. In fact, this is so hard to write that I didn't even take the trouble to write it. It would have taken me forever. This was just twenty 0's right here. A hundred 0's I would have filled up this screen and you have found it boring. So I didn't even write it. So clearly, this is easier to write."
    },
    {
        "Q": "At 0:18 how do you know the difference between a large and a very large number?\n",
        "A": "One is BIG and the other is SMALL",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            18
        ],
        "3min_transcript": "I don't think it's any secret that if one were to do any kind of science, they're going to be dealing with a lot of numbers. It doesn't matter whether you do biology, or chemistry, or physics, numbers are involved. And in many cases, the numbers are very large. They are very, very large numbers. Very large numbers. Or, they're very, small, very small numbers. Very small numbers. You could imagine some very large numbers. If I were to ask you, how many atoms are there in the human body? Or how cells are in the human body? Or the mass of the Earth, in kilograms, those are very large numbers. If I were to ask you the mass of an electron, that would be a very, very small number. So any kind of science, you're going to be dealing with these. And just as an example, let me show you one of the most common numbers you're going to see, in especially chemistry. It's called Avogadro's number. Avogadro's number. And if I were write it in just the standard way of writing as-- do it in a new color. It would be 6022-- and then another 20 zeroes. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. And even I were to throw some commas in here, it's not going to really help the situation to make it more readable. Let me throw some commas in here. This is still a huge number. If I have to write this on a piece of paper or if I were to publish some paper on using Avogadro's number, it would take me forever to write this. And even more, it's hard to tell if I forgot to write a zero or if I maybe wrote too many zeroes. So there's a problem here. Is there a better way to write this? So is there a better way to write this than to write it all out like this? To write literally the 6 followed by the 23 digits, or the 6022 followed by the 20 zeroes there? you're curious, Avogadro's number, if you had 12 grams of carbon, especially 12 grams of carbon-12, this is how many atoms you would have in that. And just so you know, 12 grams is like a 50th of a pound. So that just gives you an idea of how many atoms are laying around at any point in time. This is a huge number. The point of here isn't to teach you some chemistry. The point of here is to talk about an easier way to write this. And the easier way to write this we call scientific notation. Scientific notation. And take my word for it, although it might be a little unnatural for you at this video. It really is an easier way to write things like things like that. Before I show you how to do it, let me show you the underlying theory behind scientific notation. If I were to tell you, what is 10 to the 0 power? We know that's equal to 1."
    },
    {
        "Q": "\nat 18:36 sal added two powers 10^-3 and 10^-4 = 10^-7. should it not be 10^7 ?",
        "A": "No, Sal is correct. -3 + -4 = -7. A negative plus a negative equals a bigger negative. (You may be getting confused with multiplication; a negative TIMES a negative is a positive.)",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            1116
        ],
        "3min_transcript": "And especially if you're dealing with twenty or thirty 0's on either sides of the decimal point. Put a 0 here to make my wife happy. But when you do it in scientific notation, it will actually simplify it. This guy can be rewritten as 5 times 10 to the what? I have 1, 2, 3 spaces behind the decimal. 10 to the third. And then this is 8, so this is times 8 times 10 to the-- sorry, this is 5 times 10 to the minus 3. That's very important. 5 times 10 to the 3 would have been 5,000. Be very careful about that. Now, what is this guy equal to? This is 1, 2, 3, 4 places behind the decimal. So it's 8 times 10 to the minus 4. If we're multiplying these two things, this is the same thing as 5 times 10 to the minus 3 times 8 times 10 to the minus 4. There's nothing special about the scientific notation. It literally means what it's saying. So for multiplying, you could write it out like this. And multiplication, order doesn't matter. 10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have So this is a really small number. But the multiplication, when you do it in scientific notation, is actually fairly straightforward. This is going to be equal to 6.0-- let me write it properly. 6.022 times 10 to the 23rd times 7.23 times 10 to the minus 22. We can change the order, so it's equal to 6.022 times 7.23. That's that part. So you can view it as these first parts of our scientific notation times 10 to the 23rd times 10 to the minus 22. And now, this is-- you're going to do some little decimal multiplication right here. It's going to be-- some number-- 40 something, I think. I can't do this one in my head. But this part is pretty easy to calculate. I'll just leave this the way it is. But this part right here, this will be times. 10 to the 23rd times 10 to the minus 22."
    },
    {
        "Q": "\nIn problem #10, minute 1:56, wouldn't the triangles be called RTP or PTR or just T for the one in purple, and the other to be called APT or TPA or just P. I thought that the major/main angle of the triangle needed to be in-between the other two letters when naming it?",
        "A": "I m not aware of defining a triangle based solely on the reference to the major angle.",
        "video_name": "bWTtHKSEcdI",
        "timestamps": [
            116
        ],
        "3min_transcript": "Actually, just right after stopping that video, I realized a very simple way of showing you that RP is congruent to TA, a little bit more of a rigorous definition. If we can show that this triangle right there, that one I drew in purple, and this triangle right here are congruent, then we could make a fairly reasonable argument that RP is going to be congruent to TA, because they're essentially the corresponding sides of the two congruent triangles. This congruent triangles would be kind of flipping each other. So how can we make that argument? Well, on the purple triangle, this angle is going to be equal to this angle on the yellow triangle. Actually, we got that from the fact that this is an isosceles trapezoid, so the base angles are going to be the same. know that this side, right there, is going to be congruent to this side. And then finally, they both share this side right here. So we could use the argument-- once again, side, angle, side-- that the side, angle, and side are congruent to this side, angle, and side. So you could say by SAS, triangle TRP is congruent to triangle TAP. And if they're congruent, then all of the corresponding sides are equal, so then TA is congruent to RP. Once again, you didn't have to do all that. It's a multiple choice test. But I wanted to give you that. I felt bad I wasn't giving you a more rigorous definition. A more rigorous proof. So anyway, problem number 11. If a quadrilateral has perpendicular diagonals, then it is a rhombus. Fair enough. Which of the following is a counterexample to the statement above? So they're saying, if it's perpendicular diagonals, then it's a rhombus. So if we could find something that has perpendicular diagonals that is not a rhombus, then we have a counterexample. Then this would not be true. So let's find something with perpendicular diagonals that is not a rhombus. Well, this one has perpendicular diagonals. The diagonals are perpendicular to each other, all 90-degree angles. And this is clearly not a rhombus. This is like a kite. This is not parallel to this and that is not parallel, so this is not a rhombus. So this is definitely a counterexample."
    },
    {
        "Q": "at about 3:30, the instructor states that (a^b)^d=a^bd. How is it possible for the exponent to shift down a degree?\n",
        "A": "This is how exponents work. It doesn t shift it down a degree. Here are a few cases and examples: (2^3)^2=8^2=64 (2^3)^2=2^(3*2)=2^6=64 (2^1=2,2^2=2,2^3=8,2^4=16,2^5=32,2^6=64) (3^2)^3=9^3=729 (3^2)^3=3^(2*3)=3^6=729 (3,9,27,81,243,729) This is not unlike what happens when you multiply two numbers with the same base raise to different powers. In that case you add the exponents: 2^2*2^4=2^(2+4)=2^6=64 2^2*2^4=4*16=64 Good question. It is confusing at first, but when you get the hang of it, it s not too hard.",
        "video_name": "Pb9V374iOas",
        "timestamps": [
            210
        ],
        "3min_transcript": "Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as-- let me do this in a different color-- that 3 is the same thing-- we could put it out front-- that's the same thing as 3 times the logarithm base 5 of x. This is just another way of writing it using this property. And so you could argue that this is a what-- maybe this is a simplification because you took the exponent outside of the logarithm, and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that-- and I'll just pick some arbitrary letters here-- let's say that we know that a to the b power is equal to c. And so if we know that-- that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here, and raise it to the dth power. So let me make it-- so let's raise-- take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting this statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a B. Actually, let's say I'm using all lowercase letters. This is a lowercase c. So let me write it this way, a to the-- so I'm going to raise this to the dth power, and I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now, what's interesting over here is we can use what we know about exponent properties. Say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. This is-- let me do that in a different color. I've already used that green. This right over here, using what we know about exponent properties, this is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we would write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to get to c to the dth power?"
    },
    {
        "Q": "I understood the video, but one part confused me. When he's talking about B = -B at 1:12 through 1:53, that confused me. Wouldn't B equal -B because it is after the zero? Thanks in advance!\n",
        "A": "No. If you assume the scale is one then B = -1. -B would then be - -1. Two negatives is a positive so B= -1 and -B = 1. 1 is not equal to -1 so B=-B is not true. Just because there is a negative in front of the variable does not mean it is negative. It just means that it is multiplied by -1. I hope this clears up your confusion.",
        "video_name": "vRa6XxykfbY",
        "timestamps": [
            72,
            113
        ],
        "3min_transcript": "- [Voiceover] So let's do a couple of examples of the number opposites exercise on Khan Academy. Just to make sure that we've really digested what it means to be the opposite of a number. So they have a number line here, and then they've marked some of the numbers with letters. And this is A, B, C, D, E. And then they say, \"What can we say about point B?\" And point B is one hash mark to the left of zero, all right. \"Select all that applies.\" So the first thing, they say is, \"B is equal to negative D.\" So that means that B needs to be the opposite of D. Let's see if that's true. D is here. D is one hash mark to the right of zero. The opposite of D, or negative D, should be one hash mark to the left of zero. Which is B. B is the opposite of D. B is negative D, so this is true. The next statement is, \"B is the opposite of E.\" Let's see, E is one, two, three hash marks to the right of zero. So the opposite of E should be one, two, three hash marks to the left of zero. B is not the opposite of E. A is the opposite of E, but that isn't what they write here so we're not going to check that. And then they say, \"B is equal to negative B.\" That's saying that B is the same thing as the opposite of B. That is not true. The opposite of B is going to go on the other side of the number line. That's D, which is not B. So this is also not true. This would have been true if they said, \"C is equal to negative C.\" Because C is at zero, it's zero away from zero. C is zero. The opposite of zero is going to be zero away on the other side of the number line. Well, that's still zero away. So zero is negative zero. Or we could have said C is equal to negative C. But the only number for which that is true is zero. It's not going to be true for B, because B is clearly not zero. So let's do a couple of here. \"What can we say about negative C?\" All right, so now C is zero. \"Select all that apply.\" Yeah, the negative of zero is going to still be zero. The negative of zero, the opposite of zero, is still going to be zero. The opposite of zero, that's not going to be this number. This is some non-zero number. So that's that. All right. \"What can we say about point T?\" T once again is at zero. So we say, \"T is zero.\" \"Zero is the opposite of zero.\" It is equal to negative zero. Both of them -- one is zero, you could say zero to the left, one is zero to the right. Well, that still means that they're still on zero. And they're right, they are zero. \"T is equal to the opposite of the opposite of T.\" Well, this would actually be true for any number. For any number, the opposite of the opposite is going to be that number. We could have put the Q here, the R here. Really for any number the opposite of the opposite is going to be that number again. And so it's definitely going to be true for zero as well. So that's definitely going to be true. So hopefully that kind of gives you a little bit"
    },
    {
        "Q": "\nAt 5:38, does Sal mean Law of Sines?",
        "A": "yes you are definitely correct that is a funny mess up on Sal part",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            338
        ],
        "3min_transcript": "that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees. If we wanted actual numerical value, we could just write this as two square roots of two. But let's actually figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out everything else about it. Or if you actually had two sides and an angle, you also would be able to figure out everything else about the triangle."
    },
    {
        "Q": "At 3:36, you said that we should take a reciprocal to the both sides of the equation. But can't we just cross multiply the equation to get a=4xsin105 degrees\n",
        "A": "Either way works. The answer is the same.",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            216
        ],
        "3min_transcript": "So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees."
    },
    {
        "Q": "\nI calculated the sides using the Pythagorean Theorem, and I came up with a = 3.47, which is .4 off the answer they said at 4:14. When I calculated the side of b from the given answer (3.86) I got 3.3, also about .4 off the listed answer at 5:14 (2.83). Is this just because of rounding, or was something wrong?",
        "A": "Hello GoMcLucky, The Pythagorean Theorem doesn t apply to this problem as we do not have right triangle. Instead we want to set up a ratio: SIN(105) / a = sin(30)/2 a = 3.86 Regards, APD",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            254,
            314
        ],
        "3min_transcript": "So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees."
    },
    {
        "Q": "At 1:04 why is Sal Subtracting those numbers?\n",
        "A": "Remember that the sum of a triangle s three angles is 180 degrees. You can see that we know two of the angles already, 45 and 30. To find the final unknown angle, Sal simply subtracts the other two angles from 180, the sum of a triangle s angles.",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            64
        ],
        "3min_transcript": "Voiceover:We've got a triangle here where we know two of the angles and one of the sides. And what I claim, is that I can figure out everything else about this triangle just with this information. You give me two angles and a side, and I can figure out what the other two sides are going to be. And I can, of course, figure out the third angle. So, let's try to figure that out. And the way that we're going to do it, we're going to use something called the Law of Sines. In a future video, I will prove the Law of Sines. But here, I am just going to show you how we can actually apply it. And it's a fairly straightforward idea. The Law of Sines just tells us that the ratio between the sine of an angle, and the side opposite to it, is going to be constant for any of the angles in a triangle. So for example, for this triangle right over here. This is a 30 degree angle, This is a 45 degree angle. They have to add up to 180. So this right over here has to be a, let's see, it's going to be 180 minus 45 minus 30. That's 180 minus 75, so this is going to equal And so applying the Law of Sines, actually let me label the different sides. Let's call this side right over here, side A or has length A. And let's call this side, right over here, has length B. So the Law of Sines tells us that the ratio between the sine of an angle, and that the opposite side is going to be constant through this triangle. So it tells us that sine of this angle, sine of 30 degrees over the length of the side opposite, is going to be equal to sine of a 105 degrees, over the length of the side opposite to it. Which is going to be equal to sine of 45 degrees. So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say,"
    },
    {
        "Q": "At around 5:03 how does he go from 4*\u00e2\u0088\u009a2/2 to 2\u00e2\u0088\u009a2?\nSolving both problems on the calculator give the same answer so they are both the same thing, but how did he simplify it from 4*\u00e2\u0088\u009a2/2 to 2\u00e2\u0088\u009a2?\n\nEDIT: Ok... its pretty obvious It just didn't click for some reason. He just simplified the 2 in 4*\u00e2\u0088\u009a2/2 to convert it to 2\u00e2\u0088\u009a2.. idk how I didn't see that.\n",
        "A": "that got me as well",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            303
        ],
        "3min_transcript": "that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees. If we wanted actual numerical value, we could just write this as two square roots of two. But let's actually figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out everything else about it. Or if you actually had two sides and an angle, you also would be able to figure out everything else about the triangle."
    },
    {
        "Q": "At 8:34, does it matter where you place your coordinate letter points? Like for example, can you put point T at the top left of the isosceles trapezoid instead of the bottom left?\n",
        "A": "Yes, as long the the diagonals correspond and the letters are in order going around the trapezoid. The answer would still be the same!",
        "video_name": "4PPMnI8-Zsc",
        "timestamps": [
            514
        ],
        "3min_transcript": "Given, TRAP, that already makes me worried. Given TRAP is an isosceles trapezoid with diagonals RP and TA, which of the following must be true? OK, let's see what we can do here. So an isosceles trapezoid means that the two sides that lead up from the base to the top side are equal. Kind of like an isosceles triangle. So let me draw that. Actually, I'm kind of guessing that. I haven't seen the definition of an isosceles triangle anytime in the recent past. An isosceles trapezoid. But it sounds right. So I'll go with it. And that's a good skill in life. So I think what they say when they say an isosceles trapezoid, they are essentially saying that this to be equal to that. They're saying that this side is equal to that side. An isosceles trapezoid. And they say RP and TA are diagonals of it. Let me draw that. So let me actually write the whole TRAP. So this is T R A P is a trapezoid. Let me draw the diagonals. RP is that diagonal. And TA is this diagonal right here. OK. All right, let's see what we can do. Which of the following must be true? RP is perpendicular to TA. Well, I can already tell you that that's not going to be true. And you don't even have to prove it. Because you can even visualize it. If you squeezed the top part down. If you were to squeeze the top down, they didn't tell us how high it is. Then these angles, let me see if I can draw it. That angle and that angle, which are opposite or vertical angles, which we know is the U.S. word for it. Those are going to get smaller and smaller if we squeeze it down. And in order for both of these to be perpendicular those would have to be 90 degree angles. And we already can see that that's definitely not the case. All right. RP is parallel to TA. Well that's clearly not the case, they intersect. All right, they're the diagonals. RP is congruent to TA. Well, that looks pretty good to me. Because it's an isosceles trapezoid. If we drew a line of symmetry here, everything you see on this side is going to be kind of congruent to its mirror"
    },
    {
        "Q": "At 0:13, what does 'hone in' mean?\n",
        "A": "To hone in means to sharpen or make something more precise",
        "video_name": "ojFuf9RYmzI",
        "timestamps": [
            13
        ],
        "3min_transcript": "Voiceover:So we've got 3 Y squared plus 6 X to the third and we're raising this whole to the fifth power and we could clearly use a binomial theorem or pascal's triangle in order to find the expansion of that. But what I want to do is really as an exercise is to try to hone in on just one of the terms and in particular I want to hone in on the term that has some coefficient times X to the sixth, Y to the sixth. So in this expansion some term is going to have X to the sixth, Y to sixth and I want to figure out what the coefficient on that term is and I encourage you to pause this video and try to figure it out on your own. So I'm assuming you've had a go at it and you might have at first found this to be a little bit confusing. I'm only raising it to the fifth power, how do I get X to the sixth, Y to the sixth? But then when you look at the actual terms of the binomial it starts to jump out at you. Okay, I have a Y squared term, I have an X to the third term, so when I raise these to powers I'm going to get, I could have powers higher than the fifth power. But to actually think about which of these terms has the X to the sixth, Y to the sixth, in I guess the actual expansion without even thinking about its coefficients. And we've seen this multiple times before where you could take your first term in your binomial and you could start it off it's going to start of at a, at the power we're taking the whole binomial to and then in each term it's going to have a lower and lower power. So let me actually just copy and paste this. So this is going to be, so copy and so that's first term, second term, let me make sure I have enough space here. Second term, third term, fourth term, fourth term, fifth term, and sixth term it's going to have 6 terms to it, you always have one more term than the exponent. And then, actually before I throw the exponents on it, let's focus on the second term. So the second term, actually I'll write it like this. let me copy and paste that, whoops. So let me copy and paste that. So we're going to put that there. That there. And that there. And that there. And then over to off your screen. I wrote it over there. We'll see if we have to go there. And then let's put the exponents. So this exponent, this is going to be the fifth power, fourth power, third power, second power, first power and zeroeth power. And for the blue expression, for 6 X to the third, this is going to be the zeroeth power, first power, first power, second power, third power, fourth power, and then we're going to have the fifth power right over here."
    },
    {
        "Q": "\nAt 1:34, why are the points 2,0 and 0,2?",
        "A": "how does that help",
        "video_name": "wyTjyQMVvc4",
        "timestamps": [
            94
        ],
        "3min_transcript": "So let's say I have a function of x and y; f of x and y is equal to x plus y squared. If I try to draw that, let's see if I can have a good attempt at it. That is my y axis-- I'm going to do a little perspective here --this is my x axis-- I make do the negative x and y axis, could do it in that direction --this is my x axis here. And if I were to graph this when y is 0, it's going to be just a-- let me draw it in yellow --is going to be just a straight line that looks something like that. And then for any given actually, we're going to have a parabola in y. y is going to look something like that. I'm just going to it in the positive quadrant. It's going to look something like that. It'll actually, when you go into the negative y, you're going to see the other half of the parabola, but I'm not going to worry about it too much. So you're going to have this surface. it looks something like that. Maybe I'll do to another attempt at drawing it. But this is our ceiling we're going to deal with again. And then I'm going to have a path in the xy plane. equal to 2, y is 0. And I'm going to travel, just like we did in the last video, I'm going to travel along a circle, but this time the circle's going to have of radius 2. Move counter clockwise in that circle. This is on the xy plane, just to be able to visualize it properly. So this right here's a point 0, 2. And I'm going to come back along the y axis. This is my path; I'm going to come back along the y access and then so I look a left here, and then I'm going to take another left here in and come back along the x axis. I drew it in these two shades of green. That is my contour. And what I want to do is I want to evaluate the surface area of essentially this little building that has the roof of f of xy is equal to x plus y squared, and I want to find the surface area of its walls. Then you're going to have this wall, which is along the curve; it's going to look something like kind of funky wall on that curved side right there. I'll try my best effort to try to-- it's going to be curving way up like that and then along the y axis. It's going to have like a half a parabolic wall right there. I'll do that back wall along the y axis. I'll do that in orange, I'll use magenta. That is the back wall along the y axis. Then you have this front wall along the x axis. And then you have this weird curvy curtain or wall-- do that maybe in blue --that goes along this curve right here, this part of a circle of radius 2. So hopefully you get that visualization. It's a little harder; I'm not using any graphic program at this time."
    },
    {
        "Q": "\nAt 8:14, when Sal says this curve actually has another point where the slope is equal to the average slope, I was wondering whether there are as many such points as the number of inflection points in the graph, or atleast in the closed interval we are concerned about.\nIs this true? Are there really as many such points as the number of inflection points?",
        "A": "You re almost right. If n = the number of inflection points, then there will be n + 1 points where the slope is equal to the average slope. Except straight lines, of course, where the slope of every point is equal to the average slope.",
        "video_name": "bGNMXfaNR5Q",
        "timestamps": [
            494
        ],
        "3min_transcript": "Let me do it in a blue. So that's the average slope between those two points, right? So what does the mean value theorem tell us? It says, if f of x is defined over this closed interval from a to b, and f of x is continuous, and it's differentiable, that you could take the derivative at any point, that there must be some points c f prime of c is equal to this thing. So is equal to f prime of c. I shouldn't have written it here. So what is that telling us? So all that's telling us, is if we're continuous, differentiable, defined over the closed interval, that there's some point c, oh, and c has to be between a and b, of the points, but there's some point c where the derivative at c, or the slope at c, the instantaneous slope at c, is exactly equal to the average slope over that interval. So what does that mean? So we can look at it visually. Is there any point along this curve where the slope looks very similar to this average slope that we calculated? Well, sure, let's see. It looks like, maybe, this point, right here? This is very inexact. But that point looks like the slope, you know, I could say the slope is something like that, right there. So we don't know what, analytically, this function is, but visually, you could see that at this point c, the derivative, so I just picked that point. So this could be our point c. And how do we just say that? to the average slope. So f prime of c is this thing, and it's going to be equal to the average slope over the whole thing. And this curve actually probably has another point where the slope is equal to the average slope. Let's see. This one looks, like, right around there. Just the way I drew it. Looks like the slope there could look something like, could be parallel as well. These lines should be parallel. The tangent lines should be parallel. So hopefully that makes a little sense to you. Another way to think about it is that your average, actually, let me draw a graph just to make sure that we hit the point home. Let's draw my position as a function of time. So this is something, this'll make it applicable to the real world. So that's my x-axis, or the time axis, that's my position axis. This is going back to our original intuition of what"
    },
    {
        "Q": "why was it that one corner was 30* and one was 49* at 8:44? I still don't understand :(\n",
        "A": "It was because when you add up 101 (the measure of corner 1) and 30 (measure of corner 2), you get 131. In a triangle, when you add up all 3 of the angles, you get 180*, no matter what. So, 180-131=49. Therefore, the measure of the other angle would be 49*. Hope that helped! ^_^",
        "video_name": "kqU_ymV581c",
        "timestamps": [
            524
        ],
        "3min_transcript": "What is this angle? This is a good time to pause because I will now show you the solution. So what can we do here? So this angle, well jeez, I just like to just mess around and see what I can figure out. So, if this angle here is 101 degrees, what other angles can we figure out? We could figure out -- well, we could figure out this angle. We could figure out a bunch of angles. We could figure out that -- let me switch the color, these are my figure out angles. So that's 101, then this is supplementary, that's 79 degrees, right? That's also 79 degrees because this is also supplementary. This angle right here is opposite to it, so this angle right here is going to be 101 degrees. What else can figure out? We could figure out this angle because it's supplementary, we could figure out this angle. We could also figure out this angle because we see this triangle right here. So let's call this angle b, b for blue. So b plus 75 plus 75 is going to equal 180. And I'm just using this triangle right here. So b plus 150 is equal to 180, or b is equal to 30 degrees. So we're able to figure this out. Now, what will you do if I told you that we are now ready to figure out this yellow angle? It might not be obvious to you. You kind of have to look at the triangle in the right way, and the SAT will do this to you all the time, all the time. That's why I'm testing you this way. Well, let me give you a little hint. Look at this triangle. Non-ideal color, let me do it in red so it really stands out. Look at this triangle. just looking at the right triangle and kind of seeing that oh wow, I actually can figure out something. Look at this triangle right here. We know this angle of it, 101 degrees. We know this angle, we just figured it out, it was 30 degrees. So all we have left is to figure out this yellow angle, call it x. So x plus 101 plus 30 is equal to 180 degrees because the angles in a triangle add up to 180 degrees. So x plus 131 is equal to 180. x is equal to what? 49 degrees. There you go. We've done the second problem in the angle game. I think that's all of the time I have now in this video. In the next video maybe I'll do a couple more of these angle game problems. See you soon."
    },
    {
        "Q": "At 9:55 shouldn't the sqrt of 1 be +1 and -1 and then say that its because its in the first quadrant we can drop -1?\n",
        "A": "Sal took the positive square root of 1 because ds(the length of a small segment of the curve) is always positive because it is a length.",
        "video_name": "uXjQ8yc9Pdg",
        "timestamps": [
            595
        ],
        "3min_transcript": "curtain that has our curve here as kind of its base, and has this function, this surface as it's ceiling. So we go back down here, and let me rewrite this whole thing. So this becomes the integral from t is equal to o to t is equal to pi over 2-- I don't like this color --of cosine of t, sine of t, cosine times sine-- that's just the xy --times ds, which is this expression right here. And now we can write this as-- I'll go switch back to that color I don't like --the derivative of x with respect to t is minus sine of t, and we're going to square it, plus the derivative of y with respect to t, that's cosine of t, and we're going to square it-- let me make my radical a little bit bigger --and then all of that times dt. you realize that this right here, and when you take a negative number and you squared it, this is the same thing. Let me rewrite, do this in the side right here. Minus sine of t squared plus the cosine of t squared, this is equivalent to sine of t squared plus cosine of t squared. You lose the sign information when you square something; it just becomes a positive. So these two things are equivalent. And this is the most basic trig identity. This comes straight out of the unit circle definition: sine squared plus cosine squared, this is just equal to 1. So all this stuff under the radical sign is just equal to 1. And we're taking the square root of 1 which is just 1. So all of this stuff right here will just become 1. And so this whole crazy integral simplifies a good bit pi over 2 of-- and I'm going to switch these around just because it will make it a little easier in the next step --of sine of t times cosine of t, dt. All I did, this whole thing equals 1, got rid of it, and I just switched the order of that. It'll make the next up a little bit easier to explain. Now this integral-- You say sine times cosine, what's the antiderivative of that? And the first thing you should recognize is, hey, I have a function or an expression here, and I have its derivative. The derivative of sine is cosine of t. So you might be able to a u substitution in your head; it's a good skill to be able to do in your head. But I'll do it very explicitly here. So if you have something that's derivative, you define that something as u. So you say u is equal to sine of t and then du, dt, the derivative of u with respect to t is equal to cosine of t."
    },
    {
        "Q": "\nAt 1:55 isn't it supposed to be 3.14159",
        "A": "Sal used 3.14 as an approximate value for \u00cf\u0080. (3.14159 is also an approximate, but a little more precise.)",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            115
        ],
        "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi."
    },
    {
        "Q": "why did you divide by 2 at 1:10\n",
        "A": "because the diameter is twice as big as radius. so you divide 2 from diameter to find the radius",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            70
        ],
        "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi."
    },
    {
        "Q": "\nat 1:29, why is it 64 pi not 16 pi because dont you multiply pi and the diameter to get the answer",
        "A": "pi * diameter (same as 2*pi*radius) would be the circumference of the wafer which is 16 pi. The question wants us to compute the area so we need to use pi*r^2.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            89
        ],
        "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi."
    },
    {
        "Q": "How do you get 64 at 1:27? Do you just multiply the number that's the radius by itself or something? I'm confused with this.\n",
        "A": "Yes. pi*r^2 means radius in second power. What s power? How many times the number needs to be multiplied by itself. If power is 2 than the number needs to be multiplied twice. If radius is 8 than r^2 is 8^2 which is 8*8 (multiply by itself twice) which equals 64.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            87
        ],
        "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi."
    },
    {
        "Q": "\nAt 3:50, How does 8*8*8 simplify to just 8?",
        "A": "good question. this simplifies to 8 because all those 8s are inside of a cube root symbol. so we had \u00e2\u0088\u009a8*8*8. this simplifies to just 8. sry if this doesn t help :( its kinda hard to explain in words.",
        "video_name": "DKh16Th8x6o",
        "timestamps": [
            230
        ],
        "3min_transcript": "And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512."
    },
    {
        "Q": "0:32 You need to use prime factorization to help you with finding cube roots?\n",
        "A": "The prime factorization of the base is rather important in finding any type of root (whether square root, cube root, 5th root, or whatever). Do you need instruction on how to do this?",
        "video_name": "DKh16Th8x6o",
        "timestamps": [
            32
        ],
        "3min_transcript": "We are asked to find the cube root of negative 512. Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1. Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups."
    },
    {
        "Q": "At 3:48, I don't get it how Sal got the square root of 8*8*8 to get the answer 8.....Can someone explain to me what happened there?\n",
        "A": "well he was answering the cubed root of -512 thats how he got 8 so -8*-8*-8 is -512",
        "video_name": "DKh16Th8x6o",
        "timestamps": [
            228
        ],
        "3min_transcript": "And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512."
    },
    {
        "Q": "\nAt 4:33 was Sal meant to write \"X2 +2\" or am I confused ?",
        "A": "You re right, he should have written X2 +2 instead of X1 +2",
        "video_name": "r4bH66vYjss",
        "timestamps": [
            273
        ],
        "3min_transcript": "It's just the way that we do it. The way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes. The first point I go along the horizontal, what we traditionally call our x-axis. And I go 1 in that direction. And then convention is, the second point I go 1 in the vertical direction. So the point 1, 1. Oh, sorry, let me be very clear. This is 2 and 2, so one is right here, and one is right there. So the point 1, 1 would be right there. That's just the standard convention. Now our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that. I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1, and x2. To represent the vector, what we do is we draw a line from that point to the point x1. And let me call this, let's say that we wanted to draw a. So x1 minus 1. So this is, I'm representing a. So this is, I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point. That one right there. That's my starting point. So minus 4, 4. Now if I want to represent my vector a, what I just said is So x1 plus minus 1 or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, let's see, I'm starting at the point minus 4 comma 4. If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1. And then 4 plus the second term. 4 plus 2. And so this is what? This is minus 5 comma 6. So I go to minus 5 comma 6. So I go to that point right there and I just draw a line. So my vector will look like this. I draw a line from there to there. And I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually let me do it a little bit better. Because minus 5 is actually more, a little closer to right here. Minus 5 comma 6 Is right there, so I draw"
    },
    {
        "Q": "At 5:09, shouldn't it be (x1-1, x2+2)?\n",
        "A": "Yes it should",
        "video_name": "r4bH66vYjss",
        "timestamps": [
            309
        ],
        "3min_transcript": "It's just the way that we do it. The way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes. The first point I go along the horizontal, what we traditionally call our x-axis. And I go 1 in that direction. And then convention is, the second point I go 1 in the vertical direction. So the point 1, 1. Oh, sorry, let me be very clear. This is 2 and 2, so one is right here, and one is right there. So the point 1, 1 would be right there. That's just the standard convention. Now our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that. I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1, and x2. To represent the vector, what we do is we draw a line from that point to the point x1. And let me call this, let's say that we wanted to draw a. So x1 minus 1. So this is, I'm representing a. So this is, I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point. That one right there. That's my starting point. So minus 4, 4. Now if I want to represent my vector a, what I just said is So x1 plus minus 1 or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, let's see, I'm starting at the point minus 4 comma 4. If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1. And then 4 plus the second term. 4 plus 2. And so this is what? This is minus 5 comma 6. So I go to minus 5 comma 6. So I go to that point right there and I just draw a line. So my vector will look like this. I draw a line from there to there. And I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually let me do it a little bit better. Because minus 5 is actually more, a little closer to right here. Minus 5 comma 6 Is right there, so I draw"
    },
    {
        "Q": "\nAt 21:32, Khan says that the vector \"y-x\" is equal to (-6, -5) and graphs it by going six points down and five points to the left. Shouldn't he graph it by going six points to the left and five points down?",
        "A": "Good catch. You are correct.",
        "video_name": "r4bH66vYjss",
        "timestamps": [
            1292
        ],
        "3min_transcript": "x minus y. That's the difference between the two vectors. You can view the difference as, how do you get from one vector to another vector, right? Like if, you know, let's go back to our kind of second grade world of just scalars. If I say what 7 minus 5 is, and you say it's equal to 2, well that just tells you that 5 plus 2 is equal to 7. Or the difference between 5 and 7 is 2. And here you're saying, look the difference between x and y is this vector right there. It's equal to that vector right there. Or you could say look, if I take 5 and add 2 I get 7. Or you could say, look, if I take vector y, and I add vector x minus y, then I get vector x. Now let's do something else that's interesting. Let's do what y minus x is equal to. y minus x. What is that equal to? Do it in another color right here. Well we'll take minus 4, minus 2 which is minus 6. It's minus 5. So y minus x is going to be, let's see, if we start here we're going to go down 6. 1, 2, 3, 4, 5, 6. And then back 5. So back 2, 4, 5. So y minus x looks like this. It's really the exact same vector. Remember, it doesn't matter where we start. It's just pointing in the opposite direction. So if we shifted it here. I could draw it right on top of this. It would be the exact as x minus y, but just in the opposite direction. Which is just a general good thing to know. So you can kind of do them as the negatives of each other. And actually let me make that point very clear. You know we drew y. Actually let me draw x, x we could draw as 2, 3. So you go to the right 2 and then up 3. I've done this before. This is x in non standard position. What is negative x? Negative x is minus 2 minus 3. So if I were to start here, I'd go to minus 2, then I'd go minus 3. So minus x would look just like this. Minus x. It looks just like x. It's parallel. It has the same magnitude. It's just pointing in the exact opposite direction. And this is just a good thing to kind of really get seared into your brain is to have an intuition for these things. Now just to kind of finish up this kind of idea of adding and subtracting vectors. Everything I did so far was in R2. But I want to show you that we can generalize them. And we can even generalize them to vector spaces that aren't normally intuitive for us to actually visualize. So let me define a couple of vectors. Let me define vector a to be equal to 0, minus 1, 2, and 3."
    },
    {
        "Q": "At 25:07, it looks like he accidentally enabled the smudge tool on the tablet. Do vectors like these have anything to do with matrix algebra?\n",
        "A": "Yes, you could actually see the columns of matrices as vectors (if I am not mistaken).",
        "video_name": "r4bH66vYjss",
        "timestamps": [
            1507
        ],
        "3min_transcript": "We can do the same addition and subtraction operations with them. It's just it'll be hard to visualize. We can keep them in just vector form. So that it's still useful to think in four dimensions. So if I were to say 4 times a. This is the vector a minus 2 times b. What is this going to be equal to? This is a vector. What is this going to be equal to? Well we could rewrite this as 4 times this whole column vector, 0, minus 1, 2, and 3. Minus 2 times b. Minus 2 times 4, minus 2, 0, 5. And what is this going to be equal to? This term right here, 4 times this, you're going to get, the 4 times this, you're going to get 4 times 0, 0, minus 4, 8. 4 times 3 is 12. And then minus, I'll do it in yellow, minus 2 times 4 is 8. 2 times minus 2 is minus 4. 2 times 0 is 0. 2 times 5 is 10. This isn't a good part of my board, so let me just. It doesn't write well right over there. I haven't figured out the problem, but if I were just right it over here, what do we get? With 0 minus 8? Minus 8. Minus 4, minus 4. Minus negative 4. So that's minus 4 plus 4, so that's 0. 8 minus 0 is 8. 12 minus, what was this? Oh, this is a 10. Now you can see it again. Something is very bizarre. 2 times 5 is 10. So it's 12 minus 10, so it's 2. So when we take this vector and multiply it by 4, and subtract 2 times this vector, we just get this vector. And even though you can't represent this in kind of an easy kind of graph-able format, this is a useful concept. And we're going to see this later when we apply some of these vectors to multi-dimensional spaces."
    },
    {
        "Q": "At 5:35 how did Sal know that the two areas are equal? The problem doesn't say that they function has x intercept at 1.5, and picture isn't necessarily to scale\n",
        "A": "looking at the graph and where the coordinates are given and marked, we can clearly see that the two areas are equal",
        "video_name": "OvMBNVi5bLY",
        "timestamps": [
            335
        ],
        "3min_transcript": "So it's going to be pi, and then we're going to divide it by four, because it's only one fourth of that circle. And we're going to subtract that. So we have negative pi and we were multiplying it times 1/4 out here, because it's just a quarter circle in either case. And we're subtracting it, because the area is below the x-axis. And so this simplifies to-- this is equal to 1/4 times 8 pi, which is the same thing as 2 pi. Did I do that right? 1/4 times 8 pi. So this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly it is a negative number here. More negative than negative 8. So let's try the other bounds. So let's see what g of positive 3 is. So g of positive 3, when x is equal to 3, that-- we go back to our definition-- that is 2 times 3 plus the integral from 0 to 3 f of t dt. And this is going to be equal to 2 times 3 is 6. And the integral from 0 to 3 f of t dt, that's this entire area. So we have positive area over here. And then we have an equal negative area right over here because it's below the x-axis. So the integral from 0 to 3 is just going to be 0, you're going to have this positive area, and then this negative area right over here is going to completely cancel it out. Because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6. So we already know that our starting point, g of negative 4-- that when x is equal to negative 4-- that is not where g hits a global maximum. Because that's a negative number. And we already found the end point, where g hits a positive value. So negative 4 is definitely not a candidate. where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points in there where it's either undifferentiable or its derivative is equal to 0. So let's look at this derivative. So g prime of x-- we just take the derivative of this business up here. Derivative of 2x is 2. Derivative of this definition going from 0 to x of f of t dt-- we did that in part a, this is just the fundamental theorem of calculus-- this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere, over the interval. So you'll get a number here, and obviously two is just two,"
    },
    {
        "Q": "\nAt 1:38 ,does he have to write out all of the possibility's like that . i tried that and it took a long time (not to mention how tricky it was). Can't he use a possibility diagram/probability space diagram?",
        "A": "It is just a way to show every possible outcome, like with truth tables. A shorter way to represent this is to determine the rule governing the outcomes and simply calculate mathematically. I think Sal is trying to show all possibilities so that even students witj no prior exposure to probabiliy can understand.",
        "video_name": "xSc4oLA9e8o",
        "timestamps": [
            98
        ],
        "3min_transcript": "Let's think about the situation where we have a completely fair coin here. So let me draw it. I'll assume it's a quarter or something. So this is a quarter. Let me draw my best attempt at a profile of George Washington. Well, that'll do. So it's a fair coin. And we're going to flip it a bunch of times and figure out the different probabilities. So let's start with a straightforward one. Let's just flip it once. So with one flip of the coin, what's the probability of getting heads? Well, there's two equally likely possibilities. And the one with heads is one of those two equally likely possibilities, so there's a 1/2 chance. Same thing if we were to ask what is the probability of getting tails? There are two equally likely possibilities, and one of those gives us tails, so 1/2. And this is one thing to realize. If you take the probabilities of heads plus the probabilities of tails, you get 1/2 plus 1/2, which is 1. And this is generally true. The sum of the probabilities of all of the possible events should be equal to 1. all of these fractions, and the numerator will then add up to all of the possible events. The denominator is always all the possible events. So you have all the possible events over all the possible events when you add all of these things up. Now let's take it up a notch. Let's figure out the probability of-- I'm going to take this coin, and I'm going to flip it twice-- the probability of getting heads and then getting another heads. There's two ways to think about it. One way is to just think about all of the different possibilities. I could get a head on the first flip and a head on the second flip, head on the first flip, tail on the second flip. I could get tails on the first flip, heads on the second flip. Or I could get tails on both flips. So there's four distinct, equally likely possibilities. And one way to think about is on the first flip, I have two possibilities. On the second flip, I have another two possibilities. And so I have four possibilities. For each of these possibilities, for each of these two, I have two possibilities here. So either way, I have four equally likely possibilities. And how many of those meet our constraints? Well, we have it right over here, this one right over here-- having two heads meets our constraints. And there's only one of those possibilities. I've only circled one of the four scenarios. So there's a 1/4 chance of that happening. Another way you could think about this-- and this is because these are independent events. And this is a very important idea to understand in probability, and we'll also study scenarios that are not independent. But these are independent events. What happens in the first flip in no way affects what happens in the second flip. And this is actually one thing that many people don't realize. There's something called the gambler's fallacy, where someone thinks if I got a bunch of heads in a row, then all of a sudden, it becomes more likely on the next flip to get a tails. That is not the case."
    },
    {
        "Q": "At 2:32 you said 240x^2. Why and How is it squared?\n",
        "A": "It s squared because you are multiplying 30x and 8x. When multiplying variables, you write the variable, (x), and add what powers that they are each to together, (x^1 times x^1).You will end up with x^2. Here is an example. x^3 times x^2. First, write the variable, (x), and then, add their powers together, (^3+^2)=^5. Your answer will be x^5. :)",
        "video_name": "bamcYQDzVTw",
        "timestamps": [
            152
        ],
        "3min_transcript": "So the volume is going to be the area of the base, which is 3x squared plus 30x plus 5, times the height, times 8x minus 5. Now, to multiply something like this out, it might seem really complicated, but we once again, we just have to do the distributive property. If you viewed this big thing in pink here as just a number, if this was like the number 7, you'd just say, well, this is going to be 7 times 8x minus 7 times 5 or minus 5 times 7. You would just distribute it out. You would just multiply this entire thing times each of the terms. That's what the distributive property tells us when you first learned it. So let's do that. So it's going to be this entire thing times 8x, or we can view it as 8x times this entire thing. So 8x times this entire thing: 3x squared plus 30x plus 5, times minus 5. So once again you get 3x squared plus 30x plus 5. And now we just multiply these out. We just distribute out the 8x over this whole thing and we distribute out the negative 5 over that whole thing. So then we get 8x times 3x squared is 24x to the third power. 8x times 30x is what? That's 240x squared, so plus 240x squared. 8x times 5 is plus 40x. And then we multiply this negative 5 out. Negative 5 times 3x squared is negative 15x squared. Negative 5 times 30x is negative 150x. And then we just have to simplify it from here. So we only have one third-degree term, one thing that has an x to the third in it. We have this term right here, so we'll write that as 24x to the third. And then what are our x squared terms? We have 240x squared minus 15x squared. So what's two 240 minus 15? It's 225x squared. So plus 225x squared. That's adding that term to that term right over there. And then we have 40x minus 150x. That would be negative 110x. And then finally we have just this negative 25 out here. That's the only constant term. And we're done! We found the volume of the tank."
    },
    {
        "Q": "I think he said that 3-dimensional object volumes were the products of the base and the height, somewhere around 1:00. That probably is a mistake, or I heard incorrectly, because pyramid volume has a different formula. So, are we just supposed to assume that the tank is not a pyramid? I've never heard of a pyramid-shaped tank, so that might be it.\n",
        "A": "He said a three-dimensional object like this referring to a cylinder, and in this case, and with similar 3D objects (like prisms), base x height applies.",
        "video_name": "bamcYQDzVTw",
        "timestamps": [
            60
        ],
        "3min_transcript": "Find the volume of a tank whose base has an area of 3x squared plus 30x plus 5 square feet and whose height is 8x minus 5. So let me draw a potential tank here. Maybe it's a cylinder, so let me draw a cylindrical tank here, just like that. It's supposed to be a cylinder, and maybe if it was transparent, you would see this back side over there. And these two are the same. These are supposed to be straight lines. And the volume of a-- let me just label it first. So the area of the base, which is the same as the area of the top here or of the base here, the area of that is 3x squared plus 30x plus 5. And they tell us the height is 8x minus 5. The height of this tank is 8x minus 5. And if you want to find the volume of a three-dimensional object like this, you just multiply the area of the base So the volume is going to be the area of the base, which is 3x squared plus 30x plus 5, times the height, times 8x minus 5. Now, to multiply something like this out, it might seem really complicated, but we once again, we just have to do the distributive property. If you viewed this big thing in pink here as just a number, if this was like the number 7, you'd just say, well, this is going to be 7 times 8x minus 7 times 5 or minus 5 times 7. You would just distribute it out. You would just multiply this entire thing times each of the terms. That's what the distributive property tells us when you first learned it. So let's do that. So it's going to be this entire thing times 8x, or we can view it as 8x times this entire thing. So 8x times this entire thing: 3x squared plus 30x plus 5, times minus 5. So once again you get 3x squared plus 30x plus 5. And now we just multiply these out. We just distribute out the 8x over this whole thing and we distribute out the negative 5 over that whole thing. So then we get 8x times 3x squared is 24x to the third power. 8x times 30x is what? That's 240x squared, so plus 240x squared. 8x times 5 is plus 40x. And then we multiply this negative 5 out. Negative 5 times 3x squared is negative 15x squared. Negative 5 times 30x is negative 150x."
    },
    {
        "Q": "(3:33)\nThe C(A) is (1, 2, 3) and (1, 1, 4). We know this because X1 and X2 are the pivot variables?\nand columns 3 and 4 are not part of the column space because X3 and X4 are free variables?\n",
        "A": "C(A) is the span of v1 = (1, 2, 3) and v2 = (1, 1, 4). Why? Because by setting x3 and x4 in turn to 0 we can show that v3 and v4 are each in the span of v1 and v2. Then clearly all 4 column vectors are in the span of the 1st two, and by closure the span of any vectors within a span is in the span.",
        "video_name": "EGNlXtjYABw",
        "timestamps": [
            213
        ],
        "3min_transcript": "So this is x1. Let me scroll up a little bit. This is associated with x1, right? When you multiply this times x1, you get this column times x1, this column times x2, this column times x3, this column times x4 like that. When you look at just regular A, when you look at just your matrix A, it's the same thing. If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3 and x4 like that. What you can do is you put it in reduced row echelon form. You say which columns have my pivot entries or are associated with pivot variables? You say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns, and so those first two columns would be a basis for the column space. How did I get this? Am I just making up things on the fly? Well, no! a situation where the vectors associated with the free variables you can write them as linear combinations of the vectors associated with the pivot variables, and we did a special case of that last time. But a very quick and dirty way of doing it, and I don't know if it's actually dirty, is just take your matrix, put it in it reduced row echelon form, and you say, look, this column and this column are associated with my free variables. Therefore, this column and this column must be associated with my free variables. The solution sets are the same to Ax equal to 0 or the reduced row echelon form of Ax is equal to 0. So they're the same. So if this column and this column are associated with free variables, so are this column and this column, which means that they can be expressed just by judiciously selecting your values of your free variables as linear variables, with the pivot entries, which are that column and that column. So this column and this column would be a basis for A, and we see that. We see that all the way down here. 1, 2, 3 and 1, 1, 4, We did a lot of work and we got all the way there, and we said this is a basis for the column span of A. Now, doing all of that work, let's see if we can actually visualize what the column space of A looks like. I have a strange feeling I might have said column span a couple of times, but the column space, what does it look like? Well, there's a couple of ways to think about what it looks like. One way is we can say, look, this span this is 2-- that's a member of R3. That's a vector in R3 and this is a vector in R3. Let me draw my x, z and-- well, normally it's drawn--"
    },
    {
        "Q": "\nAt 2:25 Why isn't y = 2sin(t)?\nThen again at 7:12 Why isn't x = 2cos(t)?\nWe were given these values in the beginning of the question (previous video), and yet Sal made these up? How?",
        "A": "It could be, if you wanted it to be. The point is that on the c2 path y needs to go from 2 down to 0. On the c3 path x needs to go from 0 up to 2. How it gets there isn t that important, and doesn t have to be the same or even related to the c1 path parametrization, so Sal just picked the simplest possible way to do what he wanted.",
        "video_name": "Qqanbd3gLhw",
        "timestamps": [
            145,
            432
        ],
        "3min_transcript": "In the last video, we set out to figure out the surface area of the walls of this weird-looking building, where the ceiling of the walls was defined by the function f of xy is equal to x plus y squared, and then the base of this building, or the contour of its walls, was defined by the path where we have a circle of radius 2 along here, then we go down along the y-axis, and then we take another left, and we go along the x-axis, and that was our building. And in the last video, we figured out this first wall's surface area. In fact, you can think of it, our original problem is, we wanted to figure out the line integral along the closed path, so it was a closed line integral, along the closed path c of f of xy, and we're always multiplying f of xy times a little bit, a little, small distance of our path, ds. We're writing this in the most abstract way possible. And what we saw in the last video is, the easiest way to do this is to break this up into multiple paths, or into So you can imagine, this whole contour, this whole path we call c, but we could call this part, we figured out in the last video, c1. This part we can call, let me make a point, c2, and this point right here is c3. So we could redefine, or we can break up, this line integral, this closed-line integral, into 3 non-closed line integrals. This will be equal to the line integral along the path c1 of f of xy ds, plus the line integral along c2 of f of x y ds plus the line integral, you might have guessed it, along c3 of f of xy ds, and in the last video, we got as far as figuring out this first part, this first curvy wall all right here. Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other 2 parts. And in order to do it, we need to do another parameterization of x and y. It's going to be different than what we did for this part. We're no longer along this circle, we're just along the y-axis. So as long as we're there, x is definitely going to be equal to 0. So that's my parameterization, x is equal to 0. If we're along the y-axis, x is definitely equal to 0. And then y, we could say it starts off at y is equal to 2. Maybe we'll say y is equal to 2 minus t, for t is between 0, t is greater than or equal to 0, less than or equal to 2. And that should work. When t is equal to 0, we're at this point right there, and then as t increases towards 2, we move down the y-axis, until eventually when t is equal to 2, we're at that point right there. So that's our parameterization. And so let's evaluate this line, and we could do our derivatives, too, if we like. What's the derivative, I'll write it over here. What's dx dt?"
    },
    {
        "Q": "\nAt 12:18, when Sal is simplifying x^4 = 81, he straightaway puts x^2 = 9. It could have been -9 as well, but wouldn't have worked in the end, since it would yield complex values for x.\nWhat do you think - was that unintentional (a mistake), or he knew it would not be helpful in the end?",
        "A": "Could have been either of those, you can t really know for sure. You re right that it wouldn t have been useful though, as at 6:30 Sal even mentions that we re assuming that we re dealing with real numbers. Had he put x^2 = +- 9, the only thing that would have changed would be the length of the video, as he d have probably explained what you ve just covered in your post. :)",
        "video_name": "zC_dTaEY2AY",
        "timestamps": [
            738
        ],
        "3min_transcript": "if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself. If we add x to the fourth to both sides, you get x to the fourth is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus three. So these are our candidate inflection points, x is equal to 0, x is equal to plus 3, or x is equal to minus 3. So what we have to do now, is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. So what happens when x is slightly below 0? So let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them, necessarily, but if x is like 0.1. What is the second derivative going to be doing?"
    },
    {
        "Q": "\nat 12:00 did he factor out 4x^2 correctly? it doesn't look right for some reason.",
        "A": "It s correct. 4x\u00c2\u00b2*81 - 4x\u00c2\u00b2*x^4 = 27 * 12 * x\u00c2\u00b2 - 4x^6",
        "video_name": "zC_dTaEY2AY",
        "timestamps": [
            720
        ],
        "3min_transcript": "if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself. If we add x to the fourth to both sides, you get x to the fourth is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus three. So these are our candidate inflection points, x is equal to 0, x is equal to plus 3, or x is equal to minus 3. So what we have to do now, is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. So what happens when x is slightly below 0? So let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them, necessarily, but if x is like 0.1. What is the second derivative going to be doing?"
    },
    {
        "Q": "\nAt 2:36, Sal said that the function of the graph is y = | x - 3 |. Can someone please explain this?",
        "A": "Two things can make the graph of functions shift, a change in the x or a change in the y. For absolute value functions, the shifts are defined by y = | x - h | + k, so a y shift is outside the absolute value and follows the sign. Inside for the x shift, however, the graph shifts in the opposite direction, so a -3 shifts it in the positive x direction. The other way to think about is that if x = 3, then the value inside the absolute value is 0.",
        "video_name": "Wri26sPEBoI",
        "timestamps": [
            156
        ],
        "3min_transcript": "and think about how that might change the equation. So let's just visualize what we're even talking about. So if we're gonna shift three to the right, it would look like, it would look like this. So that's what we first wanna figure out the equation for and so how would we think about it? Well, one way to think about it is, well, something interesting is happening right over here at x equals three. Before, that interesting thing was happening at x equals zero. Now, it's happening at x equals three. And the interesting thing that happens here is that you switch signs inside the absolute value. Instead of taking an absolute value of a negative, you're now taking the absolute value as you cross this point of a positive and that's why we see a switch in direction here of this line and so you see the same thing happening right over here. So at this point right over here, we know that our equation needs to evaluate out to zero and this is where it's going to switch signs and so we say, okay, well, this looks like an absolute value so it's going to have the form, y is equal to the absolute value of something and so you say, okay, if x is three, how do I make that equal to zero? Well, I can subtract three from it. If I say y is equal to the absolute value of x minus three, well, let's try it out. Let's see if it makes sense. So when x is equal to three, three minus three is zero, absolute value of that is zero. That works out. When x is equal to four, four minus three is one, absolute value of one is indeed, is indeed one. And if x is equal to two, well, two minus three is negative one So once again, I'm showing you this by really trying out numbers, trying to give you a little bit of an intuition because that wasn't obvious to me when I first learned this that if I'm shifting to the right which it looks like I'm increasing an x value but what I would really do is replace my x with an x minus the amount that I'm shifting to the right but I encourage you to try numbers and think about what's happening here. At this vertex right over here, whatever was in the absolute value sign was equaling zero. It's when whatever was in the absolute value sign is switching from negative signs to positive signs. So once again, if you shift three to the right, that has to happen at x equals three. So whatever is inside the absolute value sign has to be equal to zero at x equals three and this, pause this video and really think about this if it isn't making sense and even as you get more and more familiar with this,"
    },
    {
        "Q": "At around 2:30 in the video Sal invokes the chain rule: d/dx(ln(e^x))=d/dx(e^x)*d/d e^x(ln(e^x))=d/dx(e^x)*(1/e^x)=1. The derivative of the composite function ln(e^x) is the derivative of the inner function w.r.t. x multiplied by the derivative of the outer function w.r.t. e^x. I don't recognize this chain rule. I always thought the chain rule was: d/dx(e^x)*ln(e^x)+d/d e^x(ln(e^x))*e^x. Furthermore, is there a more direct proof; without starting from the composite function d/dx(ln(e^x))?\n",
        "A": "Well, you could define e^x such that d/dx e^x = d/dx e^x, and then show that the e in question actually is the e. However, in rigorous calculus they first define a function ln x to be \u00e2\u0088\u00ab{0,x}(dt/t), such that its derivative clearly is 1/x (and d/dx ln(f(x)) = f\u00c2\u00b4(x)/f(x)) and then define e^x to be its inverse function.",
        "video_name": "sSE6_fK3mu0",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's prove with the derivative of e to the x's, and I think that this is one of the most amazing things, depending on how you view it about either calculus or math or the universe. Well we're essentially going to prove-- I've already told you before that the derivative of e to the x is equal to e to the x, which is amazing. The slope at any point of that line is equal to the x value-- is equal to the function at that point, not the x value. The slope at any point is equal to e. That is mind boggling. And that also means that the second derivative at any point is equal to the function of that value or the third derivative, or the infinite derivative, and that never ceases to amaze me. But anyway back to work. So how are we going to prove this? Well we already proved-- I actually just did it right before starting this video-- that the derivative-- and some people actually call this the definition of e. They go the other way around. They say there is some number for which this is true, So it could almost be viewed as a little bit circular, but be we said that e is equal to the limit as n approaches infinity of 1 over 1 plus n to the end. And then using this we actually proved that derivative of ln of x is equal to 1/x. The derivative of log base e of x is equal to 1/x. So now that we prove this out, let's use this to prove this. Let me keep switching colors to keep it interesting. Let's take the derivative of ln of e to the x. This is almost trivial. This is equal to the logarithm of a to the b is equal to b times the logarithm of a, so this is equal to the derivative of x ln of e. Well, to the first power, right? So this just equals the derivative of x, which we have shown as equal to 1. I think we have shown it, hopefully we've shown it. If we haven't, that's actually a very easy one to prove. OK fair enough. We did that. But let's do this another way. Let's use the chain rule. So what doe the chain rule say? If we have f of g of x, where we have one function embedded in another one, the chain rule say we take the derivative of the inside function, so d/dx of e to the x. And then we take the derivative of the outside function or the derivative of the outside function with respect to the inner function. You can almost view it that way. So the derivative ln of e to the x with"
    },
    {
        "Q": "At 0:56 I don't know why he added the four there, but I see how it works with the equation. Where does the four come from?\nThanks for your help!\n",
        "A": "Hi, I would be happy to help you; is the 0:56 time reference that you made correct? (That part of the video is long division)",
        "video_name": "Y2-tz27nKoQ",
        "timestamps": [
            56
        ],
        "3min_transcript": "PROBLEM: \"Express the rational number 19/27 (or 19 27ths) as a terminating decimal or a decimal that eventually repeats. Include only the first six digits of the decimal in your answer.\" Let me give this a shot. So we want to express 19/27 \u2013 which is the same thing as 19 \u00f7 27 \u2013 as a decimal. So let's divide 27 into 19. So 27 going into 19. And we know it's going to involve some decimals over here, because 27 is larger than 19, and it doesn't divide perfectly. So let's get into this. So 27 doesn't go into 1. It doesn't go into 19. It does go into 190. And it looks like 27 is roughly 30. It's a little less than 30. 30 times 6 would be 180. So let's go with it going 6 times. Let's see if that works out. Well, 6 \u00d7 7 is 42. 6 \u00d7 2 is 12, + 4 is 16. Actually, we could've had another 27 in there. Because when we subtract \u2013 So we get a 10 from the 10's place. So that becomes 8 10's. This became 28. So we could have put one more 27 in there. So let's do that. So let's put one more 27 in there. So 7 27's. 7 \u00d7 7 is 49. 7 \u00d7 2 is 14, + 4 is 18. And now our remainder is 1. We can bring down another 0. 27 goes into 10 0 times. 0 \u00d7 27 is 0. [Not \"10,\" as Sal states by mistake.] Subtract \u2013 we have a remainder of 10. But now, we have to bring down another 0. So let's bring down this 0 right over here. So now, 27 goes into 100 3 times. 3 \u00d7 27 is 60 + 21, is 81. Well, we could take 100 from the 100's place, and make it 10 10's. And then we could take 1 of those 10's from the 10's place and turn it into 10 1's. And so 9 10's minus 8 10's is equal to 1 10. And then 10 -1 is 9. So it's equal to 19. You probably \u2013 You might have been able to do that in your head. And then we have \u2013 And I see something interesting here \u2013 because when we bring down our next 0, we see 190 again. We saw 190 up here. But let's just keep going. So 27 goes into 190 \u2013 And we already played this game. It goes into it 7 times. 7 \u00d7 27 \u2013 we already figured out \u2013 was 189. We subtracted. We had a remainder of 1. Then we brought down another 0. We said 27 goes into 10 0 times. 0 \u00d7 27 is 0. Subtract. Then you have \u2013"
    },
    {
        "Q": "What does Sal mean at 2:01 when he says \"Any of those - well that's just going to give you 3.\"? I also don't understand how he got 15 either. I've been stuck on this lesson re-watching it trying to figure out what he is saying, but it doesn't make sense!\n",
        "A": "The square root of 3*3 is 3. 3*3 = 9, and the square root of nine is three.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            121
        ],
        "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can."
    },
    {
        "Q": "in 2:06 to 2:14 -- why wasn't the square root of 3*3 simplified to the square root of 9 to get 81\n",
        "A": "You are taking the square root of 9 not squaring 9. When you do 9 square or 9^2 you get 81. When your taking square root , you finding which number you have to multiply twice to get to 9. By which your answer will be 3, since 3*3 = 9",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            126,
            134
        ],
        "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can."
    },
    {
        "Q": "At 1:55 why did he write 13 separately?\n",
        "A": "The \u00e2\u0088\u009a(117) is split into 2 square roots: \u00e2\u0088\u009a(9) * \u00e2\u0088\u009a(13). The reason we do this is we can simplify \u00e2\u0088\u009a(9). It = 3. So, the radical will go away on that piece. We can not simplify \u00e2\u0088\u009a(13), so it stays in that form. Hope this helps.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            115
        ],
        "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can."
    },
    {
        "Q": "\nKahn takes the derivative at 2:17, but what is the purpose of that? Does that signify the rate of change of the area or is that the area from (x,x+\u00e2\u0088\u0086x)?\n\nPlease answer both questions.",
        "A": "It would indeed signify the rate of change of the area, but that is not why he is doing it. Sal is demonstrating that the integral (in terms of area under a curve) happens to coincide with a sort-of inverse operation of taking a derivative. This concept is at the core of the Fundamental Theorem of Calculus.",
        "video_name": "pWtt0AvU0KA",
        "timestamps": [
            137
        ],
        "3min_transcript": "Let's say that we've got some function f that is continuous on the interval a to b. So let's try to see if we can visualize that. So this is my y-axis. This right over here, I want to make it my t-axis. We'll use x little bit later. So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a. This is t is equal to b. So we're saying that it is continuous over this whole interval. Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a is a lower bound to x of f of t-- x is in this interval, where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, the definite integral, this has to do with differentiation an antiderivatives and all that. But we don't know that yet. All we know right now is that this is the area under the curve f between a and x, so between a, and let's say this right over here is x. So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So the derivative, f prime of x-- well this definition of derivatives, it's the limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to? Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of-- what's f of x plus delta x? Well, put x in right over here. You're going to get the definite integral from a to x plus delta"
    },
    {
        "Q": "I don't understand the factoring. At 1:09 when Sal starts simplifying and factoring everything out, how did he get that (x2+3x+2) was the same thing as (x+2) (x+1) and that x2-4 was the same as (x+2)(x-2)?\n",
        "A": "He noticed that the factors of 2 (1 and 2) add up to 3, and that (x^2-4) is a difference of squares, which has a simple factoring rule. If you re having trouble with factoring, you may want to go back and review your algebra before going further. If your algebra is rusty, calculus will be very difficult.",
        "video_name": "oUgDaEwMbiU",
        "timestamps": [
            69
        ],
        "3min_transcript": "The function, f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4, is not defined at x is equal to positive or negative 2. And we see why that is, if x is equal to positive or negative 2 then x squared is going to be equal to positive 4, and 4 minus 4 is 0, and then we're going to have a 0 in the denominator. And that's not defined. We don't know what that happens when you divide-- well we've never defined what happens when you divide by 0. So they say, what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x-- I'll just rewrite it-- is equal to-- Actually let me just start simplifying right from the get go. So in the numerator I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over-- and the denominator, this is a difference of squares. This is x plus 2 times x minus 2. So this is going to be equal to 6 times-- let me do it a different color. So we think of two numbers and if I take their product If I take their sum I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1. When you take the product there you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2. Then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2 then x plus 2 is going to be equal to 0. And you won't be able to do that. We don't know what it means divide something by 0. So we could say that this is going to be equal to-- so we can divide the numerator equal to negative 2. So this is equal to 6 times-- we're going divided by x plus 2 in the numerator, x plus 2 in the denominator-- so it's going to be 6 times x plus 1 over x minus 2. And we have to put the constraint here because now we've changed it. Now this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0. So you could say, for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what could we assign f of negative 2 to make the function continuous at the point?"
    },
    {
        "Q": "\nAt 1:58. isn't the question mark supposed to be on top of the equals sign, to indicate that we are not sure that the equation is equal to -11, instead of next to -11?",
        "A": "You could do either the way he does it in this is more informal",
        "video_name": "SkMNREAMNvc",
        "timestamps": [
            118
        ],
        "3min_transcript": "Is negative 1 comma 7 a solution for the system of linear equations below? And they give us the first equation is x plus 2y is equal to 13. Second equation is 3x minus y is equal to negative 11. In order for negative 1 comma 7 to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and y is equal to 7, need to satisfy both of these equations in order for it to be a Solution. So let's try it out. Let's try it out with the first equation. So we have x plus 2y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative 1, and y is equal to 7, will x plus 2y equals 13? So we have negative 1 plus 2 times 7-- y should be 7-- this needs to be equal to 13. because we don't know whether it does. So this is the same thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least, satisfy this first equation. This point does sit on the graph of this first equation, or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue. We have 3 times negative 1 minus y, so minus 7, needs to be equal to negative 11. I'll put a question mark here because we don't know whether it's true or not. So let's see, we have 3 times negative 1 is negative 3. And then we have minus 7 needs to be equal to negative 11-- I put the question mark there. Negative 3 minus 7, that's negative 10. No, negative 10 does not equal a negative 11. So x equaling negative 1, and y equaling 7 does not satisfy the second equation. So it does not sit on its graph. So this over here is not a solution for the system. So the answer is no. It satisfies the first equation, but it doesn't satisfy the second. In order to be a solution for the system, it has to satisfy both equations."
    },
    {
        "Q": "\nAt 0:33, how did he know that i squared equals -1?",
        "A": "Hi Anushka, Recall that i is defined as sqrt(-1). if we square this we get liberate radicand yielding -1. Here is a challenge for you. Calculate i squared, i cubed, and i squared. Look for the pattern. It will help you in the future. Regards, APD",
        "video_name": "SP-YJe7Vldo",
        "timestamps": [
            33
        ],
        "3min_transcript": "Voiceover:Most of your mathematical lives you've been studying real numbers. Real numbers include things like zero, and one, and zero point three repeating, and pi, and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. Then we explored something interesting. We explored the notion of what if there was a number that if I squared it I would get negative one. We defined that thing that if we squared it we got negative one, we defined that thing as i. So we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i, and pi times i, and e times i. This might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real or imaginary numbers? Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to, is equal to the real number five plus the imaginary number three times i. So this thing right over here we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. They won't make any sense. These are kind of going in different, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary, imaginary. A number like this we call a complex number, a complex number. It has a real part and an imaginary part. or someone will say what's the real part? What's the real part of our complex number, z? Well, that would be the five right over there. Then they might say, \"Well, what's the imaginary part? \"What's the imaginary part of our complex number, z? And then typically the way that this function is defined they really want to know what multiple of i is this imaginary part right over here. In this case it is going to be, it is going to be three. We can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we on the vertical axis we plot the imaginary part, so that's the imaginary part. On the horizontal axis we plot the real part. We plot the real part just like that."
    },
    {
        "Q": "\nat 0:51 y would u cross out the division and turn it into 2/3\nPLEASE ANSWER THIS QUESTION I WANT TO STUDY THIS TOPIC",
        "A": "what i dont understand what that means can u be more specific",
        "video_name": "Io9i1JkKgN4",
        "timestamps": [
            51
        ],
        "3min_transcript": "Determine whether 30/45 and 54/81 are equivalent fractions. Well, the easiest way I can think of doing this is to put both of these fractions into lowest possible terms, and then if they're the same fraction, then they're equivalent. So 30/45, what's the largest factor of both 30 and 45? 15 will go into 30. It'll also go into 45. So this is the same thing. 30 is 2 times 15 and 45 is 3 times 15. So we can divide both the numerator and the denominator by 15. So if we divide both the numerator and the denominator by 15, what happens? Well, this 15 divided by 15, they cancel out, this 15 divided by 15 cancel out, and we'll just be left with 2/3. So 30/45 is the same thing as 2/3. It's equivalent to 2/3. 2/3 is in lowest possible terms, or simplified form, Now, let's try to do 54/81. Now, let's see. Nothing really jumps out at me. Let's see, 9 is divisible into both of these. We could write 54 as being 6 times 9, and 81 is the same thing as 9 times 9. You can divide the numerator and the denominator by 9. So we could divide both of them by 9. 9 divided by 9 is 1, 9 divided by 9 is 1, so we get this as being equal to 6/9. Now, let's see. 6 is the same thing as 2 times 3. 9 is the same thing as 3 times 3. We could just cancel these 3's out, or you could imagine this is the same thing as dividing both the numerator and the denominator by 3, or multiplying both the numerator and the denominator by 1/3. These are all equivalent. I could write divide by 3 or multiply by 1/3. Let me write divide by 3 for now. I don't want to assume you know how to multiply fractions, because we're going to learn that in the future. So we're going to divide by 3. 3 divided by 3 is just 1. 3 divided by 3 is 1, and you're left with 2/3. So both of these fractions, when you simplify them, when you put them in simplified form, both end up being 2/3, so they are equivalent fractions."
    },
    {
        "Q": "could somebody explain to me which formula is Sal using at 6:00? is it the Heron's formula?\n",
        "A": "it is the distance formula",
        "video_name": "GiGLhXFBtRg",
        "timestamps": [
            360
        ],
        "3min_transcript": "The centroid of a triangle is just going to be the average of the coordinates of the vertices. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here. You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center is just the average of these coordinates. Now, what we want to do is use this information. Let's just use this coordinate right here and then compare just using the distance formula. Let's compare this distance up here in orange to this distance down here in yellow. And remember, this point right over here-- this is the median of this bottom side right over here. It's just going to be the average of these two points. And so the x-coordinate-- 0 plus a over 2 is going to be a over 2. b plus 0 over 2 is going to be b over 2. And then it has no z-coordinates, so it's just going to be 0. 0 plus 0 over 2 is 0. So we know the coordinates for this point that point and that point. So we can calculate the yellow distance and we can calculate the orange distance. So let's calculate the orange distance. So that is going to be equal to the square root of-- of these points squared. So it's a over 3 minus 0 squared. So that's going to be a squared over 9, plus b over 3 minus 0 squared. So that's b squared over 9. Plus c over 3 minus c, which is negative 2/3. And we want to square that. So we're going to have positive 4 over 9c squared. Did I do that right? c over 3, so 1/3 minus 1 is negative 2/3. So this is negative 2/3 c. You're going to get 4/9 c squared. So that's the orange distance. Now, let's calculate-- and if we want to do it, we can express this-- let me express it a little bit simpler than this. This is the same thing as the square root of a squared plus b"
    },
    {
        "Q": "At 1:41, why are angles C and B 90 degrees? I know it was explained in the video, but I didn't really understand it.\n",
        "A": "AB and AC are tangents, meaning that they intersect the circle at one and only one point. One property of tangents are that if you draw a radius to the tangent point, the tangent will be perpendicular to that radius.",
        "video_name": "ZiqHJwzv_HI",
        "timestamps": [
            101
        ],
        "3min_transcript": "Angle A is a circumscribed angle on circle O. So this is angle A right over here. Then when they say it's a circumscribed angle, that means that the two sides of the angle are tangent to the circle. So AC is tangent to the circle at point C. AB is tangent to the circle at point B. What is the measure of angle A? Now, I encourage you to pause the video now and to try this out on your own. And I'll give you a hint. It will leverage the fact that this is a circumscribed angle as you could imagine. So I'm assuming you've given a go at it. So the other piece of information they give us is that angle D, which is an inscribed angle, is 48 degrees and it intercepts the same arc-- so this is the arc that it intercepts, arc CB I guess you could call it-- it intercepts this arc right over here. It's the inscribed angle. The central angle that intersects that same arc So this is going to be 96 degrees. I could put three markers here just because we've already used the double marker. Notice, they both intercept arc CB so some people would say the measure of arc CB is 96 degrees, the central angle is 96 degrees, the inscribed angle is going to be half of that, 48 degrees. So how does this help us? Well, a key clue is that angle is a circumscribed angle. So that means AC and AB are each tangent to the circle. Well, a line that is tangent to the circle is going to be perpendicular to the radius of the circle that intersects the circle at the same point. So this right over here is going to be a 90-degree angle, and this right over here is going to be a 90-degree angle. OC is perpendicular to CA. OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right We have a quadrilateral going on here. ABOC is a quadrilateral, so its sides are going to add up to 360 degrees. So we could know, we could write it this way. We could write the measure of angle A plus 90 degrees plus another 90 degrees plus 96 degrees is going to be equal to 360 degrees. Or another way of thinking about it, if we subtract 180 from both sides, if we subtract that from both sides, we get the measure of angle A plus 96 degrees is going to be equal to 180 degrees."
    },
    {
        "Q": "\nJust a quick question, at 9:38 you cannot cancel the top vector v and the bottom vector v right? Is this because they are dot products and not multiplication signs?",
        "A": "You get a different answer (a vector divided by a vector, not a scalar), and the answer you get isn t defined.",
        "video_name": "27vT-NWuw0M",
        "timestamps": [
            578
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 2:08, he says \"sin^2(theta)\" why didn't he say \"sin(theta)^2\"?",
        "A": "sin ^2 theta is the more formal way to write out the equation, as it avoids the possibility of mistaking the theta as theta^2 .",
        "video_name": "HnDvUaVjQ1I",
        "timestamps": [
            128
        ],
        "3min_transcript": "So we've got a right triangle drawn over here where this base's length is a, the height here is b, and the length of the hypotenuse is c. And we already know when we see something like this, we know from the Pythagorean theorem, the relationship between a, b, and c, we know there's a squared plus b squared is going to be equal to the hypotenuse squared, is going to be equal to c squared. What I want to do in this video is explore how we can relate trig functions to, essentially, the Pythagorean theorem. And to do that, let's pick one of these non-right angles. So let's pick this angle right over here as theta, and let's just think about this what the sine of theta is and what the cosine of theta is, and see if we can mess with them a little bit to somehow leverage the Pythagorean theorem. So before we do that, let's just write down sohcahtoa just so we remember the definitions of these trig functions. So sine is opposite over hypotenuse. Cah, cosine is adjacent over hypotenuse. And toa, tan is opposite over adjacent, So let's think about sine of theta. I will do it, I'll do it in this blue color. So sine of theta is what? It is opposite over hypotenuse, so it is equal to the length of b or it is equal to b-- b is the length-- b over the length of the hypotenuse, which is c. Now what is cosine of theta? Well, the adjacent side, the side of this angle that is not the hypotenuse, it has length a. So it's the length of the adjacent side over the length of the hypotenuse. Now how could I relate these things? Well it seems like, if I square sine of theta, then I'm going to have sine squared theta is equal to b squared over c squared, and cosine squared theta is going to be a squared over c squared. Seems like I might be able to add them to get something that's pretty close to the Pythagorean theorem here. So let's try that out. So sine squared theta is equal to b squared over c squared. Cosine squared theta is equal to a squared over c squared. So what's this sum? What's sine squared theta plus cosine squared theta? Is going to be equal to what? Sine squared theta is b squared over c squared, plus a squared over c squared, which is going to be equal to-- Well we have a common denominator of c squared. And the numerator, we have b squared plus a squared. Now, what is b squared plus a squared?"
    },
    {
        "Q": "At 0:35, Sal said a greatest common factor and a greatest common divisor are \"kinda\" the same thing. Does he mean exactly the same?\n",
        "A": "Yes GCF & GCD are the same thing",
        "video_name": "jFd-6EPfnec",
        "timestamps": [
            35
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n4:59.",
        "A": "i get it",
        "video_name": "jFd-6EPfnec",
        "timestamps": [
            299
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 2:37 arent the base and height souposed to be switched? or they can stay the same?",
        "A": "well, at 2:37, the base is supposed to be one of the sides, and the height should be inside the triangle.",
        "video_name": "ukPjc3Oyad4",
        "timestamps": [
            157
        ],
        "3min_transcript": "And now the height, I guess you could say this, if you were to drop a penny from here, it's sitting outside the triangle. So it looks different from this one. But this would still be the height. This would still be the height. Right over here, ya do the same thing. One half times base times height would give you the area of this triangle. So how can we apply that over here? Well, this triangle is on this grid, but it's kind of at an angle. With this grid it's hard to pick out the base and the height. for this triangle as a whole. But what we could do, there's actually several ways that we can approach this. Is we can break this triangle up into two or more triangles where we can figure out the base and the height for each of them. So for example, I can break this one. Let's see I could, I'm picking this point and this point because it breaks it up into two triangles where I can figure out the base and the height. Well what am I talking about? Well this triangle over here, that I am shading in blue. I've rotated 90 degrees. But if you view this yellow. If you view this yellow as the base of this triangle, you see that the base is three. So, let me write the base is equal to three units. And what's the height here? Well the height here is going to be The height here is going to be this distance right over here, which is four. Height is equal to four. So the area, the area of that triangle right over there, is going to be one half times three times four, which is equal to six. So this part right over here, the area is six. And now we can do a similar thing with this other triangle. 'Cause once again, we can view this yellow line, or now I have this yellow and blue line, as the base. The base is equal to three. So I could write that base is equal to three. And once again, I've rotated. So now the base is on the side. And then the height here, the height of this triangle is two. If this is the base, remember if this is the base here we've just rotated it. Then this right over here is the height Height would be equal to two. So what's the area of this one? The area of this one is going to be one half times the base, three, times the height which is two. One half times two is one times three. This is going to be equal to three. So the area of the whole thing is gonna be this area of three plus this area of six. It's gonna be the area of nine. Area is equal to nine. Now that's one way you could do it. Is you could just break it up into triangles where you could figure out the base and the height. Another way, and this is you can kinda view it as a maybe a trickier way. Or ya kinda have to think a little bit outside of the box, or maybe outside of the triangle to do it this way. Is to, instead of doing it this way, visualize this triangle. And actually, let me, let me clean this up a little bit. Let me undo all this work that I just did. Let me undo this to show you the other approach."
    },
    {
        "Q": "\nWhat is happening at 4:27 when Sal states that Cos(theta) = -1 can be viewed as (2n + 1)pi? Why add the one to that function, when the previous function \"2n(pi)\" does the same thing?",
        "A": "(2n+1) where n is an integer, is the standard way of specifying any odd integer. Whereas 2n is even. So, cos \u00ce\u00b8 = \u00e2\u0088\u00921 only when \u00ce\u00b8 is an odd integer multiplied by \u00cf\u0080.",
        "video_name": "SdHwokUU8xI",
        "timestamps": [
            267
        ],
        "3min_transcript": "on and on, and it makes sense. Theta equaled, or sorry, cosine of theta, the x-coordinate on this unit circle equaled one right when we were at zero angle, and we had to go all the way around the circle to get back to that point, two pi radians. But then it'll be again when we get to four pi radians, and then six pi radians, so two pi, four pi, six pi, and I guess you could see the pattern here. We're gonna keep hitting cosine of theta equals one every two pi, so you could really kind of view this as every multiple of two pi. Two pi n, where n is an integer, n is integer... is an integer. And that applies also for negative values. If you're going the other way around, if we're going the other way around, we don't get back until we get to negative two pi. Notice we were at zero, and then the next time we're at one again is at negative two pi, and then negative four pi, and then over and over and over again. But this applies, if n is an integer, n can be to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well we can just look at this graph right over here. Well when theta is equal to pi, when theta is equal to pi, and let's see, well, it kinda goes off this graph, but this graph would keep going like this, would keep going like this, and you'd see it would also be at three pi. And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians, and then it won't happen again until we get to two pi, three pi radians, three pi radians. And it won't happen again until we go to two pi, until we add another two pi, until we make one entire revolution, And you can keep going on and on and on, and that's also true in the negative direction, so if we take two pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case, and you actually see it right over here on the graph. So you could think about this as two pi, two pi n plus pi, or you could view it as two n plus one, or two n plus one times pi, where pi is, sorry, where n is an integer. Let me right that a little bit neater, n is integer. At every one of those points, cosine, or for every one of these thetas, cosine of theta is going to keep hitting negative one over and over again. And you see it, it goes it goes from one bottom, where you can kind of valley to the next valley, it takes two pi to get to the next valley,"
    },
    {
        "Q": "at 3:50, why is it that it goes pi, 3 pi, 5 pi for cos (-1) rather than o, 2, 4 pi...?\n",
        "A": "The cosine is just the x-value of a point on the unit circle. At 0, 2pi, 4pi, etc, the x-value is 1. But at pi, 3pi, 5pi, etc, the x-value is -1.",
        "video_name": "SdHwokUU8xI",
        "timestamps": [
            230
        ],
        "3min_transcript": "on and on, and it makes sense. Theta equaled, or sorry, cosine of theta, the x-coordinate on this unit circle equaled one right when we were at zero angle, and we had to go all the way around the circle to get back to that point, two pi radians. But then it'll be again when we get to four pi radians, and then six pi radians, so two pi, four pi, six pi, and I guess you could see the pattern here. We're gonna keep hitting cosine of theta equals one every two pi, so you could really kind of view this as every multiple of two pi. Two pi n, where n is an integer, n is integer... is an integer. And that applies also for negative values. If you're going the other way around, if we're going the other way around, we don't get back until we get to negative two pi. Notice we were at zero, and then the next time we're at one again is at negative two pi, and then negative four pi, and then over and over and over again. But this applies, if n is an integer, n can be to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well we can just look at this graph right over here. Well when theta is equal to pi, when theta is equal to pi, and let's see, well, it kinda goes off this graph, but this graph would keep going like this, would keep going like this, and you'd see it would also be at three pi. And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians, and then it won't happen again until we get to two pi, three pi radians, three pi radians. And it won't happen again until we go to two pi, until we add another two pi, until we make one entire revolution, And you can keep going on and on and on, and that's also true in the negative direction, so if we take two pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case, and you actually see it right over here on the graph. So you could think about this as two pi, two pi n plus pi, or you could view it as two n plus one, or two n plus one times pi, where pi is, sorry, where n is an integer. Let me right that a little bit neater, n is integer. At every one of those points, cosine, or for every one of these thetas, cosine of theta is going to keep hitting negative one over and over again. And you see it, it goes it goes from one bottom, where you can kind of valley to the next valley, it takes two pi to get to the next valley,"
    },
    {
        "Q": "\nI don't get it... At 4:08 5/9 was given as the answer for cleaning the entire bathroom. I just cannot wrap my head around it because I feel like 5/9 is for cleaning 3/5 of the bathroom still? Please help. Thanks",
        "A": "Anon is right... just break it down on a scratchpad",
        "video_name": "2DBBKArGfus",
        "timestamps": [
            248
        ],
        "3min_transcript": "He's able to take one third of a bottle so I could write it here 13 of a bottle to clean to clean three-fifths of a bathroom three-fifths of a bathroom three-fifths of a bathroom So hopefully it's clear now. Why this was helpful? We say okay? We want to take how many bottles it takes to clean a certain number of bathrooms one third of a bottle take can clean? three-fifths of a bathroom and makes it clear that we need to take the one-third and Divide it by three-fifths because then we're going to get bottles per bathroom. We're going to get the rate We're not going to get bathrooms per bottle we're going to get bottles per bathroom Which is what we care about what fraction of the bottle well? It will it take to clean entire bathroom to clean one entire bathroom? So now we just have to take one-third and divide it by three-fifths 13 three-fifths Divided by three fifths and the units are going to be bottles per bathroom. I'll write it like that Bottles or since if we know that that's going to be a fraction. We could just say rate bottle bathroom bottle Bathroom and so this is going to be the same thing as well we're going to have one third divided by three fifths is the same thing as One-Third times the reciprocal of three fifths so one-third times five thirds one-Third times five thirds Bottle bathroom, so let me write that down bottle bathroom Bottle bathroom and what is this going to be well? We multiply the numerators 1 times 5 is going to give us 5 and the denominator three times three is nights? bathroom Bathroom I wrote these a little bit further apart than I would want to let me write that a little bit closer together 59 59 of a bottle bathroom So just as a reminder of what we did here It could see him a little bit daunting But we said okay look he we want how much I'm what fraction of the bottle to clean his entire bathroom So we care about bottles per bathroom or since we're talking about a fraction I guess we could just say bottle some fraction of a bottle per bathroom so it's bottle Bathrooms that's going to give us the correct rate So it took one third of a bottle to clean Three-fifths of a bathroom so you divide one third by three-fifths? You get five ninths of a bottle per bathroom, so this? Right over here that is 59 it takes five ninths of a bottle of cleaning solution to clean his entire bathroom"
    },
    {
        "Q": "at 1:56 what do you mean by bottles an bathroom?\n",
        "A": "at 1:56 he is just using this ratio as an example nothing too important",
        "video_name": "2DBBKArGfus",
        "timestamps": [
            116
        ],
        "3min_transcript": "It's a little daunting because it has fractions, but then when we work through it step-by-step Hopefully, it'll feel a little bit more intuitive so it says Calvin cleans three-fifths of his bathroom with one third of a bottle of cleaning solution at this rate What fraction of the bottle of cleaning Solution will Calvin used to clean his entire bathroom? And like always encourage you to pause the video and try to take a stab at this yourself So as [I] mentioned it's a little bit You know it's a three-fifths of his bathroom with one third of a bottle How do we [think] about this and what my brain does is I'd like to say well How would we like to answer the question what fraction of the bottle of cleaning solution will Calvin used to clean his entire bathroom? So we want to figure out is we want to figure out how many bottles so?? bottles bottles Per let me write it this way So we want to figure out he uses a certain number up with that? bottles per Bathroom if we knew this then we have the answer to the question if this would this might be I don't know [two-fifths] Bottles two-fifths of a bottle bathroom it might be two Bottles per bathroom, but if we know whatever this? Is and we know the answer How many bottles doesn't need to take to clean a bathroom or what fraction of a bottle? We don't know that they're hinting that it's a fraction of a bottle but how much of a bottle or how many bottles per bathroom or another way to think about this if we want to know if We wanted to express it as a as a rate more mathematically. We could say? bottles bottles per bathroom Bathroom and the Reason why this is helpful It makes it clear that look we want to figure out we want to take our Units tell us that we want to divide the number of bottles or the fraction of a bottle it takes? To clean the number of bathrooms or certain fraction of the bathrooms and they tell us over here they tell us that He's able to take one third of a bottle so I could write it here 13 of a bottle to clean to clean three-fifths of a bathroom three-fifths of a bathroom three-fifths of a bathroom So hopefully it's clear now. Why this was helpful? We say okay? We want to take how many bottles it takes to clean a certain number of bathrooms one third of a bottle take can clean? three-fifths of a bathroom and makes it clear that we need to take the one-third and Divide it by three-fifths because then we're going to get bottles per bathroom. We're going to get the rate We're not going to get bathrooms per bottle we're going to get bottles per bathroom Which is what we care about what fraction of the bottle well? It will it take to clean entire bathroom to clean one entire bathroom? So now we just have to take one-third and divide it by three-fifths"
    },
    {
        "Q": "\nAt 2:13 Sal said lets see how they look an a ___ diagram. Does anyone know what he said? \"Argan\"\nThanks!\nI am assuming it is named after a mathematician.",
        "A": "It s called an Argand Diagram.",
        "video_name": "BZxZ_eEuJBM",
        "timestamps": [
            133
        ],
        "3min_transcript": "I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z, and it's equal to a plus bi, I used a word. And I have to be careful about that word, because I used in the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I did make it clear that if you were to see the function the real part of z, this would spit out the a. And the function the imaginary part of z, this would spit out-- and we talked the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here."
    },
    {
        "Q": "At 2:34 Why does Sal draw the number Z with an arrow like a vector?\n",
        "A": "It is not an arrow, but simply a line on top of Z, it stands for conjugate of Z in this case.",
        "video_name": "BZxZ_eEuJBM",
        "timestamps": [
            154
        ],
        "3min_transcript": "the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here. to visually add the complex number and its conjugate. So we said these are just like position vectors. So if we were to add z and its conjugate, we could essentially just take this vector, shift it up here, do heads to tails. So this right here, we are adding z to its conjugate. And so this point right here, or the vector that specifies that point, is z plus z's conjugate. And you can see right here, just visually, this is going to be 2a. And to do that algebraically. If we were to add z-- that's a plus bi-- and add that to its conjugate, so plus a minus bi, what are we going to get? These two guys cancel out. We're just going to have 2a. Or another way to think about it-- and really, we're just playing around with math-- if I take any complex number, and to it I add its conjugate,"
    },
    {
        "Q": "\nAt 2:14, he referenced a z*. What does that mean? Additionally, is an Argand diagram just an imaginary coordinate plane?",
        "A": "z is the conventional variable for a complex number. a z with a bar over it, or a z with an asterisk means the conjugate of z, which is what the video is all about.",
        "video_name": "BZxZ_eEuJBM",
        "timestamps": [
            134
        ],
        "3min_transcript": "I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z, and it's equal to a plus bi, I used a word. And I have to be careful about that word, because I used in the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I did make it clear that if you were to see the function the real part of z, this would spit out the a. And the function the imaginary part of z, this would spit out-- and we talked the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here."
    },
    {
        "Q": "\nAt 8:15 is the reason that ||a-b||=||b-a|| because the length is always a positive value?",
        "A": "Yes. Remember that when you subtract two things (x - y) it is equal to -1 * (y - x). Since the magnitude only deals with the absolute value, the magnitudes will be the same, but the direction will be exactly opposite.",
        "video_name": "5AWob_z74Ks",
        "timestamps": [
            495
        ],
        "3min_transcript": "But here now, if I put little parentheses here, now I can apply the triangle inequality. And I say, well, you know what? This is going to be, by the triangle inequality, which we've proved, it's going to be less than or equal to the lengths of each of these vectors. Vector b plus the length of vector a minus b. So we know that the length of a is less than the sum of that one and that one. So we don't have to worry about this being our problem. We know that that is not true. Now let's look at b. So is there any way that I can rewrite b as a sum of two Well sure. I can write it as a sum of a plus, let me put it this way. If that vector right there is a minus b, the same vector in the reverse direction is going to be the vector b minus a. That's the same thing as b. And you can see it right here. The a's would cancel out and you're just left with the b there. Now by the triangle inequality, we know that this is less than or equal to the length of vector a plus the length of vector b minus a. Now you're saying hey, Sal, you're dealing with b minus a. This is the length of a minus b. And I can leave this for you to prove it based on our definition of vector lengths, but the length of b minus a is equal to minus 1 times a minus b. And I'll leave it to you to say that look, these lengths are equal. Because essentially-- I could leave that, but I think you can take that based on just the visual depiction of them And I have to be careful with length because it's not just in two dimensions. But I think you get the idea and I'll leave that for you to prove that these lengths are the same thing. So we know that b is less than the length of those two things. So we don't have to worry about that one right there. Finally, a minus b. The magnitude or the length of vector a minus b. Well I can write that as the length of-- or I can write that as vector a plus vector minus b. If we just put a minus b right there and go in the other directions, we could say minus b, which would be in that direction plus a would give us our vector a minus b."
    },
    {
        "Q": "\nThere is one but. In the video (2:43) it said that after the price for topping became 2$, the equation must be 8+2(p+8)/p+8. While I was doing the exercises I got the same problem, but in the hints the answer was that the second part should be 24+2p/p+8. Where did they get 24 and how?",
        "A": "The answer is (24+2p)/(p+8), because when you simplify [8+2(p+8)]/(p+8), that s what you get. Here s the process: [8+2(p+8)]/(p+8) (8+2p+16)/(p+8) -- used distributive property (24+2p)/(p+8) -- added together; 8+16=24",
        "video_name": "jQ15tkoXZoA",
        "timestamps": [
            163
        ],
        "3min_transcript": "and then her total cost per pizza after if she bakes 8 more pizzas. So before, we're going to use p to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients So 1.5, or $1.50, times the number of pizzas. This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients. So $2 per pizza. And instead of saying that she's baking p pizzas, So it's going to be p plus 8. And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making p plus 8 pizzas, you would divide by p plus 8. And the problem tells us that these two things are equivalent. Here you had a higher cost in ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what p has to be. p has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made p, is going to be the same as her total cost per pizza when she's making p plus 8 pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked And we used p to represent the number of pizzas. But now for fun, let's actually just solve for p. So let's just simplify things a little bit. So this part right over here. Actually, let's just cross multiply this on both sides. Or another way of thinking is multiply both sides times p plus 8 and multiply both sides times p. So if we multiply by p plus 8, and we multiply by p, we multiply by p plus 8, and we multiply by p, that cancels with that. That cancels with that. On the left-hand side-- so let's see. We have to just do the distributive property twice right over here. What is p times 8 plus 1.5p? Well, that's going to be 8p. I'm just multiplying the p times this stuff first. Plus 1.5p squared. And now let's multiply the 8 times both of these terms."
    },
    {
        "Q": "At 1:14, why is it the square root of x+2? I know that the square root of x looks like that, but how can you determine if it is x + 2?\n",
        "A": "If x+2 is substituted for x, then the graph is centered at x= -2. That s because the general format is x-c, where c is where the graph is centered. For example, If x-2 was substituted in, then the graph would be centered at x=2.",
        "video_name": "IJWDyPFXGyM",
        "timestamps": [
            74
        ],
        "3min_transcript": "Select the piecewise function whose graph is shown below. Or I guess we should say to the right. I copied and pasted it so it's on the right now. So we have this piecewise continuous function. So it's not defined for x being negative 2 or lower. But then starting at x greater than negative 2, it starts being defined. It's continuous all the way until we get to the point x equals 2 and then we have a discontinuity. And then it starts getting it defined again down here. And then it is continuous for a little while all the way. And then when x is greater than 6, it's once again undefined. So let's think about which of these functions describe this one over here. So this one looks like a radical function shifted. So square root of x would look like this. Let me do it in a color that you could see. Square root of x would look like this. And this just looks like square root of x shifted 2 to the left. So this looks like square root of x plus 2. This one right over here looks like square root of x plus 2. And you could verify that. When x is negative 2, negative 2 plus 2, square root of that is going to be 0. And it's not defined there, but we see that if we were to continue it would have been defined there. Let's try some other points. When x is negative 1, negative 1 plus 2 is 1. Principal root of 1 is 1. Let's try 2. When x is equal to 2, 2 plus 2 is 4. Principal root of 4 is positive 2. So this just looks like a pretty good candidate. So it looks like our function. So let's see, it looks like our function would be-- and I'm not going to call it anything because it could be p, h, g, or f. But our function, if I were to write it out, So it looks like it's the square root of x plus 2 for negative 2 being less than x-- it's not defined at negative 2, but as long as x is greater than negative 2 and x is less than or equal to 2. And x is less than or equal to 2. So that's this part of the function. And then it jumps down. Now, this looks like x to the third. x to the third looks something like this. So x to the third power looks something like that. So let's see, negative 2 to the third power is negative 8. So it looks something like that. That's what x to the third looks like. So 2 to the third power is 8. So x to the third looks something like that. This looks like x to the third shifted over 4. So I'm guessing that this is x minus 4 to the third power."
    },
    {
        "Q": "\nisnt the slope of the first graph on the left corner negative..and how at 2:17, does SAL say that the slope is close to 1?.?",
        "A": "it s positive because as x increases y increases, too",
        "video_name": "nZrxs-U9d8o",
        "timestamps": [
            137
        ],
        "3min_transcript": "Let's say that this right over here is the graph of lowercase f of x. That's lowercase f of x there. And let's say that we have some other function capital F of x. And if you were to take its derivative, so, capital F prime of x, that's equal to lowercase f of x, lower case f of x. So given that, which of these, which of these, could be the graph of capital, of capital F of x? And I encourage you to pause this video, and try and think about it on your own before we work through it. Well if, if this curve is going to be the derivative of one of them. That means that any, for any x value it's describing what the instantaneous rate of change. Or what the slope of the tangent line is, of which ever one of these is the possible capital F of x. So, let's just look at a couple of things right here so what, what do we know about lower case f of x? Well, one thing we know is it's always positive. It, it has as we go to negative infinity, it asymptotes towards 0. But it's always positive. So since this is describing the slope of one of these. That means that the slope of one of these always, or out of the candidates, has to always be positive. And, if we look at this, the slope of the tangent line here is, indeed, always positive. The slope of the tangent line here does look like it's positive. Every time we increase an x, we're increasing by y. Here it's positive. But here it's negative. When we increase by x, we decrease by y. So, we can rule, we can rule this one out. Now, what else, what else do we know? Well, this is the derivative. This is telling us the slope of the tangent line. So, for example, when x is equal to, when x is equal to negative 4, f of, f of negative 4 is pretty close to 0. It's pretty close to 0. So, it's slightly, slightly more than 0. of x has to be pretty close to 0, when x equals negative 4. So, let's see, when x equals negative 4, the slope of tangent line, here, isn't close to 0, this actually looks closer to 1. So, we could rule this one out. Over here, when x is equal to negative 4, the slope of the tangent line, yeah, that actually does look pretty close to 0. So I won't rule that one out. And over here, the slope of the tangent line, when x is equal to negative 4, that also looks pretty close to 0. So these are still both in the running. So let's see how we can think of it different. So let's just pick another point. When x is equal to, when x is equal to 0, f of 0 looks like it's pretty close to 1. I don't know if it's exactly to 1. Actually it looks almost exactly. Almost exactly equal to 1. So when capital F of, so at capital F of 0, the"
    },
    {
        "Q": "\nAt 11:26 he forgets to write a line above the x2 (the standard deviation of the difference of the sample 'means'). Right?",
        "A": "Yes, he does, as commented in the next video and as you can see from his calculation of the confidence intervals, he gets a little confused sometimes in videos that are longer than 10 minutes ;D After all, its not that easy to talk about so serious stuff for quiet long, without making any small mistakes :)",
        "video_name": "hxZ6uooEJOk",
        "timestamps": [
            686
        ],
        "3min_transcript": "variances of each of those distributions. Let me write it this way. So the variance, I'll kind of re-prove it. The variance of our distribution is going to be equal to the sum of the variances of each of these sampling distributions. And we know that the variance of each of these sampling distributions is equal to the variance of this sampling distribution, is equal to the variance of the population distribution, divided by our sample size. And the variance of this sampling distribution, for our control, is going to be equal to the variance of the population distribution for the control divided by its sample size. And since we don't know what these are, we can approximate them. Especially, because our n is greater than 30 for both circumstances. We can approximate these with our sample variances for each of these distributions. So let me make this clear. Our sample variances for each of these distributions. So this is going to be our sample standard deviation one squared, which is the sample variance for that distribution, over 100. Plus my sample standard deviation for the control squared, which is the sample variance. variance divided by 100. And this will give us the variance for this distribution. And if we want the standard deviation, we just take the square roots of both sides. If we want the standard deviation of this distribution right here, this is the variance right now, so we just need to take the square roots. Let's calculate this. We actually know these values. S1, our sample standard deviation for group one is 4.67. We wrote it right here, as well. It's 4.76 and 4.04. The S is 4.67, we're going to have to square it. And the S2 is 4.04, we're going to have to square it. So let's calculate that. So we're going to take the square root of 4.67 squared"
    },
    {
        "Q": "7:45 Sal says there's a 95% chance that 1.91 is within 1.96 standard deviations. Shouldn't the interpretation be that \"we are 95% confident...etc.?\" Because this isn't really about probability, right?\n",
        "A": "Yes, because he is talking about the specific value that was observed (1.91). If we re still talking theoretically, we can talk about probability. So we can say there is a 95% chance that xbar will be within 1.96\u00cf\u0083 of \u00ce\u00bc. When we start plugging in numbers and get, say, that xbar is 9.31, then we no longer have a random variable, and the observed mean either is or is not (100% or 0%) within 1.96\u00cf\u0083 of \u00ce\u00bc.",
        "video_name": "hxZ6uooEJOk",
        "timestamps": [
            465
        ],
        "3min_transcript": "So this critical Z value right here is 1.96 standard deviations. This is 1.96 times the standard deviation of x1 minus x2. And then this right here is going to be negative 1.96 times the same thing. Let me write that. So this right here, it's symmetric. This distance is going to be the same as that distance. So this is negative 1.96 times the standard deviation of this distribution. So let's put it this way, there's a 95% chance that our sample as a difference of these other samples. There's a 95% chance that 1.91 lies within 1.96 times the standard deviation of that distribution. So you could view it as a standard error of this statistic. So x1 minus x2. Let me finish that sentence. There's a 95% chance that 1.91, which is the sample statistic, or the statistic that we got, is within 1.96 times the standard deviation of this distribution of the true mean of of the distribution. There's a 95% chance that the true mean of the distribution is within 1.96 times the standard deviation of the distribution of 1.91. These are equivalent statements. If I say I'm within three feet of you, that's equivalent to saying you're within three feet of me. That's all that's saying. But when we construct it this way, it becomes pretty clear, how do we actually construct the confidence interval? We just have to figure out what this distance right over here is. And to figure out what that distance is, we're going to have to figure out what the standard deviation of this distribution is. Well the standard deviation of the differences of the sample means is going to be equal to, and we saw this in the last video-- in fact, I think I have it right at the bottom"
    },
    {
        "Q": "\nAt 3:34, Sal drew a distribution of the difference of sample means. This looks like it's a normal distribution? In the last video, the distribution of the difference of sample means he drew looked normal, too. How do we know this is normally distributed? If you add two normal r.v.s together, is the resulting distribution normal, and why?\n\nThanks! Beth",
        "A": "Central Limit Theorem. Sal has vids on it.",
        "video_name": "hxZ6uooEJOk",
        "timestamps": [
            214
        ],
        "3min_transcript": "So just based on what we see, maybe you lose an incremental 1.91 pounds every four months if you are on this diet. And what we want to do in this video is to get a 95% confidence interval around this number. To see that in that 95% confidence interval, maybe, do we always lose weight? Or is there a chance that we can actually go the other way with the low-fat diet? So in this video, 95% confidence interval. In the next video, we'll actually do a hypothesis test using this same data. And now to do a 95% confidence interval, let's think about the distribution that we're thinking about. So let's look at the distribution. Of course we're going to think about the distribution that we're thinking about. We want to think about the distribution of the difference of the means. So it's going to have some true mean here. of the sample means. Let me write that. It's not a y, it's an x1 and x2. So it's the sample mean of x1 minus the sample mean of x2. And then this distribution right here is going to have some standard deviation. So it's the standard deviation of the distribution of the mean of x1 minus the sample mean of x2. It's going to have some standard deviation here. And we want to make an inference about this. Or I guess, the best way to think about it, we want to get a 95% confidence interval. Based on our sample, we want to create an interval around this, where we're confident that there's a 95% chance that this true mean, the true mean of the differences, lies within that interval. And to do that let's just think of it the other way. are 95% sure that any sample from this distribution, and this is one of those samples, that there is a 95% chance that we will select from this region right over here. So we care about a 95% region right over here. So how many standard deviations do we have to go in each direction? And to do that we just have to look at a Z table. And just remember, if we have 95% in the middle right over here, we're going to have 2.5% over here and we're going to have 2.5% over here. We have to have 5% split between these two symmetric tails. So when we look at a Z table, we want the critical Z value that they give right over here. And we have to be careful here. We're not going to look up 95%, because a Z table gives us the cumulative probability up to that critical Z value."
    },
    {
        "Q": "at 4:23, how do you find the height and acquaint as srt2 over 2? I cannot understand.\n",
        "A": "What do you mean by acquaint ? Numbers don t get acquainted with each other, people do. Regarding how did Sal find the sine and cosine of -45 degrees... 45 degree angles are a very common occurrence. Sal has memorized these values from the unit circle. You might want to search the internet and find a good picture of the unit circle to use as reference.",
        "video_name": "Idxeo49szW0",
        "timestamps": [
            263
        ],
        "3min_transcript": "And so if you draw a line-- Let me draw a little unit circle here. So if I have a unit circle like that. And let's say I'm at some angle. Let's say that's my angle theta. And this is my y-- my coordinates x, y. We know already that the y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x-value is the cosine of theta. So what's the tangent going to be? It's going to be this distance divided by this distance. Or from your algebra I, this might ring a bell, because we're starting at the origin from the point 0, 0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So this has to be a 45 45 90 triangle. This is an isosceles triangle. These two have to add up to 90 and they have to be the same. So this is 45 45 90. And if you know your 45 45 90-- Actually, you don't even have to know the sides of it. In the previous video, we saw that this is going to be-- Right here. This distance is going to be square root of 2 over 2. So this coordinate in the y-direction is minus square root of 2 over 2. And then this coordinate right here on the x-direction is square root of 2 over 2 because this length right there is that. So the square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45 45 90 triangle. So this angle right here is-- Well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x-axis, we'll call"
    },
    {
        "Q": "\nat \"9:00\", Why do you include the first quadrant in your restriction of Theta? Why wouldn't you restrict your range of Theta to only the fourth quadrant?",
        "A": "You need to have all the positive and all the negative possibilities. If you exclude Q1 you can t have any positive answers. Q1 covers the positive angles and Q4 covers the negative angles.",
        "video_name": "Idxeo49szW0",
        "timestamps": [
            540
        ],
        "3min_transcript": "It would be-- I'll just say 2 pi minus pi over 4. Or 4 pi minus pi. It can't map to all of these different things. So I have to constrict the range on the inverse tan function. And we'll restrict it very similarly to the way we restricted the sine-- the inverse sine range. We're going to restrict it to the first and fourth quadrants. So the answer to your inverse tangent is always going to be something in these quadrants. But it can't be this point and that point. Because a tangent function becomes undefined at pi over 2 and at minus pi ever 2. Because your slope goes vertical. You start dividing-- Your change in x is 0. You're dividing-- Your cosine of theta goes to 0. So if you divide by that, it's undefined. So your range-- So if I-- Let me write this down. So if I have an inverse tangent of x, I'm going to-- Well, So if I have the tangent of theta is equal to x, what are all the different values that x could take on? These are all the possible values for the slope. And that slope can take on anything. So x could be anywhere between minus infinity and positive infinity. x could pretty much take on any value. But what about theta? Theta, you can only go from minus pi over 2 all the way to pi over 2. And you can't even include pi over 2 or minus pi over 2 because then you'd be vertical. So then you say your-- So if I'm just dealing with vanilla tangent. Not the inverse. The domain-- Well the domain of tangent can go multiple times around, so let me not make that statement. But if I want to do inverse tangent so I don't have a 1 to many mapping. I want to cross out all of these. I'm going to restrict theta, or my range, to be greater than And so if I restrict my range to this right here and I exclude that point and that point. Then I can only get one answer. When I say tangent of what gives me a slope of minus 1? And that's the question I'm asking right there. There's only one answer. Because if I keep-- This one falls out of it. And obviously as I go around and around, those fall out of that valid range for theta that I was giving you. And then just to kind of make sure we did it right. Our answer was pi over 4. Let's see if we get that when we use our calculator. So the inverse tangent of minus 1 is equal to that. Let's see if that's the same thing as minus pi over 4. Minus pi over 4 is equal to that. So it is minus pi over 4. But it was good that we solved it without a calculator because"
    },
    {
        "Q": "\nThe restrictions on this test seem redundant.\n\nfrom 0:53 to 1:33, Sal gives three restrictions on the series:\n\n1) Bn \u00e2\u0089\u00a5 0 for all relevant n (namely positive integers n).\n2) lim as Bn\u00e2\u0086\u0092\u00e2\u0088\u009e = 0\n3) {Bn} is a decreasing sequence.\n\nDon't the first two rules imply the third?",
        "A": "Consider the function f(n) = x*e^(-n). This will be our Bn. What do you get at n= 0, n=1, n=2? f(0) = 0, f(1) = e^-1 = 0.37, f(2) = 2/e^-2 = 0.27 So the first 3 terms of the sequence are: [0, 0.37, 0.27]. Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n\u00e2\u0086\u0092\u00e2\u0088\u009e = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.",
        "video_name": "91qVGeyTl44",
        "timestamps": [
            53,
            93
        ],
        "3min_transcript": "- [Voiceover] Let's now expose ourselves to another test of conversions, and that's the alternating series test. And I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete. So let's say that I have some series, some infinite series. Let's say it goes from n equals k to infinity of a sub n. Let's say I can write it as, or I can rewrite a sub n. So a sub n is equal to negative one to the n, times b sub n, or a sub n is equal to negative one to the n plus one, times b sub n, where b sub n is greater than or equal to zero for all the n's we care about. So for all of these integer n's greater than or equal to k. and we know two more things. And we know number one, the limit as n approaches infinity of b sub n is equal to zero. And number two, b sub n is a decreasing sequence. Then that lets us know that the original infinite series is going to converge. So this might seem a little bit abstract right now. Let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series from n equals one to infinity And we can write it out just to make this series a little bit more concrete. When n is equal to one, this is going to be negative one to the one power. Actually, let's just make this a little bit, let's make this a little bit more interesting. Let's make this negative one to the n plus one. So when n is equal to one, this is going to be negative one squared over one, which is going to be one. And then when n is two, it's going to be negative one to the third power, which is going to be negative one half. So it's minus one half, plus one third, minus one fourth, plus, minus, and it just keeps going on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite our a sub n, so let me do that. So a sub n, which is equal to negative one,"
    },
    {
        "Q": "At 0:38-0:48: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?\n",
        "A": "We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.",
        "video_name": "91qVGeyTl44",
        "timestamps": [
            38,
            48
        ],
        "3min_transcript": "- [Voiceover] Let's now expose ourselves to another test of conversions, and that's the alternating series test. And I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete. So let's say that I have some series, some infinite series. Let's say it goes from n equals k to infinity of a sub n. Let's say I can write it as, or I can rewrite a sub n. So a sub n is equal to negative one to the n, times b sub n, or a sub n is equal to negative one to the n plus one, times b sub n, where b sub n is greater than or equal to zero for all the n's we care about. So for all of these integer n's greater than or equal to k. and we know two more things. And we know number one, the limit as n approaches infinity of b sub n is equal to zero. And number two, b sub n is a decreasing sequence. Then that lets us know that the original infinite series is going to converge. So this might seem a little bit abstract right now. Let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series from n equals one to infinity And we can write it out just to make this series a little bit more concrete. When n is equal to one, this is going to be negative one to the one power. Actually, let's just make this a little bit, let's make this a little bit more interesting. Let's make this negative one to the n plus one. So when n is equal to one, this is going to be negative one squared over one, which is going to be one. And then when n is two, it's going to be negative one to the third power, which is going to be negative one half. So it's minus one half, plus one third, minus one fourth, plus, minus, and it just keeps going on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite our a sub n, so let me do that. So a sub n, which is equal to negative one,"
    },
    {
        "Q": "\nat 3:25 he said twice,but ment two times",
        "A": "Twice is the same as two times. Once, twice, thrice are still used occasionally.",
        "video_name": "N1X0vf5PUz4",
        "timestamps": [
            205
        ],
        "3min_transcript": "How many of these 4 equal pieces would I have to eat. I encourage you to pause the video and think about that. Well, I would eat this piece. You could imagine me eating this piece and this piece right over here. I've eaten the same amount of the pizza. Each of these pieces you could imagine got cut into 2 pieces when I cut the whole pizza this way. And so now I have to eat 2 slices of the 4, as opposed to 1 slice of the 2. So I just ate 2 out of the 4 slices. I'm using different numbers here. Here I'm using a 1 in the numerator and 2 in the denominator. Here, I'm using a 2 in the numerator and a 4 in the denominator. These two fractions represent the same quantity. I ate the same amount of pizza. If I eat 2/4 of a pizza, if I eat 2 out of 4 equal pieces, that's the same fraction of the pizza So we would say that these two things are equivalent fractions. Now let's do another one like this. Instead of just dividing it into 4 equal pieces, let's divide it into 8 equal pieces. So now we could cut once like this. So now we have 2 equal pieces. Cut once like this. Now we have 4 equal pieces. And then divide each of those 4 pieces into 2 pieces. So I'll cut those in-- So let's see. I want to make them equal pieces. Those don't look as equal as I would like. So that looks more equal, and that looks reasonably equal. So now how many equal pieces do I have? I have 8 equal pieces. But let's say I wanted to eat the same fraction of the pizza. Well, how many of those 8 equal pieces have I eaten? Well, I've eaten 1, 2, 3, 4 of those 8 equal pieces. And so once again, this fraction, 4 of 8, or 4/8, is equivalent to 2/4, which is equivalent to 1/2. And you might see a little bit of a pattern here. Going from this scenario to this scenario, I got twice as many equal slices. Because I had twice as many equal slices, I needed to eat two times the number of slices. So I multiply the denominator by 2, and I multiply the numerator by 2. If I multiply the numerator and the denominator by the same number, then I'm not changing what that fraction represents. And you see that over here. Going from 4 slices to 8 slices, I cut every slice, I turned every slice into 2 more slices, so I had twice as many slices. And then if I want to eat the same amount,"
    },
    {
        "Q": "2:41 alligator eats the bigger number\n",
        "A": "Im finally done this thing!",
        "video_name": "nFsQA2Zvy1o",
        "timestamps": [
            161
        ],
        "3min_transcript": "These are both equal quantities. What I have here on the left hand side, this is 1 plus 1 minus 1 is 1 and this right over here is 1. These are both equal quantities. Now I will introduce you to other ways of comparing numbers. The equal sign is when I have the exact same quantity on both sides. Now we'll think about what we can do when we have different quantities on both sides. So let's say I have the number 3 and I have the number 1 and I want to compare them. So clearly 3 and 1 are not equal. In fact, I could make that statement with a not equal sign. So I could say 3 does not equal 1. But let's say I want to figure out which one is a larger So if I want to have some symbol where I can compare them, where I can tell, where I can state which of these is larger. And the symbol for doing that is the greater than symbol. This literally would be read as 3 is greater than 1. 3 is a larger quantity. And if you have trouble remembering what this means-- greater than-- the larger quantity is on the opening. I guess if you could view this as some type of an arrow, or some type of symbol, but this is the bigger side. Here, you have this little teeny, tiny point and here you have the big side, so the larger quantity is on the big side. This would literally be read as 3 is greater than-- so let me write that down-- greater than, 3 is greater than 1. I could write an expression. I could write 1 plus 1 plus 1 is greater than, let's say, well, just one 1 right over there. This is making a comparison. But what if we had things the other way around. What if I wanted to make a comparison between 5 and, let's say, 19. So now the greater than symbol wouldn't apply. It's not true that 5 is greater than 19. I could say that 5 is not equal to 19. So I could still make this statement. But what if I wanted to make a statement about which one is larger and which one is smaller? Well, as in plain English, I would want to say 5 is less than 19. So I would want to say-- let me write that down-- I want to write 5 is less than 19."
    },
    {
        "Q": "\nAt 3:41 can you explain how you went from 4(y^2-4y+4) to 4(y-2)^2?",
        "A": "Sal factored the trinomial. Find 2 numbers that multiply to +4 but also add to -4. The numbers are -2 and -2 This creates the factors 4(y -2)(y - 2). If you write it using exponents, you get: 4 (y-2)^2 Hope this helps.",
        "video_name": "cvA4VN1dpuY",
        "timestamps": [
            221
        ],
        "3min_transcript": "Is equal to 109. And the things we're going to add, those are what complete the square. Make these things a perfect square. So, if I take this, you have a minus 4 here. I take half of that number. This is just completing the square, I encourage you to watch the video on completing the square where I explain why this works. But I think I have a minus 4. I take half of that, it's minus 2. And then minus 2 squared is plus 4. Now, I can't do one thing to one side of the equation without doing it to the other. And I didn't add a 4 to the left-hand side of the equation. I actually added a 4 times 4, right? Because you have this 4 multiplying it out front. So I added at 16 to the left side of the equation, so I have to also add it to the right-hand side of the equation. This is equivalent to also having a plus 16 here. That might make a little bit clearer, right? When you factor it out, and it becomes a 4. And we would have added a 16 up here as well. Likewise, if we take half of this number here. 1 squared is 1. We didn't add a 1 to the left-hand side of the equation, we added a 1 times minus 25. So we want to put a minus 25 here. And, likewise, this would have been the same thing as adding a minus 25 up here. And you do a minus 25 over here. And now, what does this become? The y terms become 4 times y minus 2 squared. y minus 2 squared. Might want to review factoring a polynomial, if you found that a little confusing, that step. Minus 25 times x plus 1 squared. That's that, right there. x plus 1 squared, is equal to, let's see, 109 plus 16 is 25 minus 25, it equals 100. We're almost there. So we want a 1 here, so let's divide both sides So, you will get y minus 2 squared. 4 divided by 100 is the same thing as 1/25, so this becomes over 25. Minus, let's see, 25/100 is the same thing as 1/4, so this becomes x plus 1 squared over 4 is equal to 1. And there you have it. We have it in standard form and, yes, indeed, we do have a hyperbola. Now, let's graph this hyperbola. So the first thing we know is where the center of this hyperbola is. Is the center of this hyperbola is at the point x is equal to minus one. So it's an x is equal to minus 1. y is equal to 2. And let's figure out the asymptotes of this hyperbola. So if this was -- this is the way I always do it, because I always forget the actual formula. If this was centered at 0 and it looked something like this. y squared over 25 minus x squared over 4 is equal to 1."
    },
    {
        "Q": "\nAt about 1:14 he starts writing the \"number tree\". Do you always put the larger number on the right?",
        "A": "The order of the numbers don t change the actual math. Usually people will use one side or the other for the prime numbers which makes a clear picture where it is easy to find all the primes. However, it is not fixed and when working with students I ll break it up by whatever factors they can think of and circle the primes as we get them. (For example, when factoring 48 a student may say 6*8 initially)",
        "video_name": "znmPfDfsir8",
        "timestamps": [
            74
        ],
        "3min_transcript": "What is the least common multiple of 36 and 12? So another way to say this is LCM, in parentheses, 36 to 12. And this is literally saying what's the least common multiple of 36 and 12? Well, this one might pop out at you, because 36 itself is a multiple of 12. And 36 is also a multiple of 36. It's 1 times 36. So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. There we go. Let's do a couple more of these. That one was too easy. What is the least common multiple of 18 and 12? And they just state this with a different notation. The least common multiple of 18 and 12 is equal to question mark. So let's think about this a little bit. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. We care about 18, and we care about 12. So there's two ways that we could approach this. One is the prime factorization approach. of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. So let's do that. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. 9 is 3 times 3. So we could write 18 is equal to 2 times 3 times 3. That's its prime factorization. 12 is 2 times 6. 6 is 2 times 3. So 12 is equal to 2 times 2 times 3. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover because we want the least common multiple or the smallest common multiple. So let's think about it. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. So let's write that down. So we have to have a 2 times 3 times 3. This makes it divisible by 18. If you multiply this out, you actually get 18. And now let's look at the 12. So this part right over here-- let me make it clear. This part right over here is the part that makes up 18, makes it divisible by 18. And then let's see. 12, we need two 2's and a 3. Well, we already have one 3, so our 3 is taken care of. We have one 2, so this 2 is taken care of. But we don't have two 2s's. So we need another 2 here. So, notice, now this number right over here has a 2 times 2 times 3 in it, or it has a 12 in it, and it has a 2 times 3 times 3, or an 18 in it. So this right over here is the least common multiple"
    },
    {
        "Q": "\n@ 7:43 why and how did you guess 'y' to be = Ae^(2x) ?",
        "A": "Because we knew that after some algebra with both the second derivative, the first derivative and the original function, the result must be 3e\u00c2\u00b2\u00cb\u00a3, so a good candidate that will maintain that exponent but with enough room to play with the coefficients was Ae\u00c2\u00b2\u00cb\u00a3",
        "video_name": "znE4Nq9NJCQ",
        "timestamps": [
            463
        ],
        "3min_transcript": "And so our general solution-- I'll call that h. Well, let's call that y general. y sub g. So our general solution is equal to-- and we've done this many times-- C1 e to the 4x plus C2 e to the minus 1x, or minus x. Fair enough. So we solved the homogeneous equation. So how do we get, in that last example, a j of x that will give us a particular solution, so on the right-hand side we get this. Well here we just have to think a little bit. And this method is called The Method of Undetermined And you have to say, well, if I want some function where I take a second derivative and add that or subtracted some multiple of its first derivative minus some multiple of the function, I get e to the 2x. That function and its derivatives and its second times e to the 2x. So essentially we take a guess. We say well what does it look like when we take the various derivatives and the functions and we multiply multiples of it plus each other? And all of that. We would get e the to 2x or some multiple of e to the 2x. Well, a good guess could just be that j-- well I'll call it y particular. Our particular solution here could be that-- and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution. A solution that gives us this on the right-hand side. So let's say that the one I pick is some constant A times e to the 2x. If that's my guess, then the derivative of that is equal to 2Ae the to 2x. And the second derivative of that, of my particular And now I can substitute in here, and let's see if I can solve for A, and then I'll have my particular solution. So the second derivitive, that's this. So I get 4Ae to the 2x minus 3 times the first derivitive. So minus 3 times this. So that's minus 6Ae to the 2x minus 4 times the function. So minus 4Ae to the 2x, and all of that is going to be equal to 3e to the 2x. Well we know e to the 2x equal 0, so we can divide both sides by that. Just factor it out, really. Get rid of all of the e's to the 2x. On the left-hand side, we have 4A and a minus 4A. Well, those cancel out. And then lo and behold, we have minus 6A is equal to 3. Divide both sides by 6 and get A is equal to minus 1/2."
    },
    {
        "Q": "@ 8:38 ish, what happens when the left side all adds up to 0?\nex: y''(x)-6y'(x)+9y(x)=5e^3x is the equation and my guess would be Ae^3x\ni'd end up with coefficients of 9-18+9=5e^3x\n",
        "A": "In that case, you ve proved your guess wrong by contradiction. That means you must change your guess accordingly. For example, I would change y(x) = Ae^3x to y(x) = Ae^Bx instead. This gives y(x) = Ae^Bx, y (x) = ABe^Bx and y (x) = AB^2e^3x y (x) - 6y (x) + 9y(x) = AB^2e^3x - 6ABe^3x + 9Ae^3x = 5e^3x Therefore, AB^2 - 6AB + 9A = 5. Solving for A, we have A = 5/(B - 3)^2; B =/= 3 That s it. Hope that helps!",
        "video_name": "znE4Nq9NJCQ",
        "timestamps": [
            518
        ],
        "3min_transcript": "times e to the 2x. So essentially we take a guess. We say well what does it look like when we take the various derivatives and the functions and we multiply multiples of it plus each other? And all of that. We would get e the to 2x or some multiple of e to the 2x. Well, a good guess could just be that j-- well I'll call it y particular. Our particular solution here could be that-- and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution. A solution that gives us this on the right-hand side. So let's say that the one I pick is some constant A times e to the 2x. If that's my guess, then the derivative of that is equal to 2Ae the to 2x. And the second derivative of that, of my particular And now I can substitute in here, and let's see if I can solve for A, and then I'll have my particular solution. So the second derivitive, that's this. So I get 4Ae to the 2x minus 3 times the first derivitive. So minus 3 times this. So that's minus 6Ae to the 2x minus 4 times the function. So minus 4Ae to the 2x, and all of that is going to be equal to 3e to the 2x. Well we know e to the 2x equal 0, so we can divide both sides by that. Just factor it out, really. Get rid of all of the e's to the 2x. On the left-hand side, we have 4A and a minus 4A. Well, those cancel out. And then lo and behold, we have minus 6A is equal to 3. Divide both sides by 6 and get A is equal to minus 1/2. It is equal to minus 1/2 e to the 2x. And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogeneous equation. So we can call this the most general solution-- I don't know. I'll just call it y. It is our general solution C1e to the 4x plus C2e to the minus x plus our particular solution we found. So that's minus 1/2e to the 2x. Pretty neat. Anyway, I'll do a couple more examples of this. And I think you'll get the hang of it. In the next examples, we'll do something other than an e to the 2x or an e function here."
    },
    {
        "Q": "\nAt 2:34, could you replace the outermost parentheses with brackets so it can be 4+[2(7- 2x)] instead of 4+(2(7-2x)?\nAre both expressions valid or is one expression more correct than the other?",
        "A": "Yes, they both mean the same thing. The brackets would be understood to be parentheses; they are usually used to set the groups of things you multiply first apart.",
        "video_name": "xLYVo_k0_us",
        "timestamps": [
            154
        ],
        "3min_transcript": "First consider the expression for negative 5 plus the quantity of 4 times x. Now, take the product of negative 8 and that expression and then add 6. So let's do it step by step. First, we're going to have this expression-- negative 5 plus something. So it's going to be negative 5 plus the quantity of 4 times x. Well, that's just going to be 4x. So it's going to be negative 5 plus 4x. So that's this expression up here. Now, take the product of negative 8, so were going to just take negative 8, and we're going to multiply the product of negative 8 and that expression. So we're going to take negative 8 and multiply it so that expression is this thing right over here. So if we say the product of negative 8 in that expression is going to be negative 8 times that expression, that expression is negative 5 plus 4x, so that's negative 8. The product of the two, so we could put a multiplication sign there, or we could just leave that out and implicitly it would mean multiplication, take the product of negative 8 and that expression and then add 6. So that would be then adding 6 right over here. So we could write it as negative 8 open parentheses negative 5 plus 4x and then add 6. Let's do one more. First, consider the expression the sum of 7 and-- so that's going to be 7 plus something-- and the product of negative 2 and x. The product of negative 2 and x is negative 2x. So it's 7 plus negative 2x. We could write that as 7 minus 2x. So this is equal to 7 minus 2x. These are the same expression. So now we're saying 4 plus the quantity of 2 times that expression? So it's going to be 4 plus some quantity. I'll put that in parentheses. The quantity of 2 times-- I'll do this in magenta or in yellow. 2 times that expression-- let me do this in blue-- that expression is this thing right over here. So 4 plus the quantity of 2 times that expression, 2 times 7 minus 2x. And we are done."
    },
    {
        "Q": "I don't understand at 1:25 what falls out of the pythagorean theorem\n",
        "A": "The video states that the distance formula falls out of the the Pythagorean Theorem. The distance formula is A squared plus B squared equals C squared. C represents the length of the hypotenuse of a right triangle (The longest side). A or B can represent the length of one of the sides that make a right angle in a right triangle.",
        "video_name": "9ASWQi14FlE",
        "timestamps": [
            85
        ],
        "3min_transcript": "Point A is at negative 5 comma 5. So this is negative 5 right over here. This is 1, 2, 3, 4, 5. That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then, it's a center of circle A, which I won't draw just yet, because I don't know the radius of circle A. Point B is at-- let me underline these in the appropriate color-- point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there, so center of circle B. Point P is at 0, 0, so it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information. Because that tells us, if this is on both circles, then that means that this is B's radius away from the point B from the center. And this tells us that it is circle A's radius away from its center, which is at point A. And so we can imagine-- let me draw a radius or the radius for circle A-- we now know since P sits on it, that this could be considered the radius for circle A. And you could use a distance formula, but what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius or the distance between those two points squared is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. That's our change in x, negative 5 minus 0 squared, plus our change in y, 5 minus 0 squared, which is the length of the radius squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50, we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root Now, I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 Or you could say, this is 0 minus negative 5. This distance right over here is 5."
    },
    {
        "Q": "At 8:15, when he proves T(V) is a subspace, is he proving simultaneously that R^m is a subspace as well??\n",
        "A": "If R^n is a vector space (which it is, as it is easy to show), then yes, because T(x) = Ax maps R^n to R^m.",
        "video_name": "hZ827mfh1Jo",
        "timestamps": [
            495
        ],
        "3min_transcript": "So what is this equal to? Well, we know from our properties, our definition of linear transformations, the sum of the transformations of two vectors is equal to the transformation of the sum of their of vectors. Now, is the transformation of a plus b, is this a member of TV? Is it a member of our image? Well, a plus b is a member of V, and the image contains the transformation of all of the members of V. So the image contains the transformation of this guy. This guy, a plus b is a member of V. So you're taking a transformation of a member of V which, by definition, is in your image of V under T. So this is definitely true. Now, let's ask the next question. If I take a scalar multiple of some member of my image of V If I take the sum scalar, what is this equal to? By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector. Now is this going to be a member of our image of V under T? Well we know that ca is definitely in V, right? That's from the definition of a subspace. This is definitely in V. And so, if this is in V, the transformation of this has to be in V's image under T. So this is in -- this is also a member of V. And obviously, you can set this equal to 0. The zero vector is a member of V, so any transformation of -- if you just put a 0 here, you'll get the zero vector. So the zero vector is definitely -- I don't care what this is, if you multiply it times 0, you are going to get the zero vector. also a member of TV. So we come on the result that T -- the image of V under T, is a subspace. Which is a useful result which we will be able to use later on. But this, I guess, might naturally lead to the question, what if we go -- everything we have been dealing with so far have been subsets, with the case of this triangle, or subspaces, in the case of V. But what if I were to take the image Rn under T, right? This is the image of Rn under T. Let's think about what this means. This means, what do we get when we take any member of Rn, what is the set of all of the vectors? Then when we take the transformation of all of the members of Rn, let me write this."
    },
    {
        "Q": "AT 8:00 shouldn't it be PLUS zero not MINUS?\n",
        "A": "Actually during 5:43 and 5:53",
        "video_name": "fyJkXBvcA2Q",
        "timestamps": [
            480
        ],
        "3min_transcript": "Let's figure out what the Laplace transform of t squared is. And I'll do this one in green. And maybe we'll see a pattern emerge. The Laplace transform of t squared. Well, it equals 1/s times the Laplace transform of it's So what's it's derivative? Times the Laplace transform of 2t plus this evaluated at 0. Well, that's just 0. So this is equal to-- well we can just take this constant out. This is equal to 2/s times the Laplace transform of t. Well, what does that equal? We just solved it. So it's 2/s times 1/s squared. So it's equal to 2/s to the third. Well, let me just do t the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of t to the third is 1/s times the Laplace transform of it's derivative, which is 3t squared. Which is-- take the constant out because it's a linear operator. 3/s times the Laplace transform of t squared. So it equals what? What's the Laplace transform of t squared? It's 2/s to the third. So this equals 3 times 2 over what? s to the fourth. And you can put a t/n here and use an inductive argument to And that general formula is-- I think you see the pattern here. Whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. And then the numerator is the factorial of my exponent. So in general, and this is one more entry in our Laplace transform table. The Laplace transform of t to the nth power is equal to n factorial over s to the n plus 1. That's a parenthesis. I guess I didn't have to write those parenthesis. That just confuses it. But anyway, when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all of that. But it's just saying with this pattern we showed, t to the"
    },
    {
        "Q": "At 7:24, Sal said g(x) = 3 (1/3)^x\nCould this be simplified down to 1^x?\n",
        "A": "No, you need to retain the base of 3, but you could simplify it to 3^(1-x) Like this: 3*(1/3)^x = (3^1)*(3^(-1))^x = (3^1)*(3^(-x)) = 3^(1-x) But it s harder to see what this will do than in the earlier form.",
        "video_name": "gFdh_rE2XgU",
        "timestamps": [
            444
        ],
        "3min_transcript": "would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here. We could just take this a and substitute it in right over here for a, and so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing an a here, I'll write 9r, times r is equal to one. Or we could write, let me scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the positive and negative square root of both sides, but they tell us that r is greater than zero, so we can just take the principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to nine times r. So our exponential function could be written as g of x is equal to a, which is three, times r, which is 1/3, 1/3 to the x power."
    },
    {
        "Q": "\nAt 5:07 in the equation ar^-1, does -1 apply to r only or both a and r. I'm not sure because Sal didn't put any parentheses in.",
        "A": "if no parentheses => applies only to one symbol which it is written after, (r)",
        "video_name": "gFdh_rE2XgU",
        "timestamps": [
            307
        ],
        "3min_transcript": "I gotta be careful here, I got a little bit. Every time you increase your x by one, you're decreasing your y. And here on the x axis, we're marking off every half. So every time you increase your x by one, you are decreasing your y. You are decreasing your y by four there, so that makes sense that the slope is negative four. So now let's think about what b is. So to figure out b, we could use either one of these points to figure out, given an x, what f of x is, and then we can solve for by. Let's try one, 'cause one is a nice, simple number. So we could write f of one, which would be negative four times one plus b. And they tell us that f of one is one, is equal to one. And so this part right over here, we could write that as negative four plus b is equal to one, and then we get b is equal to five. So we get f of x is equal to negative 4x plus five. Now, does that make sense that the y-intercept here is five? Well, you see that right over here. By inspection, you could have guessed, actually, that the y-intercept here is five, but now we've solved it. Maybe this was 5.00001 or something, but now we know for sure it's negative 4x plus five. Or another way you could've said it, if the slope is negative four, if this right over here is nine, you increase one in the x direction, you're gonna decrease four in the y direction, and that will get you to y is equal to five, so that is the y-intercept. But either way, we have figured out the linear function. Now let's figure out the exponential function. So here we could just use the two points to figure out these two unknowns. So, for example, let's try this first point. So g of negative one, which if we look would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here."
    },
    {
        "Q": "At 6:40 how to get from r^2=1/9 to r=1/3. What was done to get this answer. I don't get it. I'm confused. Could someone explain please?\n",
        "A": "You need to take the square root of both sides. The square root of r^2 = r. The square root of 1/9 = 1/3.",
        "video_name": "gFdh_rE2XgU",
        "timestamps": [
            400
        ],
        "3min_transcript": "would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here. We could just take this a and substitute it in right over here for a, and so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing an a here, I'll write 9r, times r is equal to one. Or we could write, let me scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the positive and negative square root of both sides, but they tell us that r is greater than zero, so we can just take the principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to nine times r. So our exponential function could be written as g of x is equal to a, which is three, times r, which is 1/3, 1/3 to the x power."
    },
    {
        "Q": "\nAt 0:23 Sal says \"literally\". Is that a math term I am not aware of or something?",
        "A": "He means that it s condensed to basic numbers. Like a regular equation is 6743-2340 but a literal equation is (6000+700+40+3) - (2000+300+40)",
        "video_name": "RRk5qLd__Ro",
        "timestamps": [
            23
        ],
        "3min_transcript": "We've got 6,798 minus 3,359. So let's see how far we can get with the subtraction. So immediately, when we go into the ones place, we're going to try to subtract a 9 from an 8. So we immediately reach a little bit of a stumbling block. And to see what our options are here in terms of regrouping, I'm going to rewrite both of these numbers. So I'm going to rewrite 6,798 literally. So this is equal to 6,000. That's this right over here. That's 6,000, plus 700, plus 90, plus 8. Minus all of this. So I could subtract each of the places. So I could say this is going to be minus 3,000 minus 300 minus 50-- a 5 in a tens place So here, we're just explicitly showing what those place values represent. A 6 in the thousands place is 6,000. A 3 in the hundreds place is 300. Let's go back to our problem. We wanted to subtract a 9 from an 8. Well, that's a little bit of a stumbling block. But what if we could take some of the value from some of these and give it to the 8? In particular, we could go one place value up to the 90. And what if we were to take 10 from the 90? So let's do that. If we were to take 10 from the 90, then 90 becomes 80. But we don't want to change the value of the entire number. So we're going to give that 10 to this 8. We're essentially regrouping right over here. And then that 8 can become an 18. Notice, I did not change the value of the number. Instead of saying it's 6,000 plus 700 plus 90 plus 8, I'm just saying that it's 6,000 plus 700 plus 80 plus 18. Those are both going to give you 6,798. But now it becomes a little bit easier for us to actually subtract. Now, if we subtract, I have 18 minus 9, which is 9. I have 80-- not 90, now. I have 80 minus 50, which is 30. And these are all positive, so this is plus 9. This is a positive 30. 80 minus 50 is 30. I have 700 minus 300, which is 400. And I have 6,000 minus 3,000, which is 3,000. So this is literally going to be 3,000 plus 400 plus 30 plus 9, or 3,439. Now, how would you do it if you didn't want to write it out like this? And this is where you'll see a kind of shorthand notation."
    },
    {
        "Q": "In 0:32, How did he get 200cm squared for the height of the cereal box?\n",
        "A": "It was the area of the front of the cereal box, so he did 20cm(height of box) times 10cm (width of front of box) which equals 200 cm which is the area of the front of the box.",
        "video_name": "1iSBNSYhvIU",
        "timestamps": [
            32
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can figure out the surface area of this cereal box. And there's a couple of ways to tackle it. The first way is, well let's figure out the surface area of the sides that we can see, and then think about what the surface area of the sides that we can't see are and how they might relate, and then add them all together. So let's do that. So the front of the box is 20 centimeters tall and 10 centimeters wide. It's a rectangle, so to figure out its area we can just multiply 20 centimeters times 10 centimeters, and that's going to give us 200 centimeters. 200 centimeters, or 200 square centimeters, I should say. 200 square centimeters, that's the area of the front. And let me write it right over here as well, 200. Now we also know there's another side that has the exact same area as the front of the box, and that's the back of the box. And so let's write another 200 square centimeters for the back of the box. Now let's figure out the area of the top of the box. The top of the box, we see the box is three centimeters deep so this right over here is three centimeters. It's three centimeters deep and it's 10 centimeters wide. So the top of the box is gonna be three centimeters times 10 centimeters, which is 30 square centimeters of area. So that's the top of the box, 30 square centimeters. Well, the bottom of the box is gonna have the exact same area, we just can't see it right now, so that's gonna be another 30. Then we have two more sides, cuz this box has six sides, We have this side panel that is 20 centimeters tall, we see that the height of the box is 20 centimeters and three centimeters deep. So three times 20, let me write that a little bit neater. Three times 20, that's 20 centimeters right there. Three centimeters times 20 centimeters is going to give us 60 square centimeters. Now that's this side panel, but there's another side panel that has the exact same area that's on the other side of the box, so it's 60 centimeters squared for this side, and then another 60 for the corresponding side And now we can just add up all of these together. And so we get zero, this is going to be carry the one, or regroup the one, it's a 100, and then we have 500. So we get 580 square centimeters is the surface area of this box."
    },
    {
        "Q": "At 0:00, is the bead counter actually an 'Abacus'? Thank you if you answer.\n\n-TheAmericanBerserker\n",
        "A": "Yes , at 0:00 , the bead counter is an abacus.-facepalm- Stop doing this just for badges.",
        "video_name": "v3gdX07Q6qE",
        "timestamps": [
            0
        ],
        "3min_transcript": "Hey, Sal Hello, Brit I picked this up at a garage sale and I know you like colours. I love colours! You wear colourful shirts everyday and I thought you might like this. I do. This is very kind. I like it, yeah. So what are you hoping to do with this thing? Well at minimum maybe just represent numbers. Like counts? Keep track of numbers? Yeah you know counting the number of days or? 1, 2, 3. Yeah I could imagine doing that. Oh ok. So moving the bead down is a number, great. And that there's, there's 10. 10 there, so maybe that'd be 20, 30, 40, 50, 60, 70, 80, 90, 100. Yeah you could count 100 beads to keep track of things. What if I need to count to 105 or 6? Do I need to buy another one? I don't even know where to get more than these. Um, that would be an option. I guess maybe that's not an option if you don't know where to get it. Let's see. Um. Well, you know the different colours, you know, just like we have different forms of currency. Maybe we can have each of these, each of these colours or maybe the columns they represent a different amount. But what if we had one of these red beads represent 10 of these blue beads on this first column. So then you could go, this would be 10. Or you could say that is 10? So there's two representations of 10 here. On, yeah the way we just worked it out here, yeah. Okay. And then 11 would be that. Okay, and this is 21. Yeah, you have two 10s and a 1. The wooden colour or whatever colour. The skintone bead is going to represent all of the red beads. Yeah if each of one of these. Yeah that's a good system. If each column represents 10 of the beads to the column to its right that could be interesting. Because if this represents 10 red beads. That's ten 10s That'd be equivalent, this is equivalent to 100 blue beads. Which would be 100 of the red beads. Which would be 1,000 of the blue beads. And this would be 10,000 of the blue beads. And this would be 100,000 of the other blue beads. This would be 1,000,000 of the blue beads. And we're going to be able to repesent all the numbers in between? 1 and say 1,000,000? I think we can. Let me just give you a number and see how you do. What about 15,003? So let's think a little bit about this. So let's try the big numbers first. So see each of these is 1 and each of these is 10. Each of these is 100. Each of these is 1,000. I don't have 15 of these. But these are 10,000. Each of these are 10,000 So this is one 10,000 And then I could do 5 thousands. 1, 2, 3, 4, 5. So this is 15 thousand. So 1 ten thousand, 5 thousands. Then 0 hundreds, 0 tens, and then throw a 3 there. So 15,003."
    },
    {
        "Q": "In discussing reflections at 4:38, why does not Sal place the reflection line through the origin? This approach would better help novice learners of the concept.\n",
        "A": "I agree especially if you do not have a computer program to do these reflections - along the line y=x makes easy reflections, the line on the graph does not appear to be precisely defined, but it sort of looks like y = x + 1/2.",
        "video_name": "XiAoUDfrar0",
        "timestamps": [
            278
        ],
        "3min_transcript": "The point of rotation, actually, since D is actually the point of rotation that one actually has not shifted, and just 'til you get some terminology, the set of points after you apply the transformation this is called the image of the transformation. So, I had quadrilateral BCDE, I applied a 90-degree counterclockwise rotation around the point D, and so this new set of points this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set that are on our quadrilateral, I could rotate around, I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation. I could rotate around any point. Now let's look at another transformation, and that would be the notion of a reflection, You imagine the reflection of an image in a mirror or on the water, and that's exactly what we're going to do over here. If we reflect, we reflect across a line, so let me do that. This, what is this one, two, three, four, five, this not-irregular pentagon, let's reflect it. To reflect it, let me actually, let me actually make a line like this. I could reflect it across a whole series of lines. Woops, let me see if I can, so let's reflect it across this. Now, what does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image, and you imagine this as the line of symmetry that the image and the original shape they should be mirror images across this line we could see that. Let's do the reflection. There you go, and you see we have a mirror image. This is this far away from the line. This, its corresponding point in the image This point over here is this distance from the line, and this point over here is the same distance but on the other side. Now, all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again you could just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't stretch or scale up or scale down it kind of maintains its shape, and that's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here the distance between this point and this point, between points T and R, and the difference between their corresponding image points, that distance is the same. The angle here, angle R, T, Y, the measure of this angle over here,"
    },
    {
        "Q": "Do units in this case kilometers (km) cancel each other out? At 3:06 Sal mentions km cancel km to just get hours.\n",
        "A": "Yes, the km units cancel creating a final unit in hours.",
        "video_name": "Q0tTfe2lKIc",
        "timestamps": [
            186
        ],
        "3min_transcript": "I'll draw a bigger building, that's his school, and we know that this distance is 35, 35 kilometers and we also know that it took one and a half hours. One and a half hours, now he traveled at different rates for different distances. So he traveled some distance to the school bus so this is, or to the bus stop, so that's the bus stop, right over there, and we're seeing this distance to the bus stop, that's how much he covered by walking. So this right over here, this distance, right over here, that is W and the rest of the distance, he covered by the bus, so the rest of this distance, all of this distance right over there, that is going to be B. So what do we know? We know the distance covered by walking plus the distance covered by bus is going to be 35 kilometers, 35 kilometers here, this is the entire, that is the entire distance from home to school, so we know that W plus B, is equal to 35 kilometers and just with one equation, we're not going to be able to figure out what W and B are but we have another constraint. We know the total amount of time. So the total amount of time is going to be one and half hours, so we'll just write that over here. This is going to be 1.5, so what's the time traveled by, what's the time he walks? Let me write this over here, time time walking, we'll that's going to be the distance walking divided by the rate walking. So the distance walking is W kilometers W kilometers divided by his rate, the distance divided by your rate is gonna give you your time, so let's see, his rate is five kilometers per hour, five kilometers per hour and so you're gonna and if you divide by or if you have one over hours in the denominator, that's going to be the same thing, this is gonna be W over five hours, so the units work out. So his time walking is W over five, W over five and by that same logic, his time on the bus is going to be the distance on the bus divided by, divided by the average speed of the school bus, so this is going to be 60. This is all going to be in hours and now we can solve this system of equations. We have two linear equations with two unknowns. We should be able to find W and B that satisfy both of these. Now what's an easy thing to do? Let's see, if I can multiply this second equation by negative five, and I'm gonna, this is going to be a negative W here so it'll cancel out with this W up there. So let's do that, let's multiply the second equation by, I'm just gonna switch to one color here,"
    },
    {
        "Q": "\nMr Sal At 3:43\nhow can square root of X-3 be greater or equal to 0\nif i plug 0 in place of x i would -3 which is wrong",
        "A": "If x=0, then it would be sqrt(-3). Then would be an imaginary number. That s why domain comes in handy.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            223
        ],
        "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the"
    },
    {
        "Q": "\nHey there! I'm new to Khan Academy and I've got a doubt regarding inequalities involving modulo, i.e, |x|.\n\nAt 5:58, Sal says that when |x|>=3 ---> x<=(-3) or x>=3, But what if x was positive, how then would x<=(-3)?\n\nThanks!",
        "A": "First |x| represent the absolute value not modulo (modulo is a very different thing). As for your question, the solution is that x<=3 OR x>=3 so if x is positive, it satisfies the right half of the inequality and since it is an OR it doesn t need to satisfy the other side, likewise for a negative number <-3. If the solution was instead x<=-3 AND x>=3 then we would have a problem since anything that satisfied one half could not satisfy the other, thus resulting in no valid solution.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            358
        ],
        "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator,"
    },
    {
        "Q": "At 6:04 when hes discussing that x can < or = to -3, im lost. If x was -4 and then you subtracted 3 youd get -7. If x has to be = or > than 0 then i dont understand how it can be -3 or less. Some body please explain this.\n",
        "A": "i understand now thank you",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            364
        ],
        "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator,"
    },
    {
        "Q": "\nAt 6:21, Sal asks, \"Is 'such that' the colon or the line?\" I've always used those interchangeably. Is there actually a difference?",
        "A": "Well I ve always used the pipe ( | ) symbol there (e.g { x in R | x \u00e2\u0089\u00a0 -2 } ) out of laziness, but Sal s comment in this video made me question that practice...",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            381
        ],
        "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator,"
    },
    {
        "Q": "\n4:28 It is to my understanding that if f(x)=radical x-3 then the x cannot be any less than three, which means x greater or EQUAL TO 0* cannot be so it is just x is ONLY greater than or equal to zero. If I am missing something please let me know.\n*just putting an emphasis",
        "A": "Sal said it was greater than or equal to zero, so I don t know what you mean",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            268
        ],
        "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the"
    },
    {
        "Q": "\nAround 6:10, I don't undersand how x could be smaller than -3.\nSay x= -4, than we get square root of -4 -3 = square root of -7, which isn't possible, or at least doesn't make sense to me..",
        "A": "It is because you are taking the absolute value of x. So, for x = -4, we would get: \u00e2\u0088\u009a( |x| - 3) = \u00e2\u0088\u009a( |-4| - 3) = \u00e2\u0088\u009a( 4 - 3) = \u00e2\u0088\u009a( 1) = \u00c2\u00b1 1 It is only when x is between -3 and 3 that we would get a negative number under the square root.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            370
        ],
        "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator,"
    },
    {
        "Q": "\nAt 9:12 why can't we make the trasformation R2+R3 and then R3+R2 to recover two zero rows as in the case lambda=3? This way at the end I would get an eigenspace that is equal to the span of the vectors (-1,1,0) and (-1,0,1). And if you sum them up you actually get the same vector Sal recovered in the video! Are they simply two \"equivalent\" solutions or is there something wrong about mine?\nThanks in advance for any answer :D",
        "A": "You can replace any row with a scalar multiple of itself, or with itself minus another row (or equivalently, another row minus itself ). There are different paths to the same result. If you don t interchange any columns, you should end up with the same equation Sal got, otherwise you made an error.",
        "video_name": "3Md5KCCQX-0",
        "timestamps": [
            552
        ],
        "3min_transcript": "And all the other things don't change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors in the eigenspace that corresponds to lambda is equal to minus 3, is going to be equal to 0. I'm just applying this equation right here which we just derived from that one over there. So, the eigenspace that corresponds to lambda is equal to minus 3, is the null space, this matrix right here, are all the vectors that satisfy this equation. So what is-- the null space of this is the same thing as the null space of this in reduced row echelon form So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So minus 2, minus 2, minus 2. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0 minus 5 plus minus-- or let me Let me replace it with the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And then let me do the last row in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5. So that is 3. Now let me replace-- and I'll do it in two steps. So this is 1, 1, 1. I'll just keep it like that. And actually, well let me just keep it like that. And then let me replace my third row with my third row plus my second row. It'll just zero out. If you add these terms, these all just become 0. That guy got zeroed out. And let me take my second row and divide it by 3. So this becomes 0, 1, minus 1. And I'm almost there. I'll do it in orange. So let me replace my first row with my first row minus my second row. So this becomes 1, 0, and then 1 minus minus 1 is 2. 1 minus minus 1 is 2. And then in the second row is 0, 1, minus 1."
    },
    {
        "Q": "At 2:56 while dividing 54 by 4, why did 4 went to twenty while it is written as 2.0? I see that it is number two because the decimal is just next to it on the right hand side. In the other words 2.0 is equals to 2 x 10^0. While the number twenty is equals to 2 x 10^1. Also, the number twenty is equals to 20.0, which looks different than the twenty that is written when we added a zero at 2:54.\n",
        "A": "Ah, I believe that Sal was carrying the decimal point down so you can see where it would be.",
        "video_name": "jTCZfMMcHBo",
        "timestamps": [
            176,
            174
        ],
        "3min_transcript": "So how much does the amount of money I make change when I work a certain number of hours, when my hours worked change by a certain amount. So let's just take some data points here. We could take really any of these data points, I'll take some of the smaller numbers. So let's say if when I go from four to eight hours, so my change in x is going to be what? If I go from four to eight, might change in x is going to be eight minus four, four hours, right? So this is going to be my change in x. I'm just picking these two points, I could have picked four and forty if I wanted, but the math would become more complicated. But how much does the amount of money earn change if I go from four hours to eight hours? Well, I go from $54 to $108, so the difference in the amount of money I make is $108 minus $54. Well, that's going to be $108 minus $54, that's just $54. And then what was the change in the amount of hours I worked? Well, the change in the hours I worked was four hours. So, if I work four more hours, I make 54 more dollars. Let me put a little equal sign there. So what is 54 divided by four? So four goes into 54-- looks like there's going to be decimal here-- four goes into five one time, one times four is four. Subtract, you get five minus four is one, bring down this four you get 14. Four goes into 14 three times, three times four is 12. Fourteen minus 12 is two, bring down a 0 right here, four goes into 20 five times. And of course you have this decimal right here. Five times four is 20. So this is equal to 13.5, but since we're talking in terms of dollars, maybe say $13.50, because that's our numerator, right? This is money earned, dollars per hour, because that's our denominator, dollars per hour. So that essentially answers our question. What does the slope represent in this situation? It represents the hourly wage for working at wherever this might be. Frankly, for this problem, you didn't even have to take two We could have said hey, if you work four hours and make $54, 54 divided by four is 13.50. Or we could have said hey, if we work eight hours, we get $108, 108 divided by eight is 13.50. So you didn't even have to take two data points here, you could have just taken any of these numbers divided by any of these numbers. But hopefully we also learned a little bit about what slope is."
    },
    {
        "Q": "\nAt 4:24, 3=2 has no solutions but what about if x =0? Then 0=0 would be one solution.",
        "A": "If you are asking if X=0 can be a solution to the middle equation, the answer is no. It really has no solution. To verify whether or not X=0 is solution, substitute into the equation. -7x + 3 = 2x + 2 - 9x -7(0) + 3 = 2(0) + 2 - 9(0) 0 + 3 = 0 + 2 + 0 3 = 2 This is false, so X=0 is not a solution. Hope this helps.",
        "video_name": "qsL_5Y8uWPU",
        "timestamps": [
            264
        ],
        "3min_transcript": "If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side, going to be left with a 2 there. So all I did is I added 7x. I added 7x to both sides of that equation. And now we've got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides, just to get rid of this constant term. So we're going to get negative 7x on the left hand side. On the right hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides. To subtract 2x from both sides, you're going to get-- so subtracting 2x,"
    },
    {
        "Q": "At 3:20 the symbol Sal used, is that the sign of infinite?\n",
        "A": "Yes \u00e2\u0088\u009e is the infinite sign. Great educated guess!",
        "video_name": "qsL_5Y8uWPU",
        "timestamps": [
            200
        ],
        "3min_transcript": "and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no. No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5-- and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for. Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side,"
    },
    {
        "Q": "5:40 Why that line is called secant line?\n",
        "A": "A secant line is a line that intersects a curve of some sort, at two points. A secant line is what we use to find average rates of change.",
        "video_name": "f4MYCepzLyQ",
        "timestamps": [
            340
        ],
        "3min_transcript": "it still is positive, but it seems a lot less positive. So it looks like the rate of change. It looks like the rate of change is actually changing here. In this case it looks like the line is getting steeper and steeper and steeper as time goes by. Or another way to think about it is, our rate of change of distance with respect to time is Increasing its not just constant. It's not a constant slope. The slope is increasing as time increases. So for something like this, how do we think about rate of change? How do we think about rate of change of one variable with respect to another, in particular distance with respect to time? Well later on in your math careers you'll find out that this is actually what most of differential calculus is all about and you will get to differential calculus. But for our purposes, we have another tool at our disposal and this is actually a good foundation for the calculus that you will learn in the future. This is the notion of average rate of change. Let me write this down average rate of change I already hopefully gave you an argument Why it's very hard you'll need actually calculus to figure out the Instantaneous rate of change the rate of change Right at that point the slope of the tangent line that you're going to need a calculus But to figure out the average rate of change between two points We can use very similar tools that we use to figure out the slope of a line. So for example: We could figure out the average rate of change between any two points on this curve. So for example: We could say the average rate of change from when we go from t equals 0 to t equals 3 That's going to be the slope of this secant line So let me draw that. So let me do this in a more fun color. I'll do it in this color All right So as I said the actual Instantaneous rate of change is constantly changing. In this case it's increasing, and we'll need calculus for that. But now we could think about average rate of change which would be the slope of the line that connects these two points. an average rate of change. As we see, the actual curve its rate of change is lower earlier on, and then it's rate of change is higher as we get closer and closer to 3 seconds. But the average rate of change is going to be the slope of this line right over here. And we could think about that. This is going to be our change. So the slope of this line, is going to be a change in distance over change in time. An average rate of change especially when you're talking about a curve like this it depends on what starting and ending point This is the average rate of change for the first 3 seconds, is going to be- Well what is our distance right at the third second? Well, it's going to be d of 3. Now what was our initial distance? It's going to be d of zero. So this expression right over here. It's going to give me my change in distance. It's going to give me my change in distance right over here Let me write. It actually maybe I could do it this way So I could do it over here. So our change in distance is this."
    },
    {
        "Q": "\nAt 0:50 what does sal mean by with respect to time?",
        "A": "Sal meant . what will be the change in distance with respect to time . Example :- you car moves with a speed of 80/kmph , so what do you mean by this, your car moves with a speed of 80 km, per hour and that is your speed . if you want to write it in meter/ sec it would be 22.2/sec . =)",
        "video_name": "f4MYCepzLyQ",
        "timestamps": [
            50
        ],
        "3min_transcript": "definitions for the function D over here over here d of t is equal to 3 t plus 1. We could imagine the d could represent distance as a function of time. Distance measured in meters and time measured in seconds. So over here when time is 0 right when we're starting our distance is going to be 1 meter. After 1 second has gone by our distance is now going to be, so 3 times 1 plus 1 is going to be 4. Our distance now is going to be 4 is going to be 4 meters. After 2 seconds our distance is going to be 3 times two is 6 plus 1 it's going to be 7 it's going to be 7 meters. So given this definition of D of T this function definition. What is the rate of change of distance with respect to time and let me write it this way. What is the change in distance the rate of change of distance with respect to time which people sometimes called speed. Well, what is this going to be? Let's just take two points Let's just say the change in distance over change in time when time goes from time equals 0 to time equals 1 So over here our change in time is equal to 1 our change in time is 1 and what's our change in distance? Well our change in distance when our time increased by 1 our distance increases by 3 it goes from 1 meters to 4 meters. So our change in distance is equal to 3. So it's going to be equal to 3 over 1 or just 3. If we wanted the units it would be 3 meters, every 1 second. Now let's think about it, does that change if we pick any other two points? what if we were to say between 1 second and 2 seconds so between 1 second and 2 seconds our change in time is 1 second and then our change in distance is So once again our change in distance over change in time is 3 meters per second. This is all review and you might recognize, we pick any two points on this line here and we're going to have the same rate of change of distance with respect to time. In fact that's what defines a line. Or one of the ways to think about a line or a linear function is that the rate of change of one variable with respect to the other one, is constant. In this particular one we're talking about the rate of change of the vertical variable with respect to the horizontal one. We're talking about the slope of the line. This is the slope. This line has a slope of 3. That's what defines a line or one of the things that defines a line is the slope between any two points is going to be exactly 3. Just as a little bit of review from other Algebra you've seen before. You can even pick it out in the function definition."
    },
    {
        "Q": "\nI still don't understand why at 1:36 you have to multiply both sides by Pi to simplify the equation.",
        "A": "He multiplied both sides by the reciprocal.",
        "video_name": "tVcasOt55Lc",
        "timestamps": [
            96
        ],
        "3min_transcript": "I have a circle here whose circumference is 18 pi. So if we were to measure all the way around the circle, we would get 18 pi. And we also have a central angle here. So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference-- let me write this down-- should be the same as the ratio of the central angle to the total number of angles if you were So let's just think about that. We know the circumference is 18 pi. We're looking for the arc length. I'm just going to call that a. a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. So we could simplify this by multiplying both sides by 18 pi. And we get that our arc length is equal to-- well, 10/360 is the same thing as 1/36. So it's equal to 1/36 times 18 pi, so it's 18 pi over 36, which is the same thing as pi/2. to be pi/2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine the same circle. So it's the same circle here. Our circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. But now I'm going to make the central angle an obtuse angle. So let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length-- so all of this."
    },
    {
        "Q": "\nAt 0:51 he said 9/5 is equals to 1 4/5, but why?",
        "A": "9/5 = 5/5 + 4/5 = 1 + 4/5 = 1 4/5. Have a blessed, wonderful new year!",
        "video_name": "-lUEWEEpmIo",
        "timestamps": [
            51
        ],
        "3min_transcript": "Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%-- this literally means 30 per 100. So you could view this as 30/100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. Now we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30/100-- and you could view this times 6/1-- this is equal to 180/100. We can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9/5, which is the same thing as 1 and 4/5. And then if we wanted to write this as a decimal, 4/5 is 0.8. And if you want to verify that, you and there's going to be a decimal. So let's throw some decimals in there. It goes into 4 zero times. So we don't have to worry about that. It goes into 40 eight times. 8 times 5 is 40. Subtract. You have no remainder, and you just have 0's left here. So 4/5 is 0.8. You've got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would've gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6-- let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal. So I'm only going to have one number to the right of the decimal here. So I just count one number. It's going to be 1.8. So either way you think about it or calculate it, 30% of 6"
    },
    {
        "Q": "\nAt 0:19 why does Sal say that we can get a 1 coefficient? Why do we need to get a 1 coefficient?",
        "A": "Your 1st step in factoring should always be to look for and remove the greatest common factor. The reason this should be your 1st step is that it makes doing any other factoring technique easier to do. Think of it has cleaning out the clutter, all the other numbers get smaller and easier to work with. And, you have to factor it out sometime, so do it as your 1st step and make the rest of the work easier.",
        "video_name": "R-rhSQzFJL0",
        "timestamps": [
            19
        ],
        "3min_transcript": "We're asked to factor 35k squared plus 100k, minus 15. And because we have a non-1 coefficient out here, the best thing to do is probably to factor this by grouping. But before we even do that, let's see if there's a common factor across all of these terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. And if we look at all of these numbers, they all look divisible by 5. In fact their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here, so we're still going to have to factor by grouping. But at least the numbers here are smaller so it'll be easier to think about it in terms of finding numbers whose product sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better if I were to take their product, I get 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs, so when you add numbers of different signs, you could view it as you're taking the difference of the positive versions. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21, and 1 will be the negative, because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups."
    },
    {
        "Q": "\nI'm really confused. Where does Sal get the 21 from @1:18? O_O",
        "A": "He multiplies 7 by -3. He factored the polynomial, then applied the product/sum theorem for quadratic polynomials to get the answers for their roots.",
        "video_name": "R-rhSQzFJL0",
        "timestamps": [
            78
        ],
        "3min_transcript": "We're asked to factor 35k squared plus 100k, minus 15. And because we have a non-1 coefficient out here, the best thing to do is probably to factor this by grouping. But before we even do that, let's see if there's a common factor across all of these terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. And if we look at all of these numbers, they all look divisible by 5. In fact their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here, so we're still going to have to factor by grouping. But at least the numbers here are smaller so it'll be easier to think about it in terms of finding numbers whose product sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better if I were to take their product, I get 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs, so when you add numbers of different signs, you could view it as you're taking the difference of the positive versions. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21, and 1 will be the negative, because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups."
    },
    {
        "Q": "What is the difference between a square root and a principle square root? 0:49 3:00\n",
        "A": "A square root can be negative or positive, but the principle square root is always positive.",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            49,
            180
        ],
        "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i."
    },
    {
        "Q": "\nAt 1:50 to 1:57, is the reason why that following that logic you would get i^infinity 2\u00e2\u0088\u009a13 or am I missing something?",
        "A": "I m confused where you got the idea of i^infinity from. Sal is explaining why sqrt(+52) cannot be sqrt(-1) \u00c3\u0097 sqrt(-52). When you simplify this, you end up with i \u00c3\u0097 i\u00c3\u0097sqrt(52), which is -sqrt(52). This is not the correct answer; therefore it doesn t work.",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            110,
            117
        ],
        "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i."
    },
    {
        "Q": "At, 2:12 Sal says the property does not work if both nos. are negative. I wanted to know why?\n",
        "A": "If both numbers are negative, you get something like what he showed in the video. To recap: if sqrt(1) = sqrt(-1) * sqrt(-1) <-- both numbers are negative then 1 = -1 <-- you already know sqrt(1) = 1, and sqrt(x) * sqrt(x) = x As you can see, it just doesn t work. Because of this, both numbers can t be negative.",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            132
        ],
        "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i."
    },
    {
        "Q": "\nAt 2:03 , Sal says u cant separate a principal square root number into the product of two numbers with the principal square root of a negative number. But you can, if you take the only the square root (not principal square root ) of the number ,say 4 , right? Cuz, u'll end up with i^2 sq.root(4) which\nalso can be written as [ -sq.root(4) ] which will give you the same two answers. Jst need a confirmation. Thanks.",
        "A": "I can understand what you are trying to say, but -2 x -2 = 4 not -4, likewise 2 x 2 is not = -4 hopefully this helps if you still need the question answered after 2 years.",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            123
        ],
        "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i."
    },
    {
        "Q": "When you are answering the question for credit on a test or something, would you write the constraints explained at 3:12 along with your answer?\n",
        "A": "yes, you would its part of the answer",
        "video_name": "XChok8XlF90",
        "timestamps": [
            192
        ],
        "3min_transcript": "factor this top expression over here\u2014which was the expression for the width\u2014this is of the form a-squared minus b-squared, where b-squared is 9. So this is going to be the same thing as a plus the square root of 9 times a minus the square root of 9. So this is a plus 3 times a minus 3 and I just recognized that from just the pattern; if you ever see something a-squared minus b-squared, it's a plus b times a minus b and you can verify that for yourself; multiply this out, you'll get a-squared minus b-squared. So, the width can be factored into a plus 3 times a minus 3. Let's see if we can do something for the denominator. So here, if we want to factor this out, we have to think of two numbers that when we add them, I get positive 6, and when I take their product, I get negative 27. Let's see, if I have positive 9 and negative 3, that would work. 9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27\u2014of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here\u2014we don't want to forget the constraints on our domain\u2014 a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined"
    },
    {
        "Q": "\nIs the constraint a \u00e2\u0089\u00a0 -9 at 3:47 really necessary? The simplified expression is already undefined at a = -9, so it looks as if it doesn't have to be restated.\n\nAdditionally, the original context of the word problem provides a rectangle whose width and length are rational expressions. Can't you not have a negative number in the simplified expression anyway?",
        "A": "1) Yes, it s obvious. 2) Set the area > 0. You get a^4+6a^3-36a^2-54a+261 > 0",
        "video_name": "XChok8XlF90",
        "timestamps": [
            227
        ],
        "3min_transcript": "9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27\u2014of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here\u2014we don't want to forget the constraints on our domain\u2014 a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined right over there; a cannot be equal to 3. Hopefully you found that useful."
    },
    {
        "Q": "what is that sound at 3:40 ?\n",
        "A": "some one laughing",
        "video_name": "X2jVap1YgwI",
        "timestamps": [
            220
        ],
        "3min_transcript": "of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake. So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%."
    },
    {
        "Q": "At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?\n",
        "A": "It s because from Earth the moon is closer than the sun therefore it looks like it is blocking it.",
        "video_name": "X2jVap1YgwI",
        "timestamps": [
            331
        ],
        "3min_transcript": "especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom x is the same thing as 1x, right? So 1x plus 0.25x, well that's just the same thing as 1 plus 0.25, plus x-- we're just doing the distributive property in reverse-- equals 100. And what's 1 plus 0.25? That's easy, it's 1.25. So we say 1.25x is equal to 100. Not too hard. And after you do a lot of these problems, you're going to intuitively say, oh, if some number grows by 25%, and it becomes 100, that means that 1.25 times that number is equal to 100. And if this doesn't make sense, sit and think about it a little bit, maybe rewatch the video, and hopefully it'll, over time, start to make a lot of sense to you. This type of math is very very useful. I actually work at a hedge fund, and I'm doing this type of math in my head day and night. So 1.25 times x is equal to 100, so x would equal"
    },
    {
        "Q": "\nAny one knows the answer for this it would be so helpful\nA Squate and a rectangle have the same area.\nThe sides of the rectangle are in the ratio 4:1. It's perimeter is 200cm.\nWhat is the length of the sides of the Square?",
        "A": "For problems like this, it is useful to write one of the unknown in terms of a symbol or letter. (i ll go with x) Here, length of shorter side of the rectangle is x. (Since the longer side is 4 times longer, i ll call it 4x) perimeter of rectangle= length+width+length+width = 200 Therefore, x+4x+x+4x =200 10x=200 x=20 Now to find the area of the rectangle, area= lengtth * width =20*80 =1600 area of square = 1600 = length*length length= sqrt(1600) =40 (fill in the units everywhere when you work on paper)",
        "video_name": "4ywTWCaLmXE",
        "timestamps": [
            241
        ],
        "3min_transcript": "is equal to question mark. And instead of writing question mark, I'll put a variable in there. Actually, let me put a question mark there just so you really understand it is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross-multiplying, but that cross-multiplying is still using a little bit of algebra. In cross-multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross-multiplication, you'll say that 2 times 9 must be equal to question mark times 3, must be equal to whatever is in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3, and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. this cross-multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x just to simplify the writing a little bit. So we have 2/3 is equal to-- instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you have to do the same thing to it. Now, what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x? So that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9,"
    },
    {
        "Q": "\nAt 4:55 , how is that multiplying 9 times 9 times cancels each other?\n9*9=81....am so confuse",
        "A": "Ah I see whats going on. Alright, theres 2/3 = x/9. Lets change this just a tiny bit to 2/3 = x * 1/9, now multiplying both sides by 9. we get 9 * 2/3 = x * 1/9 * 9. On the right side it s One Divided by 9 that gets multiplied by 9. Or if I just swapped it around. 9 * 2/3 = x * 9 * 1/9 = x * 9/9. So now multiplying and dividing it all out it is. 6 = x",
        "video_name": "4ywTWCaLmXE",
        "timestamps": [
            295
        ],
        "3min_transcript": "In cross-multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross-multiplication, you'll say that 2 times 9 must be equal to question mark times 3, must be equal to whatever is in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3, and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. this cross-multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x just to simplify the writing a little bit. So we have 2/3 is equal to-- instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you have to do the same thing to it. Now, what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x? So that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9, Otherwise they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2/3, or 9/1 times 2/3. Or this is equal to 18/3. And we know that 18/3 is the same thing as 6. So these are all legitimate ways to do it. I wanted you to understand that what I'm doing right here That's actually the reasoning why cross-multiplication works. But for a really simple problem like this, you could really just use common sense. If you're increasing the cups of oatmeal by a factor of 3, then increase the cups of flour by a factor of 3."
    },
    {
        "Q": "At 2:010 in the video above, is the pink loop twice the size of the smaller loop?\n",
        "A": "It should be, since the pink loop was going around the original loop twice.",
        "video_name": "Am-a5x9DGjg",
        "timestamps": [
            121
        ],
        "3min_transcript": "So I saw these Fruit by the Foots, or Fruit by the Feet. Maybe Fruits by the Foot. Anyway, I figured they had mathematical potential. So I decided to just record myself playing with them. The first thing that comes to mind when there's a strip of paper, even paper covered in fruit flavored sugar, is to make a Mobius strip. So I did. Thing about Mobius strips is they have one side, while Fruits by the Foot are, by nature, two-sided. A normal loop would have a paper side and a sugar side. Putting a half twist in would make a sudden transition from one to the other. But you can also do this. Wrap it around twice so the sugary part sticks to itself and the entire outside is covered in paper. Here is our Mobius strip. After confirming that Sharpie flip chart markers won't bleed through the paper, I drew a line along a single side of the Mobius strip. But that's not telling us anything that we didn't already know. Just like one line can cover both sides, except really one side, the paper does, too. So unlike a normal loop of Fruit by the Foot covered in paper, which would need two pieces, one for each side, the Mobius strip can be unwrapped by pulling off a single strip of paper. and then came back to see what else I could do. The strip of fruit-flavored gunk has these two lines going down it, perforations, dividing it into thirds. One of the go-to Mobius strip fun things is to cut it in half down the middle. But after that, cutting it into thirds is the next thing to do. And it's as if Fruit by the Foot is designed for this purpose. So I started separating it along this line. You might want to pause here and think about what you think would happen. So I continue around. And when I loop around once, now I'm on the other side. And it turns out to not have been two lines, but one line that goes around twice. Let's see what we've got. It's completely whoa. I mean, this is totally magical to me, that there's these two loops and they're linked together, and they're not even the same size. And they're made of flavored sugar gunk. But let's understand what's going on. You can do it with paper, too. I cut out a strip, twist, and tape. Now I'm coloring the edge, both to demonstrate that the Mobius Because when we cut a strip into thirds, you could also think of it like this. You're cutting the edge off. This leaves a thinner Mobius strip and a long looping edge. Because of the twist, the edge loops around the body of a Mobius strip so the two are linked. Really, though, you should just try it yourself. After making a Mobius strip and ripping it into thirds, I poked at the leftover bit of Fruit by the Foot, waiting to be inspired. It's spirally? I opened up another package so I'd have more to play with. Whoa. It's got this pattern on it. I mean, I don't remember this at all about Fruit by the Foot. But then again, it's not like I've had one in the past 15 years. And last time I had a Fruit by the Foot, I wouldn't have recognized that this is a frieze pattern, a symmetric pattern that repeats in one dimension. I mean, repeating patterns are good because they have a rolling stamp that just kind of presses this pattern into it. But it's got other symmetry, too. The two halves of this pattern are exactly the same, but not"
    },
    {
        "Q": "At 1:50, why did it divide like, into two loops? How come they aren't the same? Help me!\n",
        "A": "Because there is only one edge on a M\u00c3\u00b6bius strip, when she cut the edge off, it completely separated. Because of the twist in the M\u00c3\u00b6bius strip, the longer edge curled around the thicker center.",
        "video_name": "Am-a5x9DGjg",
        "timestamps": [
            110
        ],
        "3min_transcript": "So I saw these Fruit by the Foots, or Fruit by the Feet. Maybe Fruits by the Foot. Anyway, I figured they had mathematical potential. So I decided to just record myself playing with them. The first thing that comes to mind when there's a strip of paper, even paper covered in fruit flavored sugar, is to make a Mobius strip. So I did. Thing about Mobius strips is they have one side, while Fruits by the Foot are, by nature, two-sided. A normal loop would have a paper side and a sugar side. Putting a half twist in would make a sudden transition from one to the other. But you can also do this. Wrap it around twice so the sugary part sticks to itself and the entire outside is covered in paper. Here is our Mobius strip. After confirming that Sharpie flip chart markers won't bleed through the paper, I drew a line along a single side of the Mobius strip. But that's not telling us anything that we didn't already know. Just like one line can cover both sides, except really one side, the paper does, too. So unlike a normal loop of Fruit by the Foot covered in paper, which would need two pieces, one for each side, the Mobius strip can be unwrapped by pulling off a single strip of paper. and then came back to see what else I could do. The strip of fruit-flavored gunk has these two lines going down it, perforations, dividing it into thirds. One of the go-to Mobius strip fun things is to cut it in half down the middle. But after that, cutting it into thirds is the next thing to do. And it's as if Fruit by the Foot is designed for this purpose. So I started separating it along this line. You might want to pause here and think about what you think would happen. So I continue around. And when I loop around once, now I'm on the other side. And it turns out to not have been two lines, but one line that goes around twice. Let's see what we've got. It's completely whoa. I mean, this is totally magical to me, that there's these two loops and they're linked together, and they're not even the same size. And they're made of flavored sugar gunk. But let's understand what's going on. You can do it with paper, too. I cut out a strip, twist, and tape. Now I'm coloring the edge, both to demonstrate that the Mobius Because when we cut a strip into thirds, you could also think of it like this. You're cutting the edge off. This leaves a thinner Mobius strip and a long looping edge. Because of the twist, the edge loops around the body of a Mobius strip so the two are linked. Really, though, you should just try it yourself. After making a Mobius strip and ripping it into thirds, I poked at the leftover bit of Fruit by the Foot, waiting to be inspired. It's spirally? I opened up another package so I'd have more to play with. Whoa. It's got this pattern on it. I mean, I don't remember this at all about Fruit by the Foot. But then again, it's not like I've had one in the past 15 years. And last time I had a Fruit by the Foot, I wouldn't have recognized that this is a frieze pattern, a symmetric pattern that repeats in one dimension. I mean, repeating patterns are good because they have a rolling stamp that just kind of presses this pattern into it. But it's got other symmetry, too. The two halves of this pattern are exactly the same, but not"
    },
    {
        "Q": "At 2:04. Why does Sal say, \" third of the way around the triangle \". What does it mean?\nThanks.\n",
        "A": "A triangle s three angles measure up to 180. Since it is a equilateral triangle Sal is constructing, each angle is 60 degrees. Many people become confused when Sal says 120 degrees is one third of the triangle, but he is talking about the arc of the circle. To find the arc or the angle formed by the arc, use this equation: arcX = 2 angleX. Thus, the angle is 60 degrees and one third of a triangle, due to what I said earlier. (60 x 3 = 180)",
        "video_name": "gWMTTP58_J0",
        "timestamps": [
            124
        ],
        "3min_transcript": "Construct an equilateral triangle inscribed inside the circle. So let me construct a circle that has the exact same dimensions as our original circle. Looks pretty good. And now, let me move this center, so it sits on our original circle. So they now sit on each other. Or their centers now sit on each other. So I can make it, and that looks pretty good. And now, let's think about something. If I were to draw this segment right over here, this, of course, has the length of the radius. Now, let's do another one. And that's either of their radii, because they have the same length. Now, let's just center this at our new circle and take it out here. Now, this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here. So these two segments have the same length. Now, if I were to connect that point to that point, this is a radius of our original circle. So this right over here, I have constructed an equilateral triangle. Now, why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now, why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now, why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees. Or it's a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1/3 of the entire circle. And now, I can keep doing this. Let's move-- I'll reuse these-- let's move our circle around. And so now, I'm going to move my circle along the circle. And once again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc you could say, is 120 degrees. So this is going to be one side of our equilateral triangle."
    },
    {
        "Q": "\nAt 1:02, what does Sal mean with the fact that the sum of 1/n is unbounded, but that the sum of 1/n^2 is bounded?",
        "A": "i think an unbound sum is basically like a divergent integral and a bound sum is like a convergent integral one goes to infinity and the other has a fixed value",
        "video_name": "u1UKIljUWuc",
        "timestamps": [
            62
        ],
        "3min_transcript": "- [Voiceover] Let's explore a bit the infinite series from n equals one to infinity of one over n squared. Which of course is equal to one plus one fourth, that's one over two squared, plus one over three squared, which is one ninth, plus one sixteenth and it goes on and on and on forever. So there's a couple of things that we know about it. The first thing is that all of the terms here are positive. So all of the terms here are positive. So they're all positive and that they're decreasing. It looks like they're decreasing quite quickly here from one to one fourth to one ninth to one sixteenth, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive we know that this sum right over here, if it does converge is going to be greater than zero. So the only reason why it wouldn't converge is if which we know if this was one over n it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that will be a pretty good argument for why this thing right over here converges because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms. Or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum, none of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're going to do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. one over n squared as f of n if I were to write it this way. So why is this interesting? Well let's graph it. So that's the graph of y is equal to f of x. And notice this is a continuous, positive, decreasing function, especially over the interval that I care about right over here. I guess we could say for positive values of x, it is a continuous, positive, decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well one, this first term right over here, you could view that as the area of this block right over here."
    },
    {
        "Q": "At 6:33 -- Does Sal mean\n(1 + def-integral from 2 to inf of 1/(x^2)) instead of (1 + def-integral from 1 to inf of 1/(x^2))?\n",
        "A": "No, it is the indefinite integral from 1 to infinity. The series does start at n=2 and goes to infinity, because we took out the 1. But both the integral and the series, that starts at n = 2, start at x = 1. The only difference is that the series is an understimate (a right Reimann Sum) of the actual area of the curve (the indefinite integral).",
        "video_name": "u1UKIljUWuc",
        "timestamps": [
            393
        ],
        "3min_transcript": "So our original series from n equals one to infinity of one over n squared. It's going to be equal to this first block, the area of this first block plus the area of all the rest of the blocks, the one fourth plus one ninth plus one sixteenth, let me do this in a new color. Which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of expressed this as a sum of this plus all of that stuff. Now what's interesting is that this, what I just wrote in this blue notation that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice it's an underestimate it's always below the curve, so it's going to So we can write that this thing is going to be less than one plus instead of writing this I'm gonna write the definite integral. One plus the definite integral from one to infinity of one over x squared dx. Now why is that useful? Well we know how to evaluate this and I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of and I'll just write this as x to the negative two dx. Which is equal to the limit as t approaches infinity of negative x to the negative one, or actually I And we're going to evaluate that at t and at one, which is equal to the limit as t approaches infinity of negative one over t and then minus negative one over one so that would just be plus one. And as t approaches infinity this term right over here is going to be zero, so this is just going to simplify to one. So this whole thing evaluates to one. So just like that we were able to place an upper-bound on this series. We're able to say that the series under question or in question, so the infinite sum from n equals one to infinity of one over n squared is going to be less than one plus one or it's going to be less than two. Or another way to think about it, it's going to be the two"
    },
    {
        "Q": "\nAt 5:35, why does Sal define the range of the Volume function according to the height (x) and width (20 - 2x) of the box? Why that specific combination of dimensions and not say height (x) & length (30 - 2x) or width & length instead?",
        "A": "At that point he isn t defining the range of the volume, he s defining the realistic range for x, where x will ultimately represent the height of the box. Eventually some value of x will result in an optimum volume relative to its height (where the height is a direct function of x), width and length, all of which have been defined in terms of x.",
        "video_name": "MC0tq6fNRwU",
        "timestamps": [
            335
        ],
        "3min_transcript": "So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus x-- sorry, 20 minus 2x times the depth, which is 30 minus 2x. Now, what are possible values of x that give us a valid volume? Well, x can't be less than 0. You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to 0. So let me write this down. x is going to be greater than or equal to 0. And what does it have to be less than? Well, I can cut at most-- we can see here the length right over here, this pink color, this mauve color, So this has got to be greater than 0. This is always going to be shorter than the 30 minus 2x, but the 20 minus 2x has to be greater than or equal to 0. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to 2x, or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. That's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between 0 and 10. Otherwise we've cut too much, or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our-- essentially of our domain, of what x can be for our volume. Well, our volume when x is equal to 0 is equal to what? We can have 0 times all of this stuff. You're not going to have any height here. So you're not going to have any volume, so our volume would be 0. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be 0. So once again, we would have no volume. And this term right over here, if we just look at it algebraically would also be, equal to 0, so this whole thing would be equal to 0. So someplace in between x equals 0 and x equals 10 we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value, let me make that 0. We know that x cannot be less than 0. My maximum x-value, well, 10 seems pretty good. My minimum y-value, this is essentially"
    },
    {
        "Q": "Towards the end of the video at around 19:00, didn't Sal mean to write rank(A) = n?\n",
        "A": "I believe he did",
        "video_name": "M3FuL9qKTBs",
        "timestamps": [
            1140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 4:22, Sal said that there is variability in the data. can you please explain this?",
        "A": "Variability means that something changes. So, if your set of data has values that differ from each other or change, then there is variability in the data.",
        "video_name": "qyYSQDcSNlY",
        "timestamps": [
            262
        ],
        "3min_transcript": "Do dogs run faster than cats? Once again, there are many dogs and many cats, and they all run at different speeds. Some dogs run faster than some cats, and some cats run faster than some dogs. So we would need some statistics to get a sense of in general or on average how fast do dogs run and then maybe on average how fast do cats run. Then we could compare those averages, or we could compare the medians in some way. So this is definitely a statistical question. Once again, we're talking about, in general, a whole population of dogs, the whole species of dogs, versus cats. And there's variation in how fast dogs run and how fast cats run. If we were talking about a particular dog and a particular cat, well, then there would just be an answer. Does dog A run faster than cat B? Well, sure. That's not going to be a statistical question. You don't have to use the tools of statistics. Actually, no, this fits the pattern of the previous one. Do wolves weigh more than dogs? Once again, there are some very light dogs and some very heavy So those wolves definitely weigh more than those dogs. But there's some very, very, very heavy dogs. And so what you want to do here, because we have variability in each of these, is you might want to come with some central tendency. On average, what's the median wolf weight? What's the average, the mean wolf weight? Compare that to the mean dog's weight. So once again, since we're speaking in general about wolves, not a particular wolf, and in general about dogs, and there's variation in the data, and we're trying to glean some numbers from that to compare, this is definitely a statistical question. Does your dog weigh more than that wolf? And we're assuming that we're pointing at a particular wolf. Now this is the particular. We're comparing a particular dog to a particular wolf. and come up with an absolute answer. There's no variability in this dog's weight, at least at the moment that we weigh it, no variability in this wolf's weight at the moment that we weigh it. This is not a statistical question. I'll put an x next to the ones that are not statistical questions. Does it rain more in Seattle than Singapore? Once again, there is variation here. And we would also probably want to know, does it rain more in Seattle than Singapore in a given year, over a decade, or whatever? But regardless of those questions, however we ask it, in some years, it might rain more in Seattle. In other years, it might rain more in Singapore. Or if we just picked Seattle, it rains a different amount from year to year. In Singapore, it rains a different amount from year So how do we compare? Well, that's where the statistics could be valuable. There's variability in the data. So we can look at the data set for Seattle and come up with some type of an average, some type of a central tendency, and compare that to the average, the mean, the mode-- the mode probably wouldn't be that useful here-- to Singapore."
    },
    {
        "Q": "7:54... Shouldn't the new vector have \"m\" components? One for each row?\n",
        "A": "Yes (mistake).",
        "video_name": "7Mo4S2wyMg4",
        "timestamps": [
            474
        ],
        "3min_transcript": "m-th row, and then the m-th row will be am1. This is the m-th row first column. am1 times x1 plus -- it's hard to keep switching colors -- plus am2 times x2, all the way until we get to amn times xn. So what is this vector going to look like? It's essentially going to have -- let's say we call this vector-- Let's say it's equal to vector b. What does vector b look like? How many entries is it going to have? Well it has an entry for each row of this, right? We're taking each row and we're essentially taking the dot product of this row vector with this column vector. notation in a second. But I think you understand that this is a dot product. The first component times the first component plus the second component times the second component plus the third component times the third component, all the way to the n-th component plus the n-th component times the n-th component. So this is essentially the dot product of this row vector. We've been writing all of our vectors as columns, so we could call them column vectors, you're just writing them as rows. And we can be a little bit more specific with the notation in a second, but what's this going look like? Well we're doing this m times, so we're going to have m entries. You're going to b1 b2 all the way to bn. If you viewed these all as matrices, you can kind of view it as -- and this will eventually work for the matrix math we're going to learn -- this is an m by n matrix and we're multiplying it by -- how many rows does this guy have? He has n rows. So m by n times an n by 1, you essentially can ignore these middle two terms, and they'll result with -- how many rows does this guy have? He has m rows, and 1 column. These middle two terms have to be equal to each other just for the multiplication to be defined, and then you're left with an m by 1 matrix. So this was all abstract, let me actually apply it to some actual numbers. But it's important to actually set the definition. Now that we have the definition we can apply it to some actual matrices and vectors. So let's say we have the matrix. Let's say I want to multiply the matrix minus 3, 0, 3, 2. Now I'll do this one in yellow. 1, 7, minus 1, 9. And I want to multiply that by the vector."
    },
    {
        "Q": "At 0:36 could the sequence be written instead as the sum from n=0 to infinity of (1/2)^n? and if so which one is used most commonly?\n",
        "A": "Sure - both versions are widely used.",
        "video_name": "rcRg_gO7-7E",
        "timestamps": [
            36
        ],
        "3min_transcript": "Let's say that I have a geometric series. A geometric sequence, I should say. We'll talk about series in a second. So a geometric series, let's say it starts at 1, and then our common ratio is 1/2. So the common ratio is the number that we keep multiplying by. So 1 times 1/2 is 1/2, 1/2 times 1/2 is 1/4, 1/4 times 1/2 is 1/8, and we can keep going on and on and on forever. This is an infinite geometric sequence. And we can denote this. We can say that this is equal to the sequence of a sub n from n equals 1 to infinity, with a sub n equaling 1 times our common ratio to the n minus 1. So it's going to be our first term, which is just 1, times our common ratio, which is 1/2. 1/2 to the n minus 1. And you can verify it. This right over here you can view as 1/2 to the 0 power. This is 1/2 half to the first power, this is 1/2 squared. So the first term is 1/2 to the 0. The second term is 1/2 to the 1. The third term is 1/2 squared. So the nth term is going to be 1/2 to the n minus 1. So this is just really 1/2 to the n minus 1 power. Fair enough. Now, let's say we don't just care about looking at the sequence. We actually care about the sum of the sequence. So we actually care about not just looking at each of these terms, see what happens as I keep multiplying by 1/2, but I actually care about summing 1 plus 1/2 plus 1/4 plus 1/8, and keep going on and on and on forever. So this we would now call a geometric series. And because I keep adding an infinite number of terms, this is an infinite geometric series. So this right over here would be the infinite geometric series. A series you can just view as the sum of a sequence. Now, how would we denote this? Well, we can use summing notation. We could say that this is equal to the sum. Let me make sure I'm not falling off the page. Let me just scroll over to the left a bit. The sum from n equals 1 to infinity of a sub n. And a sub n is just 1/2 to the n minus 1. 1/2 to the n minus 1 power. So you just say OK, when n equals 1, it's 1/2 to the 0, which is 1. Then I'm going to sum that to when n equals 2, which is 1/2, when n equals 3, it's 1/4. On and on, and on, and on. So all I want to do in this video is to really clarify differences between sequences and series, and make you a little bit comfortable with the notation. In the next few videos, we'll actually try to take sums of geometric series and see if we actually get a finite value."
    },
    {
        "Q": "\n3:04 she says that circles can be defined by three points, can't a circle be defined by more than three?",
        "A": "No. A circle is only defined by three points. If you have three points, then there is only one circle that can pass through all three (unless they are collinear in which case no circle can). A fourth point would either be redundant or make the circle previously defined invalid.",
        "video_name": "DK5Z709J2eo",
        "timestamps": [
            184
        ],
        "3min_transcript": "and it's cool that that's possible. But I'm not really interested in doing calculations. So we'll come back to camels. Here's a fractal. You start with these circles in a circle, and then keep drawing the biggest circle that fits in the spaces between. This is called an Apollonian Gasket. And you can choose a different starting set of circles, and it still works nicely. It's well known in some circles because it has some very interesting properties involving the relative curvature of the circles, which is neat, But it also looks cool and suggests an awesome doodle Step 1, draw any shape. Step 2, draw the biggest circle you can within this shape. Step 3, draw the biggest circle you can in the space left. Step 4, see step 3. As long as there is space left over after the first circle, meaning don't start with a circle, this method turns any shape into a fractal. You can do this with triangles. You can do this with stars. And don't forget to embellish. You can do this with elephants, or snakes, or a profile of one I choose Abraham Lincoln. Awesome. OK, but what about other shapes besides circles? For example, equilateral triangles, say, filling this other triangle, which works because the filler triangles, and orientation matters. This yields our friend, Sierpinski's triangle, which, by the way, you can also make out of Abraham Lincoln. But triangles seem to work beautifully in this case. But that's a special case. And the problem with triangles is that they don't always fit snugly. For example, with this blobby shape, the biggest equilateral triangle has this lonely hanging corner. And sure, you don't have to let that stop you, and it's a fun doodle game. But I think it lacks some of the beauty of the circle game. Or what if you could change the orientation of the triangle to get the biggest possible one? What if you didn't have to keep it equilateral? Well, for polygonal shapes, the game runs out pretty quickly, so that's no good. But in curvy, complicated shapes, the process itself becomes difficult. How do you find the biggest triangle? It's not always obvious which triangle has more area, especially when you're starting shape is not very well defined. This is an interesting sort of question because there is a correct answer, but if you were going to write a computer program that filled a given shape with another shape, following even the simpler version of the rules, you might need to learn some computational geometry. And certainly, we can move beyond triangles to squares, or even elephants. But the circle is great because it's just so fantastically round. A circle can be defined by three points. So draw three, arbitrary points, and then try and find the circle they belong to. So one of the things that intrigues me about the circle game is that, whenever you have one of these sorts of corners, you know there's going to be an infinite number of circles Thing is, for every one of those infinite circles, you create a few more little corners that are going to need an infinite number of circles. And for every one of those, and so on. You just get an incredible number of circles breeding more circles. And you can see just how dense infinity can be. Though the astounding thing is that this kind of infinity is still the smallest, countable kind of infinity. And there are kinds of infinity that are just mind bogglingly infiniter. But wait, here's an interesting thing. If you call this distance 1 arbitrary length unit, then this distance plus this, dot, dot, dot, is an infinite series that approaches 1. And this is another, different, series that still approaches 1. And here's another, and another. And as long as the outside shape is well defined, so will the series be. But if you want the simple kind of series, where each circle's diameter is a certain percentage of the one before it,"
    },
    {
        "Q": "\nat 0:56, vi says a camel is only a third of the page, but if so, couldn't you just mark 1/6 of the page, and 1/12, 1/24, & so on?",
        "A": "Yeah, she could ve just acted like the 2/3 left was 1 page, then made the camels half of that, then half of that, and so on. She could ve also made the camels bigger or only used 2/3 of the page for camels, making the 1/2 of 1/2 of 1/2\u00e2\u0080\u00a6 again and used the extra for a circle game shape.",
        "video_name": "DK5Z709J2eo",
        "timestamps": [
            56
        ],
        "3min_transcript": "So you mean you're in math class, yet again, because they make you go every single day. And you're learning about, I don't know, the sums of infinite series. That's a high school topic, right? Which is odd, because it's a cool topic. But they somehow manage to ruin it anyway. So I guess that's why they allow infinite serieses in the curriculum. So, in a quite understandable need for distraction, you're doodling and thinking more about what the plural of series should be than about the topic at hand. Serieses, serises, seriesen, seri? Or is it that the singular should be changed? One serie, or seris, or serum? Just like the singular of sheep should be shoop. But the whole concept of things like 1/2 plus 1/4 plus 1/8 plus 1/16, and so on approaches 1 is useful if, say, you want to draw a line of elephants, each holding the tail of the next one. Normal elephant, young elephant, baby elephant, dog-sized elephant, puppy-sized elephant, all the way down to Mr. Tusks, and beyond. Which is at least a tiny bit awesome because you can get an infinite number of elephants in a line and still have it fit across a single notebook page. But there's questions, like what if you started with a camel, which, being smaller than an elephant, only goes across a third of the page. How big should the next camel be in order to properly approach the end of the page? and it's cool that that's possible. But I'm not really interested in doing calculations. So we'll come back to camels. Here's a fractal. You start with these circles in a circle, and then keep drawing the biggest circle that fits in the spaces between. This is called an Apollonian Gasket. And you can choose a different starting set of circles, and it still works nicely. It's well known in some circles because it has some very interesting properties involving the relative curvature of the circles, which is neat, But it also looks cool and suggests an awesome doodle Step 1, draw any shape. Step 2, draw the biggest circle you can within this shape. Step 3, draw the biggest circle you can in the space left. Step 4, see step 3. As long as there is space left over after the first circle, meaning don't start with a circle, this method turns any shape into a fractal. You can do this with triangles. You can do this with stars. And don't forget to embellish. You can do this with elephants, or snakes, or a profile of one I choose Abraham Lincoln. Awesome. OK, but what about other shapes besides circles? For example, equilateral triangles, say, filling this other triangle, which works because the filler triangles, and orientation matters. This yields our friend, Sierpinski's triangle, which, by the way, you can also make out of Abraham Lincoln. But triangles seem to work beautifully in this case. But that's a special case. And the problem with triangles is that they don't always fit snugly. For example, with this blobby shape, the biggest equilateral triangle has this lonely hanging corner. And sure, you don't have to let that stop you, and it's a fun doodle game. But I think it lacks some of the beauty of the circle game. Or what if you could change the orientation of the triangle to get the biggest possible one? What if you didn't have to keep it equilateral? Well, for polygonal shapes, the game runs out pretty quickly, so that's no good. But in curvy, complicated shapes, the process itself becomes difficult. How do you find the biggest triangle? It's not always obvious which triangle has more area, especially when you're starting shape is not very well defined. This is an interesting sort of question because there is a correct answer, but if you were going to write a computer program that filled a given shape with another shape, following even the simpler version of the rules, you might need to learn some computational geometry. And certainly, we can move beyond triangles to squares, or even elephants. But the circle is great because it's just so fantastically round."
    },
    {
        "Q": "At 3:57, couldn't you just make the intervals on 0 and 3 and not include the numbers -1 and 4?\n",
        "A": "Remember, there are lots of numbers (fractions and decimals) that are located between the integers. If you make the interval 0 to 3, you lose all the real numbers from -1 to 0 and 3 to 4 (numbers like -0.5, -1/4, -0.999999, 3.6; 3.999, etc.). So, you have a completely different set of numbers and a much smaller set of numbers. Thus, the interval needs to be (-1, 4)",
        "video_name": "UJQkqV2zGv0",
        "timestamps": [
            237
        ],
        "3min_transcript": "that we include, this bracket on the left says that we include negative three, and this bracket on the right says that we include positive two in our interval. Sometimes you might see things written a little bit more math-y. You might see x is a member of the real numbers such that... And I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying that the set of all x's that are a member of the real number, so this is just fancy math notation, it's a member of the real numbers. I'm using the Greek letter epsilon right over here. It's a member of the real numbers such that. This vertical line here means \"such that,\" negative three is less x is less than-- negative three is less than or equal to x, is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. So let's-- Let me draw a number line again. So, a number line. And now let me do-- Let me just do an open interval. An open interval just so that we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Let me use a different color. So the values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four, and I'm not going to include negative one. Notice I have open circles here. Over here had closed circles, the closed circles told me that I included negative three and two. it's all the values in between negative one and four. Negative .999999 is going to be included, but negative one is not going to be included. And 3.9999999 is going to be included, but four is not going to be included. So how would we-- What would be the notation for this? Well, here we could say x is going to be a member of the real numbers such that negative one-- I'm not going to say less than or equal to because x can't be equal to negative one, so negative one is strictly less than x, is strictly less than four. Notice not less than or equal, because I can't be equal to four, four is not included. So that's one way to say it. Another way I could write it like this. x is a member of the real numbers such that x is a member of... Now the interval is from negative one to four but I'm not gonna use these brackets."
    },
    {
        "Q": "\nAt 8:13 could it also be expressed as {X \u00cf\u00b5 R | 1 < X < 1 } ?",
        "A": "Sorry, but no. 1 < X < 1 is the notation for and , not or . 1 < X < 1 is the null set since both 1 < X and x < 1 must be true. There is no number that is both greater than one and less than one.",
        "video_name": "UJQkqV2zGv0",
        "timestamps": [
            493
        ],
        "3min_transcript": "So we're not going to include negative four. Negative four is strictly less than, not less than or equal to, so x can't be equal to negative four, open circle there. But x could be equal to negative one. It has to be less than or equal to negative one. It could be equal to negative one so I'm going to fill that in right over there. And it's everything in between. If I want to write it with this notation I could write x is a member of the real numbers such that x is a member of the interval, so it's going to go between negative four and negative one, but we're not including negative four. We have an open circle here so I'm gonna put a parentheses on that side, but we are including negative one. We are including negative one. So we put a bracket on that side. That right over there would be the notation. Now there's other things that you could do You could say, well hey, everything except for some values. Let me give another example. Let's get another example here. Let's say that we wanna talk about all the real numbers except for one. We want to include all of the real numbers. All of the real numbers except for one. Except for one, so we're gonna exclude one right over here, open circle, but it can be any other real number. So how would we denote this? Well, we could write x is a member of the real numbers such that x does not equal one. So here I'm saying x can be a member of the real numbers but x cannot be equal to one. It can be anything else, but it cannot be equal to one. You could say x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that. Or you could do something interesting. This is the one that I would use, this is the shortest and it makes it very clear. You say hey, everything except for one. But you could even do something fancy, like you could say x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from-- or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when we're talking about negative infinity or positive infinity, you always put a parentheses."
    },
    {
        "Q": "When you work out the problem at 3:55, wouldn't it be f(x+1)/f(x)=1/4*2^x+1/1/4*2^x=1? Because 1/4 and x's cancel and you are left with 2/2 which is equal to 1.\n",
        "A": "First off, be careful how you write it, you need extra sets of parentheses or your equation is totally different so f(x+1)/f(x) = 1/4 * 2^ (x+1) / (1/4*2^x). The 1/4 cancel so you are left with 2^(x+1)/2^x x s do not cancel, we have to use exponent rules where dividing same bases require us to subtract the exponents, so as Sal noted x + 1 - x is just 1, so we have 2^1 or just 2",
        "video_name": "G2WybA4Hf7Y",
        "timestamps": [
            235
        ],
        "3min_transcript": "Five times one is just five. So the initial value is once again, that. So if you have exponential functions of this form, it makes sense. Your initial value, well if you put a zero in for the exponent, then the number raised to the exponent is just going to be one, and you're just going to be left with that thing that you're multiplying by that. Hopefully that makes sense, but since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio. The common common ratio. And in my brain, we say well why is it called a common ratio? Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color, h of zero is equal to, we already established one-fourth. Now, what is h of one going to be equal to? two to the first power. So it's going to be one-fourth two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one it would be two. That is the common ratio between successive whole number inputs into our function. So, h of I could say plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one one-fourth two to the n. Two to the n plus one, divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential function look like? And they're telling you this is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five."
    },
    {
        "Q": "\nAt 2:43, what does the word arithmetic mean in general? I've heard people use the word a lot.",
        "A": "The worldwide definition of arithmetic is: the branch of mathematics relating to the manipulation of numbers. Fractions, addition and subtraction, division and multiplication fall into this category.",
        "video_name": "uhxtUt_-GyM",
        "timestamps": [
            163
        ],
        "3min_transcript": "and we'll talk a lot about samples versus populations, but I think just a basic sense of what that is. If I survey three people who are going to vote for president, I clearly haven't surveyed the entire population. I've surveyed a sample. But what inferential statistics are all about are if we can do some math on the samples, maybe we can make inferences or conclusions about the population as a whole. Anyway, that's just the big picture of what statistics is all about. So let's just get into the meat of it. We'll start with the descriptive. So the first thing that I would want to do it, or I think most people would want to do when they're given a whole set of numbers and they're told to describe it. It's like well, maybe I can come up with some number that is most indicative of all of the numbers in that set, or some number that represents the central tendency. This is a word you'll see a lot in statistics books. The central tendency of a set of numbers. And I'll be a little bit more exact here than I normally am with the word average. When I talk about in this context, it just means that the average is a number that somehow is giving us a sense of the central tendency, or maybe a number that is most representative of a set. I know that sounds all very abstract, but let's do a couple of examples. So there's a bunch of ways you can actually measure of the central tendency, or the average, of a set of numbers, and you've probably seen these before. They are the mean-- actually, there are types of means, but we'll stick with the arithmetic mean. Later when we talk about stock returns and things, we'll do geometric means. And maybe we'll cover the harmonic mean one day. There's a mean, the median, and the mode. can kind of be representative of a data set, or a population's central tendency, or a sample central tendency. And they all are collectively-- they can all be forms of an average. And I think when we see examples, it'll make a little bit more sense. In every day speak when people talk about an average-- I think you've already computed averages in your life-- they're usually talking about the arithmetic mean. So normally when someone says, let's take the average of these numbers and they expect you to do something, they want you to figure out the arithmetic mean. They don't want you to figure out the median or the mode. But before we go any further, let's figure out what these things are. So let me make up a set of numbers. Let's say I have the number 1, let's say I have another 1, 2, 3, let's say I have a 4. That's good enough. We just wanted a simple example. So the mean, or the arithmetic mean,"
    },
    {
        "Q": "In this video about Statistics in 8:04 and 10:42 when Sal was talking about The median does this type of rule apply When there is odd numbers there is a middle number but when there is an even number you have to divde it by the two middle numbers that is your median.Does that rule apply in Statistics? Just want to know.\n",
        "A": "Thanks Andrew for letting me know!",
        "video_name": "uhxtUt_-GyM",
        "timestamps": [
            484,
            642
        ],
        "3min_transcript": "And I know what you might be thinking-- well, that was easy enough when we had five numbers. But what if we had six numbers? What if it was like this? What if this was our set of numbers? 1, 1, 2, 3, let's add another 4 there. So now there's no middle number. 2 is not the middle number because there's two less than it and three larger than it. And then 3 is not the middle number because there's two larger and three smaller than it. So there's no middle number. So when you have a set with even numbers and someone tells you to figure out the median, what you do is you take the middle two numbers. And then you take the arithmetic mean of those two numbers. So in this case, of this set, the median would be 2.5. Fair enough. But let's put this aside, because I want to compare the median and the means and the modes for the same set of numbers. But that's a good thing to know, because sometimes it can be a little confusing. These are all mathematical tools for getting our heads around numbers. It's not like one day someone saw one of these formulas on the face of the sun and says, oh, that's part of the universe. That is how the average should be calculated. These are human constructs to just get our heads around large sets of data. I mean, this isn't a large set of data. But instead of five numbers, if we had five million numbers, you could imagine that you don't like thinking about every number individually. Anyway, before I talk more about that, let me tell you what the mode is. And the mode, to some degree, it's the one that I think most people forget or never learn. And when they see it on an exam, it confuses them because they're like, oh, that sounds very advanced. But in some ways, it is the easiest of all of the measures of central tendency or of average. The mode is essentially what number is most common in a set. So in this example, there's two ones and then there's one of everything else. So the mode here is 1. So mode you can say is the most common number. 1, 1, 2, 3, 4, 4. Here, I have two 1s and I have two 4s. And this is where the mode gets a little bit tricky, because either of these would have been a decent answer for the mode. You could have actually said the mode of this is 1, or the mode of this is 4. And it gets a little bit ambiguous, and you probably want a little clarity from the person asking you. Most times on a test when they ask you, there's not going to be this ambiguity. There will be a most common number in the set. Now you're saying, oh, why wasn't just one of these good enough? Why did we learn averages? Why don't we just use arithmetic mean all the time? What's median and mode good for? Well, I'll try to do one example of that and see if it rings true with you. And then you could think a little bit more. Let's say I had this set of numbers-- 3, 3, 3, 3, 3, and 100."
    },
    {
        "Q": "\nAt 11:10 sal, u wrote 115 /6 what is that supposed to mean were talking about arithmatic not fractions and also your example for the mode did not make sense may you clarify that for me",
        "A": "He was calculating the mean of the data set {3,3,3,3,3,100}, which is found by taking the sum of all those numbers (=115) and dividing by the number of elements in the set (6). Hence 115/6, which is, as he says, 19 1/6. You could also write it as 19.7, but 115/6 is more precise, as you don t have to round. For that same data set, the mode is pretty straightforward. The number 3 is clearly the most common number in that set (thee are five 3 s but only one 100), so the mode is 3.",
        "video_name": "uhxtUt_-GyM",
        "timestamps": [
            670
        ],
        "3min_transcript": "What's the mean here? So one, two, three, four, five 3s and 100. So it would be 115 divided by 6. Because I have one, two, three, four, five, six numbers. 115 is just the sum of all of these. So that's equal to-- how many times does 6 go into 115? 6 goes into it one time. 1 times 6 is 6. 55 goes into it 9 times. 9 times 6 is 54. So it's equal to 19 and 1/6. Fair enough. I just added all the numbers and divided by how many there are. But my question is, is this really representative of this set? I mean, I have a ton of 3s and then I have 100 all of a sudden. And we're saying that the central tendency is 19 and 1/6. 19 and 1/6 doesn't really seem indicative of this set. I mean, maybe it does, depending on your application. But it just seems a little bit off. I mean, my intuition would be that central tendency is something closer to 3. So what the median tell us? We already put these numbers in order. If I had given you out of order, you'd And you'd say, what's the middle number? Let's see. The middle two numbers, since I have an even number, are 3 and 3. So if I take the average of 3 and 3-- I should be particular with my language-- if I take the arithmetic mean of 3 and 3, I get 3. And this is maybe a better measurement of the central tendency or of the average of this set of numbers. Essentially, what it does is by taking the median, I wasn't so much affected by this really large number that's very different than the others. In statistics, they call that an outlier. If you talked about average home prices, maybe every house in the city is $100,000 and then there's one house that costs a trillion dollars. And then if someone told you the average house price was a million dollars, you might have a very wrong perception of that city. But the median house price would be $100,000, are like. So similarly, this median gives you a better sense of what the numbers in this set are like. The arithmetic mean was skewed by what they'd call an outlier. Being able to tell what an outlier is is one of those things that a statistician will say, well, I know it when I see it. There isn't really a formal definition for it, but it tends to be a number that really sticks out. And sometimes it's due to a measurement error, or whatever. And then finally, the mode. What is the most common number in this set? Well, there's five 3s and there's one 100. So the most common number is, once again, it's a 3. So in this case, when you had this outlier, the median and the mode tend to be maybe a little bit better about giving you an indication of what these numbers represent. Maybe this was just a measurement error. We don't actually know what these represent. If these are house prices, then I would argue that these are probably more indicative measures of what the houses in an area cost."
    },
    {
        "Q": "At 18:10 when Sal says/writes \"set X spans subspace V and X has 5 elements you now know that no set than spans the subspace V can have fewer than 5 elements\", isn't this statement incorrect? The set X didn't claim to be a BASIS for V, just span V. (I know Sal continues with \"even better if X is a basis...\" but it sounds like he's going on to make a separate statement).\n",
        "A": "Indeed. This threw me for a loop. Sal needs to correct this with a pop-up.",
        "video_name": "Zn2K8UIT8r4",
        "timestamps": [
            1090
        ],
        "3min_transcript": "Remember we said that n is greater than m. Or when we defined B, we said that m is less than n. Same thing, that this was a smaller set. Now we're saying that this spans V, but at the same time we said this was a basis. This is just our starting fact, that this is a basis for V. Basis means two things: it means it spans V and it means it's linearly independent. Now we just got this result by assuming that we had some set B that's smaller than this set here that spans V. We were able to construct this by saying that a1 through am also spans V. The result we got is that this spans V. But if this subset of A spans V, then A becomes linearly Because if this subset spans V, that means that an can be represented as some linear combination of these guys. contradiction with our original statement that set A is a basis for V, because that means it's linearly independent. If you're able to do this, then this means that there was some smaller spanning set, you get the result that A has to be linearly dependant, even though we said it was linearly independent. So we now know, we get our contradiction, we say that there cannot be [typing] a spanning set B that has fewer elements than A. And this is a pretty neat outcome, because now, if I the subspace V again. Then you know that X has five elements. You now know that no set that spans the subspace V can have fewer than five elements. Even better, if I told you that X is a basis for V, and I told you it has five elements, and Y is a basis for V. [typing] You know that Y also has to have exactly five elements."
    },
    {
        "Q": "At 4:41, why are you using the last equation?\n",
        "A": "It doesn t matter what equation of the 3 is used to find out the value z. Any of the three equations can be used.",
        "video_name": "f7cX-Ar2cEM",
        "timestamps": [
            281
        ],
        "3min_transcript": "and add the two equations. So let's do that. So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to 7 times negative 10, is equal to negative 70. And now we can add these two equations. I'm now trying to eliminate the y's. So we have 16x plus 56x. That is 72x. So we have 72x, these guys eliminate, equal to negative 72, negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1. And now we just have to substitute back to figure out what y and z are equal to. So we have 8x minus y is equal to negative 10. If x is equal to negative 1, that means 8 times negative 1, or negative 8 minus y is equal to negative 10. We can add 8 to both sides. And so we have negative y is equal to negative 2, or multiplying both sides by negative 1, y is equal to 2. let me square that off. So x is equal to negative 1, y is equal to 2. We now just have to worry about z, and we can go back to any of these up here. So I'll just use this last one. The numbers seem lower. So if we substitute back into this last equation right over here, we have 3 times x, which is 3 times negative 1 plus y, which is 2, minus z is equal to 3. So it's negative 3 plus 2 minus z is equal to 3. And this is negative 1 minus z is equal to 3. negative z is equal to 4. Multiplying both sides by negative 1, you get z is equal to negative 4. So we're done. Let's verify that these solutions, or the solution of x is negative 1, y is equal to 2, z is equal to negative 4, actually satisfies all three of these constraints. So let's substitute into this first one. So you have x plus 2y plus 5z. So x is negative 1 plus 2 times y, so plus 4, plus 5z. So minus 20 has to be equal to negative 17. And this is negative. This right here is positive 3 minus 20 is indeed equal to negative 17. So it satisfies the first constraint. Let's look at the second one. 2 times x, 2 times negative 1. That's negative 2 minus 3 times y."
    },
    {
        "Q": "at 2:30 why is he adding all the numbers when he could be subtracting?\n",
        "A": "so that he can eliminate the z variable.",
        "video_name": "f7cX-Ar2cEM",
        "timestamps": [
            150
        ],
        "3min_transcript": "Solve this system. And once again, we have three equations with three unknowns. So this is essentially trying to figure out where three different planes would intersect in three dimensions. And to do this, if we want to do it by elimination, if we want to be able to eliminate variables, it looks like, well, it looks like we have a negative z here. We have a plus 2z. We have a 5z over here. If we were to scale up this third equation by positive 2, then you would have a negative 2z here, and it would cancel out with this 2z there. And then if you were to scale it up by 5, you'd have a negative 5z here, and then that could cancel out with that 5z over there. So let's try to cancel out. Let's try to eliminate the z's first. So let me start with this equation up here. I'll just rewrite it. So we have-- I'll draw an arrow over here-- we have x plus 2y plus 5z is equal to negative 17. And then to cancel out or to eliminate the z's, I'll multiply this equation here times 5. So I'm going to multiply this equation times 5. So 3x times 5 is 15x, y times 5 is plus 5y, and then negative z times 5 is negative 5z-- that's the whole point and why we're multiplying it by 5-- is equal to 3 times 5, which is equal to 15. And so if we add these two equations, we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z, those are going to cancel out. And that is going to be equal to negative 17 plus 15 is negative 2. So we were able to use the constraints in that equation and that equation, and now we have an equation in just x and y. So let's try to do the same thing. Let's trying to eliminate the z's. But now I'll use this equation and this equation. So this equation-- let me just rewrite it over here. We have 2x minus 3y plus 2z is equal to negative 16. And now, so that this 2z gets eliminated, let's multiply this equation times 2. So let's multiply it times 2, so we'll have a negative 2z here to eliminate with the positive 2z. So 2 times 3x is 6x, 2 times y is plus 2y, and then 2 times negative z is negative 2z is equal to 2 times 3 is equal to 6. And now we can add these two equations. 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get canceled out. And then that is equal to negative 16 plus 6 is negative 10. So now we have two equations with two unknowns. We've eliminated the z's. And let's see, if we want to eliminate again, we have a negative y over here. We have a positive 7y."
    },
    {
        "Q": "at around 15:40. Are you allowed to just say dx*(1/dt) = dx/dt and have it be the derivative of x with respect to t??\n",
        "A": "I m not sure what you are asking, but remember that since dx and dt were on different sides of the radical, you d have to deal with them slightly differently.",
        "video_name": "_60sKaoRmhU",
        "timestamps": [
            940
        ],
        "3min_transcript": "algebraically manipulate differentials, what we can do is let us multiply and divide by dt. So one way to think about it, you could rewrite, so let me just do this orange part right here. Let's do a little side right here. So if you take this orange part, and write it in pink, and you have dx squared, and then you have plus dy squared, and let's say you just multiply it times dt over dt, right? That's a small change in t, divided by a small change in t. That's 1, so of course you can multiply it by that. If we're to bring in this part inside of the square root sign, right, so let me rewrite this. This is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt. I just wanted to write it this way to show you I'm just multiplying by 1. And here, I'm just taking this dt, writing it there, and And now if I wanted to bring this into the square root sign, this is the same thing, this is equal to, and I'll do it very slowly, just to make sure, I'll allow you to believe that I'm not doing anything shady with the algebra. This is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and all of that times dt, right? I didn't do anything, you could just take the square root of this and you'd get 1 over dt. And if I just distribute this, this is equal to the square root, and we have our dt at the end, of dx squared, or we could even write, dx over dt squared, plus dy over dt squared. Right? dx squared over dt squared is just dx over dt squared, same thing with the y's. And now all of a sudden, this starts to look We said that these are equivalent. And I'll switch colors, just for the sake of it. So we have the integral. From t is equal to a. Let me get our drawing back, if I-- from t is equal to a to t is equal to b of f of x of t times, or f of x of t and f of, or and y of t, they're both functions of t, and now instead of this expression, we can write the square root of, well, what's dx, what's the change in x with respect to, whatever this parameter is? What is dx dt? dx dt is the same thing as g prime of t. Right? x is a function of t. The function I wrote is g prime of t. And then dy dt is same thing as h prime of t."
    },
    {
        "Q": "Okay, I guess Sal is using Synthetic Division at 10:20, but I don't see how he's doing it.\nFor example, when he uses the 1:\n\u00e2\u0080\u00a2 We write down the 1\n1, ...\n\u00e2\u0080\u00a2 1 x 1 + (-3) = -2\n1, -3, ...\n\u00e2\u0080\u00a2 1 x (-2) + (-9) = -11\n1, -3, -11, ...\n\u00e2\u0080\u00a2 1x (-11) + 27 = 16\n1, -3, -11, 16\nNow this doesn't add up to zero, but it still doesn't make any sense to me.\nUsing 3 I get:\n1, 0, -9, 0\nWhat am I doing wrong?\n",
        "A": "I don t use synthetic division. It s quicker when you get used to it, but harder to understand. Long division (what he used in this video) always works. You don t have to relearn it whenever you haven t used it for a while.",
        "video_name": "11dNghWC4HI",
        "timestamps": [
            620
        ],
        "3min_transcript": "it times that guy. So plus lambda squared. Minus 4 lambda plus 4. And now of course, we have these terms over here. So we're going to have to simplify it again. So what are all of our constant terms? We have a 23 and we have a plus 4. So we have a 27. Plus 27. And then, what are all of our lambda terms? We have a minus 9 lambda and then we have a-- let's see. We have a minus 9 lambda, we have a plus 4 lambda, and then we have a minus 4 lambda. So these two cancel out. So I just have a minus 9 lambda. And then, what are my lambda squared terms? I have a plus lambda squared and I have a minus 4 lambda squared. So if you add those two that's going to be minus 3 lambda squared. And then finally, I have only one lambda cubed term, that So this is the characteristic polynomial for our matrix. So this is the characteristic polynomial and this represents the determinant for any lambda. The determinant of this matrix for any lambda. And we said that this has to be equal to 0 if any only if lambda is truly an eigenvalue. So we're going to set this equal to 0. And unlucky or lucky for us, there is no real trivial-- there is no quadratic. Well there is, actually, but it's very complicated. And so it's usually a waste of time. So we're going to have to do kind of the art of factoring a quadratic polynomial. I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class or really, in an algebra class generally-- it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions. And if you are dealing with integer solutions, then your Especially if you have a 1 coefficient out here. So your potential roots-- in this case, what are the factors of 27? So 1, 3, 9 and 27. So all these are potential roots. So we can just try them out. 1 cubed is 1 minus 3. So let me try 1. So if we try a 1, it's 1 minus 3 minus 9 plus 27. That does not equal 0. It's minus 2 minus 9 is minus 11. Plus 16. That does not equal 0. So 1 is not a root. If we try 3 we get 3 cubed, which is 27. Minus 3 times 3 squared is minus 3 times 3, which is minus 27. Minus 9 times 3, which is minus 27. Plus 27. That does equal 0. So lucky for us, on our second try we were able to find one 0 for this."
    },
    {
        "Q": "At 2:18, I didn't get what the speaker meant by \"that multiplies the rest of the term\". Can anyone explain?\n",
        "A": "A number is a coefficient, which defines how much there is. In maths , especially algebra, numbers get an x or another letter. When you put a 5 before a x you get 5x. Whereby 5 is the coefficient and x the rest of the term . It s get tricky when you add multiple variables to one coefficient. Like this 4xy or 57abc.",
        "video_name": "9_VCk9tWT0Y",
        "timestamps": [
            138
        ],
        "3min_transcript": "What I want to do in this video is think about how expressions are formed and the words we use to describe the different parts of an expression. And the reason why this is useful is when you hear other people refer to some expression and say, oh, I don't agree with the second term, or the third term has four factors, or why is the coefficient on that term 6, you'll know what they're talking about, and you can communicate in the same way. So let's think about what those words actually mean. So we have an expression here. And the first thing I want to think about are the terms of an expression or what a term is. And one way to think about it is the terms are the things that are getting added and subtracted. So, for example, in this expression right over here, you have three things that are getting added and subtracted. The first thing, you're taking 2 times 3. You're adding that to 4. And then from that, you're subtracting 7y. So in this example, you have three terms. The first term is 2 times 3. The second term is just the number 4. Now, let's think about the term \"factor.\" And when people are talking about a factor, especially in terms of an expression, they're talking about the things that are getting multiplied in each term. So, for example, if you said, what are the factors of the first term? The first term refers to this one right over here-- 2 times And there's two factors. There's a 2 and a 3, and they are being multiplied by each other. So here you have two factors in the first term. What about the second term? This was the first term. The second term here has only one factor, just the 4. It's not being multiplied by anything. And the third term here, once again, has two factors. It's the product of 7 times y. So we have two factors here. We have a 7 and a y. And this constant factor here, this number 7 also has a special name. It is called the coefficient of this term-- coefficient. And the coefficient is the nonvariable that multiplies the rest of the term. That's one way of thinking about it. So here's 7y. Even if it was 7xy or 7xyz or 7xyz squared, that nonvariable that's multiplying everything else, we would consider to be the coefficient. Now, let's do a few more examples. And in each of these-- I encourage you actually right now to pause the video-- think about what the terms are. How many terms are there in each expression, how many factors in each term, and what are the coefficients? So let's look at this first one. It's clear that we have three things being added together. This is the first term. This is the second term. And this is the third term. So this is the first term. This is the second term. This is the third term. And they each have two factors. This first one has the factors 3 and x."
    },
    {
        "Q": "\nAt 1:03, Sal says that the term is 7y, but isn't it -7y ?",
        "A": "Whether it s negative (7y) or not, it really depends on what the value of y is. : If y is negative, then 7y will be a negative number too : However if y is positive, then this term will be positive",
        "video_name": "9_VCk9tWT0Y",
        "timestamps": [
            63
        ],
        "3min_transcript": "What I want to do in this video is think about how expressions are formed and the words we use to describe the different parts of an expression. And the reason why this is useful is when you hear other people refer to some expression and say, oh, I don't agree with the second term, or the third term has four factors, or why is the coefficient on that term 6, you'll know what they're talking about, and you can communicate in the same way. So let's think about what those words actually mean. So we have an expression here. And the first thing I want to think about are the terms of an expression or what a term is. And one way to think about it is the terms are the things that are getting added and subtracted. So, for example, in this expression right over here, you have three things that are getting added and subtracted. The first thing, you're taking 2 times 3. You're adding that to 4. And then from that, you're subtracting 7y. So in this example, you have three terms. The first term is 2 times 3. The second term is just the number 4. Now, let's think about the term \"factor.\" And when people are talking about a factor, especially in terms of an expression, they're talking about the things that are getting multiplied in each term. So, for example, if you said, what are the factors of the first term? The first term refers to this one right over here-- 2 times And there's two factors. There's a 2 and a 3, and they are being multiplied by each other. So here you have two factors in the first term. What about the second term? This was the first term. The second term here has only one factor, just the 4. It's not being multiplied by anything. And the third term here, once again, has two factors. It's the product of 7 times y. So we have two factors here. We have a 7 and a y. And this constant factor here, this number 7 also has a special name. It is called the coefficient of this term-- coefficient. And the coefficient is the nonvariable that multiplies the rest of the term. That's one way of thinking about it. So here's 7y. Even if it was 7xy or 7xyz or 7xyz squared, that nonvariable that's multiplying everything else, we would consider to be the coefficient. Now, let's do a few more examples. And in each of these-- I encourage you actually right now to pause the video-- think about what the terms are. How many terms are there in each expression, how many factors in each term, and what are the coefficients? So let's look at this first one. It's clear that we have three things being added together. This is the first term. This is the second term. And this is the third term. So this is the first term. This is the second term. This is the third term. And they each have two factors. This first one has the factors 3 and x."
    },
    {
        "Q": "\non 5:34 that expression the last 3 factors are x, y, z what would be the coefficients of that term?",
        "A": "The coefficient would be 1. It doesn t have to be stated that 1*x*y^2*z^5 for 1 to be the coefficient as the 1 multiplies way to x*y^2*z^5. Also the 1 is not a factor of that term.",
        "video_name": "9_VCk9tWT0Y",
        "timestamps": [
            334
        ],
        "3min_transcript": "And if you look at that level, if you look at the first term, and you say, well, how many factors does that have? Well, you would say that it has three factors-- x, y, and z. How many factors does the second term have? Well, you could say, well, it has two factors. One factor is x plus y, and then the other factor is y. The first factor is x plus 1. And the second one is y. It's multiplying this expression. This smaller expression itself is one of the factors. And the other one is y. And then this third one also has two factors, a 4 and an x. And if someone said, hey, what's the coefficient on this term? You would say, hey, look, the coefficient is the 4. Now, let's look at this one over here. Actually, before I look at that one, what was interesting about this is that here you had a little smaller expression itself acting as one of the factors. So then you can go and then zoom in on this expression right over here. And you can ask the same question. On this smaller expression, how many terms does it have? Well, it has two terms-- an x and a 1. And each of them have exactly one factor. So when we're giving these, you can keep nesting these expressions to think about when you talk about terms or factors or factors of terms, you have to really specify what part of the nesting you're thinking about. If you're talking about the terms of this whole expression, there's one, two, three. But then you could look at this subexpression, which itself is a factor of a term, and say, oh, well, there's only two terms in this one. Now, let's look at this one. How many terms? Well, once again, there's clearly three. Actually, let me add one more, because I'm tired of expressions with three terms. So I'm just going to add a 1 here. So now, we clearly have four terms. This is the first term, second term, third term, fourth term. And how many factors are in each of them? Well, this is interesting. You might say, well, the factors are the things that are being multiplied. But here I'm dividing by y. Remember, dividing by y is the same thing as multiplying by its reciprocal. to have three factors here, where the factors are 3x and 1/y. If you multiplied 3 times x times 1/y, you're going to get exactly what you have right over here. So you would say this has three factors. If someone asks, what's the coefficient here? Well, you'd say, well, that 3 is the coefficient. Here how many factors do you have? And this is a little bit tricky, because you might say, well, isn't 5x squared times y, isn't that equal to 5 times x times x times y? And you'd be right. So it would be very tempting to say that you have four factors. But the convention, the tradition that most people use, is that they consider the exponent with x as a base as just one factor, this as just one factor. So traditionally, people will say this has three factors. It has a 5x squared and a y. x squared is just considered a factor. And once again, what's the coefficient? It's the 5."
    },
    {
        "Q": "At 4:08, Sal's final expression (2rp - 4\u00cf\u0080r^2) had three terms : r, p, and r^2. But didn't the question ask for a binomial, which has two terms? I'm confused....\n",
        "A": "This expression has two terms. I think you counted the types of variables in the expression instead of the terms. A term is an expression that doesn t have the operations of addition and subtraction. 2rp<---One term 4\u00cf\u0080r^2<---Second term There are two terms in this expression; therefore, it s a binomial. I hope this helped!",
        "video_name": "EvvxBdNIUeQ",
        "timestamps": [
            248
        ],
        "3min_transcript": "If p is greater than 7r, then 2-- let me write it this way. We know that p is greater than 7r. So if we're going to multiply both sides of this equation by 2rr-- and 2r is positive, we're dealing with positive distances, positive lengths-- so if we multiply both sides of this equation by 2r, it shouldn't change the equation. So multiply that by 2r, and then multiply this by 2r. And then our equation becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even multiply this by 2r? Well, that's so that this becomes the same as the area of the rectangle. So this is the area of the rectangle. And what's 14r squared? Well, 4 times pi, is going to get us something less than 14. So this is 4 pi is less than 14. 14 is 4 times 3 and 2-- let me put it this way. 4 times 3.5 is equal to 14. So 4 times pi, which is less than 3.5, is going to be less than 14. So we know that this over here is larger than this quantity over here. It's larger than 4 pi r squared. And so we know that this rectangle has a larger area than the circle. So we can just subtract the circle's area from the rectangle's area to find the difference. So the difference is going to be the area of the rectangle, which we already figured out is 2rp. And we're going to subtract from that the area of the circle. The area of the circle is 4 pi r squared. And one point I want to clarify. I gave the equation of the area of a circle to be pi r squared. And then we said that the radius is actually 2r in this case. So I substituted 2r for r. Hopefully that doesn't confuse you. This r is the general term for any radius. They later told us that the actual radius is 2 times some letter r. So I substitute that into the formula. Anyway, hopefully you found that useful."
    },
    {
        "Q": "5:42: How did he get 2x squared?\n",
        "A": "during the Pythagorean theorem he had x^2 + X^2 = c^2 right? Well when u put x^2 and x^2 together u get 2x^2",
        "video_name": "McINBOFCGH8",
        "timestamps": [
            342
        ],
        "3min_transcript": "Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. in a 45-45-90 triangle or a right isosceles triangle, the ratio of the sides are one of the legs can be 1. Then the other leg is going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2. So this is 45-45-90. That's the ratios. And just as a review, if you have a 30-60-90, the ratios were 1 to square root of 3 to 2. And now we'll apply this in a bunch of problems."
    },
    {
        "Q": "Why do you not express the ratio of the 30,60,90 as 3:4:5? I seems a lot less confusing.\n",
        "A": "Just because 3:4:5 immediately relates to pytha. thereom... which isn t always the case with 30.60.90 s.",
        "video_name": "McINBOFCGH8",
        "timestamps": [
            184
        ],
        "3min_transcript": "on this ratio right over here. Just an example, if you see a triangle that looks like this, where the sides are 2, 2 square root of 3, and 4. Once again, the ratio of 2 to 2 square root of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry. And this is a 45-45-90 triangle. Or another way to think about is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees. But you can have a right angle, you can have a right triangle, that is isosceles. And isosceles-- let me write this-- this is a right isosceles triangle. are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2"
    },
    {
        "Q": "What does delta mean like at 5:08?\n",
        "A": "Delta means the change in a quantity. It is a Greek letter that looks like a triangle. It is used in Math, Science, and others to show that you re looking at how something is changing without having to write out: the change in... . Kinda like when you re laughing out loud and you text someone lol .",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            308
        ],
        "3min_transcript": "One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens We go up by 3. delta x. delta y. Our change in y is 3. So delta y over delta x, When we go to the right, our change in x is 1. Our change in y is positive 3. So our slope is equal to 3. What is our y-intercept? Well, when x is equal to 0, y is equal to 1. So b is equal to 1. So this was a lot easier. Here the equation is y is equal to 3x plus 1. Let's do that last line there. Line C Let's do the y-intercept first. You see immediately the y-intercept-- when x is equal to 0, y is negative 2. So b is equal to negative 2. And then what is the slope? m is equal to change in y over change in x."
    },
    {
        "Q": "\nI dont get it. Any equation on how 0.2 becomes 1/5? at 8:03?",
        "A": "0.2 is equal to 1/5. A simple way to explain it is how many 0.2 s in 1.0. There are 5. 0.2 is just 1 of those 5, so 1/5.",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            483
        ],
        "3min_transcript": "If we go over to the right by one, two, three, four. So our change in x is equal to 4. What is our change in y? Our change in y is positive 2. So change in y is 2 when change in x is 4. So the slope is equal to 1/2, 2/4. So the equation here is y is equal to 1/2 x, that's our slope, minus 2. And we're done. Now let's go the other way. Let's look at some equations of lines knowing that this is the slope and this is the y-intercept-- that's the m, that's the b-- and actually graph them. Let's do this first line. I already started circling it in orange. The y-intercept is 5. When x is equal to 0, y is equal to 5. You can verify that on the equation. So when x is equal to 0, y is equal to one, two, three, four, five. That's the y-intercept and the slope is 2. in the y-direction. If I move 1 in the x-direction, I move up 2 in If I move 1 in the x-direction, I move up 2 in the y-direction. If I move back 1 in the x-direction, I move down 2 in If I move back 1 in the x-direction, I move down 2 in I keep doing that. So this line is going to look-- I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. It'll just keep going on, on and on and on. So that's our first line. I can just keep going down like that. Let's do this second line. y is equal to negative 0.2x plus 7. Let me write that. y is equal to negative 0.2x plus 7. It's always easier to think in fractions. So 0.2 is the same thing as 1/5. We could write y is equal to negative 1/5 x plus 7. So it's one, two, three, four, five, six. That's our y-intercept when x is equal to 0. This tells us that for every 5 we move to the right, we move down 1. We can view this as negative 1/5. The delta y over delta x is equal to negative 1/5. For every 5 we move to the right, we move down 1. So every 5. One, two, three, four, five. We moved 5 to the right. That means we must move down 1. We move 5 to the right. One, two, three, four, five. We must move down 1. If you go backwards, if you move 5 backwards-- instead of this, if you view this as 1 over negative 5. These are obviously equivalent numbers. If you go back 5-- that's negative 5. One, two, three, four, five. Then you move up 1. If you go back 5-- one, two, three, four, five-- you move up 1."
    },
    {
        "Q": "I still don't get how to find x and how you know that it is automatically zero based on something else at 9:12 and why is it that when y=-x that equeals M but when y=3.75 that equeals b. is x always zero?\n",
        "A": "i think you subtract,add, divide,or multiply 3.75 to both sides of the equation depending on what the sign is you do the opposite :)",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            552
        ],
        "3min_transcript": "in the y-direction. If I move 1 in the x-direction, I move up 2 in If I move 1 in the x-direction, I move up 2 in the y-direction. If I move back 1 in the x-direction, I move down 2 in If I move back 1 in the x-direction, I move down 2 in I keep doing that. So this line is going to look-- I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. It'll just keep going on, on and on and on. So that's our first line. I can just keep going down like that. Let's do this second line. y is equal to negative 0.2x plus 7. Let me write that. y is equal to negative 0.2x plus 7. It's always easier to think in fractions. So 0.2 is the same thing as 1/5. We could write y is equal to negative 1/5 x plus 7. So it's one, two, three, four, five, six. That's our y-intercept when x is equal to 0. This tells us that for every 5 we move to the right, we move down 1. We can view this as negative 1/5. The delta y over delta x is equal to negative 1/5. For every 5 we move to the right, we move down 1. So every 5. One, two, three, four, five. We moved 5 to the right. That means we must move down 1. We move 5 to the right. One, two, three, four, five. We must move down 1. If you go backwards, if you move 5 backwards-- instead of this, if you view this as 1 over negative 5. These are obviously equivalent numbers. If you go back 5-- that's negative 5. One, two, three, four, five. Then you move up 1. If you go back 5-- one, two, three, four, five-- you move up 1. I have to just connect the dots. I think you get the idea. I just have to connect those dots. I could've drawn it a little bit straighter. Now let's do this one, y is equal to negative x. Where's the b term? I don't see any b term. You remember we're saying y is equal to mx plus b. Where is the b? Well, the b is 0. You could view this as plus 0. Here is b is 0. When x is 0, y is 0. That's our y-intercept, right there at the origin. And then the slope-- once again you see a negative sign. You could view that as negative 1x plus 0. So slope is negative 1. When you move to the right by 1, when change in x is 1, change in y is negative 1. When you move up by 1 in x, you go down by 1 in y. Or if you go down by 1 in x, you're going to go up by 1 in y. x and y are going to have opposite signs. They go in opposite directions. So the line is going to look like that."
    },
    {
        "Q": "\n5:20 Why is it written n greater than or equal to 2?",
        "A": "In the formula being used, the nth term = the (n-1)th term + a common difference. In other words, the value of the nth term depends on the value of the term before it If n were 1, the term before it would be the 0th term --- which doesn t exist. So n is restricted to 2 and above in order that the term before will be no less than the first term.",
        "video_name": "_cooC3yG_p0",
        "timestamps": [
            320
        ],
        "3min_transcript": "And we could write that this is the sequence a sub n, n going from 1 to infinity of-- and we could just say a sub n, if we want to define it explicitly, is equal to 100 plus we're adding 7 every time. And then each term-- the second term we added 7 once. Third term-- we add 7 twice. So for the nth term, we're going to add 7 n minus 1 times. So this is an explicit definition of it, but we could also do it recursively. So just to be clear, this is one definition where we write it like this, or we could write a sub n, from n equals 1 to infinity. And in either case I should write with. And if I want to define it recursively, I could say a sub 1 is equal to 100. And then, for anything larger than 1, for any index above 1, And so we're done. This is another way of defining it. So in general, if you wanted a generalizable way to spot or define an arithmetic sequence, you could say an arithmetic sequence is going to be of the form a sub n-- if we're talking about an infinite one-- from n equals 1 to infinity. If you want to define it explicitly, you could say a sub n is equal to some constant, which would essentially the first term. It would be some constant plus some number that your incrementing-- or I guess this could be a negative number, or decrementing by-- times n minus 1. So this is one way to define an arithmetic sequence. In this case, d was 2. In this case, d is 7. That's how much you're adding by each time. And in this case, k is negative 5, and in this case, k is 100. the recursive way of defining an arithmetic sequence generally, you could say a sub 1 is equal to k, and then a sub n is equal to a sub n minus 1. A given term is equal to the previous term plus d for n greater than or equal to 2. So once again, this is explicit. This is the recursive way of defining it. And we would just write with there. Now the last question I have is is this one right over here an arithmetic sequence? Well, let's check it out. We start at 1. Then we add 2. Then we add 3. So this is an immediate giveaway that this is not an arithmetic sequence. Now we are adding 4. We're adding a different amount every time. So this, first of all, this is not arithmetic. This is not an arithmetic sequence. But how could we define this, since we're trying to define our sequences?"
    },
    {
        "Q": "At 4:19, Sal pronounces \"arithmetic\" wrong.\n",
        "A": "That was defiantly Worth pointing out",
        "video_name": "_cooC3yG_p0",
        "timestamps": [
            259
        ],
        "3min_transcript": "I could say a sub 1 is equal to negative 5. And then each successive term, for a sub 2 and greater-- so I could say a sub n is equal to a sub n minus 1 plus 3. Each term is equal to the previous term-- oh, not 3-- plus 2. So this is for n is greater than or equal to 2. So either of these are completely legitimate ways of defining the arithmetic sequence that we have here. We can either define it explicitly, or we could define it recursively. Now let's look at this sequence. Is this one arithmetic? Well, we're going from 100. We add 7. 107 to 114, we're adding 7. 114 to 121, we are adding 7. So this is indeed an arithmetic sequence. So just to be clear, this is one, And we could write that this is the sequence a sub n, n going from 1 to infinity of-- and we could just say a sub n, if we want to define it explicitly, is equal to 100 plus we're adding 7 every time. And then each term-- the second term we added 7 once. Third term-- we add 7 twice. So for the nth term, we're going to add 7 n minus 1 times. So this is an explicit definition of it, but we could also do it recursively. So just to be clear, this is one definition where we write it like this, or we could write a sub n, from n equals 1 to infinity. And in either case I should write with. And if I want to define it recursively, I could say a sub 1 is equal to 100. And then, for anything larger than 1, for any index above 1, And so we're done. This is another way of defining it. So in general, if you wanted a generalizable way to spot or define an arithmetic sequence, you could say an arithmetic sequence is going to be of the form a sub n-- if we're talking about an infinite one-- from n equals 1 to infinity. If you want to define it explicitly, you could say a sub n is equal to some constant, which would essentially the first term. It would be some constant plus some number that your incrementing-- or I guess this could be a negative number, or decrementing by-- times n minus 1. So this is one way to define an arithmetic sequence. In this case, d was 2. In this case, d is 7. That's how much you're adding by each time. And in this case, k is negative 5, and in this case, k is 100."
    },
    {
        "Q": "\nat 6:41, is there not a way to write the third question in an explicit from? If there is, can someone show me how?",
        "A": "Actually you re asking for the common formula that is taught in the next lessons: a_n = (n * (n + 1))/2",
        "video_name": "_cooC3yG_p0",
        "timestamps": [
            401
        ],
        "3min_transcript": "the recursive way of defining an arithmetic sequence generally, you could say a sub 1 is equal to k, and then a sub n is equal to a sub n minus 1. A given term is equal to the previous term plus d for n greater than or equal to 2. So once again, this is explicit. This is the recursive way of defining it. And we would just write with there. Now the last question I have is is this one right over here an arithmetic sequence? Well, let's check it out. We start at 1. Then we add 2. Then we add 3. So this is an immediate giveaway that this is not an arithmetic sequence. Now we are adding 4. We're adding a different amount every time. So this, first of all, this is not arithmetic. This is not an arithmetic sequence. But how could we define this, since we're trying to define our sequences? So we could say, this is equal to a sub n, where n is starting at 1 and it's going to infinity, with-- we'll say our base case-- a sub 1 is equal to 1. And then for n is 2 or greater, a sub n is going to be equal to what? So a sub 2 is the previous term plus 2. a sub 3 is the previous term plus 3. a sub 4 is the previous term plus 4. So it's going to be the previous term plus whatever your index is. So this looks close, but notice here we're changing the amount that we're adding based on what our index is. We're adding the amount of index to the previous term. And so this is for n is greater than or equal to 2. Well for an arithmetic sequence, we're adding the same amount regardless of what our index is. Here we're adding the index itself. an interesting sequence nonetheless."
    },
    {
        "Q": "On 0:08, is it first first quadratic formula? Yes or no?\n\nAnd on 2:27-2:28, what's binomial? Anyone tell me??\n",
        "A": "The equation shown at 0:08 is the standard form of a quadratic expression or equation, not the quadratic formula. A binomial is a polynomial having only two terms. Hope that helped.",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            8,
            147,
            148
        ],
        "3min_transcript": "In the last video, I told you that if you had a quadratic equation of the form ax squared plus bx, plus c is equal to zero, you could use the quadratic formula to find the solutions to this equation. And the quadratic formula was x. The solutions would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers, because you have a plus or a minus right there. What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do, so that I can start completing the square from this point right here, is-- let me rewrite the equation right here-- so we have ax-- bx, plus c is equal to 0. So the first I want to do is divide everything by a, so I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax, plus c over a, is equal to 0 over a, which is still just 0. Now we want to-- well, let me get the c over a term on to the right-hand side, so let's subtract c over a from both sides. And we get x squared plus b over a x, plus-- well, I'll just leave it blank there, because this is gone now; we subtracted it from both sides-- is equal to negative c over a I left a space there so that we can complete the square. And you saw in the completing the square video, you and you square it. So what is b over a divided by 2? Or what is 1/2 times b over a? Well, that is just b over 2a, and, of course, we are going to square it. You take 1/2 of this and you square it. That's what we do in completing a square, so that we can turn this into the perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left-hand side. We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well, this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out."
    },
    {
        "Q": "At 1:17 Sal divides \"everything by a\" but writes ... x(b/a) instead of (b * x) / a. Why does x not get divided by a?\n",
        "A": "x(b/a) is exactly the same as (b * x) / a Here are the steps: x (b / a) = x/1 * (b / a) = (b * x) / (a * 1) = (b * x) / a",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            77
        ],
        "3min_transcript": "In the last video, I told you that if you had a quadratic equation of the form ax squared plus bx, plus c is equal to zero, you could use the quadratic formula to find the solutions to this equation. And the quadratic formula was x. The solutions would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers, because you have a plus or a minus right there. What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do, so that I can start completing the square from this point right here, is-- let me rewrite the equation right here-- so we have ax-- bx, plus c is equal to 0. So the first I want to do is divide everything by a, so I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax, plus c over a, is equal to 0 over a, which is still just 0. Now we want to-- well, let me get the c over a term on to the right-hand side, so let's subtract c over a from both sides. And we get x squared plus b over a x, plus-- well, I'll just leave it blank there, because this is gone now; we subtracted it from both sides-- is equal to negative c over a I left a space there so that we can complete the square. And you saw in the completing the square video, you and you square it. So what is b over a divided by 2? Or what is 1/2 times b over a? Well, that is just b over 2a, and, of course, we are going to square it. You take 1/2 of this and you square it. That's what we do in completing a square, so that we can turn this into the perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left-hand side. We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well, this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out."
    },
    {
        "Q": "At 6:30, why couldn't you take \"b^2\" out of the radical? Because the square root of b^2 is just plus/minus b ? Or am I missing something?\n",
        "A": "You can only take factors out of root signs if they re factors, but in this case it s being added not multiplied so you can t split it up.",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            390
        ],
        "3min_transcript": "So the left-hand side simplifies to this. The right-hand side, maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit. So the right-hand side, we can rewrite it. This is going to be equal to-- well, this is going to be b squared. I'll write that term first. This is b-- let me do it in green so we can follow along. So that term right there can be written as b squared over 4a square. And what's this term? What would that become? This would become-- in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that. denominator by 4a. In fact, the 4's cancel out and then this a cancels out and you just have a c over a. So these, this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right here, I can rewrite as b squared minus 4ac, all of that over 4a squared. This is looking very close. Notice, b squared minus 4ac, it's already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation, so let's do that. So if you take the square root of both sides, the left-hand side will just become x plus-- let me scroll down a little bit-- x plus b over 2a is going to be equal to the plus or minus square root of this thing. numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now, what is the square root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now, this is looking very close to the quadratic. We have a b squared minus 4ac over 2a, now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a, plus or minus the square root of b squared minus 4ac over 2a, common"
    },
    {
        "Q": "\nAt 5:00, How do you already know that the sin^2(a) plus the cos^2(a) equals one?",
        "A": "let us imagine a right triangle \u00ef\u00bc\u008co=opposite\u00ef\u00bc\u008ch=hypotenuse\u00ef\u00bc\u008ca= adjacent. sin\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=o/h\u00ef\u00bc\u008ccos\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=a/h\u00ef\u00bc\u008cright\u00ef\u00bc\u009fso \u00ef\u00bc\u009a sin \u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=o\u00c2\u00b2/h\u00c2\u00b2 and cos\u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=a\u00c2\u00b2/h\u00c2\u00b2 we can add up these\u00ef\u00bc\u008cand get\u00ef\u00bc\u009a sin \u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089+ cos\u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=\u00ef\u00bc\u0088o\u00c2\u00b2+a\u00c2\u00b2\u00ef\u00bc\u0089/h\u00c2\u00b2 \u00ef\u00bc\u008c we know that \u00ef\u00bc\u009a\u00ef\u00bc\u0088o\u00c2\u00b2+a\u00c2\u00b2\u00ef\u00bc\u0089=h\u00c2\u00b2 \u00ef\u00bc\u008cthen we can get the answer\u00ef\u00bc\u008chope that could help you\u00ef\u00bc\u0081",
        "video_name": "a70-dYvDJZY",
        "timestamps": [
            300
        ],
        "3min_transcript": "well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared Or you could write that-- let me think of the best way to do this. You could write that the sine squared of a is equal to 1 minus the cosine sign squared of a. And then we could take this and substitute it right here. So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a. But the sine squared of a is this right there. So minus-- I'll do it in a different color. Minus 1 minus cosine squared of a. That's what I just substituted for the sine squared of a. And so this is equal to the cosine squared of a minus 1 plus the cosine squared of a. Which is equal to-- we're just adding. I'll just continue on the right. We have 1 cosine squared of a plus another cosine squared of a, so it's 2 cosine squared of a minus 1."
    },
    {
        "Q": "At 12:06pm, How do i verify csc^4 (x) - cot^4 (x) = 1 + 2 cot^2 (x)\ni got to (csc^2 (x) + cot^2 (x))1 = 1 + 2cot^2 (x) (i think the right side can convert to 2cscx but im stuck)\n",
        "A": "Identity to use: 1 + cot\u00c2\u00b2 x = csc\u00c2\u00b2 x We need to square this. Thus, [1 + cot\u00c2\u00b2 x]\u00c2\u00b2 = [csc\u00c2\u00b2 x ]\u00c2\u00b2 1 + 2cot\u00c2\u00b2 x + cot\u00e2\u0081\u00b4 x = csc\u00e2\u0081\u00b4 x Let us then use this to substitute for csc\u00e2\u0081\u00b4 x in your equation: csc\u00e2\u0081\u00b4 (x) - cot\u00e2\u0081\u00b4 (x) = 1 + 2 cot\u00c2\u00b2 (x) (1 + 2cot\u00c2\u00b2 x + cot\u00e2\u0081\u00b4 x) - cot\u00e2\u0081\u00b4 (x) = 1 + 2 cot\u00c2\u00b2 (x) 1 + 2cot\u00c2\u00b2 x = 1 + 2 cot\u00c2\u00b2 (x) As, both sides of the equation are now the same thing, we have proved the original.",
        "video_name": "a70-dYvDJZY",
        "timestamps": [
            726
        ],
        "3min_transcript": "squared of a is equal to 1/2 times 1 minus cosine of 2a. And we have our other discovery I guess we could call it. Our finding. And it's interesting. It's always interesting to look at the symmetry. Cosine squared-- they're identical except for you have a plus 2a here for the cosine squared and you have a minus cosine of 2a here for the sine squared. So we've already found a lot of interesting things. Let's see if we can do anything with the sine of 2a. Let me pick a new color here that I haven't used. Well, I've pretty much used all my colors. So if I want to figure out the sine of 2a, this is equal to the sine of a plus a. Which is equal to the sine of a times the co-- well, I don't Times the cosine of a plus-- and this cosine of a, that's the second a. Actually, you could view it that way. Plus the sine-- I'm just using the sine of a plus b. Plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just people to 2 sine of a, cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result. I know I'm a little bit tired by playing with all of these sine and cosines. And I was able to get all the results that I needed for my calculus problem, so hopefully this was a good review for you because it was a good review for me. You can write these things down. You can memorize them if you want, but the really important take away is to realize that you really can derive all of these formulas really from these initial formulas that we just had. And even these, I also have proofs to show you how to get trig functions."
    },
    {
        "Q": "\nat 2:01, vi said \"hamentashe\" or something. What are those?!",
        "A": "Hamantash is a triangular cookie or pastry, often associated with the Jewish holiday Purim.",
        "video_name": "o6KlpIWhbcw",
        "timestamps": [
            121
        ],
        "3min_transcript": "So you're me and you're in math class and-- triangles, triangles, triangles-- I don't know. The teacher keeps saying words, and you're supposed to be doing something with trigons, whatever those are. But you're bored and-- triangles, triangles, triangles. Sure you could draw your triangles separately, They're happiest when snuggled up together into a triangle party. Everybody knows triangles love parties. Sometimes they get together and do these triangle congo lines. If you keep adding new triangles on the same side, it gets all curvy and spirally. Or you can alternate and it goes pretty straight. In fact, since all the sides of the triangle are supposed to be straight lines, and since they're all lying on top of some previous straight line, then this whole line would have to be straight if these were actually triangles. Since it's not, it's proof that these aren't quite triangles. Maybe they've been partying a little too hard, but hey, at least you're not doing math. Speaking of which, the teacher is still going on about types of trigons, and you're supposed to be taking notes. But you're more interesting in types of triangles, which you already know all about. There are fat triangles, and pointy triangles, and perfect triangles, and cheese slice triangles which are a kind of pointy triangle but are symmetric like a slice of cheese or cake. Super pointy triangles are fun to stack into triangle stacks. You can put all the points facing one direction, but the stack starts to wobble too much towards that direction. So it's good to put some facing the other direction before you go too far. You'll notice pretty quickly, that the skinnier the triangle, the less wobble it adds to the stack. can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180,"
    },
    {
        "Q": "\nat 1:09 did she say ''beacaus you have to navigate your triangle stack around your nose''?\nor ''beacaus you have to navigate your triangle stack around your notes''?",
        "A": "I heard nose but notes is probably more correct",
        "video_name": "o6KlpIWhbcw",
        "timestamps": [
            69
        ],
        "3min_transcript": "So you're me and you're in math class and-- triangles, triangles, triangles-- I don't know. The teacher keeps saying words, and you're supposed to be doing something with trigons, whatever those are. But you're bored and-- triangles, triangles, triangles. Sure you could draw your triangles separately, They're happiest when snuggled up together into a triangle party. Everybody knows triangles love parties. Sometimes they get together and do these triangle congo lines. If you keep adding new triangles on the same side, it gets all curvy and spirally. Or you can alternate and it goes pretty straight. In fact, since all the sides of the triangle are supposed to be straight lines, and since they're all lying on top of some previous straight line, then this whole line would have to be straight if these were actually triangles. Since it's not, it's proof that these aren't quite triangles. Maybe they've been partying a little too hard, but hey, at least you're not doing math. Speaking of which, the teacher is still going on about types of trigons, and you're supposed to be taking notes. But you're more interesting in types of triangles, which you already know all about. There are fat triangles, and pointy triangles, and perfect triangles, and cheese slice triangles which are a kind of pointy triangle but are symmetric like a slice of cheese or cake. Super pointy triangles are fun to stack into triangle stacks. You can put all the points facing one direction, but the stack starts to wobble too much towards that direction. So it's good to put some facing the other direction before you go too far. You'll notice pretty quickly, that the skinnier the triangle, the less wobble it adds to the stack. can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180,"
    },
    {
        "Q": "At 2:49, how did she make the triangle look upset when she didn't even change sides?\n",
        "A": "it was a short smile she just made a big arch and the she got a frown",
        "video_name": "o6KlpIWhbcw",
        "timestamps": [
            169
        ],
        "3min_transcript": "can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180, Which is fine, unless you're afraid your triangle might get eaten by sea monsters. Anyway, stacking triangles into a curve is nice, and you probably want to make a spiral. But if you're not careful, it'll crash into itself. So you'd better think about your angles. Though, if you do it just right, instead of a crash disaster, you'll get a wreath thing. Or you can get a different triangle circle by starting with a polygon, extending the sides in one direction, and then triangling around it, to get this sort of aperture shape. And then you should probably add more triangles, triangles, triangles-- One last game. Start with some sort of asterisk, then go around a triangle it up. Shade out from the obtusest angles, and it'll look pretty neat. You can then extend it with another layer of triangles, and another, and if you shade the inner parts of these triangles, it's guaranteed to be an awesome triangle party. And there's lots of other kinds of triangle parties just waiting to happen. Ah, the triangle. So simple, yet so beautiful. The essence of two-dimensionality, the fundamental object of Euclidean geometry, the three points that define a plane. You can have your fancy complex shapes, they're just made up of triangles. Dissect a square into triangles, make symmetric arrangements, some reminiscent of spherical and hyperbolic geometry. Triangles branches into binary fractal trees."
    },
    {
        "Q": "at 5:17 do you multiply positive numbers the same way you multiply regular numbers..? if so do the positive numbers cancel out and the awnser become negative..?\n",
        "A": "ok but what I don t understand is how negative x negative =positive but positive x positive =positive how does that work..? shouldn t it be positive x positive =negative...?",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            317
        ],
        "3min_transcript": "This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, the negatives are going to cancel out. And that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples. I'm encouraging you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one. And we have a negative times a negative. They cancel out. Negative times a negative give me a positive. So this is going to be positive one. I can just write one, or I can literally write a plus sign there to emphasize. This is a positive one. What happened if I did negative one times zero? Now this might seem, this doesn't fit into any of these circumstances, zero is neither positive nor negative. And here you just have to remember anything times zero So negative one times zero is going to be zero. Or I could've said zero times negative seven hundred and eighty-three, that is also going to be zero. Now what about two--let me do some interesting ones. What about--I'm looking a new color. Twelve times negative four. Well, once again, twelve times positive four would be fourty-eight. And we are in the circumstance where one of these two numbers, right over here, is negative. This one right here. If exactly one of the two numbers is negative, then the product is going to be negative. We are in this circumstance, right over here. We have one negative, so the product is negative. You could imagine this as repeatedly adding negative four twelve times And so you will get to negative fourty-eight. Let's do another one. What is seven times three? Well, this is a bit of a trick. There are no negative numbers here. Positive seven times positive three. The first circumstance, which you already knew how to do before this video. This would just be equal to twenty-one. Let's do one more. So if I were to say negative five times negative ten-- well, once again, negative times a negative. The negatives cancel out. You are just left with a positive product. So it's going to be five times ten. It's going to be fifty. The negative and the negative cancel out. Your product is going to be positive. That's this situation right over there."
    },
    {
        "Q": "\nAt 2:48 why does pos times neg have to be neg? is there a reason for that?",
        "A": "If you tell yourself that Multiplication is the same as repeated addition, then: 4x -9 is the same as saying four times -9 , or: -9 + -9 + -9 + -9 -> -9-9-9-9 = -36",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            168
        ],
        "3min_transcript": "if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here. This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, the negatives are going to cancel out. And that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples. I'm encouraging you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one. And we have a negative times a negative. They cancel out. Negative times a negative give me a positive. So this is going to be positive one. I can just write one, or I can literally write a plus sign there to emphasize. This is a positive one. What happened if I did negative one times zero? Now this might seem, this doesn't fit into any of these circumstances, zero is neither positive nor negative. And here you just have to remember anything times zero"
    },
    {
        "Q": "\nAt 0:25 - what's the point of writing down \"a^2 + -ab...\"? Why not just \"a^2 - ab\"? The teachers I've had would have marked the way you'd used as wrong.",
        "A": "You could do that, and you re correct, it is proper form. Sal just did that to keep everyone on the same page, and solve the problem step by step. Later in the video, it did get simplified down to a^2 - ab, and as long as your final answer is simplified properly, it won t really matter where you do it.",
        "video_name": "YahJQvY396o",
        "timestamps": [
            25
        ],
        "3min_transcript": "We need to factor 49x squared minus 49y squared. Now here there's a pattern that you might already be familiar with. But just to make sure you are, let's think about what happens if we multiply a plus b-- where these are just two terms in a binomial-- times a minus b. If you multiply this out, you have a times a, which is a squared, plus a times negative b, which is negative ab-- that's a times negative b-- plus b times a, which is the same thing as ab. And then you have b times negative b, which is negative b squared. So when you do that, you have a negative ab and a positive ab, they cancel out. And you're just going to be left with an a squared minus a b squared. Now, this thing that we have here is exactly that pattern. 49x squared is a perfect square. 49y squared is a perfect square. We can rewrite it like that. We could rewrite this over here as 7x squared minus-- and And so you see it's a pattern. It's a squared minus b squared. So if you wanted to factor this-- if you would just use this pattern that we just derived-- you would say that this is the same thing as a, 7x plus b plus 7y times 7x minus b, minus 7y. And you'd be done. Now there's one alternate way that you could factor this and it'd be completely legitimate. You could start from the beginning and say, you know what? 49 is a common factor here, so let me just factor that out. So you could say it's equivalent to 49 times x squared minus y squared. And you say, oh, this fits the pattern of-- this is a squared minus b squared. So this will be x plus y times x minus y. times x minus y. And to see that this, right here, is the exact same thing as this right over here, you could just factor 7 out of You'd factor out a 7 out of that term, a factor 7 out of that term. And when you multiply them, you'd get the 49. So these are-- this or this-- these are both ways to factor this expression."
    },
    {
        "Q": "\nAround 9:13, what type of function would it be if there was an x value that wasn't mapped to a y value?",
        "A": "Such a scenario doesn t really exist. Say we had f: {a, b, c} -> {1, 2}, defined by f(a)=1, f(b)=2, and f(c)=7. Then either it is a typo where (whoever gave such a problem) forgot to put 7 in {1, 2}, or the object f is not a function, because a function f: A->B must be a subset of AxB. Since 7 is not in B, such an f wouldn t be a subset of AxB, so it wouldn t be a function.",
        "video_name": "xKNX8BUWR0g",
        "timestamps": [
            553
        ],
        "3min_transcript": "most one x that maps to it. Or another way to say it is that for any y that's a member of y-- let me write it this way --for any y that is a member y, there is at most one-- let me write most in capital --at most one x, such that f of x is equal to y. There might be no x's that map to it. So for example, you could have a little member of y right here that just never gets mapped to. Everyone else in y gets mapped to, but that guy never gets mapped to. So this would be a case where we don't have a surjective function. This is not onto because this guy, he's a member of the co-domain, but he's not a member of the He doesn't get mapped to. But this would still be an injective function as long as every x gets mapped to a unique y. Now, how can a function not be injective or one-to-one? And I think you get the idea when someone says one-to-one. Well, if two x's here get mapped to the same y, or three get mapped to the same y, this would mean that we're not dealing with an injective or a one-to-one function. So that's all it means. Let me draw another example here. Let's actually go back to this example right here. When I added this e here, we said this is not surjective anymore because every one of these guys is not being mapped to. Is this an injective function? Well, no, because I have f of 5 and f of 4 both mapped to d. injectiveness. This is what breaks it's surjectiveness. Now if I wanted to make this a surjective and an injective function, I would delete that mapping and I would change f of 5 to be e. Now everything is one-to-one. I don't have the mapping from two elements of x, going to the same element of y anymore. And everything in y now gets mapped to. So this is both onto and one-to-one."
    },
    {
        "Q": "\nAt 0:40, Sal starts thinking about the question. Maybe that makes sense but what I was thinking was different. What I did was, x^5 simplify with x^5, x^2 with x^2 and then I tried solving the question. I got a very different answer.....-0.03......That is way off 2/3! I know 2/3 is an approximation but having a difference of signs is a big difference! Why?",
        "A": "Ack! NEVER, EVER, EVER, cancel out anything next to a plus or minus sign! This is a very bad algebra mistake. Here s an example so you can see why this is not okay: Consider (x + 1) / (x + 2). If you could just cancel out the x s, you would be left with 1/2. But, try plugging in some values for x and you will see that this can t be right. For instance, if you plug in x = 1, you will get (1+1)/(1+2) = 2/3, not 1/2. Different values of x would give you different ratios.",
        "video_name": "gv9ogppphso",
        "timestamps": [
            40
        ],
        "3min_transcript": "So we have f of x equaling 4x to the fifth minus 3x squared plus 3, all of that over 6x to the fifth minus 100x squared minus 10. Now, what I want to think about is-- what is the limit of f of x, as x approaches infinity? And there are several ways that you could do this. You could actually try to plug in larger and larger numbers for x and see if it seems to be approaching some value. Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is, as x gets very, very large-- let's just focus on the numerator. As x gets very, very large, this term right over here in the numerator-- 4x to the fifth-- is going to become a much, much more significant than any of these other things. Something squaring gets large. But something being raised to the fifth power gets raised that much, much faster. Similarly, in the denominator, this term right over here-,, is going to grow much, much, much faster than any of these Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the fifth power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the fifth over 6x to the fifth for a very large, large x Or we could say as x approaches infinity. Now, what could this be simplified to? Well, you have x to the fifth divided by x to the fifth. These are going to grow together. So these you can think of them as canceling out. And so you are left with 2/3. So what you could say is-- the limit of f of x, as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much. And so it's going to approach 2/3. if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2/3. So lets look at the graph. So right here is the graph. Got it from Wolfram Alpha. And we see, indeed, as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2/3. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2/3. So let me draw it as neatly as I can. So this right over here is y is equal to 2/3. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2/3. And when we just look at the graph here, it seems like the same thing is happening from the bottom direction, when x approaches negative infinity."
    },
    {
        "Q": "i didn't understand at 2:40 why x squared was x. I thought this was radicals not square roots. Or is x just 1\nHelp Please :/\n",
        "A": "In the video, it actually shows the square root of x squared, not just x squared. So, if you think of the equation as the square root of x*x, you get x. Hope this helps!",
        "video_name": "egNq4tSfi1I",
        "timestamps": [
            160
        ],
        "3min_transcript": "plus three \"a\"'s which will give you four \"a\"'s, in this case \"a\" is all of this business right over here so we added those terms, and then we wanted to think about we have four principle roots of \"a\" and we have one more principle roots of \"a\", so same idea you have four of these things I am circling in magenta and you have one more of these things that I am circling in magenta, that one co-efficient is implicit so if I have four of something plus one more of something it becomes five of that something so plus plus five times the square root, plus five times the square root of eight and now we'll see if we can simplify this anymore, we have four of something and we have five of something else, so you can't just add these two things together, but maybe we can simplify this a little bit so we know that the principle root of two x squared, this is the same thing as, so let me write the four out front, so we have the four, and the principle root of two x squared is the same thing as the principle and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have"
    },
    {
        "Q": "\nAt 2:47, Sal says that sqrt(x^2) is the absolute value of x, and I get that. How come at 3:14, he says that sqrt(4) would be 2? Shouldn't he also consider -2?",
        "A": "He qualified his answer by saying that the principal square root of 4 is 2. The principal square root is the unique nonnegative square root of a nonnegative real number.",
        "video_name": "egNq4tSfi1I",
        "timestamps": [
            167,
            194
        ],
        "3min_transcript": "plus three \"a\"'s which will give you four \"a\"'s, in this case \"a\" is all of this business right over here so we added those terms, and then we wanted to think about we have four principle roots of \"a\" and we have one more principle roots of \"a\", so same idea you have four of these things I am circling in magenta and you have one more of these things that I am circling in magenta, that one co-efficient is implicit so if I have four of something plus one more of something it becomes five of that something so plus plus five times the square root, plus five times the square root of eight and now we'll see if we can simplify this anymore, we have four of something and we have five of something else, so you can't just add these two things together, but maybe we can simplify this a little bit so we know that the principle root of two x squared, this is the same thing as, so let me write the four out front, so we have the four, and the principle root of two x squared is the same thing as the principle and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have"
    },
    {
        "Q": "Why at 9:45 do we use n! in the denominator instead of just the n. I understood up to the second derivative why we introduce 1/2 but then on the third it I do not understand why we have 1/3! as this is 1/6 and not 1/3 which means that it will not cancel out. I hope you understand my questions.\nThank you very much for your help :)\n",
        "A": "For the second derivative, it s not 1 / 2 it s really 1 / 2! It s just that 2! happens to be equal to 2. The factorial works well for repeated differentiation: d/dx x^5 / 5! = x^4 / 4! d/dx x^4 / 4! = x^3 / 3! d/dx x^3 / 3! = x^2 / 2! d/dx x^2 / 2! = x / 1! d/dx x / 1! = 1 / 0! d/dx 1 / 0! = 0",
        "video_name": "epgwuzzDHsQ",
        "timestamps": [
            585
        ],
        "3min_transcript": "And let's see how it does on its third derivative, or I should say the second derivative. So p prime prime of x is equal to-- this is a constant, so its derivative is 0. So you just take the coefficient on the second term is equal to f prime prime of 0. So what's the second derivative of p evaluated at 0? Well, it's just going to be this constant value. It's going to be f prime prime of 0. So notice, by adding this term, now, not only is our polynomial value the same thing as our function value at 0, its derivative at 0 is the same thing as the derivative of the function at 0. And its second derivative at 0 is the same thing as the second derivative of the function at 0. So we're getting pretty good at this. And you might guess that there's a pattern here. Every term we add, it'll allow us to set up the situation so that the n-th derivative of our approximation at 0 of our function at 0. So in general, if we wanted to keep doing this, if we had a lot of time on our hands and we wanted to just keep adding terms to our polynomial, we could-- and let me do this in a new color. Maybe I'll do it in a color I already used. We could make our polynomial approximation. So the first term, the constant term, will just be f of 0. Then the next term will be f prime of 0 times x. Then the next term will be f prime prime of 0 times 1/2 times x squared. I just rewrote that in a slightly different order. Then the next term, if we want to make their third derivative the same at 0, would be f prime prime prime of 0. The third derivative of the function at 0, times 1/2 times 1/3, so 1 over 2 times 3 times x to the third. And we can keep going. Plus, if we want to make their fourth derivatives at 0 coincide, it would be the fourth derivative of the function. I could put a 4 up there, but this is really emphasizing-- it's the fourth derivative at 0 times 1 over-- and I'll change the order. Instead of writing it in increasing order, I'll write it as 4 times 3 times 2 times x to the fourth. And you can verify it for yourself. If we just had this only, and if you were to take the fourth derivative of this, evaluate it at 0, it'll be the same thing as the fourth derivative of the function evaluated at 0. And in general, you can keep adding terms where the n-th term will look like this. The n-th derivative of your function evaluated at 0 times x to the n over n factorial. Notice this is the same thing as 4 factorial. 4 factorial is equal to 4 times 3 times 2 times 1."
    },
    {
        "Q": "at about 3:09, Sal adds an \"x\" after f '(0). Why?\n",
        "A": "It s so p (0) still equals f (0) after he takes the derivative of his expression f(0)+f (0)x. If the x wasn t there, then the f (0) would end up equaling zero when he took the derivative, just as the f(0) did. Reviewing power rule may be helpful in clarifying this.",
        "video_name": "epgwuzzDHsQ",
        "timestamps": [
            189
        ],
        "3min_transcript": "So at first, maybe we just want p of 0, where p is the polynomial that we're going to construct, we want p of 0 to be equal to f of 0. So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0. So if I were to graph it, it would look like this. It would just be a horizontal line at f of 0. And you could say, Sal, that's a horrible approximation. It only approximates the function at this point. Looks like we got lucky at a couple of other points, but it's really bad everywhere else. And now I would tell you, well, try to do any better using a horizontal line. At least we got it right at f of 0. So this is about as good as we can do with just a constant. And even though-- I just want to remind you-- this might not look like a constant, but we're assuming that given the function, will just give us a number. So whatever number that was, we would put it right over here. We'd say p of x is equal to that number. It would just be a horizontal line right there at f of 0. But that obviously is not so great. So let's add some more constraints. Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0. Let me do this in a new color. So we also want, in the new color, we also want-- that's not a new color. We also want p prime. We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0. And we don't want to lose this right over here. So what if we set p of x as being equal to f of 0? So we're taking our old p of x, but now we're Plus f prime of times x. So let's think about this a little bit. If we use this as our new polynomial, what happens? What is p is 0? p of 0 is going to be equal to-- you're going to have f of 0 plus whatever this f prime of 0 is times 0. If you put a 0 in for x, this term is just going to be 0. So you're going to be left with p of 0 is equal to f of 0. That's cool. That's just as good as our first version. Now what's the derivative over here? So the derivative is p prime of x is equal to-- you take the derivative of this. This is just a constant, so its derivative is 0. The derivative of a coefficient times x is just going to be the coefficient. So it's going to be f prime of 0. So if you evaluate it at 0-- so p prime of 0."
    },
    {
        "Q": "\nWhat is the function approaching as x=1? At 3:30 could anyone explain or phrase that in a different way or ways?",
        "A": "What is the y-value of the line (or function) approaching as x gets closer to 1? This example is simple as y is always 1 except on the exact point of (1,1) since we cannot divide by zero. when x=0, y=1. when x=0.5, y=1. when x=0.9, y=1.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            210
        ],
        "3min_transcript": "which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing."
    },
    {
        "Q": "At 5:06 Sal drew a curve like thing after writing g(x).What is that curve?\n",
        "A": "Braces ( curly braces ), it is the same of using { .",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            306
        ],
        "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this."
    },
    {
        "Q": "\nAt 10:34, Sals calculator said that 1.999999999999^2 was 4. Well it is actually 3.999999996...but close enough.",
        "A": "Listen again, he said his calculator rounded the result to 4 but that the result wasn t exactly 4, but it was really really really really close to 4. And that is the whole deal with limits. What is the value of a function as you get really really really really close to the limit value. Keep Studying!",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            634
        ],
        "3min_transcript": "as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. so 1.99, and once again, let me square that. Well now I'm at 3.96. What if I do 1.999, and I square that? I'm going to have 3.996. Notice I'm going closer, and closer, and closer to our point. And if I did, if I got really close, 1.9999999999 squared, what am I going to get to. It's not actually going to be exactly 4, this calculator just rounded things up, but going to get to a number really, really, really, really, really, really, really, really, really close to 4. And we can do something from the positive direction too. And it actually has to be the same number when we approach from the below what we're trying to approach, and above what we're trying to approach. So if we try to 2.1 squared, we get 4.4. let me go a couple of steps ahead, Now we are getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4. So once again, that's a numeric way of saying that the limit, as x approaches 2 from either direction of g of x, even though right at 2, the function is equal to 1, because it's discontinuous. The limit as we're approaching 2, we're getting closer, and closer, and closer to 4."
    },
    {
        "Q": "Hey @4:15 This might sound stupid but why is the answer 1 and not zero?\n",
        "A": "I think I was thinking about slopes and derivatives. I was a little rusty on my calculus at the time I watched this video.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            255
        ],
        "3min_transcript": "And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety,"
    },
    {
        "Q": "at 5:22, Sal says \"it\" what is he referring to?\n",
        "A": "His it is referring to g(x) , and yes he is trying to say g(2) = 1 , or at the x-value 2, g(x) outputs a 1.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            322
        ],
        "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this."
    },
    {
        "Q": "\nAt 6:39 he says the discontinuity is when X=2, why can't it also be at X=-2?",
        "A": "That is the whole point of limits, x cannot equal 2 because that is your limit!",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            399
        ],
        "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function"
    },
    {
        "Q": "at 7:08 on the y=f(x) axis you put an open circle at 4 the top why\n",
        "A": "Putting an empty circle in a graph of this kind indicates that the function does not have a value on the curve at that point. Often this is because the function is undefined at that point, but in this case it is because the function has a quirky definition stating that the function has a different value, not on this curve, at that particular point.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            428
        ],
        "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function"
    },
    {
        "Q": "\nAt 2:44 why did he draw an open circle and not an asymptote?",
        "A": "Good question. It s because there is a finite limit as x approaches 1. He uses the open circle to show that the function is undefined at that point. If you graph something like 1/(x-1) you ll notice that it goes to +infiniti as you approach 1+ and -infiniti as you approach 1-. So there is an asymptote at 1. In the example of (x-1)/(x-1) the function is 1 at every x value except 1 so there is no asymptote.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            164
        ],
        "3min_transcript": "which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing."
    },
    {
        "Q": "\nAt 0:10, Sal said that limits are the ideas that calculus is based upon. How important are limits for anyone more familiar with calculus? (just curious)",
        "A": "Limits are very important to calculus. Derivatives and integrals use calculus and many modeling applications in calculus use limits. Limits are very important in calculus because many concepts are based on that sole aspect.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            10
        ],
        "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here."
    },
    {
        "Q": "At 4:40 , f(x) must tend to one but why it is equal to one?\n",
        "A": "It is the two-sided limit of the function as x approaches 1.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            280
        ],
        "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this."
    },
    {
        "Q": "At 6:27, why is he using g(x)? Can't it be something else?\n",
        "A": "What do you mean? It is just used to write a function.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            387
        ],
        "3min_transcript": "familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that."
    },
    {
        "Q": "At 0:35, Sal explains another way of finding the determinant, by taking each value of the first row terms in the original matrix and multiplying it by the corresponding terms in the cofactor matrix, and taking the sum of those products. Why does this work?\n",
        "A": "This is called rule of Sarrus, only applies to the 3 x 3 matrices. It s just a coincidence of the traditional minor and cofactor method.",
        "video_name": "ArcrdMkEmKo",
        "timestamps": [
            35
        ],
        "3min_transcript": "We're nearing the home stretch of our quest to find the inverse of this three-by-three matrix here. And the next thing that we can do is find the determinant of it, which we already have a good bit of practice doing. So the determinant of C, of our matrix-- I'll do that same color-- C, there are several ways that you could do it. You could take this top row of the matrix and take the value of each of those terms times the cofactor-- times the corresponding cofactor-- and take the sum there. That's one technique. Or you could do the technique where you rewrite these first two columns, and then you take the product of the top to left diagonals, sum those up, and then subtract out the top right to the bottom left. I'll do the second one just so that you can see that you get the same result. The determinant is going to be equal to-- I'll 1, negative 2, 2, 2, 1, 1, 3, 4, 5. And let me now, just to make it a little bit simpler, rewrite these first two columns. So negative 1, negative 2, 2, 1, 3, 4. So the determinant is going to be equal to-- so let me write this down. So you have negative 1 times 1 times 5. Well that's just going to be negative 5, taking that product. Then you have negative 2 times 1 times 3. Well that's negative 6. So we'll have negative 6. Or you could say plus negative 6 there. And then you have 2 times 2 times 4. Well that's just 4 times 4, which is just 16. So we have plus 16. And then we do the top right to the bottom left. Well that's negative 4 times 5. So that is negative 20. But we're going to subtract negative 20. So that's negative 4 times 5, negative 20, but we're going to subtract negative 20. Obviously that's going to turn into adding positive 20. Then you have negative 1 times 1 times 4, which is negative 4. But we're going to subtract these products. We're going to subtract negative 4. And then you have 2 times 1 times 3, which is 6. But we have to subtract it. So we have subtracting 6. And so this simplifies to negative 5 minus 6 is negative 11, plus 16 gets us to positive 5. So all of this simplifies to positive 5. And then we have plus 20 plus 4."
    },
    {
        "Q": "At 0:57, 2i-7i is -5i. Based on this, is i a coefficient?\n",
        "A": "No, the coefficients are the 2, -7 and -5. i is like a variable. But, it is unique because we know its value. i always = sqrt(-1).",
        "video_name": "SfbjqVyQljk",
        "timestamps": [
            57
        ],
        "3min_transcript": "We're asked to add the complex number 5 plus 2i to the other complex number 3 minus 7i. And as we'll see, when we're adding complex numbers, you can only add the real parts to each other and you can only add the imaginary parts to each other. So let's add the real parts. So we have a 5 plus a 3. And then the imaginary parts-- we have a 2i. So plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3-- that's pretty straightforward. That's just going to be 8. And then if I have two of something and from that I subtract seven of that something-- and in this case, the something is the imaginary unit, the number i. If I have two i's and I take away seven i's, then I have negative five i's. 2 minus 7 is negative 5. So then I have negative 5i. So when you add these two complex numbers, You get another complex number. It has a real part and an imaginary part."
    },
    {
        "Q": "\n6*3 is 3 times larger than 2*3. how is this in-between as Sal says @1:30\nI can perform the computation, I just don't know why it works. Can someone help.",
        "A": "Note that if we multiply 2 \u00e2\u0080\u00a2 3 by 3, we can have 3 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 3. Then, we can use the associativity of multiplication to find that this is equivalent to (3 \u00e2\u0080\u00a2 2) \u00e2\u0080\u00a2 3 = 6 \u00e2\u0080\u00a2 3. Therefore, 6 times 3 is thrice 2 times 3.",
        "video_name": "j3-XYLnxJDY",
        "timestamps": [
            90
        ],
        "3min_transcript": "So right here, we have a four-sided figure, or a quadrilateral, where two of the sides are parallel to each other. And so this, by definition, is a trapezoid. And what we want to do is, given the dimensions that they've given us, what is the area of this trapezoid. So let's just think through it. So what would we get if we multiplied this long base 6 times the height 3? So what do we get if we multiply 6 times 3? Well, that would be the area of a rectangle that is 6 units wide and 3 units high. So that would give us the area of a figure that looked like-- let me do it in this pink color. The area of a figure that looked like this would be 6 times 3. So it would give us this entire area right over there. Now, the trapezoid is clearly less than that, but let's just go with the thought experiment. Now, what would happen if we went with 2 times 3? has a width of 2 and a height of 3. So you could imagine that being this rectangle right over here. So that is this rectangle right over here. So that's the 2 times 3 rectangle. Now, it looks like the area of the trapezoid should be in between these two numbers. Maybe it should be exactly halfway in between, because when you look at the area difference between the two rectangles-- and let me color that in. So this is the area difference on the left-hand side. And this is the area difference on the right-hand side. If we focus on the trapezoid, you see that if we start with the yellow, the smaller rectangle, it reclaims half of the area, half of the difference between the smaller rectangle It gets exactly half of it on the left-hand side. And it gets half the difference between the smaller and the larger on the right-hand side. So it completely makes sense that the area of the trapezoid, this entire area right over here, should really just be the average. It should exactly be halfway between the areas of the smaller rectangle and the larger rectangle. So let's take the average of those two numbers. It's going to be 6 times 3 plus 2 times 3, all of that over 2. So when you think about an area of a trapezoid, you look at the two bases, the long base and the short base. Multiply each of those times the height, and then you could take the average of them. Or you could also think of it as this is the same thing as 6 plus 2. And I'm just factoring out a 3 here. 6 plus 2 times 3, and then all of that over 2,"
    },
    {
        "Q": "At 1:36,why the point (1,-1) is the center?\n",
        "A": "you set each numerator equal to zero. so x-1=0 ==> x=1 and y+1=0 ==> y=-1 If you want to know why setting the each numerator finds the center... I think about it like this. (x-1)^2 is symmetric about x=1. That is if you add or subtract the same number from x, say 1 (x=2, x=0), then you will get the same output. (2-1)^2=(0-1)^2. The same goes for (y+1)^2. ***this may not be the exact terminology but is a way to think about it.",
        "video_name": "lGQw-W1PxBE",
        "timestamps": [
            96
        ],
        "3min_transcript": "Let's see if we can tackle a slightly more difficult hyperbola graphing problem. Let's add the hyperbola. Make this up on the fly x minus 1 squared over 16 minus y plus 1 squared over 4 is equal to 1. So the first thing to recognize is that this is a hyperbola and we'll in a few videos, do a bunch of problems where the first point is just to identify what type of conic section we have and then the second step is actually graph the conic section. I already told you that we're going to be doing a hyperbola problem, so you know it's a hyperbola. But the way to recognize that is that you have this minus of the y squared term and then we actually have it shifted. The classic or the standard non-shifted form of a hyperbola or a hyperbola centered at 0 would look something like this. Especially if it has the same asymptotes just shifted, but minus y squared over 4 is equal to 1. And the difference between this hyperbola and this hyperbola the center of this hyperbola is at the point x is equal to 1 y is equal to minus 1. And the way to think about it is x equals 1 makes this whole term 0, and so that's why it's the center. And y equal to minus 1 makes this whole term 0. And on here, of course, the center is the origin. Center is 0, 0. So the easy way to graph this is to really graph this one, but you shift it so you use the center being 1 minus 1 instead of the center being 0, 0. So let's do that. So let's figure out the slope of the two asymptotes here and for this hyperbola right here. So if we go with this one, let's just solve for y. That's what I always like to do whenever I'm graphing a hyperbola. So we get minus y squared over 4. Subtracting x squared over 16 from both sides minus x squared over 16 plus 1. I'm working on this hyperbola right here, not this one, and then I'm going to just shift it later. And then let's say multiply both sides by minus 4 and you get y squared is equal to-- see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared"
    },
    {
        "Q": "\nIn 3:35, why didn't Sal factor x^2/4-4 in (x/2+2)(x/2-2)?",
        "A": "He s taking the square root of everything except the constant anyway, so it makes more sense and is easier to square root in the end.",
        "video_name": "lGQw-W1PxBE",
        "timestamps": [
            215
        ],
        "3min_transcript": "for this hyperbola right here. So if we go with this one, let's just solve for y. That's what I always like to do whenever I'm graphing a hyperbola. So we get minus y squared over 4. Subtracting x squared over 16 from both sides minus x squared over 16 plus 1. I'm working on this hyperbola right here, not this one, and then I'm going to just shift it later. And then let's say multiply both sides by minus 4 and you get y squared is equal to-- see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared And to figure out the asymptotes you just have to think about well what happens as x approaches positive or negative infinity. As x gets really positive or x gets really negative. And we've done this a bunch of times already. This is more important than just memorizing the formula, because it gives you an intuition of where those equations for the lines of the asymptote actually come from. Because these are what this graph or this equation or this function approaches as x approaches positive or negative infinity. As x approaches positive or negative infinity, what is y approximately equal to, in this case? Well once again, this term is going to dominate. This is just a 4 right here. You could imagine when x is like a trillion or a negative trillion, this is going to be huge number and this is going to be just like you know you almost view it like is going to dominate. So as you approach positive or negative infinity, y is going to be approximately equal to the square root, the positive and negative square root, of x squared over 4. So y would be approximately equal to positive or negative x over 2, or 1/2x. Let's do that. Let's draw our asymptotes. And remember, these are the asymptotes for this situation. But now of course, we're centered at 1 negative 1. So I'm going to draw two lines with these slopes, with positive 1/2 and negative 1/2 slopes, but they're going to be centered at this point. I just got rid of the shift just so I could figure out the asymptotes but of course this is the real thing that we're trying to graph, so let me do that. This is my y-axis this is my x-axis and the center of"
    },
    {
        "Q": "At ~6:20, shouldn't the second multiple choice said \"If \u00e2\u0088\u00a0AOC is rotated...\", not \"If Ray OA and Ray OC are rotated...\"? Because you can't rotate points, rays, lines, or line segments due to them not having a vertex, right?\n",
        "A": "Yes, you can rotate those around a point/origin, actually.",
        "video_name": "uYXhga17q1g",
        "timestamps": [
            380
        ],
        "3min_transcript": "but it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about. So and it's good because we felt good about the middle choice. Let's do one more of these. So they are telling us that line AOB, and they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are colinear. And COD is our straight lines, all right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example angle AOC, and angle DOB, are vertical angles. And if we wanted to prove that they are equal, we would say well their measures are gonna be equal, so theta should be equal to phi. So this one says segment OA is congruent to OD. OA is congruent to OD. We don't know that, they never even told us that. So I don't even have to read the rest of it, this is already saying, I don't know how far D is away from O, I don't know if it's the same distance A is from O. So we can just rule this first choice out. I can stop reading, this started with a statement that we don't know, based on the information they gave us. So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change, so phi must be equal to theta. So this is interesting, so let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, if you rotate it 180 degrees, it's gonna go all the way around and point in the other direction, it's going to become, it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB, and ray OC, if you rotate it 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order. That ray OA maps to ray OB, and that ray OC maps to ray OD. And we saw that, ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB, and then OC if you rotate 180 degrees, will map to OD."
    },
    {
        "Q": "\ni don't understand 2:20",
        "A": "its just saying that 70% is equal to 100% - 30%",
        "video_name": "d1oNF88SAgg",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip"
    },
    {
        "Q": "\nWhy are we subtracting 0.30 from x at 1:59?\nI don't get it\nThis algebra is a bit confusing",
        "A": "x would be considered the full price of the guavas, and you are subtracting 0.30 from it, as you are finding 30% of the full price. 0.30 is 30% in decimal notation, so by subtracting x by 0.30, you are finding the price after the 30% discount",
        "video_name": "d1oNF88SAgg",
        "timestamps": [
            119
        ],
        "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip"
    },
    {
        "Q": "\nI noticed Sal uses x in his previous video, and then he uses it here at 1:40. I know x is the same as 1x, but why is 1 used instead of 2 or 3? Is 1 supposed to be the opposite of 0? Because in a number line, the numbers between 0 and 1 would be 0.01 to 0.99. Does 1 represent 100% in this case?",
        "A": "Yes, x here represent 100%, that s why x is the same as 1x. Sal is just using x as a variable. You can use whatever letter you like, y, z, a, b, etc...",
        "video_name": "d1oNF88SAgg",
        "timestamps": [
            100
        ],
        "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip"
    },
    {
        "Q": "At 3:21, why can we divide 12.60 by 0.7 to get the full price?\n",
        "A": "The equation has 0.7 times x . To isolate x , we use the opposite operation. The opposite of multiplication is division. This is why Sal is dividing both sides of the equation by 0.7",
        "video_name": "d1oNF88SAgg",
        "timestamps": [
            201
        ],
        "3min_transcript": "That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip Where you say, 70% of the full price is equal to my sale price, right? I took 30% off. This is 70% of the full price. You might just skip to this step once you get used to these problems in a little bit. And now we just have to solve for x. Divide both sides by 0.7, so you get x is equal to 12.60 divided by 0.7. We could use a calculator, but it's always good to get a little bit of practice dividing decimals. So we get 0.7 goes into 12.60. Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right. So the 0.7 becomes a 7. Ignore that right there. The 12.60 becomes 126, put the decimal right there. Decimal right there. And we're ready to just do straight up long division. So this is now a 7, not a .7. So 7 goes into 12 1 time."
    },
    {
        "Q": "\nIn 1:13, why is sal putting it like -400w+1,100 instead of subtracting it like 1,100-400w?",
        "A": "the first looks like slope intercept form, y = mx+ b",
        "video_name": "2EwPpga_XPw",
        "timestamps": [
            73
        ],
        "3min_transcript": "Just as you were solving the potato chip conundrum in the last video, the king's favorite magical bird comes flying along and starts whispering into the king's ear. And this makes you a little bit self-conscious, a little bit insecure, so you tell the king, what is the bird talking about. And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds. And so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, OK, well I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the bird continues to whisper in the king's ear. And the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's us this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400 w from both sides. If we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And then if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100. And then we get m is equal to negative 400 divided by 100, is negative 4w. 1,100 divided by 100 is 11. Plus 11. So now we've constrained m in terms of w. This is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation? And then we will have one equation with one unknown. 200, so he's looking at that first equation now, he says 200. Instead of putting an m there, the bird says well, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute for m the expression negative 4w plus 11. And then we have the rest of it, plus 300w, is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here, in that first equation. So the first thing, you start to scratch your head. And you say, is this a legitimate thing to do. Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second."
    },
    {
        "Q": "At about 2:19, what is the word \"constraint\" used for? What exactly does that mean for this problem?\n",
        "A": "A constraint is simply another word for a limitation or restriction of the possible set of solutions satisfying both equalities. While either equality will represent visually a different line on a graph or in other words a different relationship to each variable given a set of constants, the solution set is limited to the intersection of the equalities. Thus in a system of equations, additional equalities or equations represent further restriction of the possible set of all solutions.",
        "video_name": "2EwPpga_XPw",
        "timestamps": [
            139
        ],
        "3min_transcript": "Just as you were solving the potato chip conundrum in the last video, the king's favorite magical bird comes flying along and starts whispering into the king's ear. And this makes you a little bit self-conscious, a little bit insecure, so you tell the king, what is the bird talking about. And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds. And so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, OK, well I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the bird continues to whisper in the king's ear. And the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's us this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400 w from both sides. If we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And then if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100. And then we get m is equal to negative 400 divided by 100, is negative 4w. 1,100 divided by 100 is 11. Plus 11. So now we've constrained m in terms of w. This is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation? And then we will have one equation with one unknown. 200, so he's looking at that first equation now, he says 200. Instead of putting an m there, the bird says well, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute for m the expression negative 4w plus 11. And then we have the rest of it, plus 300w, is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here, in that first equation. So the first thing, you start to scratch your head. And you say, is this a legitimate thing to do. Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second."
    },
    {
        "Q": "\nAt 0:33 if he did regroup instead of leaving it, would the outcome be the same or would it be different?",
        "A": "If by outcome you mean the solution to 65*78 then yes, it would be the same.",
        "video_name": "p0jCw2sqZgs",
        "timestamps": [
            33
        ],
        "3min_transcript": "I'm going to multiply 78 times 65 in a little less than standard way, but hopefully it'll make some sense, and you'll realize that there's multiple ways that you can multiply. And this is actually the way that I multiply numbers in my head. So 78 times 65. And then we're actually going to think about what the different parts of this process represent on this area model. So 78 times 65. So I'm going to start just the way we normally start when we multiply. I'm going to start with this 5 in this ones place, and I'm going to say 5 times 8 is 40. And instead of just writing a 0 and carrying a 4 right over here, I'm just going to write the number 40. So this was the 5 times the 8. Now, I'm going to multiply the 5 times the 7. And we have to be a little bit careful here because 5 times-- this isn't just any 7, this is a 70. So what is 5 times 70? Well, 5 times 7 would be 35, so five times 70 is 350. So I'll write that down, 350. If you add these two together, this is going to be 5 times 78. Now let's go over to the 6. So let's multiply the 6 times the 8. Now we have to be careful again. This 6 is not just a regular 6, it's in the tens place. This is a 60. 60 times 8. Well, 6 times 8 is 48, so 60 times 8 is going to be 480. So it's going to be 480. And then 6 times 7. Well, that would be 42, but we have to be careful. This is 60 times 70, so we're going to have two zeroes at the end. This is 4,200, not just 42. So 6 times 7 is 4,200. And now we can add everything together. And this is a very similar process to what we do when we do the traditional method of multiplying. I just made it a little bit more explicit But we can add everything together. In the ones place, we have a 0. In the tens place, we have 4 plus 5 is 9. 9 plus 8 is 17. And now we can carry a 1. 1 plus 3 is 4. 4 plus 4 is 8. 8 plus 2 is 10. Carry a 1, regroup of 1 even, and then you have a 5 right over there. So you get 5,070. Now, I want to think about-- I want to visualize what was going on here using this area model. So once again, we had 78. So I'm going to make this vertical length represent 78. So this distance right over here represents the 70. That's the 70, and then we'll make this distance right over here represent the 8. Let me make that a little bit cleaner so you see what I'm talking about. So this distance right over here represents the 8 and then we're going to multiply that times 65."
    },
    {
        "Q": "At 0:24 why is that shape a parallelogram because I thought it was scalene how come it is a paralleogram I don't get it\n",
        "A": "Sal says like a parallelogram . He doesn t mean to say that it IS a parallelogram. Hope that helps you understand! :)",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            24
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 2:14 what does adjacent mean?",
        "A": "Adjacent means that the two sides are directly next to each other. Not opposite in any way but connected.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            134
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:10 Sal said that mathmeticians have looked at the way kites are drawn in cartoons, why would they do that? I thought that the kite that we fly and draw in cartoons was copied off the math kite.",
        "A": "They tried to compare kites as a shape to something used in cartoons so they could see the new shape.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            10
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Using the kite in the video, preferably looking around 2:00 , are the top left and bottom right sides parallel?\n",
        "A": "Kites do not usually have any parallel sides. The only exception I know of is the rhombus (which, depending on how kite is defined is a special kind of kite).",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            120
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nsorry for got time 5:01",
        "A": "He said it can be an arbitrarily large number as long as it s greater",
        "video_name": "UTs4uZhu5t8",
        "timestamps": [
            301
        ],
        "3min_transcript": "right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater Let's do a couple more. Let's do x minus 2 is less than or equal to 1. Once again, we want to get just our x on the left-hand side. Get rid of this negative 2. Let's add 2 to both sides of this equation. Plus 2. The left-hand side just becomes an x. You have a less than or equal sign. That won't change by adding or subtracting the same thing to both sides of the inequality. And then 1 plus 2 is 3. So x needs to be less than or equal to 3. Any x that is less than or equal to 3 will satisfy this equation. So let's plot it. And I'd try out any x that's less than or equal to 3 and verify for yourself that it does indeed satisfy this I shouldn't call it an equation. This inequality. So let me graph the solution set. So let's say this is 0, 1, 2, 3, 4."
    },
    {
        "Q": "Starting at 3:35 sal subtracts to isolate the variable, but in previous videos, any time you subtract from both sides you have to change the inequality sign. Is that only done in a multiply or divide situation or did he just not switch the sign?\n",
        "A": "Correct, do not change the direction of the inequality when you add or subtract a negative. Always change the inequality when you multiply or divide by a negative. 3 < 4 -3 > -4",
        "video_name": "UTs4uZhu5t8",
        "timestamps": [
            215
        ],
        "3min_transcript": "And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well. right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater"
    },
    {
        "Q": "\nAt 4:19 Sal says that sqrt(2) * sqrt(2) is 2. How does that work?",
        "A": "We can multiply square roots. sqrt(2) * sqrt(2) = sqrt(4) What is the sqrt(4)? It = 2.",
        "video_name": "s9ppnjgmiyk",
        "timestamps": [
            259
        ],
        "3min_transcript": "And to identify the perfect squares you would say, Alright, are there any factors where I have at least two of them? Well I have two times two here. And I also have five times five here. So I can rewrite the square root of 200 as being equal to the square root of two times two. Let me just write it all out. Actually I think I'm going to run out of space. So the square root, give myself more space under the radical, square root of two times two times five times five times two. And I wrote it in this order so you can see the perfect squares here. Well this is going to be the same thing as the square root of two times two. This second method is a little bit more monotonous, (laughing) I guess is one way to think about it. And they really, they boil down to the same method. We're still going to get to the same answer. So square root of two times two times the square root times the square root of five times five, times the square root of two. Well the square root of two times two is just going to be, this is just two. Square root of five times five, well that's just going to be five. So you have two times five times the square root of two, which is 10 times the square root of two. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator and if you wanted to get rid of that, you could multiply both 'Cause notice we're just multiplying by one, we're expressing one as square root of two over square root of two, and then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two which is 20. So it could also be written like that. So hopefully you found that helpful. In fact, even this one, you could write if you want to visualize it slightly differently, you could view it as one twentieth times the square root of two. So these are all the same thing."
    },
    {
        "Q": "\nAt 3:40 Why does the the square root of 2 times 2 times the square root of 5 times 5 times the the square root of 2 equal 10 times the square root of 2? Shouldn't , sense, the square root of 2 times 2 is 2 and the square root of 5 times 5 is, it be 7 times the square root of 2?",
        "A": "you re multiplying the square roots not adding them.",
        "video_name": "s9ppnjgmiyk",
        "timestamps": [
            220
        ],
        "3min_transcript": "And to identify the perfect squares you would say, Alright, are there any factors where I have at least two of them? Well I have two times two here. And I also have five times five here. So I can rewrite the square root of 200 as being equal to the square root of two times two. Let me just write it all out. Actually I think I'm going to run out of space. So the square root, give myself more space under the radical, square root of two times two times five times five times two. And I wrote it in this order so you can see the perfect squares here. Well this is going to be the same thing as the square root of two times two. This second method is a little bit more monotonous, (laughing) I guess is one way to think about it. And they really, they boil down to the same method. We're still going to get to the same answer. So square root of two times two times the square root times the square root of five times five, times the square root of two. Well the square root of two times two is just going to be, this is just two. Square root of five times five, well that's just going to be five. So you have two times five times the square root of two, which is 10 times the square root of two. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator and if you wanted to get rid of that, you could multiply both 'Cause notice we're just multiplying by one, we're expressing one as square root of two over square root of two, and then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two which is 20. So it could also be written like that. So hopefully you found that helpful. In fact, even this one, you could write if you want to visualize it slightly differently, you could view it as one twentieth times the square root of two. So these are all the same thing."
    },
    {
        "Q": "\nAt 3:18 what does Sal mean when he says that each triangle has 180 degrees?",
        "A": "All of the internal angles of a triangle will always add up to 180\u00c2\u00b0.",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            198
        ],
        "3min_transcript": "So the measure of this angle is also x plus 40. Because they're corresponding angles, and you could see that by inspection, and if you moved around the transversal, it would make sense that that's the case. So this is x plus 40 and this is minus 40, and they're clearly supplements of each other. They're supplementary angles. Then the sum of these two angles have to be equal to 180. So let's figure it out. So x minus 40 plus x plus 40 is equal to 180, because they're supplementary. The 40's cancel out. So you just minus 40, plus 40 adds up to zero. So you're left with 2x is equal to 180. x is equal to 90. So it's D. 47: The measures of the interior angles of a pentagon are 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right?"
    },
    {
        "Q": "\nAt 10:28, why isn't the exterior angle 240 degrees, like a major arc in a circle?",
        "A": "all exterior angles add up to 360 in any polygon. in regular polygons all exterior angles are equal. thus one angle=360/6=60",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            628
        ],
        "3min_transcript": "equal to 180 minus 100 which equals 80 degrees. So this angle right here is equal to 80 degrees. And the angle they want us to figure out is the opposite of this angle, or in the U.S., I guess, they say vertical angles. And so opposite or vertical angles are equal or they're congruent, so this is going to be 80 degrees as well. And that is choice A. Problem 50: What is the measure of an exterior angle of a regular hexagon? A regular hexagon tells us that all of the sides are the same, it's equilateral, and all of the angles are the same, equiangular. So if we just knew what's the total degree measure of the interior angles, we could just divide that by 6, and then and then we could use that information to figure out the Let's just do it. So once again, I like to just draw a hexagon. Let's just draw a hexagon and count the triangles in it. Two sides, three sides, four sides, five sides and six sides. And how many triangles do I have here? One, two, three. So I have one, two, three, four triangles. The sum of the interior angles of this hexagon, of any hexagon, whether it's regular or not, are going to be 4 times 180 and that's 720 degrees. And it's a regular hexagon, so all the interior angles are going to be the same. And there's six of them. So each of them are going to be 720 divided by 6. So each of the interior angles are going to be 120 degrees. And I didn't draw it that regular, but we can assume that all of these are each 120 degrees. Fair enough. Now, if all of those are each 120 degrees, what is the measure of an exterior angle? Well, we could just extend one of these sides out a little bit. We could say, OK, if this is 120 degrees, what is its supplement? Well, these have to add up to 180, so 180 minus 120 is 60 degrees. I could do it on any side. I could extend that line out there, and I'd say, oh, that's 60 degrees. So any of the exterior angles are 60 degrees. B. All right, do I have time for one more? I'll wait for the next one in the next video. See you soon."
    },
    {
        "Q": "At 2:44, Sal says there is a formula, but he doesn't reveal it. What is the formula?\n",
        "A": "The sum of interior angles in an n-sided convex polygon is: 180(n - 2)",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            164
        ],
        "3min_transcript": "So the measure of this angle is also x plus 40. Because they're corresponding angles, and you could see that by inspection, and if you moved around the transversal, it would make sense that that's the case. So this is x plus 40 and this is minus 40, and they're clearly supplements of each other. They're supplementary angles. Then the sum of these two angles have to be equal to 180. So let's figure it out. So x minus 40 plus x plus 40 is equal to 180, because they're supplementary. The 40's cancel out. So you just minus 40, plus 40 adds up to zero. So you're left with 2x is equal to 180. x is equal to 90. So it's D. 47: The measures of the interior angles of a pentagon are 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right?"
    },
    {
        "Q": "\nAt 0:37, Sal gives a negative number as common ratio, so the terms are alternatively positive and negative. But a sequence usually has terms which keep increasing or decreasing, right?",
        "A": "Sequences might usually keep increasing or decreasing, but there s no rule that says they have to. As Sal s example illustrates.",
        "video_name": "CecgFWTg9pQ",
        "timestamps": [
            37
        ],
        "3min_transcript": "In the last video we saw that a geometric progression, or a geometric sequence, is just a sequence where each successive term is the previous term multiplied by a fixed value. And we call that fixed value the common ratio. So, for example, in this sequence right over here, each term is the previous term multiplied by 2. So 2 is our common ratio. And any non-zero value can be our common ratio. It can even be a negative value. So, for example, you could have a geometric sequence that looks like this. Maybe start at one, and maybe our common ratio, let's say it's negative 3. So 1 times negative 3 is negative 3. Negative 3 times negative 3 is positive 9. Positive 9 times negative 3 is negative 27. And then negative 27 times negative 3 is positive 81. And you could keep going on and on and on. What I now want to focus on in this video is the sum of a geometric progression or a geometric sequence, and we would call that Let's scroll down a little bit. So now we're going to talk about geometric series, which is really just the sum of a geometric sequence. So, for example, a geometric series would just be a sum of this sequence. So if we just said 1 plus negative 3, plus 9, plus negative 27, plus 81, and we were to go on, and on, and on, this would be a geometric series. And we could do it with this one up here just to really make it clear of what we're doing. So if we said 3 plus 6, plus 12, plus 24, plus 48, this once again is a geometric series, just the sum of a geometric sequence or a geometric progression. So how would we represent this in general terms and maybe Well, we'll start with whatever our first term is. And over here if we want to speak in general terms we could call that a, our first term. So we'll start with our first term, a, and then each successive term that we're going to add is going to be a times our common ratio. And we'll call that common ratio r. So the second term is a times r. Then the third term, we're just going to multiply this one times r. So it's going to be a times r squared. And then we can keep going, plus a times r to the third power. And let's say we're going to do a finite geometric series. So we're not going to just keep on going forever. Let's say we keep going all the way until we get to some a times r to the n. a times r to the n-th power."
    },
    {
        "Q": "3:28 is there a symbol for not existing or for a false statement ?\n",
        "A": "The not-equal-to sign, =/=, means that something does not equal something else. Another one to use is the naught sign, meaning that no possible solutions exist. It is denoted by a zero with a diagonal slash, like (/), only more circular.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            208
        ],
        "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that."
    },
    {
        "Q": "3:27 If both limits must be equal for the whole limit to exist, what about in instances where there is a piecewise function like this, but with the filled-in dot continuous on one of the lines? As it is approached from either side, you will get different values.\n\nTherefore, for all piecewise functions, is the limit nonexistent at the discontinuity?\n",
        "A": "Yes, if there is a discontinuity, or if the graph asymptotically approaches infinity, there is not limit. However, as you will soon learn (or maybe already have), there is a such thing as a one sided limit. For example the one sided limit of sqrt(x) approaching 0 from the right is 0, even though the whole limit does not exist at that point. I hope this helped.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            207
        ],
        "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that."
    },
    {
        "Q": "\nAt 5:30, why does he say that the limit as x approaches 4 is 5 instead of -5?",
        "A": "Because he made a mistake.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            330
        ],
        "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8."
    },
    {
        "Q": "At 5:30 I think Sal made a speed mistake. The limit as x approaches 4 is equal to -5, not 5. Am I right? :)\n",
        "A": "Yes, of course, and the correction is on there now :)",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            330
        ],
        "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8."
    },
    {
        "Q": "\nAt 3:13, sal says the limit does not exist. Does it mean that the graph is not possible, or that f(X) is not defined for that graph?",
        "A": "Another way to think of the limit failing to exist at x=2 is that f(2) cannot be re-defined in such a way that the graph becomes continuous at x=2. In other words, the discontinuity in the graph at x=2 cannot be removed even if the value of f(2) is changed! Have a blessed, wonderful day!",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            193
        ],
        "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that."
    },
    {
        "Q": "\nwhy is it the matrix vector product and not the dot product at 13:00",
        "A": "Because the dot products are for just one row and he s talking about the whole system.",
        "video_name": "qvyboGryeA8",
        "timestamps": [
            780
        ],
        "3min_transcript": "We represented our system of equation like this and we put it into reduced row echelon form, so this is A and this is 0. This right here is, let me make sure I have some space, let me put it right here. That right there is the reduced row echelon form of A. And so where essentially this equation, this is a linear equation that is trying to solve this problem. The reduced row echelon form of A times our vector x is equal to 0. So, all the solutions to this are also the solutions to our original problem, to our original ax is equal to 0. So what's the solution to this? All the x's that satisfy this, these are the null space of the reduced row echelon form of A. Right? So here are all of the x's, this is the null space, this problem, if we find all of the x's here, this is the null But we're saying that this problem is the same problem as this one, right? So we can write that the null space of A is equal to the null space of the reduced row echelon form of A. And that might seem a little bit confusing, hey, why are you even writing this out, but it's the actually very useful when you're trying to calculate null spaces. So we didn't even have to write a big augmented matrix here. We can say, take our matrix A, put it in reduced row echelon form and then figure out it's null space. We would have gone straight to this point right here. This is the reduced row echelon form of A, and then I could have immediately solved these equations, right? I would have just taken the dot product of the reduced row echelon form or, not the dot product, the matrix vector product of the reduced row echelon form of A with this vector, and I would've gotten these equations, and then these equations would immediately, I can just rewrite them in this form, and I would But anyway, hopefully you found that reasonably useful."
    },
    {
        "Q": "\nAt 3:34, how does Sal know the triangle is a 30-60-90?",
        "A": "Because of the sides. The basic 30-60-90 triangle has sides 2, 1, and sqr 3 (Watch Example: Solving a 30-60-90 triangle , Intro to 30-60-90 Triangles , 30-60-90 Triangles II ...), you can use them to find out angles and points on graphs, with this question, instead of 1 the side is 1/2, so to find the rest of the sides you simply half all the sides of the basic triangle and it is still a 30-60-90 triangle but now it fits the triangle on the graph and you can solve the problem.",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            214
        ],
        "3min_transcript": "OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles. out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle, let me do it like this. That triangle looks something like this. Where this distance right here is 1/2. That distance right there is 1/2. This distance right here is 1. Hopefully you recognize that this is going to be a 30, 60, 90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this, I don't know, just call this a. So you'd get a squared, plus 1/2 squared, which is 1/4, which is equal to 1 squared, which is 1. You get a squared is equal to 3/4, or a is equal to the So you immediately know this is a 30, 60, 90 triangle. And you know that because the sides of a 30, 60, 90 triangle, if the hypotenuse is 1, are 1/2 and square root of 3 over 2. And you also know that the side opposite the square root of 3 over 2 side is 60 degrees. That's 60, this is 90. This is the right angle, and this is 30 right up there. But this is the one we care about. This angle right here we just figured out is 60 degrees. So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees. So the arccosine, or the inverse cosine, let me write that down. The arccosine of minus 1/2 is equal to 120 degrees."
    },
    {
        "Q": "can someone explain how sal gets to 3/4 ? at 4:03?\n",
        "A": "He subtracts 1/4 from both sides.",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            243
        ],
        "3min_transcript": "OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles. out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle, let me do it like this. That triangle looks something like this. Where this distance right here is 1/2. That distance right there is 1/2. This distance right here is 1. Hopefully you recognize that this is going to be a 30, 60, 90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this, I don't know, just call this a. So you'd get a squared, plus 1/2 squared, which is 1/4, which is equal to 1 squared, which is 1. You get a squared is equal to 3/4, or a is equal to the So you immediately know this is a 30, 60, 90 triangle. And you know that because the sides of a 30, 60, 90 triangle, if the hypotenuse is 1, are 1/2 and square root of 3 over 2. And you also know that the side opposite the square root of 3 over 2 side is 60 degrees. That's 60, this is 90. This is the right angle, and this is 30 right up there. But this is the one we care about. This angle right here we just figured out is 60 degrees. So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees. So the arccosine, or the inverse cosine, let me write that down. The arccosine of minus 1/2 is equal to 120 degrees."
    },
    {
        "Q": "\nAt 6:38, Sal says that, \"We need to restrict it's range to the UPPER hemisphere\". Why do we actually need to restrict it only to UPPER hemisphere?",
        "A": "it is not mandatory that you have to restrict only the upper hemisphere. Anyways you have to restrict it to one hemisphere as there will be another value equal to it on the other hemisphere. Also as Sal said you could go one full round and have the same trig function value which would be unacceptable",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            398
        ],
        "3min_transcript": "No, it's 180 minus the 60, this whole thing is 180, so this is, right here is, 120 degrees, right? 120 plus 60 is 180. Or, if we wanted to write that in radians, you just right 120 degrees times pi radian per 180 degrees, degrees cancel out. 12 over 18 is 2/3, so it equals 2 pi over 3 radians. So this right here is equal to 2 pi over 3 radians. Now, just like we saw in the arcsine and the arctangent videos, you probably say, hey, OK, if I have 2 pi over 3 radians, that gives me a cosine of minus 1/2. And I can write that. cosine of 2 pi over 3 is equal to minus 1/2. statement up here. But I can just keep going around the unit circle. For example, I could, how about this point over here? Cosine of this angle, if I were to add, if I were to go this far, would also be minus 1/2. And then I could go 2 pi around and get back here. So there's a lot of values that if I take the cosine of those angles, I'll get this minus 1/2. So we have to restrict ourselves. We have to restrict the values that the arccosine function can take on. So we're essentially restricting it's range. We're restricting it's range. What we do is we restrict it's range to this upper hemisphere, the first and second quadrants. So if we say, if we make the statement that the arccosine of x is equal to theta, we're going to restrict our range, theta, to that top. So theta is going to be greater than or equal to 0 and less than or equal to 2 pi. Less than or equal to pi, right? Where this is also 0 degrees, or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this, this is the only point where the cosine of the angle is equal minus 1/2. We can't take this angle because it's outside of our range. And what are the valid values for x? Well any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arccosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just go check our work. Let's see if the value I got here, that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the TI-85."
    },
    {
        "Q": "\njust got confused after 9:06 .... HELP!",
        "A": "x = f(f^-1(x)) - a minor correction: f^-1 goes before f(x) Cheers",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            546
        ],
        "3min_transcript": "Less than or equal to pi, right? Where this is also 0 degrees, or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this, this is the only point where the cosine of the angle is equal minus 1/2. We can't take this angle because it's outside of our range. And what are the valid values for x? Well any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arccosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just go check our work. Let's see if the value I got here, that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the TI-85. So i need to figure out the inverse cosine, which is the same thing as the arccosine of minus 1/2, of minus 0.5. It gives me that decimal, that strange number. Let's see if that's the same thing as 2 pi over 3. 2 times pi divided by 3 is equal to, that exact same number. So the calculator gave me the same value I got. But this is kind of a useless, well, it's not a useless number. It's a valid, that is the answer. But it doesn't, it's not a nice clean answer. I didn't know that this is 2 pi over 3 radians. And so when we did it using the unit circle, we were able to get that answer. So hopefully, actually let me ask you, let me just finish this up with an interesting question. And this applies to all of them. If I were to ask you, you know, say I were to take the arccosine of x, and then I were to take the cosine of that, Well, this statement right here can be said, well, let's say that the arccosine of x is equal to theta, that means that the cosine of theta is equal to x, right? So if the arccosine of x is equal to theta, we can replace this with theta. And then the cosine of theta, well the cosine of theta is x. So this whole thing is going to be x. Hopefully I didn't get confuse you there, right? I'm saying look, arccosine of x, just call that theta. Now, by definition, this means that the cosine of theta is equal to x. These are equivalent statements. These are completely equivalent statements right here. So if we put a theta right there, we take the cosine of theta, it has to be equal to x. Now let me ask you a bonus, slightly trickier question."
    },
    {
        "Q": "\nfrom 3:30 , when sal defines the error fxn, he uses a term *'bound* how good is p(x) fitting in f(x)...\n\nBut what is meant by Bound and how is the error function bounded ?",
        "A": "Being bound simply means that you know that a value is definitely between two limits. For instance, if 10 < x < 15, then x is bound between 10 and 15. You ll actually do that bounding in another video when Sal gets to questions like, how many terms do we need in order to ensure that our approximation is good to one part in a thousand? See my other answer above for an example of when you might look for error bounds in a physical world example as opposed to pure math.",
        "video_name": "wgkRH5Uoavk",
        "timestamps": [
            210
        ],
        "3min_transcript": "Sometimes you'll see something like N comma a to say it's an Nth degree approximation centered at a. Actually, I'll write that right now. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an Nth degree polynomial centered at a. And it's going to look like this. It is going to be f of a, plus f prime of a, times x minus a, plus f prime prime of a, times x minus a squared over-- Either you could write two or two factorial, they're the same value. I'll write two factorial. You could write a divided by one factorial over here, if you like. And then plus, you go to the third derivative of f at a times x minus a to the third power, I think you see where this is going, over three factorial. And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f to the N over N factorial. And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. And you can verify that because all of these other terms have an x minus a here. So if you put an a in the polynomial, all of these other terms are going to be zero. And you'll have P of a is equal to f of a. Let me write that down. P of a is equal to f of a. And so it might look something like this. And it's going to fit the curve better the more of these terms that we actually have. So it might look something like this. I'll try my best to show what it might look like. So this is all review, I have this polynomial that's approximating this function. The more terms I have, the higher degree of this polynomial, the better that it will fit this curve But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a. So what I wanna do is define a remainder function. Or sometimes, I've seen some text books call it an error function. And I'm going to call this-- I'll just call it an error-- Just so you're consistent with all the different notations you might see in a book, some people will call this a remainder function and sometimes they'll write a remainder function for an Nth degree polynomial centered at a. Sometimes you'll see this as an error function. The error function is sometimes avoided because it looks like expected value from probability. But you'll see this often, this is E for error. E for error, R for remainder. And sometimes they'll also have the subscripts over there like that. And what we'll do is, we'll just define this function to be the difference between f of x and our approximation of f of x for any given x."
    },
    {
        "Q": "\nis 3 x = 8 i at 1:01",
        "A": "Not quite, actually 3 x = 9 at 1:01",
        "video_name": "kbqO0YTUyAY",
        "timestamps": [
            61
        ],
        "3min_transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x."
    },
    {
        "Q": "\nAt 0:57 why does Sal write down 6 with a 1 above it and say six squared-isn't it(the drawing of six squared) supposed to be a six with a 2(instead of a 1) above it?",
        "A": "it is a 2, but as the quality is low, it looks like a squished up 2 or a 1.",
        "video_name": "nMhJLn5ives",
        "timestamps": [
            57
        ],
        "3min_transcript": "I promised you that I'd give you some more Pythagorean theorem problems, so I will now give you more Pythagorean theorem problems. And once again, this is all about practice. Let's say I had a triangle-- that's an ugly looking right triangle, let me draw another one --and if I were to tell you that that side is 7, the side is 6, and I want to figure out this side. Well, we learned in the last presentation: which of these sides is the hypotenuse? Well, here's the right angle, so the side opposite the right angle is the hypotenuse. So what we want to do is actually figure out the hypotenuse. So we know that 6 squared plus 7 squared is equal to the hypotenuse squared. hypotenuse, so we'll use C here as well. And 36 plus 49 is equal to C squared. 85 is equal to C squared. Or C is equal to the square root of 85. And this is the part that most people have trouble with, is actually simplifying the radical. So the square root of 85: can I factor 85 so it's a product of a perfect square and another number? 85 isn't divisible by 4. So it won't be divisible by 16 or any of the multiples of 4. 5 goes into 85 how many times? No, that's not perfect square, either. I don't think 85 can be factored further as a So you might correct me; I might be wrong. This might be good exercise for you to do later, but as far as I can tell we have gotten our answer. The answer here is the square root of 85. And if we actually wanted to estimate what that is, let's think about it: the square root of 81 is 9, and the square root of 100 is 10 , so it's some place in between 9 and 10, and it's probably a little bit closer to 9. So it's 9 point something, something, something. And that's a good reality check; that makes sense. If this side is 6, this side is 7, 9 point something, something, something makes sense for that length. Let me give you another problem. [DRAWING] Let's say that this is 10 . This is 3. What is this side? First, let's identify our hypotenuse. We have our right angle here, so the side opposite the right angle is the hypotenuse and it's also the longest side. So it's 10."
    },
    {
        "Q": "At 1:35 you said radical is 85 the radical?\n",
        "A": "also i don t know about perfect squares but the square root of 85 can be simplified to 7 square root of 17",
        "video_name": "nMhJLn5ives",
        "timestamps": [
            95
        ],
        "3min_transcript": "I promised you that I'd give you some more Pythagorean theorem problems, so I will now give you more Pythagorean theorem problems. And once again, this is all about practice. Let's say I had a triangle-- that's an ugly looking right triangle, let me draw another one --and if I were to tell you that that side is 7, the side is 6, and I want to figure out this side. Well, we learned in the last presentation: which of these sides is the hypotenuse? Well, here's the right angle, so the side opposite the right angle is the hypotenuse. So what we want to do is actually figure out the hypotenuse. So we know that 6 squared plus 7 squared is equal to the hypotenuse squared. hypotenuse, so we'll use C here as well. And 36 plus 49 is equal to C squared. 85 is equal to C squared. Or C is equal to the square root of 85. And this is the part that most people have trouble with, is actually simplifying the radical. So the square root of 85: can I factor 85 so it's a product of a perfect square and another number? 85 isn't divisible by 4. So it won't be divisible by 16 or any of the multiples of 4. 5 goes into 85 how many times? No, that's not perfect square, either. I don't think 85 can be factored further as a So you might correct me; I might be wrong. This might be good exercise for you to do later, but as far as I can tell we have gotten our answer. The answer here is the square root of 85. And if we actually wanted to estimate what that is, let's think about it: the square root of 81 is 9, and the square root of 100 is 10 , so it's some place in between 9 and 10, and it's probably a little bit closer to 9. So it's 9 point something, something, something. And that's a good reality check; that makes sense. If this side is 6, this side is 7, 9 point something, something, something makes sense for that length. Let me give you another problem. [DRAWING] Let's say that this is 10 . This is 3. What is this side? First, let's identify our hypotenuse. We have our right angle here, so the side opposite the right angle is the hypotenuse and it's also the longest side. So it's 10."
    },
    {
        "Q": "At about 3:12, do you mean James Grime from Numberphile?\n",
        "A": "Yes, that is exactly what Vi meant.",
        "video_name": "lA6hE7NFIK0",
        "timestamps": [
            192
        ],
        "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are,"
    },
    {
        "Q": "\n@ 3:05, she mentions a type of infinity called \"aleph null\". Could someone help me out because I understand that there are multiple infinities, but what is \"aleph null\"?",
        "A": "Aleph null is the smallest infinite cardinal number. A cardinal number is the type of number we normally think of when counting things: three videos, six A s, one question, and aleph null integers. Aleph null is defined to be the number of natural numbers.",
        "video_name": "lA6hE7NFIK0",
        "timestamps": [
            185
        ],
        "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are,"
    },
    {
        "Q": "How come at 1:28 it sounds like her voice is echoing?\n",
        "A": "beacuse it is she is in an open room",
        "video_name": "lA6hE7NFIK0",
        "timestamps": [
            88
        ],
        "3min_transcript": "Voiceover: Any sequence you can come up with, whatever pattern looks fun. All your favorite celebrities birthdays lead into end followed by random numbers, whatever. All of that plus every sequence you can't come up with, each of those are the decimal places of a badly named so called real number and any of those sequences with one random digit changed is another real number. That's the thing most people don't realize about the set of all real numbers. It includes every possible combination of digits extending infinitely among aleph null decimal places. There's no last digit. The number of digits is greater than any real number, any counting number which makes it an infinite number of digits. Just barely an infinite number of digits because it's only barely greater than any finite number but even though it's only the smallest possible infinity of digits. This infinity is still no joke. It's still big enough, that for example point nine repeating is exactly precisely one and not epsilon less. You don't get that kind of point nine repeating equals one action unless your infinity really is infinite. are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to"
    },
    {
        "Q": "At 3:11, what is listable infinity and who is Jason Grime?\n",
        "A": "James Grime is a mathematician. He is often seen in several Numberphile videos, and his website is singingbanana.com. Listable infinity is his way of saying countable infinity, because the idea is that you can list the elements of a countable set, but you can t actually count them, because there are an infinite number.",
        "video_name": "lA6hE7NFIK0",
        "timestamps": [
            191
        ],
        "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are,"
    },
    {
        "Q": "\nWhat does the small hand represent in 13:11:4",
        "A": "The small hand represents the hour; in this case it is pointing to nine.",
        "video_name": "ftndEjAg6qs",
        "timestamps": [
            791
        ],
        "3min_transcript": ""
    },
    {
        "Q": "What is the elapsed time in start time 2:30 endtime 6:30\n",
        "A": "the answer is 4:00",
        "video_name": "ftndEjAg6qs",
        "timestamps": [
            150,
            390
        ],
        "3min_transcript": "We're asked, what time is it? So first, we want to look at the hour hand, which is the shorter hand, and see where it is pointing. So this right over here would have been 12 o'clock, 1 o'clock, 2 o'clock, 3 o'clock, 4 o'clock. And it looks like it's a little bit past 4 o'clock. So we are in the fourth hour. So the hour is 4. And then we have to think about the minutes. The minutes are the longer hand, and every one of these lines represent 5 minutes. We start here. This is 0 minutes past the hour, then 5 minutes past the hour, then 10 minutes past the hour. So the time is-- the minutes are 10, 10 minutes past the hour, and the hour is 4, or it's 4:10. Let's do a few more. What time is it? So first, we want to look at the hour hand. That's the shorter hand right over here. It's at-- let's see. This is 12, 1, 2, 3, 4, 5, 6, 7, 8, 9. It's just past 9. So it's still in the ninth hour. It hasn't gotten to the 10th hour yet. The ninth hour's from starting with 9 all the way until it's right almost before it gets to 10, and then it gets to the 10th hour. So the hour is 9, and then we want the minutes. Well, we can just count from 0 starting at the top of the clock. So 0, 5, 10, 15, 20, 25, 30. It's 9:30. And that also might make sense to you, because we know there are 60 minutes in an hour. And this is exactly halfway around the clock. And so half of 60 is 30. Let's do one more. What time is it? This is 12, 1-- actually, we can even count backwards. We can go 12, 11, 10. So right now we're in the 10th hour. The hour hand has passed 10, but it hasn't gotten to 11 yet. So we are in the 10th hour. So this would be 0, 5, 10, 15, 20 minutes past the hour. That's where the longer hand is pointing. It is 10:20."
    },
    {
        "Q": "\nat 2:57 how did you come up with x+y= 5?",
        "A": "Looking at the scale to the right, you can see that one side of the scale has 5 blocks(of 1). The other side has a X block and a Y block. The scale is not tipped, therefore a X and a Y have the same weight/ equal 5. So x+y=5",
        "video_name": "h9ZgZimXn2Q",
        "timestamps": [
            177
        ],
        "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is."
    },
    {
        "Q": "3:03 but how are you supposed to do this in the real world if you don't have a scale?\n",
        "A": "x+y=5 is just a given, a tool for solving the original problem. Sal simply skips over explaining that fact. On a test or worksheet the problem might be presented as something like... Solve 2x+y=8 for x if x+y=5",
        "video_name": "h9ZgZimXn2Q",
        "timestamps": [
            183
        ],
        "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is."
    },
    {
        "Q": "At about 3:00, where did he get the 5 from?\n",
        "A": "On the left side of the scale, there is an x and a y cube. That s where he gets x + y from. On the right side of the scale, there s 5 yellow cubes. That s where he gets 5 from. x + y must equal 5, because that s how the scale is balanced.",
        "video_name": "h9ZgZimXn2Q",
        "timestamps": [
            180
        ],
        "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is."
    },
    {
        "Q": "\nAt 1:40 why do you divide x^3-1 by x-1?\nWhy don't you divide it by x+1? Does it make a difference?",
        "A": "Two reasons. First, factoring by (x + 1) doesn t get us anywhere, because our problem is how to find the limit at x = 1, which means we need to get rid of a factor of (x - 1), because that s the factor that gives us a zero in the denominator at x = 1. And second, (x + 1) isn t a factor of x^3 - 1. Try dividing it and you ll see that you get a remainder.",
        "video_name": "rU222pVq520",
        "timestamps": [
            100
        ],
        "3min_transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing."
    },
    {
        "Q": "\nat 5:43, do the answers to the \"b\" value always end up as weird fractions (i.e. in the video 13/3)?",
        "A": "Not always. The y-intercept can be any real number.",
        "video_name": "XMJ72mtMn4Y",
        "timestamps": [
            343
        ],
        "3min_transcript": "y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for y-intercept to get our equation. And to do that, we can use the information that we know in fact we have several points of information We can use the fact that the line goes through the point (-1,6) you could use the other point as well. We know that when is equal to negative 1, So y is eqaul to 6. So y is equal to six when x is equal to negative 1 So negative 5 thirds times x, when x is equal to negative 1 y is equal to 6. So we literally just substitute this x and y value back into this and know we can solve for b. So let's see, this negative 1 times negative 5 thirds. So we have 6 is equal to positive five thirds plus b. And now we can subtract 5 thirds from both sides of this equation. From the left handside and subtracted from the rigth handside And then we get, what's 6 minus 5 thirds. So that's going to be, let me do it over here We take a common denominator. So 6 is the same thing as Let's do it over here. So 6 minus 5 over 3 is the same thing as 6 is the same thing as 18 over 3 minus 5 over 3 6 is 18 over 3. And this is just 13 over 3. And this is just 13 over 3. And then of course, these cancel out. So we get b is equal to 13 thirds. So we are done. We know We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept which is 13 which is 13 over 3. And we can write these as mixed numbers. if it's easier to visualize. 13 over 3 is four and 1 thirds. this y-intercept right over here. That's 0 coma 13 over 3 or 0 coma 4 and 1 thirds. And even with my very roughly drawn diagram it those looks like this. And the slope negative 5 thirds that's the same thing as negative 1 and 2 thirds. You can see here the slope is downward because the slope is negative. It's a little bit steeper than a slope of 1. It's not quite a negative 2. It's negative 1 and 2 thirds. if you write this as a negative, as a mixed number. So, hopefully, you found that entertaining."
    },
    {
        "Q": "At 4:28, would the answer be the same if you used other set of points?\n",
        "A": "I am not sure what you are asking because through any two points there is only one line. If you give different points, unless they are on the same line, then you would get a different line.",
        "video_name": "XMJ72mtMn4Y",
        "timestamps": [
            268
        ],
        "3min_transcript": "our starting point and make that our ending point. So what is our change in y? So our change in y, to go we started at y is equal to six, we started at y is equal to 6. And we go down all the way to y is equal to negative 4 So this is rigth here, that is our change in y You can look at the graph and say, oh, if I start at 6 and I go to negative 4 I went down 10. or if you just want to use this formula here it will give you the same thing We finished at negative 4, we finished at negative 4 and from that we want to subtract, we want to subtract 6. This right here is y2, our ending y and this is our beginning y This is y1. So y2, negative 4 minus y1, 6. or negative 4 minus 6. That is equal to negative 10. And all it does is tell us the change in y you go from this point to that point we have to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here. We started at x is equal to negative 1 and we go all the way to x is equal to 5. So we started at x is equal to negative 1, and we go all the way to x is equal to 5. So it takes us one to go to zero and then five more. So are change in x is 6. You can look at that visually there or you can use this formula same exact idea, our ending x-value, our ending x-value is 5 and our starting x-value is negative 1. 5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6. wich is the exact same thing as negative 5 thirds. as negative 5 over 3 I devided the numerator and the denominator by 2. y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for y-intercept to get our equation. And to do that, we can use the information that we know in fact we have several points of information We can use the fact that the line goes through the point (-1,6) you could use the other point as well. We know that when is equal to negative 1, So y is eqaul to 6. So y is equal to six when x is equal to negative 1 So negative 5 thirds times x, when x is equal to negative 1 y is equal to 6. So we literally just substitute this x and y value back into this and know we can solve for b. So let's see, this negative 1 times negative 5 thirds. So we have 6 is equal to positive five thirds plus b. And now we can subtract 5 thirds from both sides of this equation."
    },
    {
        "Q": "\nwhat did at sign that sal made mean? at 3:41?",
        "A": "That the bisector was congruent to itself...i think..",
        "video_name": "7UISwx2Mr4c",
        "timestamps": [
            221
        ],
        "3min_transcript": "So the distance from B to D is going to be the same thing as the distance-- let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. So we know that triangle ABD we know that it is congruent to triangle ACD. You have two triangles that have three sides that are congruent, or they have the same length. Then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles, these base angles, are also going to be congruent. Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. Let me draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. Let me do that in a different color. So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So they have the same exact measure. And what we want to do in this case-- we want to prove-- so let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We've proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AC is congruent to AB."
    },
    {
        "Q": "\nAt 0:57, Sal mentioned angles ABC and ACB. How do I know what order to put the letters in? For example, how do I know to write angle ABC instead of angle CBA?",
        "A": "no, order matters. the middle letter has to be the vertex",
        "video_name": "7UISwx2Mr4c",
        "timestamps": [
            57
        ],
        "3min_transcript": "So we're starting off with triangle ABC here. And we see from the drawing that we already know that the length of AB is equal to the length of AC, or line segment AB is congruent to line segment AC. And since this is a triangle and two sides of this triangle are congruent, or they have the same length, we can say that this is an isosceles triangle. Isosceles triangle, one of the hardest words for me to spell. I think I got it right. And that just means that two of the sides are equal to each other. Now what I want to do in this video is show what I want to prove. So what I want to prove here is that these two-- and they're sometimes referred to as base angles, these angles that are between one of the sides, and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it. I want to show that they're congruent. So I want to prove that angle ABC, I And so for an isosceles triangle, those two angles are often called base angles. And this might be called the vertex angle over here. And these are often called the sides or the legs of the isosceles triangle. And these are-- obviously they're sides. These are the legs of the isosceles triangle and this one down here, that isn't necessarily the same as the other two, you would call the base. So let's see if we can prove that. So there's not a lot of information here, just that these two sides are equal. But we have, in our toolkit, a lot that we know about triangle congruency. So maybe we can construct two triangles here that are congruent. And then we can use that information to figure out whether this angle is congruent to that angle there. And the first step, if we're going to use triangle congruency, is to actually construct two triangles. So one way to construct two triangles is let's set up another point right over here. Let's set up another point D. And let's just So the distance from B to D is going to be the same thing as the distance-- let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. So we know that triangle ABD we know that it is congruent to triangle ACD."
    },
    {
        "Q": "\nStarting around 2:40, I entered the same exact numbers into my calculator, 5*tan(65), and I got -7.350191288, instead of 10.7225346025. Any suggestions to what I could have done wrong?",
        "A": "Your calculator is using radians, not degrees. You can probably change it in your calculator s settings (but I can t help you with that without knowing what calculator you have). You could could convert 65 degrees into 65*\u00cf\u0080/180 radians and enter 5*tan(65*\u00cf\u0080/180) to get the correct value.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            160
        ],
        "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse."
    },
    {
        "Q": "\nAt 1:01 Sal says to use a trigonometric function. Since this is a right triangle, why can't you just use the Pythagorean Theorem?",
        "A": "In order to solve for a side using the Pythagorean theorem, you would already have to know the lengths of 2 sides of a right triangle. In this case, you only know 1 length, so you must use the trig functions to solve for a side.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            61
        ],
        "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7."
    },
    {
        "Q": "Wait a second.. what is the definition of the tangent of a number, what do you do with it? For example, at about 2:30, Mr. Khan finds the tangent of 65 degrees on his calculator. How does that work?\n",
        "A": "If you have a right triangle with a 65 degree angle in it, then the tangent of 65\u00c2\u00ba is the ratio of the side opposite the 65\u00c2\u00ba angle to the side next to it.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            150
        ],
        "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7."
    },
    {
        "Q": "At 2:50, Professor Khan used a TI-85 calculator to find the answer as 10.7. But, when I used my TI-84 calculator at home to find the answer, I got -7.4 (rounded to the nearest tenth). I also searched up on google \"5*tan 65\" and got -7.4 (rounded to the nearest tenth). Is there a reason for this?\n",
        "A": "Make sure your calculator is set to use degrees. It is likely set to radians which would cause it to produce a different result.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            170
        ],
        "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse."
    },
    {
        "Q": "\nat 2:50 he enters 5 times tangent of 65 into calculator and gets 10.72....\nwhen I enter the exact same thing.. I get -7.35....\nhave re-entered it over and over to ensure it's the same thing.\nsame result... yes it's the plain old \"tan\" function (not tan-1)",
        "A": "Make sure you aren t in radian mode.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            170
        ],
        "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse."
    },
    {
        "Q": "\nAt 5:54 , I believe Sal should have used the approximately equal sign (\u00e2\u0089\u0088) as opposed to the equal sign (=). Of course, it's a small difference so it could be my eyes, but did anyone else see it like I did?",
        "A": "actually, he is correct because the problem clearly states to round to the nearest tenth. So his answer is exact when rounded to the nearest tenth.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            354
        ],
        "3min_transcript": "And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below."
    },
    {
        "Q": "Why is it, at 2:33, that the tangent was multiplied by 5?\n",
        "A": "It was multiplied so the variable a would be isolated. It is just like in solving a regular algebraic equation- you isolate the variable first, then solve for it.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            153
        ],
        "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse."
    },
    {
        "Q": "At 1:32, why didn't Sal use the Law of Sines?\n",
        "A": "I don t know why. Using the Law of Sines would seem to make doing this problem quicker and easier. However, there are often several ways to do a problem, and maybe Sal just wanted to illustrate how to do this problem from basics (SOH CAH TOA).",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            92
        ],
        "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7."
    },
    {
        "Q": "\nat 1:30 does it matter if I use I use sin,cos,tan? in which situation would I use each one? If it does NOT matter, can I be sure that it's the same way with every other trig problem?",
        "A": "In this situation, we know the length of the adjacent side, we know the angle is 65 degrees, and we want to find the length of the opposite side. It matters very much which trig function we use in every trig problem. Here, the tan of 65 degrees = opposite / 5, so we can solve for the length of the opposite side with opposite = 5*tan 65 degrees. Which trig functions we use depends on which parts of the triangle we know.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            90
        ],
        "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7."
    },
    {
        "Q": "(at 2:38) Hi, can anyone please tell me how to plug your equation eg. 5*tan 65 into the calculator? I seem to have trouble doing it on the Khan Academy virtual calculator on the practice exercise.\n",
        "A": "First, you make sure the calculator is in degree mode. Then, you hit the buttons 5 , x , tan , 6 , 5 , ) , and = and you should get the answer.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            158
        ],
        "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse."
    },
    {
        "Q": "\nAt 5:30 you used the calculator to do 5/ (Cos 65) my calculator gets 0.422 when I do that and it is set in degrees.",
        "A": "Cos(x) is always between -1 and 1 so you should never get an absolute value below 5.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            330
        ],
        "3min_transcript": "This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out"
    },
    {
        "Q": "During 1:34 he uses tangents but insted could you have figured out the other angle and then use law of sines\n",
        "A": "He could have done that, but this lesson is using the basic trig functions. Additionally, using the Law of Sines would actually be slower.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            94
        ],
        "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7."
    },
    {
        "Q": "1:22 I dont understand how did sal said difference is -2 please help\n",
        "A": "Sal didnt mean -2, he meant -1. He said this because the whole number intervals on the graph are every other square and therefore look like its -2 but it is -1. It is negative because the slope goes down 1 unit or looking from the other way it goes left 1 unit.",
        "video_name": "hoRISaqp1Po",
        "timestamps": [
            82
        ],
        "3min_transcript": "With the graph of the function f as an aid, evaluate the following limits. So the first one is the limit as x approaches 3 of f of x minus f of 3 over x minus 3. So let's think about x minus-- x equals 3 is right over here. This right over here is f of 3, or we could say f of 3 is 1 right over here. That's the point 3 comma f of 3. And they're essentially trying to find the slope between an arbitrary x and that point as that x gets closer and closer to 3. So we can imagine an x that is above 3, that is, say, right over here. Well, if we're trying to find the slope between this x comma f of x and 3 comma f of 3, we see that it gets this exact same form. Your end point is f of x. So it's f of x minus f of 3 is your change in the vertical axis. That's this distance right over here. And we would divide by your change And that's going to be x minus 3. So that's the exact expression that we have up here when I picked this as an arbitrary x. And we see that that slope, just by looking at the line between those two intervals, seems to be negative 2. And the slope was the same thing if we go the other side. If x was less than 3, then we also would have a slope of negative 2. Either way, we have a slope of negative 2. And that's important because this limit is just the limit as x approaches 3. So it can be as x approaches 3 from the positive direction or from the negative direction. But in either case, the slope, as we get closer and closer to this point right over here, is negative 2. Now let's think about what they're asking us here. So we have 8, f of 8. So let's think. We have 8. This is 8 comma f of 8. So that's 8 comma f of 8 right over there. So our temptation might be to say, hey, 8 plus h is going to be someplace out here. It's going to be something larger than 8. But notice, they have the limit as h approaches 0 from the negative direction. So approaching 0 from the negative direction means you're coming to 0 from below. You're at negative 1, negative 0.5, negative 0.1, negative 0.0001. So h is actually going to be a negative number. So 8 plus h would actually be-- we could just pick an arbitrary point. It could be something like this right over here. So this might be the value of 8 plus h. And this would be the value of f of 8 plus h. So once again, they're finding-- or this expression is the slope between these two points. And then we are taking the limit as h approaches 0 from the negative direction. So as h gets closer and closer to 0,"
    },
    {
        "Q": "\nat 3:30 Sal says the slope for the limit from the positive direction of 8 is infinite. how does he know that?\nI thought the slop was -1.\ncan someone please help me?",
        "A": "You re correct that the slope on the positive side of 8 is clearly -1. I m not sure if Sal worded that 100% correct. I would say the limit as h approaches 0+ would be -1. The limit at 8 does not exist because limit as h ---> 0- does not equal h---> 0+ So the infinite slope he is drawing applies only AT 8.",
        "video_name": "hoRISaqp1Po",
        "timestamps": [
            210
        ],
        "3min_transcript": "And that's going to be x minus 3. So that's the exact expression that we have up here when I picked this as an arbitrary x. And we see that that slope, just by looking at the line between those two intervals, seems to be negative 2. And the slope was the same thing if we go the other side. If x was less than 3, then we also would have a slope of negative 2. Either way, we have a slope of negative 2. And that's important because this limit is just the limit as x approaches 3. So it can be as x approaches 3 from the positive direction or from the negative direction. But in either case, the slope, as we get closer and closer to this point right over here, is negative 2. Now let's think about what they're asking us here. So we have 8, f of 8. So let's think. We have 8. This is 8 comma f of 8. So that's 8 comma f of 8 right over there. So our temptation might be to say, hey, 8 plus h is going to be someplace out here. It's going to be something larger than 8. But notice, they have the limit as h approaches 0 from the negative direction. So approaching 0 from the negative direction means you're coming to 0 from below. You're at negative 1, negative 0.5, negative 0.1, negative 0.0001. So h is actually going to be a negative number. So 8 plus h would actually be-- we could just pick an arbitrary point. It could be something like this right over here. So this might be the value of 8 plus h. And this would be the value of f of 8 plus h. So once again, they're finding-- or this expression is the slope between these two points. And then we are taking the limit as h approaches 0 from the negative direction. So as h gets closer and closer to 0, And these points move closer and closer and closer together. So this is really just an expression of the slope of the line, roughly-- and we see that it's constant. So what's the slope of the line over this interval? Well, you can just eyeball it and see, well, look. Every time x changes by 1, our f of x changes by 1. So the slope of the line there is 1. It would have been a completely different thing if this said limit as h approaches 0 from the positive direction. Then we would be looking at points over here. And we would see that we would slowly approach, essentially, a vertical slope, kind of an infinite slope."
    },
    {
        "Q": "\nA the time of the video, 1:13, it tells you the inverse of the slope. What i was wondering is since you have the coordinates and the inverse slope, why don't you put in into point slope form to satisfy the question? Since the equation is looking for the equation of line B, why cant you do it this way? is it still wrong to do so? Hopefully this has made sense... :)",
        "A": "since the question does not say the answer has to be in slope-intercept form - then i guess that yes, simply putting the equation in point-slope form would be a correct answer",
        "video_name": "TsEhZRT16LU",
        "timestamps": [
            73
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 2:20 he said 9 and 9 when it was 7 and 9",
        "A": "yes, but it was actually at 2:14 hey, nobody s perfect, anyone can make a mistake",
        "video_name": "LEFE1km5ROY",
        "timestamps": [
            140
        ],
        "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place."
    },
    {
        "Q": "on 1:50 does he keep adding them all together\n",
        "A": "Sal is not adding them all together. What he is doing is using place value to determine what the tens and ones places should be. Re-watch the video from 0:32.",
        "video_name": "LEFE1km5ROY",
        "timestamps": [
            110
        ],
        "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place."
    },
    {
        "Q": "2:14 he said 9 twice\n",
        "A": "why ask me the same thing in a different way?",
        "video_name": "LEFE1km5ROY",
        "timestamps": [
            134
        ],
        "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place."
    },
    {
        "Q": "\nAt 5:00, the whole equation is written and he is describing the point-slope form. In the original equation at 4:11, is it y - y1 and x - x1 or y - y2 and x - x2?",
        "A": "Either point (x1, y1) or (x2, y2) (or any known point on the line) will work, because the slope between any 2 points on the line is the same.",
        "video_name": "LtpXvUCrgrM",
        "timestamps": [
            300,
            251
        ],
        "3min_transcript": "Again to get to this point? Well, your change in X is positive two. So your change in X is equal to two. And so what's your slope? Change in Y over change in X. Negative eight over two is equal to negative four. So now that we have a, now that we know the slope and we know a point, we know a, we actually know two points on the line, we can express this in point-slope form. And so let's do that. And the way I like to it is I always like to just take it straight from the definition of what slope is. We know that the slope between any two points on this line is going to be negative four. So if we take an arbitrary Y that sits on this line and if we find the difference between that Y and, let's focus on this point up here. So if we find the difference between that Y and this Y, and nine, and it's over the difference between This is going to be the slope between any XY on this line and this point right over here. And the slope between any two points on a line are going to have to be constant. So this is going to be equal to the slope of the line. It's going to be equal to negative four. And we're not in point-slope form or classic point-slope form just yet. To do that, we just multiply both sides times X minus four. So we get Y minus 9, we get Y minus nine is equal to our slope, negative four times X minus four. Time X minus four. And this right over here is our classic, this right over here is our classic point-slope form. We have the point, sometimes they even put parenthesis like this, but we could figure out the point from this point-slope form. The point that sits on this line with things So it would be X equals four, Y equals nine, which we have right up there, and then the slope is right over here, it's negative four. Now from this can we now express this linear equation in y-intercept form? And y-intercept form, just as a bit of a reminder, it's Y is equal to MX plus B. Where this coefficient is our slope and this constant right over here allows us to figure out our y-intercept. And to get this in this form we just have to simplify a little bit of this algebra. So you have Y minus nine. Y minus nine is equal to, well let's distribute this negative four. And I'll just switch some colors. Let's distribute this negative four. Negative four times X is negative four X. Negative four time negative four is plus 16. And now, if we just want to isolate the Y on the left hand side, we can add nine to both sides."
    },
    {
        "Q": "At 6:17 and 6:20, Sal says \"-a finite number of values,\". What does finite mean?\n",
        "A": "It means that the values are countable and not infinite. The numbers 1 to 10 are finite. Or 1 to one billion are also finite. Contrast that with an infinite number of values where any number up to infinity is allowed.",
        "video_name": "dOr0NKyD31Q",
        "timestamps": [
            377,
            380
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 08:32, does the fact that there will only be one value for Z make a difference for our discrete random variable?\n\nI ask because Y will have multiple values for the multiple students. I guess I can answer my own question because if the class contains one student then Y is equivalent to Z in that the random variable can only take on one value.",
        "A": "A discrete random variable can have more than one value but the number of values it can be is limited. In the ant example the answer can be one of many whole numbers over a range or interval. If our interval is 0 to 1,000,000 the answer could be an integer in that range (say 52). However, there there cannot be 52.6583 ants born in the universe. Either an ant is born or it isn t. To be a continuous random variable it has be able to take on ANY value in the interval.",
        "video_name": "dOr0NKyD31Q",
        "timestamps": [
            512
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:22, why does the Khan Academy guy say pre-algebra is necessary? I'm in Grade 11 summer school and I can barely pass it btw I'm 16 years old and I feel dumb :(.\n",
        "A": "You are not dumb. Sal believe s in the growth mindset, a belief that no one is dumb.",
        "video_name": "Zm0KaIw-35k",
        "timestamps": [
            142
        ],
        "3min_transcript": "Jayda takes 3 hours to deliver 189 newspapers on her paper route. What is the rate per hour at which she delivers the newspaper? So this first sentence tells us that she delivers, or she takes, 3 hours to deliver 189 newspapers. So you have 3 hours for every 189 newspapers. That's what the first sentence told us. But we want to figure out the rate per hour, or the newspapers per hour, so we can really just flip this rate So if we were to just flip it, we would have 189 newspapers for every 3 hours, which is really the same information. We're just flipping what's in the numerator and what's in Now we want to write it in as simple as possible form, and let's see if this top number is divisible by 3. 1 plus 8 is 9, plus 9 is 18. So that is divisible by 3. So let's divide this numerator and this denominator by 3 to simplify things. So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and 18 six times. 6 times 3 is 18. Subtract. Bring down the 9. 18 minus 18 was nothing. 3 goes into 9 exactly three times. 3 times 3 is 9, no remainder. So if you divide 189 by 3, you get 63, and if you divide 3 by 3, you're going to get 1. You have to divide both the numerator and the denominator by the same number. So now we have 63 newspapers for every 1 hour. Or we could write this as 63 over 1 newspapers per hour. thing as 63 newspapers per hour."
    },
    {
        "Q": "\nHow long can pi go for? 30:00",
        "A": "are you sure?",
        "video_name": "Zm0KaIw-35k",
        "timestamps": [
            1800
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nfigured out the answer at 0:30",
        "A": "that is 1:63 newspapers per hour...... ;p",
        "video_name": "Zm0KaIw-35k",
        "timestamps": [
            30
        ],
        "3min_transcript": "Jayda takes 3 hours to deliver 189 newspapers on her paper route. What is the rate per hour at which she delivers the newspaper? So this first sentence tells us that she delivers, or she takes, 3 hours to deliver 189 newspapers. So you have 3 hours for every 189 newspapers. That's what the first sentence told us. But we want to figure out the rate per hour, or the newspapers per hour, so we can really just flip this rate So if we were to just flip it, we would have 189 newspapers for every 3 hours, which is really the same information. We're just flipping what's in the numerator and what's in Now we want to write it in as simple as possible form, and let's see if this top number is divisible by 3. 1 plus 8 is 9, plus 9 is 18. So that is divisible by 3. So let's divide this numerator and this denominator by 3 to simplify things. So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and 18 six times. 6 times 3 is 18. Subtract. Bring down the 9. 18 minus 18 was nothing. 3 goes into 9 exactly three times. 3 times 3 is 9, no remainder. So if you divide 189 by 3, you get 63, and if you divide 3 by 3, you're going to get 1. You have to divide both the numerator and the denominator by the same number. So now we have 63 newspapers for every 1 hour. Or we could write this as 63 over 1 newspapers per hour. thing as 63 newspapers per hour."
    },
    {
        "Q": "\nAt 0:21, how did you conclude that the sample standard deviation was 2.98? I understood the method you used to arrive at 17.17 for the sample mean, but in in order to find the t-statistic I must know how we got \"S\".",
        "A": "The s is the sample s standard deviation.To find this take each data point subtract it by the mean(17.17) and square it. Then add it all together. finally divide that number by 10 and take the square root of that number. If you put into the calculator it would look like this: [ (15.6 - 17.17)^2 + (16.2 - 17.17)^2+.... ]/10 <-then take the square root of it and it will = 2.98.",
        "video_name": "D2sMsmL0ScQ",
        "timestamps": [
            21
        ],
        "3min_transcript": "The mean emission of all engines of a new design needs to be below 20 parts per million if the design is to meet new emission requirements. 10 engines are manufactured for testing purposes, and the emission level of each is determined. The emission data is, and they give us 10 data points for the 10 test engines, and I went ahead and calculated the mean of these data points. The sample mean of 17.17. And the standard deviation of these 10 data points right here is 2.98, the sample standard deviation. Does the data supply sufficient evidence to conclude that this type of engine meets the new standard? Assume we are willing to risk a type-1 error with a probability of 0.01. And we'll touch on this in a second. Before we do that, let's just define what our null hypothesis and our alternative hypothesis are going to be. Our null hypothesis can be that we don't meet the standards. That we just barely don't meet the standards. That the mean of our new engines is exactly 20 parts per million. And you essentially want the best possible value where we we still don't meet the standard. And then our alternative hypothesis says no, we do meet That the true mean for our new engines is below 20 parts per million. And to see if the data that we have is sufficient, what we're going to do is assume, we're going to assume that this is true. And given that this is true, if we assume this is true, and the probability of this occurring, and the probability of getting a sample mean of that is less than 1%, then we will reject the null hypothesis. So we are going to reject our null hypothesis if the probability of getting a sample mean of 17.17 given the null hypothesis is true, is less than 1%. 1% chance that we are making a type-1 error. A type-1 error is that we're rejecting it even though it's true. Here there's only a 1% chance, or less than a 1% chance that we will reject it if it is true. Now the next thing we have to think about is what type of distribution we should think about. And I guess the first thing that rings in my brain is we only have 10 samples here. We only have 10 samples. We have a small sample size right over here. So we're going to be dealing with a T-distribution and T-statistic. So with that said, so let's think of it this way. We can come up with a T-statistic that is based on these statistics right over here. So the T-statistic is going to be 17.17, our sample mean, minus the assumed population mean-- minus 20 parts per million over our sample standard deviation, 2.98--"
    },
    {
        "Q": "At 4:24, how are we able to bring the 5 to the outside of the integral?\n",
        "A": "That is a basic property of limits and an integral is a limit problem. You can factor out any constant that is convenient to factor out, though you don t have to.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            264
        ],
        "3min_transcript": "is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1"
    },
    {
        "Q": "At around 5:10, Sal applies the power rule to x^-1, making it -x^-1. Should it not actually be -x^-2, since you are supposed to subtract 1 from the exponent. Also, would the expression not turn positive, since there is a negative numerator and a negative denominator? Thanks for any help.\n",
        "A": "The power rule for integrals is the reverse of the power rule for derivatives, so you add 1 to the exponent, you don t subtract. There is no negative coefficient in the numerator, so that should be negative. The power rule for integrals, then is: \u00e2\u0088\u00ab u \u00e2\u0081\u00bf du = [ u\u00e2\u0081\u00bf\u00e2\u0081\u00ba\u00c2\u00b9 / (n+1) ] + C",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            310
        ],
        "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there."
    },
    {
        "Q": "\nwhat if at 4:35 i do not take 5 out of the antiderivative operator and then just operate it simply, it gives me a result -5x^(-1) +C . can anyone tell me why is this wrong?",
        "A": "Why would you think it is wrong. You got the same result as Sal by a slightly different method. Whether or not to factor the coefficient outside the integral sign is purely stylistic. I think it makes for less clutter, and anything that reduces the clutter of integration is welcome.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            275
        ],
        "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there."
    },
    {
        "Q": "\nAt 2:52, why is n not allowed to be equal to -1?",
        "A": "Because then you would be dividing by zero.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            172
        ],
        "3min_transcript": "And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx."
    },
    {
        "Q": "At 3:49, to check to see if the derivative of (x\u00e2\u0081\u00b6/6)+C is in fact x\u00e2\u0081\u00b5, why don't you use to quotient rule? Shouldn't the derivative be 6(x\u00e2\u0081\u00b6/6) * [(6*6x\u00e2\u0081\u00b5- 0*x\u00e2\u0081\u00b6)/ 6\u00c2\u00b2]? Which equals x^11, not x\u00e2\u0081\u00b5.\n",
        "A": "There is no reason to use the quotient method here, but the rules will always work is used correctly (unless you end up with an undefined answer) The problem with your math is that when you multiply x\u00e2\u0081\u00b6 times zero that term goes away. So you end up with 36x\u00e2\u0081\u00b5/36",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            229
        ],
        "3min_transcript": "is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1"
    },
    {
        "Q": "\nat 1:15, why did you not apply the derivative rule for division? It is supposed to be (f'(x)g(x) - f(x)g'(x))/(g(x))^2",
        "A": "keep in mind n is a constant, so n+1 is a constant f(x)=((x^(n+1))/(n+1) by quotient rule: f (x)=(x^(n)*(n+1))-(0*x^(n+1)) f (x)=((n+1)(x^n))-0 f (x)=(n+1)(x^(n) same, see?",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's take the derivative with respect to x of x to the n plus 1-th power over n plus 1 plus some constant c. And we're going to assume here, because we want this expression to be defined, we're going to assume that n does not equal negative 1. If it equaled negative 1, we'd be dividing by 0, and we haven't defined what that means. So let's take the derivative here. So this is going to be equal to-- well, the derivative of x to the n plus 1 over n plus 1, we can just use the power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x to the-- I want to use that same color. Colors are the hard part-- times x to the-- instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we were dividing by this n plus 1. And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power"
    },
    {
        "Q": "So at around 5:18, if you ended up with -5x^-1 + 5c, would that still be correct?\n",
        "A": "Yes, and I think Sal mentions this around 5:15 in the video. However, since 5 times a constant C is just another constant, we get in the habit of writing a different variable name (C_1) instead of 5C.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            318
        ],
        "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there."
    },
    {
        "Q": "\nWhen he explained about the CF is equal to 9, i did not understand that well. I did not understand how CF is equal to 9 at 6:31. Please help me as my english isn't good",
        "A": "CF corresponds to AB (which has a length of 9) on a similar triangle. Therefore, there is a proportional relationship between the two. CF is actually equal to 36/7, not nine, but it is proportional to nine.",
        "video_name": "7aGEvpHaNJ8",
        "timestamps": [
            391
        ],
        "3min_transcript": "minus x-- that's that side right there-- and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is y. So it's equal to y minus x over y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to x-- let me do that in a different color. It's going to be equal to x over this entire base right over here, so this entire BE, which once again, we know is y. So over y. And now this looks interesting, because it looks like we have three unknowns. Sorry, we know what DE is already. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between x and y. So we have three unknowns and only two equations, so it seems hard to solve at first, because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of x over y, and then we could do a substitution. So that's why this was a little tricky. So this one right here-- let me do it in that same green color. We can rewrite it as CF over 9 is equal to y minus x over y. or 1 minus x over y. All I did is, I essentially, I guess you could say, distributed the 1 over y times both of these terms. So y over y minus x over y, or 1 minus x minus y. And this is useful, because we already know what x over y is equal to. We already know that x over y is equal to CF over 12. So this right over here, I can replace with this, CF over 12. So then we get-- this is the home stretch here-- CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides, so you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here,"
    },
    {
        "Q": "\nThere is a \"Acute\" angle in 5:35. A to C and E :)",
        "A": "Rays EA and EC do make an acute angle. You would call it angle AEC. Always put the letter of the center point of the angle in the middle of the angle name.",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            335
        ],
        "3min_transcript": "We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and go through C and beyond C that is teh same thing as FE ray FE and ray FC are the same as the point E is on ray FC, then finally we have not focussed on point A you may think there is ray AE, but the line does not go beyond E, so it is not a ray, to the top of A there is no other point, so there is no ray there either that is all the rays based on the points specified. If they had given us a point over there, we could have had other rays,"
    },
    {
        "Q": "at 2:53, can the rays CE and CF also be written CEF or FEC?\n",
        "A": "No because you only use two points to identify a ray, using all three points means you are talking about an angle",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            173
        ],
        "3min_transcript": "is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays, We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and"
    },
    {
        "Q": "\nAt 2:38, can you express the ray as ray CEF?",
        "A": "No. You only use two points to express a ray. For example ray CF or ray EF or ray EC. If you use three points you are expressing an angle, in this case a straight angle.",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            158
        ],
        "3min_transcript": "is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays, We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and"
    },
    {
        "Q": "\nFrom 0:34 and onward, he starts to talk about how you can tell if it is an axis of symmetry. Can you do this with any shape? How about with a circle??",
        "A": "You don t need to worry. From 0:34, Sal started talking about that cause the topic is axis of symmetry. You always find a line of symmetry in every basic shape you find. But it is not important that there is axis of symmetry in every shape. I hope this helped!",
        "video_name": "LrTn4cvsewk",
        "timestamps": [
            34
        ],
        "3min_transcript": "For each of these diagrams, I want to think about whether this blue line represents an axis of symmetry. And the way we can tell is if on both sides of the blue line we essentially have mirror images. Let's take this top part of this polygon, the part that is above this blue line here, and let's reflect it across the blue line-- you could almost imagine that it's a reflection over some type of a lake or something-- and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go-- the same amount on the other side would get you right there. And so you immediately see we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point-- I'll do it in a different color. This point, if you were to reflect it across this blue line, it would get I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no. No, this blue line is not an axis of symmetry. Now let's look at it over here. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection. And we can even go point by point here. So this point right over here is the same distance from, if we dropped a perpendicular to this point as this one right over here; this one over here, same distance, same distance as this point right over here; and we could do that for all of the points. So in this case, the blue line does represent an axis of symmetry."
    },
    {
        "Q": "\nAt 3:04 why does it continue after that?",
        "A": "It doesn t, it ends at 3:04.",
        "video_name": "jb8mFpA1YI8",
        "timestamps": [
            184
        ],
        "3min_transcript": "Let's see if we can figure out 3 times 60. Well, there's a couple of ways you could think about it. You could literally view this as 60 three times. So you could view this as 60 plus 60 plus 60. And you might be able to compute this in your head. 60 plus 60 is 120, plus another 60 is 180. And you'd be done. But another way to think about this is that 3 times 60 is the same thing as 3 times-- instead of thinking of it as 60, you could think of 60 as 6 times 10. 3 times 6 times 10. Now, when you're multiplying three numbers like this, it doesn't matter what order you multiply them in. So we could multiply the 3 times 6 first and get 18 and then multiply that times 10. And 18 times 10 is just going to be 180. It's going to be 18 with another zero. So this is going to be 180. Now, the more practice you get here, you'll realize, but I have to worry about this 0 right over here. So I'm going to put one more zero at the end. It's going to be 180. Same answer that we got right over there. Let's do another one of these. So let's say we want to multiply 50 times 7. And I encourage you to pause the video and think about it yourself, and then unpause it and see what I do. So 50, well, there's a couple of ways you could think about it. One, you could literally try to add 50 seven times. Adding 7 fifty times would take forever, but you could literally say 50 plus 50 plus 50 plus 50-- let's see, that's four-- plus 50 plus 50. Let's see, that is six. I'll do one more right over here. 50 right over here. So this is 50 seven times. If you add together 50 plus 50 is 100, 150, 200, 250, 300, 350. So you could do it that way. You just need to realize that 50 is the same thing as 10 times 5. So we could write this as 10 times 5, and then we're multiplying that times 7. Once again, the order that we multiply does not matter. So we can multiply the 5 times the 7 first. We know that that is 35, and we're going to multiply that times 10. 10 times 35, well, we're just going to stick a zero at the end of the 35. It's going to be equal to 350. Now I want to do that zero in that same color. It's going to be 350. Now, you might realize, hey, look, I could have just looked at this 5 right over here, multiplied the 5 times the 7, and have gotten the 35. And then, not forgetting that it's actually not a 5, it's a 50. So I have to multiply by 10 again, or I have to throw that 0 at the end of it to get that 0 right over here. So 50 times 7 is 350."
    },
    {
        "Q": "\nAt 11:15, Sal mistakenly wrote 2.5, when it was supposed to be 2.9.",
        "A": "Good catch! If you look, you ll even see a little note in the lower left corner about the mistake.",
        "video_name": "i6lfVUp5RW8",
        "timestamps": [
            675
        ],
        "3min_transcript": "Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1. decimal 5 to the right. 1, 2, 3, 4, 5. So this will be 2.5, and then we're going to be left with times 10 to the negative 5. Anyway, hopefully, you found that scientific notation drill useful."
    },
    {
        "Q": "At 11:15 instead of 2.5 isn't it suppose to be 2.9\n",
        "A": "That is a known error in the video. A box pops up and tells you the error and provides the correct value of 2.9",
        "video_name": "i6lfVUp5RW8",
        "timestamps": [
            675
        ],
        "3min_transcript": "Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1. decimal 5 to the right. 1, 2, 3, 4, 5. So this will be 2.5, and then we're going to be left with times 10 to the negative 5. Anyway, hopefully, you found that scientific notation drill useful."
    },
    {
        "Q": "\nAt 0:25 I wanted to know when was scientific notation even invented?",
        "A": "not hepfull flaged",
        "video_name": "i6lfVUp5RW8",
        "timestamps": [
            25
        ],
        "3min_transcript": "There are two whole Khan Academy videos on what scientific notation is, why we even worry about it. And it also goes through a few examples. And what I want to do in this video is just use a ck12.org Algebra I book to do some more scientific notation examples. So let's take some things that are written in scientific notation. Just as a reminder, scientific notation is useful because it allows us to write really large, or really small numbers, in ways that are easy for our brains to, one, write down, and two, understand. So let's write down some numbers. So let's say I have 3.102 times 10 to the second. And I want to write it as just a numerical value. It's in scientific notation already. It's written as a product with a power of 10. So how do I write this? It's just a numeral. Well, there's a slow way and the fast way. The slow way is to say, well, this is the same thing as 3.102 times 100, which means if you multiplied 3.102 times And then we have 1, 2, 3 numbers behind the decimal point, and that'd be the right answer. This is equal to 310.2. Now, a faster way to do this is just to say, well, look, right now I have only the 3 in front of the decimal point. When I take something times 10 to the second power, I'm essentially shifting the decimal point 2 to the right. So 3.102 times 10 to the second power is the same thing as-- if I shift the decimal point 1, and then 2, because this is 10 to the second power-- it's same thing as 310.2. So this might be a faster way of doing it. the right by 1. Let's do another example. Let's say I had 7.4 times 10 to the fourth. Well, let's just do this the fast way. Let's shift the decimal 4 to the right. So 7.4 times 10 to the fourth. Times 10 to the 1, you're going to get 74. Then times 10 to the second, you're going to get 740. We're going to have to add a 0 there, because we have to shift the decimal again. 10 to the third, you're going to have 7,400. And then 10 to the fourth, you're going to have 74,000. Notice, I just took this decimal and went 1, 2, 3, 4 spaces. So this is equal to 74,000. And when I had 74, and I had to shift the decimal 1 more to the right, I had to throw in a 0 here. I'm multiplying it by 10. Another way to think about it is, I need 10 spaces between"
    },
    {
        "Q": "At 10:10 we have 5 digits. Is there any limit as to the total number of digits we can have? How does this affect significant figures?\n",
        "A": "Nope. There are no limits. The more digits you have, the bigger the exponent of 10 needs to be. The rules for significant figures aren t affected.",
        "video_name": "i6lfVUp5RW8",
        "timestamps": [
            610
        ],
        "3min_transcript": "And then we're left with this one, times 10 to the negative 3. Now, a very quick way to do it is just to say, look, let me count-- including the leading numeral-- how many spaces I have behind the decimal. 1, 2, 3. So it's going to be 2.81 times 10 to the negative 1, 2, 3 power. Let's do one more like that. Let me actually scroll up here. Let's do one more like that. Let's say I have 1, 2, 3, 4, 5, 6-- how many 0's do I have Well, I'll just make up something. 0, 2, 7. And you wanted to write that in scientific notation. Well, you count all the digits up to the 2, behind the decimal. So 1, 2, 3, 4, 5, 6, 7, 8. So this is going to be 2.7 times 10 to Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1."
    },
    {
        "Q": "\nAt 0:47, Sal says 15 to 25.\nShouldn't he say 25 to 15 ?",
        "A": "He can say 15 to 25 or 25 to 15, it depends on him, because the order in which you say ratios doesn t really matter.",
        "video_name": "jNUz0P5MG9M",
        "timestamps": [
            47
        ],
        "3min_transcript": "The following table describes the relationship between the number of servings of spaghetti bolognese-- I don't know if I'm pronouncing that-- or bolognese, and the number of tomatoes needed to prepare them. Test the ratios for equivalents, and determine whether the relationship is proportional. Well, you have a proportional relationship between the number of servings and the number of tomatoes is if the ratio of the number of servings to the number of tomatoes is always the same. Or if the ratio of the number of tomatoes to the number of servings is always the same. So let's just think about the ratio of the number of tomatoes to the number of servings. So it's 10 to 6, which is the same thing as 5 to 3. So here the ratio is 5 to 3. 15 to 9, if you divide both of these by 3, you get 5 to 3. So it's the same ratio. 15 to 25, if you divide both of these by 5, you get 5 to 3. So based on this data, it looks like the ratio between the number of tomatoes and the number of servings is always constant. So yes, this relationship is proportional."
    },
    {
        "Q": "at around 3:30, Sal siad for the second equation that why is ewual to NEGATIVE 1x +5. Why is that? Shouldn't it be positive?\n",
        "A": "No, it should not, the reason for this is that the x(1x) was already negative in the equation. The original form of the equation looked like this y<5 -x(1x) , the only thing Sal did was rewrite the equation to look like the one above it. So he moved the x(1x) to the front of the equation taking the negative sign with it y< -x(1x) +5. When you rearrange an equation the signs of numbers remain to keep the problem balanced.",
        "video_name": "CA4S7S-3Lg4",
        "timestamps": [
            210
        ],
        "3min_transcript": "So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0, if y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope of 1, for every time you move to the right 1, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line. And then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, I mean, in general I like to just think, oh, greater than, it's going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say, 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And you could try something out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey look, this is the boundary line, and we're greater than the boundary line for any given x. Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. And it has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down one because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line because we are-- so that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x."
    },
    {
        "Q": "\nAt 3:14 Sal said that y is equal to 5-x even though it said y is less then 5-x?",
        "A": "At 3:14 Sal is talking about the boundary line, not the area. That s why he says equal to.",
        "video_name": "CA4S7S-3Lg4",
        "timestamps": [
            194
        ],
        "3min_transcript": "So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0, if y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope of 1, for every time you move to the right 1, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line. And then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, I mean, in general I like to just think, oh, greater than, it's going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say, 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And you could try something out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey look, this is the boundary line, and we're greater than the boundary line for any given x. Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. And it has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down one because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line because we are-- so that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x."
    },
    {
        "Q": "At 4:30, when zero is said to be undefined, how come no one of the many mathematicians over the years were able to define it? Why is it so disputed over? Humans as a whole, over the many years of their existence weren't able to do so?\n",
        "A": "Sal says a number divided by zero is undefined, not zero itself. The reason a number divided by zero is undefined is because no such number exists. The value of a number divided by zero is infinite.",
        "video_name": "bQ-KR3clFgs",
        "timestamps": [
            270
        ],
        "3min_transcript": "we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative, If I had a negative divided by a positive, that's also going to be a negative. If I have a negative divided by a negative, that's going to give me a positive, and if obviously a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here. This is actually all multiplication, but it's interesting, because we're multiplying three (3) things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative two (-2) times negative seven (-7). Negative two (-2) times negative seven (-7). They are both negatives, and negatives cancel out, so this would give us, this part right over here, will give us positive fourteen (14). And so we're going to multiply positive fourteen (14) times this negative one (-1), times -1. Now we have a positive times a negative. Exactly one of them is negative, so this is going to be negative answer, Now let me give you a couple of more, I guess we could call these trick problems. What would happen if I had zero (0) divided by negative five (-5). Well this is zero negative fifths So zero divided by anything that's non-zero is just going to equal to zero. But what if it were the other way around? What happens if we said negative five divided by zero? Well, we don't know what happens when you divide things by zero. We haven't defined that. There's arguments for multiple ways to conceptualize this, so we traditionally do say that this is undefined. We haven't defined what happens when something is divided by zero. And similarly, even when we had zero divided by zero, this is still, this is still, undefined."
    },
    {
        "Q": "At 4:45 he says zero is undefined. Does that mean its not negative of positive?\n",
        "A": "Its neither. An integer is a whole number that can be either greater than 0, called positive, or less than 0, called negative. Zero is neither positive nor negative. Two integers that are the same distance from the origin in opposite directions are called opposites.",
        "video_name": "bQ-KR3clFgs",
        "timestamps": [
            285
        ],
        "3min_transcript": "we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative, If I had a negative divided by a positive, that's also going to be a negative. If I have a negative divided by a negative, that's going to give me a positive, and if obviously a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here. This is actually all multiplication, but it's interesting, because we're multiplying three (3) things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative two (-2) times negative seven (-7). Negative two (-2) times negative seven (-7). They are both negatives, and negatives cancel out, so this would give us, this part right over here, will give us positive fourteen (14). And so we're going to multiply positive fourteen (14) times this negative one (-1), times -1. Now we have a positive times a negative. Exactly one of them is negative, so this is going to be negative answer, Now let me give you a couple of more, I guess we could call these trick problems. What would happen if I had zero (0) divided by negative five (-5). Well this is zero negative fifths So zero divided by anything that's non-zero is just going to equal to zero. But what if it were the other way around? What happens if we said negative five divided by zero? Well, we don't know what happens when you divide things by zero. We haven't defined that. There's arguments for multiple ways to conceptualize this, so we traditionally do say that this is undefined. We haven't defined what happens when something is divided by zero. And similarly, even when we had zero divided by zero, this is still, this is still, undefined."
    },
    {
        "Q": "In algebra, why just not use the letter X so you can still use it for multiplication? 2:05\n",
        "A": "It s just that x happens to be very common, if you don t like the use of x, you can use any other variable, like j .",
        "video_name": "bQ-KR3clFgs",
        "timestamps": [
            125
        ],
        "3min_transcript": "Now that we know a little bit about multiplying positive and negative numbers, Let's think about how how we can divide them. Now what you'll see is that it's actually a very similar methodology. That if both are positive, you'll get a positive answer. If one is negative, or the other, but not both, you'll get a negative answer. And if both are negative, they'll cancel out and you'll get a positive answer. But let's apply and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So eight (8) divided by negative two (-2). So if I just said eight (8) divided by two (2), that would be a positive four (4), but since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So eight (8) divided by negative two (-2) is negative four (-4). Now negative sixteen (-16) divided by positive four (4)-- now be very careful here. If I just said positive sixteen (16) divided by positive four (4), that would just be four (4). But because one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative thirty (-30) divided by negative five (-5). If I just said thirty (30) divided by five (5), I'd get a positive six (6). And because I have a negative divided by a negative, the negatives cancel out, so my answer will still be positive six (6)! And I could even write a positive (+) out there, I don't have to, but this is a positive six (6). A negative divided by a negative, just like a negative times a negative, you're gonna get a positive answer. Eighteen (18) divided by two (2)! And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers: This is a positive divided by a positive. Which is going to be a positive. So that is going to be equal to positive nine (9). Now we start doing some interesting things, here's kind of a compound problem. We have some multiplication and some division going on. we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative,"
    },
    {
        "Q": "\nAt 0:42 why is the answer negative if the bigger number is positive?",
        "A": "It doesn t matter if the bigger number is positive or not. If you re dividing (or multiplying) and exactly one of the two numbers is negative then the answer is negative.",
        "video_name": "bQ-KR3clFgs",
        "timestamps": [
            42
        ],
        "3min_transcript": "Now that we know a little bit about multiplying positive and negative numbers, Let's think about how how we can divide them. Now what you'll see is that it's actually a very similar methodology. That if both are positive, you'll get a positive answer. If one is negative, or the other, but not both, you'll get a negative answer. And if both are negative, they'll cancel out and you'll get a positive answer. But let's apply and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So eight (8) divided by negative two (-2). So if I just said eight (8) divided by two (2), that would be a positive four (4), but since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So eight (8) divided by negative two (-2) is negative four (-4). Now negative sixteen (-16) divided by positive four (4)-- now be very careful here. If I just said positive sixteen (16) divided by positive four (4), that would just be four (4). But because one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative thirty (-30) divided by negative five (-5). If I just said thirty (30) divided by five (5), I'd get a positive six (6). And because I have a negative divided by a negative, the negatives cancel out, so my answer will still be positive six (6)! And I could even write a positive (+) out there, I don't have to, but this is a positive six (6). A negative divided by a negative, just like a negative times a negative, you're gonna get a positive answer. Eighteen (18) divided by two (2)! And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers: This is a positive divided by a positive. Which is going to be a positive. So that is going to be equal to positive nine (9). Now we start doing some interesting things, here's kind of a compound problem. We have some multiplication and some division going on. we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative,"
    },
    {
        "Q": "At 2:30, what does \"factoring out the negative sign\" mean? How do you factor out a negative sign? I also had a problem with this in \"Manipulating linear expressions with rational coefficients,\" in Algebra. I never understood how to do it.\n",
        "A": "area of each interior angle = 180 (5-2)/5 = 108, so area of each exterior angle = 180-108 = 72, so area of five exterior angle = 72*5 = 360, Am I on the right track?",
        "video_name": "95logvV8nXY",
        "timestamps": [
            150
        ],
        "3min_transcript": "Now this looks like an interesting problem. We have this polygon. It looks like a pentagon right over here, has five sides. It's an irregular pentagon. Not all the sides look to be the same length. And the sides are kind of continued on. And we have these particular exterior angles of this pentagon. And what we're asked is, what is the sum of all of these exterior angles. And it's kind of daunting, because they don't give us any other information. They don't even give us any particular angles. They don't start us off anywhere. And so what we can do, let's just think about the step by step, just based on what we do know. Well, we have these exterior angles. And these exterior angles, they're each supplementary to some interior angle. So maybe if we can express them as a function of the interior angles, we can maybe write this problem in a way that seems a little bit more doable. So let's write the interior angles over here. We already got to e. So let's call this f, this interior angle f. Let's call this interior angle g. Let's call this one i. And let's call this one j. And so this sum of these particular exterior angles, a is now the same thing as 180 minus g, because a and g are supplementary. So a is 180 minus g. And then we have plus b. But b we can write in terms of this interior angle. It's going to be 180 minus h, because these two angles once again, are supplementary. We do that in a new color. So this is going to be 180 minus h. And we could do the same thing for each of them. c, we can write as 180 minus i, so plus 180 minus i. And then d, we can write as 180 minus j, so plus 180 minus j. And then finally, e, I'm running out of colors, e, we can write as 180 minus f, so plus 180 minus f. we have 180 5 times. So this is going to be equal to 5 times 180 which is what, 900. And then you have minus g, minus h, minus i, minus j, minus f. Or we could write that as minus-- I'll try to do the same colors-- g plus h-- I'm kind of factoring out this negative sign-- g plus h-- I'll do the same color as g, that's not the same color-- g plus h, plus i, plus j, plus f. And the whole reason why did this and why this is interesting now, is that we've expressed this first thing that we need to figure out. We've expressed it in terms of sum of the interior angles."
    },
    {
        "Q": "at the time 2:02 in the video, i don't understand what he was doing? Can anyone help me please!\n",
        "A": "In this section at 2:02, he is finding the 4th root of 16 or \u00e2\u0088\u009c16 The 4th root of something is a number multiplied by itself 4 times to equal that something Well, at that time after factoring 16, he has \u00e2\u0088\u009c2\u00e2\u0088\u00992\u00e2\u0088\u00992\u00e2\u0088\u00992 written and he is talking about finding the 4th root of 2 times 2 times 2 times 2 We are looking for a factor that occurs 4 times, and there it is! We can see that 2 is a factor of 16 four different times, so \u00e2\u0088\u009c16 = \u00e2\u0088\u009c2\u00e2\u0088\u00992\u00e2\u0088\u00992\u00e2\u0088\u00992 = \u00e2\u0088\u009c2\u00e2\u0081\u00b4 = 2 And that is how he got the 2",
        "video_name": "iX7ivCww2ws",
        "timestamps": [
            122
        ],
        "3min_transcript": "So far, when we were dealing with radicals we've only been using the square root. We've seen that if I write a radical sign like this and put a 9 under it, this means the principal square root of 9, which is positive 3. Or you could view it as the positive square root of 9. Now, what's implicit when we write it like this is that I'm taking the square root. So I could have also written it like this. I could have also written the radical sign like this and written this index 2 here, which means the square root, the principal square root of 9. Find me something that if I square that something, I get 9. And the radical sign doesn't just have to apply to a square root. You can change the index here and then take an arbitrary root of a number. So for example, if I were to ask you, what-- You could imagine this is called the cube root, or you could call it the third root of 27. What is this? power, I'd get 27. Well, the only number that if you take it to the third power, you get 27 is equal to 3. 3 times 3 times 3 is equal to 27. 9 times 3, 27. So likewise, let me just do one more. So if I have 16-- I'll do it in a different color. If I have 16 and I want to take the fourth root of 16, what number times itself 4 times is equal to 16? And if it doesn't pop out at you immediately, you can actually just do a prime factorization of 16 to figure it out. 16 is 2 times 8. 8 is 2 times 4. 4 is 2 times 2. So this is equal to the fourth root of 2 times 2 times 2 times 2. You have these four 2's here. of this must be equal to 2. And you could also view this as kind of the fourth principal root because if these were all negative 2's, it would also work. Just like you have multiple square roots, you have multiple fourth roots. But the radical sign implies the principal root. Now, with that said, we've simplified traditional square roots before. Now we should hopefully be able to simplify radicals with higher power roots. So let's try a couple. Let's say I want to simplify this expression. The fifth root of 96. So like I said before, let's just factor this right here. So 96 is 2 times 48. Which is 2 times 24. Which is 2 times 12. Which is 2 times 6."
    },
    {
        "Q": "I do not entirely understand the rules in regards to a specific exponent being rewritten as a specific corresponding square root. Ex: The square root of 25= 25^1/2 power. I am absolutely lost after this point: 6:35. Are there any videos that explore this further and in greater detail? Thanks!\n",
        "A": "Basically, when a number is raised to a fractional power, it is asked for that specific root. EX: 1/2 power= square root 1/3 power = cubed root",
        "video_name": "iX7ivCww2ws",
        "timestamps": [
            395
        ],
        "3min_transcript": "Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2 to the sixth power. So this is equivalent to the sixth root of 2 to the sixth-- that's what 64 is --times x to the eighth power. Now, the sixth root of 2 to the sixth, that's pretty straightforward. So this part right here is just going to be equal to 2. That's going to be 2 times the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power, that's the same thing as x to the sixth power times x squared. This is the same thing as x to the eighth. So this is going to be equal to 2 times the sixth root of x to the sixth times x squared. And the sixth root, this part right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to 2 times x times the sixth root of x squared. Now, we can simplify this even more if you really think about. Remember, this expression right here, this is the exact same thing as x squared to the 1/6 power. And if you remember your exponent properties, when you raise something to an exponent, and then raise that to an exponent, that's equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times 1/6 power, which is the same thing-- Let me not forget to write my 2x there. And this is the same thing as 2x-- it's the same 2x there --times x to the 2/6. Or, if we want to write that in most simple form or lowest common form, you get 2x times x to the-- What do you have here? x to the 1/3. So if you want to write it in radical form, you could write this is equal to 2 times 2x times the third root of x. Or, the other way to think about it, you could just say-- So we could just go from this point right here. We could write this. We could ignore this, what we did before. And we could say, this is the same thing as 2 times x to the eighth to the 1/6 power. x to the eighth to the 1/6 power. So this is equal to 2 times x to the-- 8"
    },
    {
        "Q": "\nAt 6:20, why does the sixth root of X^6 equal X and not the absolute value of X?",
        "A": "It depends on whether or not the radical is given in the equation or if you need to insert it. If the problem has a radical shown, then you assume it is positive (what Sal did). If you are solving the problem and you introduce a radical, then you could have both the positive and negative option (so the absolute value).",
        "video_name": "iX7ivCww2ws",
        "timestamps": [
            380
        ],
        "3min_transcript": "Times 3 to the 1/5. Now I have something that's multiplied. I have 2 multiplied by itself 5 times. And I'm taking that to the 1/5. Well, the 1/5 power of this is going to be 2. Or the fifth root of this is just going to be 2. So this is going to be a 2 right here. And this is going to be 3 to the 1/5 power. 2 times 3 to the 1/5, which is this simplified about as much as you can simplify it. But if we want to keep in radical form, we could write it as 2 times the fifth root 3 just like that. Let's try another one. Let me put some variables in there. Let's say we wanted to simplify the sixth root of 64 times x to the eighth. So let's do 64 first. 64 is equal to 2 times 32, which is 2 times 16. Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2 to the sixth power. So this is equivalent to the sixth root of 2 to the sixth-- that's what 64 is --times x to the eighth power. Now, the sixth root of 2 to the sixth, that's pretty straightforward. So this part right here is just going to be equal to 2. That's going to be 2 times the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power, that's the same thing as x to the sixth power times x squared. This is the same thing as x to the eighth. So this is going to be equal to 2 times the sixth root of x to the sixth times x squared. And the sixth root, this part right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to 2 times x times the sixth root of x squared. Now, we can simplify this even more if you really think about. Remember, this expression right here, this is the exact same thing as x squared to the 1/6 power. And if you remember your exponent properties, when you raise something to an exponent, and then raise that to an exponent, that's equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times 1/6 power, which is the same thing-- Let me not forget to write my 2x there."
    },
    {
        "Q": "\nAt 3:10 Sal says that each of the objects became two groups. Exactly how are they in two groups? I'm having a hard time visualizing the two groups. I get how he divided each circle in half groups, but the 2 groups he mentions at the very end, I cannot see. Please help. Thanks.",
        "A": "Oh wait, I just saw it. The four circles were split in two.",
        "video_name": "tnkPY4UqJ44",
        "timestamps": [
            190
        ],
        "3min_transcript": "many times it goes. That'll be the whole number part of the mixed number. And then whatever's left over will be the remaining numerator over 5. So what we'll do is take 5 into 6. 5 goes into 6 one time. 1 times 5 is 5. Subtract. You have a remainder of 1. So 6/5 is equal to one whole, or 5/5, and 1/5. This 1 comes from whatever is left over. And now we're done! 3/5 divided by 1/2 is 1 and 1/5. Now, the one thing that's not obvious is why did this work? Why is dividing by 1/2 the same thing as multiplying essentially by 2. 2/1 is the same thing as 2. And to do that, I'll do a little side-- fairly simple-- example, but hopefully, it gets the point across. So we have four objects: one, two, three, four. So I have four objects, and if I were to divide into groups of two, so I want to divide it into groups of two. So that is one group of two and then that is another group of two, how many groups do I have? Well, 4 divided by 2, I have two groups of two, so that is equal to 2. Now, what if I took those same four objects: one, two, three, four. So I'm taking those same four objects. Instead of dividing them into groups of two, I want to divide them into groups of 1/2, which means each group will have half of an object in it. So let's say that would be one group right there. That is a second group. That is a third group. I think you see each group has half of a circle in it. That is the fourth. That's the fifth. That's the sixth. You have eight groups of 1/2, so this is equal to 8. And notice, now each of the objects became two groups. So you could say how many groups do you have? Well, you have four objects and each of them became two groups. I'm looking for a different color. Each of them became two groups, and so you also have eight. So dividing by 1/2 is the same thing as multiplying by 2. And you could think about it with other numbers, but hopefully, that gives you a little bit of an intuition."
    },
    {
        "Q": "\nat 12:47pm. What if you are multiplying a whole number by a fraction",
        "A": "Let s do 7 \u00c3\u00b7 3/4 Change the whole number into a fraction: 7/1 \u00c3\u00b7 3/4 Then change division to multiply by using the reciprocal of 3/4: 7/1 * 4/3 Multiply: 28/3 = 9 1/3 Hope this helps.",
        "video_name": "tnkPY4UqJ44",
        "timestamps": [
            767
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:21, cant you divide by 5?",
        "A": "At 0:21 you could divide =)",
        "video_name": "D1cKk48kz-E",
        "timestamps": [
            21
        ],
        "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like."
    },
    {
        "Q": "At 0:33, couldn't you divide by -5?\n",
        "A": "You could, however, some people consider it easier to multiply. Dividing by -5 is the same as dividing by -5/1. You can multiply the inverse -1/5 if you find multiplication easier.",
        "video_name": "D1cKk48kz-E",
        "timestamps": [
            33
        ],
        "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like."
    },
    {
        "Q": "At 1:13, why does Sal flip the inequality sign? PLEASE HELP!\n",
        "A": "We solve inequalities almost exactly the way we solve equations. The one exception is that if we multiply or divide by a negative, we have to reverse the inequality. Let s look at what happens if we don t. Start with: -4 < 12 This is current true. If we divide both sides by -2, what happens? -4/(-2) < 12/(-2) 2 < -6 This is no longer true. 2 is a larger number than -6. This is why we reverse the inequality. It is needed to maintain the integrity (truth) of the inequality. Hope this helps.",
        "video_name": "D1cKk48kz-E",
        "timestamps": [
            73
        ],
        "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like."
    },
    {
        "Q": "0:25. Why would it be one fifth and not 5?\n",
        "A": "-5 is the coefficient of c, not a subtraction from c. If it were -5+c or c - 5,, then the opposite would be to add 5. since the 5 and c are adjacent, it means multiply (which is what a coefficient is), so the opposite of multiplying by -5 is to divide by -5 which is the same as multiplying by (-1/5).",
        "video_name": "D1cKk48kz-E",
        "timestamps": [
            25
        ],
        "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like."
    },
    {
        "Q": "at 1:36 why did you shade in 2 whole halves an only 1/2 on the others?\n--------------------------------------------------------------\n",
        "A": "because, we have to add 1/2 (means half) 5 times",
        "video_name": "4PlkCiEXBQI",
        "timestamps": [
            96
        ],
        "3min_transcript": "Let's think about how or what 1/2 times 5 represents. So one way to think about it is that this could be five 1/2's added together. So you could view this as 1/2 plus 1/2 plus 1/2 plus 1/2 plus 1/2, which is the same thing as 1 plus 1 plus 1 plus 1 plus 1, over 2, which is equal to 5/2. The other way to think about this is that you start with 5 things. So let's say, that's 1 thing. Let me copy and paste that so they all look the same. So then let me paste it. So that's 2 things. That's 3 things. That's 4 things. And that's 5 things. So the other way to think about it is you start with 5 things, So what would be 1/2 of this? Well, let's see. You have 5 things, so you would get-- 5 divided by 2 would be 2 and 1/2. So you would get this far. Let me make it like this. So you would get this one. You would get this one. And you would get this one. Now, is this the same thing as 5/2? Well, what happens if we divide each of these wholes into halves? So let's do that. So if we just multiplied-- so we just divide each of these into 2. So instead of having 5 wholes, we now have 10 halves. How many of those halves have we filled in? Well, we have filled in 1, 2, 3, 4, 5. So this is also equal to 5/2. multiplication actually means. But if you said, well, how did I compute this? Well, the way you could think about it, and multiplying fractions is actually straightforward from that point of view, is as long as you can express both of them as fractions, and 5 we already know is the same thing as 5 ones, so this we can just multiply times 5/1. So now that I've expressed both of them as fractions, I can just multiply the numerator. So 1 times 5 over 2 times 1. And what's that going to be equal to? Well, 1 times 5 is 5. 2 times 1 is 2."
    },
    {
        "Q": "\nAt 3:10 he writes the value of pi = 3.14, but Sal uses it as a reference for radians, isn't the 3.14 value in degrees? I mean he says that 3 radians is close to 3.14. So basically I'm asking what measure are we using in normal arithmetic, degrees right? If so, how come he compared 3 radians to 3.14 ,what are presumebly, degrees? Please tolerate any ignorance you come across in my question, I'm simply trying to understand the concepts.",
        "A": "Pi is simply a mathematical constant. It does not have any default units attached to it. The number 5 is not in degrees or meters, it is just the number 5. The same with pi.",
        "video_name": "fYQ3GRSu4JU",
        "timestamps": [
            190
        ],
        "3min_transcript": "If we go straight up, if we rotate it, essentially, if you want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us to pi over two. That would have been a counterclockwise rotation of pi over two radians. Now is three pi over five greater or less than that? Well, three pi over five, three pi over five is greater than, or I guess another way I can say it is, three pi over six is less than three pi over five. You make the denominator smaller, making the fraction larger. Three pi over six is the same thing as pi over two. So, let me write it this way. Pi over two is less than three pi over five. It's definitely past this. We're gonna go past this. Does that get us all the way over here? If we were to go, essentially, be pointed in the opposite direction. Instead of being pointed to the right, making a full, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be five pi over five. This is less than pi radians. We are going to sit, we are going to sit someplace, someplace, and I'm just estimating it. We are gonna sit someplace like that. And so we are going to sit in the second quadrant. Let's think about two pi seven. Two pi over seven, do we even get past pi over two? Pi over two here would be 3.5 pi over seven. We don't even get to pi over two. We're gonna end up, we're gonna end up someplace, someplace over here. This thing is, it's greater than zero, so we're gonna definitely start moving counterclockwise, but we're not even gonna get to... This thing is less than pi over two. This is gonna throw us in the first quadrant. What about three radians? three is a little bit less than pi. Right? Three is less than pi but it's greater than pi over two. How do we know that? Well, pi is approximately 3.14159 and it just keeps going on and on forever. So, three is definitely closer to that than it is to half of that. It's going to be between pi over two, and pi. It's gonna be, if we start with this magenta ray, we rotate counterclockwise by three radians, we are gonna get... Actually, it's probably gonna be, it's gonna look something, it's gonna be something like this. But for the sake of this exercise, we have gotten ourselves, once again, into the second quadrant."
    },
    {
        "Q": "\nDoesn't Koch's snowflake and Mandelbrot's famous shape already address your concept of an \"infinigon/zigfinigon\" as seen around 03:09?",
        "A": "She does say something fractally at 03:35 which again Koch and Mandelbrot takcle the fractal field with their well known shapes.",
        "video_name": "D2xYjiL8yyE",
        "timestamps": [
            189
        ],
        "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0."
    },
    {
        "Q": "\nAt 3:24, your creating a pentagram, right?",
        "A": "If you mean the five-sided shape she ended up with after the Zigfinite star then that s a pentagon. If not, can you explain better what you mean?",
        "video_name": "D2xYjiL8yyE",
        "timestamps": [
            204
        ],
        "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0."
    },
    {
        "Q": "\u00e2\u0088\u009a4 is 2. But at about 2:45, it says \u00e2\u0088\u009a2 is 2. How is this possible?\n",
        "A": "This is a joke, sometimes called sophistry. In fact, angular line will never be straight, so it s proof that the square root of 2 is 2 does not work =)",
        "video_name": "D2xYjiL8yyE",
        "timestamps": [
            165
        ],
        "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0."
    },
    {
        "Q": "I did the drawing she does at about 2:50 and then divided the squares I got into right triangles and squares, and divided those squares into right triangles and squares, and so on... The top half ended up looking like Serpinsky's triangle. I encourage you to try it for yourself. Also, does anyone know why it looks like Serpinsky's triangle?\n",
        "A": "She is really good at doodling :P",
        "video_name": "D2xYjiL8yyE",
        "timestamps": [
            170
        ],
        "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0."
    },
    {
        "Q": "Why can't you just square root the whole hypotenuse formula? For example, at 1:50, instead of squaring 15, 8, and x, why can't you just use 15+8=x?\n",
        "A": "That will not get us the right answer. 15+8=23, but x=17. You have to square both numbers, add them together, and only then take the square root.",
        "video_name": "T971zHhZ3S4",
        "timestamps": [
            110
        ],
        "3min_transcript": "What I want to do in this video is attempt to find the diameter of this circle right over here. And I encourage you to pause the video and try this out on your own. Well, let's think about what's going on right over here. AB is definitely a diameter of the circle. It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know? Well, we could look at this angle right over here, angle C, and think about it is an inscribed angle. And think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle. Angle C is inscribed. If you take these two sides or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. Or another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared-- let me do this in magenta-- is going to be the length of side AB squared. So this side right over here, let me just call that x. That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64-- I want to do 225 plus 64 is 289 is equal to x squared. And then 289 is 17 squared. And you could try out a few numbers if you're unsure about that. So x is equal to 17. So the diameter of this circle right over here is 17."
    },
    {
        "Q": "\nat 4:08, why did he flip the left side?",
        "A": "He did that to get t up top . He inverted both sides of the equation which is a valid thing to do. Here is why the flip is valid: start with a / b = c / d => ad = bc (multiplied both sides by b, then by d) => d / c = b / a (divided both sides by c, then by a)",
        "video_name": "gD7A1LA4jO8",
        "timestamps": [
            248
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:08 and 5:38, how did Sal know what the graphs of the functions would look like?\n",
        "A": "Lots of practice... different types of equations create specific graphs. f(x) = x^2 is a quadratic equation. It creates a U-shaped graph. g(x) = x is a linear equation. It creates a line for a graph.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            188,
            338
        ],
        "3min_transcript": "okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\""
    },
    {
        "Q": "What are oulpuls at 1:45- 1:55\n",
        "A": "Outputs , with the crosses on the t s not drawn well.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            105,
            115
        ],
        "3min_transcript": "- As a little bit of a review, we know that if we have some function, let's call it \"f\". We don't have to call it \"f\", but \"f\" is the letter most typically used for functions, that if I give it an input, a valid input, if I give it a valid input, and I use the variable \"x\" for that valid input, it is going to map that to an output. It is going to map that, or produce, given this x, it's going to product an output that we would call \"f(x).\" And we've already talked a little bit about the notion of a domain. A domain is the set of all of the inputs over which the function is defined. So if this the domain here, if this is the domain here, and I take a value here, and I put that in for x, then the function is going to output an f(x). If I take something that's outside of the domain, let me do that in a different color... If I take something that is outside of the domain and try to input it into this function, the function will say, \"hey, wait wait,\" \"I'm not defined for that thing\" \"that's outside of the domain.\" Now another interesting thing to think about, okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here."
    },
    {
        "Q": "\nAt 4:35, Is zero a \"non-negative\" number? Is it also non-positive?",
        "A": "Yes, 0 is neither positive nor negative. If you were to add 0 to any number, let s say x, it would not increase nor decrease its value.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            275
        ],
        "3min_transcript": "What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful. in our domain, x cannot be equal to zero. If x is equal to zero, we get zero over zero, we get indeterminate form. So in order for this function to be the exact same function, we have to put that, 'cause it's not obvious now from the definition, we have to say, \"x cannot be equal to zero.\" So g(x) is equal to x for any x as long as x is not equal to zero. Now these two function definitions are equivalent. And we could even graph it. We could graph it, it's going to look, I'm gonna do a quick and dirty version of this graph. It's gonna look something like, this. It's gonna have a slope of one, but it's gonna have a hole right at zero, 'cause it's not defined at zero. So it's gonna look like this. So the domain here, the domain of g is going to be, \"x is a member of the real numbers\" \"such that x does not equal zero,\" and the range is actually going to be the same thing."
    },
    {
        "Q": "\nThe Formula Sal derives for m at 13:00min is not the same as the formula he derived in a previous video for m(Proof part 4 @ 3:00min) They seem to be negatives of each other.. Am I missing something?",
        "A": "What would happen if you multiplied both the numerator and denominator by -1?",
        "video_name": "ualmyZiPs9w",
        "timestamps": [
            780,
            180
        ],
        "3min_transcript": "to what we do with regression. The expected value of X times Y, it can be approximated by the sample mean of the products of X and Y. This is going to be the sample mean of X and Y. You take each of your XY associations, take their product, and then take the mean of all of them. So that's going to be the product of X and Y. And then this thing right over here, the expected value of Y that can be approximated by the sample mean of Y, and the expected value of X can be approximated by the sample mean of X. So what can the covariance of two random variables be approximated by? What can it be approximated by? Well this right here is the mean of their product from your sample minus the mean of your sample Y's times the mean And this should start looking familiar. This should look a little bit familiar, because what is this? This was the numerator. This right here is the numerator when we were trying to figure out the slope of the regression line. So when we tried to figure out the slope of the regression line, we had the-- let me just rewrite the formula here just to remind you-- it was literally the mean of the products of each of our data points, or the XY's, minus the mean of Y's times the mean of the X's. All of that over the mean the X squareds. And you could even view it as this, over the mean of the X times the X's. But I could just write the X squareds, over here, minus the mean of X squared. This is how we figured out the slope of our regression line. if we assume in our regression line that the points that we have were a sample from an entire universe of possible points, then you could say that we are approximating the slope of our aggression line. And you might see this little hat notation in a lot of books. I don't want you to be confused. They are saying that you're approximating the population's regression line from a sample of it. Now, this right here-- so everything we've learned right now-- this right here is the covariance, or this is an estimate of the covariance of X and Y. Now what is this over here? Well, I just said, you could rewrite this very easily as-- this bottom part right here-- you could write as the mean of X times X-- that's the same thing as X squared-- minus the mean of X times the mean of X, right? That's what the mean of X squared is. Well, what's this? Well, you could view this as the covariance of X with X."
    },
    {
        "Q": "\nWhere exactly did that x+3 go whenever it was originally at the top of the fraction at 2:09? Was is just replaced by A and B? If so how & why...?",
        "A": "The answer to your question starts at 7:20.",
        "video_name": "S-XKGBesRzk",
        "timestamps": [
            129
        ],
        "3min_transcript": "Let's see if we can learn a thing or two about partial fraction expansion, or sometimes it's called partial fraction decomposition. The whole idea is to take rational functions-- and a rational function is just a function or expression where it's one expression divided by another-- and to essentially expand them or decompose them into simpler parts. And the first thing you've got to do, before you can even start the actual partial fraction expansion process, is to make sure that the numerator has a lower degree than the denominator. In the situation, the problem, that I've drawn right here, I've written right here, that's not the case. The numerator has the same degree as the denominator. So the first step we want to do to simplify this and to get it to the point where the numerator has a lower degree than the denominator is to do a little bit of algebraic division. And I've done a video on this, but it never hurts to get a review here, so to do that, we divide the denominator into the squared minus 3x minus 40 into x squared minus 2x minus 37. So how many times? You look at the highest degree term, so x squared goes into x squared one time, one times this whole thing is x squared minus 3x minus 40, and now you want to subtract this from that to get the remainder. And see, if I'm subtracting, so I'm going to subtract, and then minus minus is a plus, a plus, and then you can add them. These cancel out. Minus 2x plus 3x, that's x. Minus 37 plus 40, that's plus 3. So this expression up here can be rewritten as-- let me scroll down a little bit-- as 1 plus x plus 3 over x squared This might seem like some type of magic thing I just did, but this is no different than what you did in the fourth or fifth grade, where you learned how to convert improper regular fractions into mixed numbers. Let me just do a little side example here. If I had 13 over 2, and I want to turn it into a mixed number, what you do-- you can probably do this in your head now-- but what you did is, you divide the denominator into the numerator, just like we did over here. 2 goes into 13. We see 2 goes into 13 six times, 6 times 2 is 12, you subtract that from that, you get a remainder of 1. 2 doesn't go into 1, so that's just the remainder. So if you wanted to rewrite this, it would be the number of times the denominator goes into the numerator, that's 6, plus the remainder over the denominator. Plus 6-- plus 1 over 2. And when you did it in elementary school, you would just write 6 1/2, but 6 1/2 is the same thing as 6 plus 1/2."
    },
    {
        "Q": "\nAt 1:20, is Sal taking the anti-derivative of the derivative of f(x)*g(x) ? So they just cancel out? Like if you had (sqrt(X))^2 and you are just left with X. Is that pretty much the same thing that is going on at that point?",
        "A": "Whether they cancel out depends on what f(x) and g(x) happen to be. Integration by parts is quite useful if you need to integrate a complicated function that would be exceedingly difficult to try to integrate altogether.",
        "video_name": "dh__n9FVKA0",
        "timestamps": [
            80
        ],
        "3min_transcript": "What we're going to do in this video is review the product rule that you probably learned a while ago. And from that, we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by parts. So let's say that I start with some function that can be expressed as the product f of x, can be expressed as a product of two other functions, f of x times g of x. Now let's take the derivative of this function, let's apply the derivative operator right over here. And this, once again, just a review of the product rule. It's going to be the derivative of the first function times the second function. So it's going to be f-- no, I'm going to do that blue color-- it's going to be f-- that's not blue-- it's going to be f prime of x times g of x times-- that's not the same color-- times g of x plus the first function times the derivative of the second, of the second. This is all a review right over here. The derivative of the first times the second function plus the first function times the derivative of the second function. Now, let's take the antiderivative of both sides of this equation. Well if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to-- well what's the antiderivative of this? This is going to be the antiderivative of f prime of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now, what I want to do is I'm going And to solve for that, I just have to subtract this business. I just have to subtract this business from both sides. And then if I subtract that from both sides, I'm left with f of x times g of x minus this, minus the antiderivative of f prime of x g of x-- let me do that in a pink color-- g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go."
    },
    {
        "Q": "Isnt integral of f'(x) is f(x) just like at 2:42\n",
        "A": "yes it is\u00e2\u0080\u00a6as d/dx of f(x) Is f (x)",
        "video_name": "dh__n9FVKA0",
        "timestamps": [
            162
        ],
        "3min_transcript": "of the second. This is all a review right over here. The derivative of the first times the second function plus the first function times the derivative of the second function. Now, let's take the antiderivative of both sides of this equation. Well if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to-- well what's the antiderivative of this? This is going to be the antiderivative of f prime of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now, what I want to do is I'm going And to solve for that, I just have to subtract this business. I just have to subtract this business from both sides. And then if I subtract that from both sides, I'm left with f of x times g of x minus this, minus the antiderivative of f prime of x g of x-- let me do that in a pink color-- g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go. So let me copy and paste it. So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book. So this is essentially the formula for integration I will square it off. You'll often see it squared off in a traditional textbook. So I will do the same. So this right over here tells us that if we have an integral or an antiderivative of the form f of x times the derivative of some other function, we can apply this right over here. And you might say, well this doesn't seem that useful. First I have to identify a function that's like this. And then still I have an integral in it. But what we'll see in the next video is that this can actually simplify a whole bunch of things that you're trying to take the antiderivative of."
    },
    {
        "Q": "\nAt 0:09, Sal said that 35 doesn't goes into 6. Did he purposely choose a number with a remainder?",
        "A": "He likely did. He s trying to demonstrate decimals. Since decimals are fractions, there needs to be a remainder to create a decimal value.",
        "video_name": "xUDlKV8lJbM",
        "timestamps": [
            9
        ],
        "3min_transcript": "Let's take 63 and divide it by 35. So the first thing that we might say is, OK, well, 35 doesn't go into 6. It does go into 63. It goes into 63 one time, because 2 times 35 is 70, so that's too big. So it goes one time. So let me write that. 1 times 35 is 35. And then if we were to subtract and we can regroup up here, we can take a 10 from the 60, so it becomes a 50, give that 10 to the 3, so it becomes a 13. 13 minus 5 is 8. 5 minus 3 is 2. So you could just say, hey, 63 divided by 35-- You could say 63 divided by 35 is equal to 1 remainder 28. But this isn't so satisfying. We know that the real answer is going to be one point something, something, something. So what I want to do is keep dividing. I want to divide this thing completely and see what type And to do that, I essentially have to add a decimal here and then just keep bringing down decimal places to the right of the decimal. So 63 is the exact same thing as 63.0, and I could add as many zeroes as I might want to add here. So what we could do is we just make sure that this decimals right over there, and we can now bring down a zero from the tenths place right over here. And you bring down that zero, and now we ask ourselves, how many times does 35 go into 280? And, as always, this is a bit of an art when you're dividing a two-digit number into a three-digit number. So let's see, it's definitely going to be-- if I were to say-- so 40 goes into 280 seven times. 30 goes into 280 about nine times. It's going to be between 7 and 9, so let's try 8. So, let's see what 35 times 8 is. 35 times 8. 5 times 8 is 40, 3 times 8 is 24, plus 4 is 28. So 35 goes into 280 exactly eight times. 8 times 5, we already figured it out. 8 times 35 is exactly 280, and we don't have any remainder now, so we don't have to bring down any more of these zeroes. So now we know exactly that 63 divided by 35 is equal to exactly 1.8."
    },
    {
        "Q": "02:29 Why did Sal use K instead of Z?\n",
        "A": "I quickly checked the timestamp out. I assume he did so as you can use any letter that isn t already used in this situation. No special reason.",
        "video_name": "lHdlHTsXbZg",
        "timestamps": [
            149
        ],
        "3min_transcript": "So, once again let's think of yourself as some type of ancient philosopher/mathematician, who is trying to extend mathematics as much as possible and try to make sure that you're not being lazy and leaving things undefined, when you might be able to define them. Whenever you start extending mathematics, especially in the realm of multiplication and division, there are few things that you hold dear to. You feel that if you define some type of division operation, that needs to be undone by multiplication; this is close to your heart. So you assume... You want to assume... You would like to assume that any type of division operation, if you start with some number and if you divide with a number over which... - division by that number is defined - so when I divide by some number and then multiply by that same number that this should get me this original number right over here, this should give me x right over here. And this happens when we just multiply and divide with regular numbers. If I get 3 divided by 2 times 2, that's gonna get me 3. If I say that's going to get me 10. The other things that I want to assume... - and this is very close to my heart - I feel that any type of definitions I make have to be constant with the idea x*0 has to be 0 or any x. So these are close to my heart. I wanna extend mathematics. These two things are things that cannot be contradicted, cannot be untrue. Now, that out of the way. You wanna start exploring the divide-by-0 question. So the first thing that you say: \"Well, let me just try to define it.\" So let's start, let's assume that I have, so this is... So let's make a further assumption... that x is some non-zero number. x divided by 0 should be, how I should divide it, let's just assume there is define, and then come up with any results that there might be, there might be a resolve for. So let's say that x divided by 0 is equal to k. Well, if this is true and if we are defining what it means to divide by zero, then we are assuming that if we multiply by zero, we'll get our original number right over here. This is something that we are not willing to contradict. So let's see what happens: x divided by 0 is equal to k. On the left hand side we have a divide by zero and than multiplied by zero. Well then if two things are equal, if I do something to one thing inorder for them to stay equal, I have to do it to the other thing. This has to be equal to that. I have to multiply the left hand AND the right hand side by zero."
    },
    {
        "Q": "\nAt 1:59 Sal says x can not be equal to zero. Why does x have to be a non-zero number? If it could be zero, couldn't 0/0 = 0? Is there any reasoning as to why x can not equal zero, or is it just to enforce other reasoning as to why 0/0 is impossible?",
        "A": "At that point it s just the boundaries of the current proof he was laying out. Later on he does another proof where 0/0=0, however that comes from multiplying the constant on the right side of the equation by 0. That constant can be any number between infinity and negative infinity, so it is indeterminate.",
        "video_name": "lHdlHTsXbZg",
        "timestamps": [
            119
        ],
        "3min_transcript": "So, once again let's think of yourself as some type of ancient philosopher/mathematician, who is trying to extend mathematics as much as possible and try to make sure that you're not being lazy and leaving things undefined, when you might be able to define them. Whenever you start extending mathematics, especially in the realm of multiplication and division, there are few things that you hold dear to. You feel that if you define some type of division operation, that needs to be undone by multiplication; this is close to your heart. So you assume... You want to assume... You would like to assume that any type of division operation, if you start with some number and if you divide with a number over which... - division by that number is defined - so when I divide by some number and then multiply by that same number that this should get me this original number right over here, this should give me x right over here. And this happens when we just multiply and divide with regular numbers. If I get 3 divided by 2 times 2, that's gonna get me 3. If I say that's going to get me 10. The other things that I want to assume... - and this is very close to my heart - I feel that any type of definitions I make have to be constant with the idea x*0 has to be 0 or any x. So these are close to my heart. I wanna extend mathematics. These two things are things that cannot be contradicted, cannot be untrue. Now, that out of the way. You wanna start exploring the divide-by-0 question. So the first thing that you say: \"Well, let me just try to define it.\" So let's start, let's assume that I have, so this is... So let's make a further assumption... that x is some non-zero number. x divided by 0 should be, how I should divide it, let's just assume there is define, and then come up with any results that there might be, there might be a resolve for. So let's say that x divided by 0 is equal to k. Well, if this is true and if we are defining what it means to divide by zero, then we are assuming that if we multiply by zero, we'll get our original number right over here. This is something that we are not willing to contradict. So let's see what happens: x divided by 0 is equal to k. On the left hand side we have a divide by zero and than multiplied by zero. Well then if two things are equal, if I do something to one thing inorder for them to stay equal, I have to do it to the other thing. This has to be equal to that. I have to multiply the left hand AND the right hand side by zero."
    },
    {
        "Q": "At @6:19 he says that he wanted to come up with a number for k, but k is already defined by 0/0 so shouldn't there be no problem since he already defined k?\n",
        "A": "It s because he wants to come up with a single number for k, but he can t seem to find one. So he still technically calls it indeterminate.",
        "video_name": "lHdlHTsXbZg",
        "timestamps": [
            379
        ],
        "3min_transcript": "Now that out of the way... OK... This was a situation when x does not equal zero. But what about when x DOES equal zero. So let's think about that a little bit. And once again, I will try to define it. So I will assume... that 0 divided by 0 is equal to some number. Well once again, so let's say it is equal to k again. And so, once again... we are trying to do the same logic, so we'll write 0/0 is equal to k. Actually, let me colourcode these zeros. This will be a magenta zero and this is a blue zero right over here. And once again, I am not willing to give up the idea that if I start with a number x, I divide it by something over which division is defined, and then I multiply by that something, I should get my original x again. for the division. So what I am gonna do - I am gonna multiply the left-hand side times 0 and by this property that I am not willing to give up, the left-hand side should simplify to this magenta zero. It should simplify to this over here. But once again, anything I do to once side of the equation, inorder for the equation to hold true, I need to the other side of the equation. And these two were equal beforehand. Any operation I do to this inorder for it to still be equal , I need to do to that. So let me multiply the right-hand side by zero. So the left I get 0, I just get this magenta 0, and on the right I could just write the zero here, but I won't simplify it. I get k times 0. Well, this I see right over here... This actually is not a contradiction. This actually is true for any k, This is one of the core assumptions that I've made in my mathematics True for any k. It's not a contradiction. But the problem here is I wanna come up with the k, I'd like a resolve for a k. It would not be nice if this turned out to be 0. if this turned out to be one, or if this turned out to be negative one But now I see, given the assumptions right here this could be ANY... this could be absolutely any k I cannot determine what k this should be This could be a hundred thousand, this could be 75, it could be anything true for any k I cannot determine what k this should be and that's why when you get a little bit more nuance in early math people will say, well 0 divided by 0, well we don't know what that's gonna be there's no consistent answer there so we're just going to call it undefined"
    },
    {
        "Q": "\nAt 5:05, why isn't the antiderivative of dw just w? The dw completely disappears, which confuses me.",
        "A": "The indefinite integral of only dw is just w (+C, of course). However, he didn t just take the indefinite integral of dw. He took the indefinite integral of (e^w)dw. The derivative of e^w+C is e^w, so the indefinite integral of (e^w)dw is e^w+C. Because there s a -1/5 in front of the integral, we must multiply the whole expression by -1/5. This results in -1/5(e^w)+C (-1/5 times a constant is still a constant). I hope this explains what Sal did!",
        "video_name": "ShpI3gPgLBA",
        "timestamps": [
            305
        ],
        "3min_transcript": "So let's try to get it in terms of the form of e to the something and not e to the negative something. So let's set. And I'm running out of colors here. Let's set w as equal to negative u. In that case, then dw, derivative of w with respect to u, is negative 1. Or if we were to write that statement in differential form, dw is equal to du times negative 1 is negative du. So this right over here would be our w. And do we have a dw here? Well, we have just a du. We don't have a negative du there. But we can create a negative du by multiplying this inside by a negative 1, but then also by multiplying the outside by a negative 1. Negative 1 times negative 1 is positive 1. We haven't changed the value. We have to do both of these in order for it to make sense. Or I could do it like this. So negative 1 over here and a negative 1 right over there. that's the same thing as negative du. This is this right over here. And so we can rewrite our integral. It's going to be equal to now it's going to be negative 1/5. Trying to use the colors as best as I can. Negative 1/5 times the indefinite integral of e to the-- well, instead of negative u, we can write w-- e to the w. And instead of du times negative 1, or negative du, we can write dw. Now this simplifies things a good bit. We know what the antiderivative of this is in terms of w. This is going to be equal to negative 1/5 e to the w. so I'll just do a plus c. And now we just have to do all of our unsubstituting. So we know that w is equal to negative u. So we could write that. So this is equal to negative 1/5. I want to stay true to my colors. Negative 1/5 e to the negative u, that's what w is equal to, plus c. But we're still not done unsubstituting. We know that u is equal to sine of 5x. u is equal to sine of 5x, so we can write this as being equal to negative 1/5 times e to the negative u, which is negative u is sine of 5x."
    },
    {
        "Q": "@ ~2:20-2:33, we have (1/5)\u00e2\u0088\u00ab(1/e^u)du,\n\nwhy can't we use w substitution on the e^u?\n",
        "A": "I tried my, and got a different answer, or maybe just another form of the answer? Thanks a bunch for your help :)",
        "video_name": "ShpI3gPgLBA",
        "timestamps": [
            140,
            153
        ],
        "3min_transcript": "Let's see if we can take the integral of cosine of 5x over e to the sine of 5x dx. And there's a crow squawking outside of my window, so I'll try to stay focused. So let's think about whether u-substitution might be appropriate. Your first temptation might have said, hey, maybe we let u equal sine of 5x. And if u is equal to sine of 5x, we have something that's pretty close to du up here. Let's verify that. So du could be equal to-- so du dx, derivative of u with respect to x. Well, we just use the chain rule. Derivative of 5x is 5. Times the derivative of sine of 5x with respect to 5x, that's just going to be cosine of 5x. If we want to write this into differential form, which is useful when we do our u-substitution, we could say that du is equal to 5 cosine 5x. And we look over here, we don't have quite du there. We have just cosine of 5x dx. So when you look over here, you have a cosine of 5x dx, but we don't have a 5 cosine of 5x dx. But we know how to solve that. We can multiply by 5 and divide by 5. 1/5 times 5 is just going to be 1, so we haven't changed the value of the expression. When we do it this way, we see pretty clearly we have our u and we have our du. Our du is 5. Let me circle that. Let me do that in that blue color-- is 5 cosine of 5x dx. So we can rewrite this entire expression as-- do that 1/5 in purple. This is going to be equal to 1/5-- I hope you don't hear that crow outside. He's getting quite obnoxious. 1/5 times the integral of all the stuff in blue is my du, So how do we take the antiderivative of this? Well, you might be tempted to well, what would you do here? Well, we're still not quite ready to simply take the antiderivative here. If I were to rewrite this, I could rewrite this as this is equal to 1/5 times the integral of e to the negative u du. And so what might jump out at you is maybe we do another substitution. We've already used the letter u, so maybe we'll use w. We'll do some w-substitution. And you might be able to do this in your head, but we'll do w-substitution just to make it a little bit clearer. This would have been really useful if this was just e to the u because we know the antiderivative of e to the u"
    },
    {
        "Q": "6:01 What is the principle root?\n",
        "A": "When you take a square root, you have two branches; a positive branch, and a negative branch. For instance, x\u00c2\u00b2 = 9 implies that x = \u00c2\u00b1 3 because (-3)\u00c2\u00b2 = 9 and 3\u00c2\u00b2 = 9 also. The principle square root is the positive branch.",
        "video_name": "q7eF5Ci944U",
        "timestamps": [
            361
        ],
        "3min_transcript": "We already know that OA is equal to OC. And we also know that OM is equal to itself. OM is clearly equal to OM. It's equal to itself. And we also know from the Pythagorean theorem that AM-- do this in a new color. We know that AM squared plus OM squared is equal to OA squared. The length of the two legs squared summed up is equal to the length of the hypotenuse squared. So we know that for the left triangle right over here, for AMO. And we can set up the same relationship for CMO. We know in CMO that-- and I'm going to try to do it corresponding-- that CM squared plus OM Now we know a few things. We know that OA is equal to OC. So, for example, right over here, where we have OA squared, we could replace this with OC. And then you can already see where this is going. You can already see that CM and AM are going to be the same. But if you want to do it a little bit more formally, you can subtract OM squared from both sides of this equation, and you'd get AM squared is equal to OC squared-- I've replaced this with OC squared-- minus OM squared. So that's on the left-hand side right over here. And then on the right-hand side, if we subtract OM squared from both sides, we have CM squared. CM squared is equal to OC squared minus OM squared. And then we could take the principal root care about the positive root because we don't want to have negative distances. So if you take the principal root of both sides, this becomes AM is equal to the principal root of that. And we also get that CM is equal to the principal root of that. Well, these two quantities are the same. So AM must be equal to CM. They both equal to this quantity right here. So AM is equal to CM, and it might be a bisector. And this is really common sense. If you know that two sides of two different right triangles are going to be congruent to each other, you can always use the Pythagorean theorem to get the third side. And that third side is uniquely constrained by the length of the other two sides, because it's a right triangle. And so these are all ways of really getting at the same thing. But now we can feel pretty good about the fact that if OD is perpendicular here, then it's definitely going to bisect this chord."
    },
    {
        "Q": "At 1:15 did anyone notice that the two smaller right triangles make a bigger right triangle\n",
        "A": "Doesn t work as a proof.",
        "video_name": "q7eF5Ci944U",
        "timestamps": [
            75
        ],
        "3min_transcript": "In a previous video, we've already shown that if we have some circle centered at O right over here, and that if OD is a radius, and it's a radius that bisects chord AC-- so bisects means that it kind of splits it in two, that AB is equal to BC-- we've proven in a previous video that OD will be perpendicular to AC. So we've proven that we can assume that is perpendicular. And that video, if you want to look it up, and you might want to prove it yourself just out of interest, but the video-- if you do a search on Khan Academy for radius is perpendicular to a chord, you'll hopefully find the proof of that. What I want to do in this video is go the other way. If we know that OD is a radius and that it is perpendicular to chord AC, what I want to do in this video is prove that it's bisecting it. So we're not assuming that it's bisecting it in this video right over here. We're just assuming that it's perpendicular. So we're essentially going to go the other way. Here, we started with the fact that it bisected it, and we established that they were perpendicular. That was in the previous video. Now we're going to start with the assumption that they're perpendicular and then prove that they bisect. we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO"
    },
    {
        "Q": "\nAt 3:00, couldnt you also say that triangle AMO is congruent to triangle CMO by aas",
        "A": "If you did that, you d either need to prove that angle A is congruent to angle C or that angle AOM is congruent to angle COM. I don t see any obvious reason to believe that either of those is true, or at least nothing that is as obvious as proving that AO is congruent to CO or that MO is congruent to itself.",
        "video_name": "q7eF5Ci944U",
        "timestamps": [
            180
        ],
        "3min_transcript": "we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO And so if we know that they're congruent, then their corresponding sides have to be congruent. So based on that, we then know that AM is a corresponding side. Let me do that in a different color. AM is a corresponding side to MC. So we know that AM must be equal to MC because they're corresponding sides. These are corresponding sides. So congruency implies that these are equal. And if those are equal, then we know that OD is bisecting AC. So we've established what we need to do. Another way that we could have proven it without RSH, is just straight up with the Pythagorean theorem. We already know, just by setting up these two radii right over here, we know that OA-- so we draw a little line here."
    },
    {
        "Q": "\nAt 5:00 and on, I still don't understand how you can get a pentagon by slicing it through once.",
        "A": "And you would need to have a blade that is bigger than the sides of the cube so there would be parts where the blade can exit the cube before you finish cutting.",
        "video_name": "aSokFEpoJFM",
        "timestamps": [
            300
        ],
        "3min_transcript": "but I think you get the idea - this would be a triangle. There's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as that upper length, or the length that intersects on this space of the cube, that's gonna be an equilateral triangle. If you pushed this point up more, actually I'd do that in a different color, you were going to have an isosceles triangle. If you were to bring this point really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle: you'd still have these angles who'd still be less than 90\u00ba, you can approach 90\u00ba. So you can't quite have an exactly a right angle, and so since you can't get to 90\u00ba, sure enough you can't get near to 91\u00ba, so actually you're not gonna be able to do But you can do an equilateral, you can do an isosceles, you can do scalene triangles. I guess you could say you could do the different types of acute triangles. But now let's do some really interesting things: Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's such a fun thing. Think about it: How can you get a pentagon by slicing a cube with a plane? All right, so here I go, this is how you can get a pentagon by slicing a cube with a plane: Imagine slicing the top - we'll do it a little bit different - so imagine slicing the top, right over there, like this, Imagine slicing this backside, like that, this back side that you can see, quite like that, now you slice this side, right over here, like this, This could be, and alike I'm gonna draw the plane - yet maybe it won't be so obvious if I try to draw the plane - but you get the actual idea: if I slice this, the right angle (not any right angle - 90\u00ba - but 'the right angle' - the proper angle. Actually I shouldn't use 'right angle', that would confuse everything.) If I slice it in the proper angle, that I'm doing with the intersection of my plane then my cube is going to be this pentagon, right over here. Now let's up the stakes something, let's up the stakes even more! What about a hexagon? Can I slice a cube in a way, with a 2D plane, to get the intersection of the plane on the cube being a hexagon? As you could imagine, I wouldn't have asked you that question unless I could. So let's see if we can do it. So if we slice this, right over there, if we slice this bottom piece,"
    },
    {
        "Q": "at 5:35 how do you make that pentigon in one slice\n",
        "A": "He uses different cuts to obtain the shape of a pentagon, the sides can be expanded to show a plane cutting into the cube. think of the solid lines as lines you can see, and think of the dotted lines as you cannot see, the front two solid lines on the pentagon you can see are lining up with the faces of the cube and the plane is tilted at an angle",
        "video_name": "aSokFEpoJFM",
        "timestamps": [
            335
        ],
        "3min_transcript": "But you can do an equilateral, you can do an isosceles, you can do scalene triangles. I guess you could say you could do the different types of acute triangles. But now let's do some really interesting things: Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's such a fun thing. Think about it: How can you get a pentagon by slicing a cube with a plane? All right, so here I go, this is how you can get a pentagon by slicing a cube with a plane: Imagine slicing the top - we'll do it a little bit different - so imagine slicing the top, right over there, like this, Imagine slicing this backside, like that, this back side that you can see, quite like that, now you slice this side, right over here, like this, This could be, and alike I'm gonna draw the plane - yet maybe it won't be so obvious if I try to draw the plane - but you get the actual idea: if I slice this, the right angle (not any right angle - 90\u00ba - but 'the right angle' - the proper angle. Actually I shouldn't use 'right angle', that would confuse everything.) If I slice it in the proper angle, that I'm doing with the intersection of my plane then my cube is going to be this pentagon, right over here. Now let's up the stakes something, let's up the stakes even more! What about a hexagon? Can I slice a cube in a way, with a 2D plane, to get the intersection of the plane on the cube being a hexagon? As you could imagine, I wouldn't have asked you that question unless I could. So let's see if we can do it. So if we slice this, right over there, if we slice this bottom piece, like that, and then you slice the side that we can see right over there, (This side, I could have made it much straighter) So hopefully you get the idea! I can slice this cube so that I can actually get a hexagon. So, hopefully, this gives you a better appreciation for what you could actually do with a cube, especially if you're busy slicing it with large planar planes - or large planar blades, in some way - There's actually more to a cube that you might have imagined in the past. When we think about it, there's six sides to a cube, and so it's six surfaces to a cube, so you can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces"
    },
    {
        "Q": "\nSal said at 2:34 that the range is limited to the first and fourth Quadrants. Why wouldn't 2.80 radians be excluded from this set?\nPerhaps he misspoke, because all positive values for y, (sin), would actually be in the first or second quadrants, correct?",
        "A": "Sal didn t misspeak. If 2.80 were in the range of arcsine, then the arcsine function would be multivalued. If we allowed values in the first and second quadrants, there would be two values that every input could map to; one in the first quadrant and one in the second.",
        "video_name": "NC7iWEQ9Kug",
        "timestamps": [
            154
        ],
        "3min_transcript": "if we do this in a color you can see, along the positive X axis. Then the other side, so let's see, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the Y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what X values, and we're assuming we're dealing in radians. What X values when if I take the sine of it are going to give me 1/3? When does Y equal 1/3 along the unit circle? That's 2/3, 1/3 right over here. We see it equals 1/3 exactly two places, here and here. There's two angles where, or at least two, if we just take one or two on each pass of the unit circle. Then we can keep adding multiples of two pi to get as many as we want. We see just on the unit circle we could have this angle. Or we could go all the way around to that angle right over there. Then we could add any multiple of two pi to those angles to get other angles that would also work where if I took the sine of them I would get 1/3. Now let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1/3. Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant if we're thinking about the unit circle right over here. We see that gave us zero point, if we round to the nearest hundredth, 34. Essentially they've given us this value. They've given us 0.34. That's this angle right over here. Well, it's a positive value. It's greater than zero, but it's less than pi over two. Pi is 3.14, so pi over two is going to be 1.57 and we can go on and on and on so this right over here is 0.34 radians. But what would this thing over here be? It's going to be whatever, if we go to the negative X axis and we subtract 0.34 so we subtracted 0.34. This is 0.34 We're going to get to this angle. It's going to be if we take pi minus the previous answer, it gets us we round to the nearest hundredth, is 2.8 radians. This is 0.34 radians, and then this one, let me do it in this purple color, this one right over here if we were to go all the way around, it's pi minus 0.34 which is 2.80 radians rounding to the nearest hundredth. Now that's not all of the values. We can add multiples of two pi to each of these."
    },
    {
        "Q": "\nAt 0:42, when Mr. Khan says \"nesting\", is he trying to say a function within a function? This looks almost like combining, but I know it really isn't. So what exactly is composing, and what is the main difference between composing and combining?",
        "A": "Composing function- applying one function to the results of another...In other words he s replacing x as the results of another function (eg. replace f(x) with f(g(2)), which is also equal to f(-3)",
        "video_name": "wUNWjd4bMmw",
        "timestamps": [
            42
        ],
        "3min_transcript": "Voiceover:So we have three different function definitions here. This is F of X in blue, here we map between different values of T and what G of T would be. So you could use this as a definition of G of T. And here we map from X to H of X. So for example, when X is equal to three, H of X is equal to zero. When X is equal to one, H of X is equal to two. And actually let me number this one, two, three, just like that. Now what I want to do in this video is introduce you to the idea of composing functions. Now what does it mean to compose functions? Well that means to build up a function by composing one function of other functions or I guess you could think of nesting them. What do I mean by that? Well, let's think about what it means to evaluate F of, not X, but we're going to evaluate F of, actually let's just start with a little warm-up. Let's evaluate F of G of two. and I encourage you to pause this video and think about it on your own. Well it seems kind of daunting at first, if you're not very familiar with the notation, but we just have to remember what a function is. A function is just a mapping from one set of numbers to another. So for example, when we're saying G of two, that means take the number two, input it into the function G and then you're going to get an output which we are going to call G of two. Now we're going to use that output, G of two, and then input it into the function F. So we're going to input it into the function F, and what we're going to get is F of the thing that we inputted, F of G of two. So let's just take it step by step. What is G of two? Well when T is equal to two, G of two is negative three. Well, I'm going to get negative three squared minus one, which is nine minus one which is going to be equal to eight. So this right over here is equal to eight. F of G of two is going to be equal to eight. Now, what would, using this same exact logic, what would F of H of two be? And once again, I encourage you to pause the video and think about it on your own. Well let's think about it this way, instead of doing it using this little diagram, here everywhere you see the input is X, whatever the input is"
    },
    {
        "Q": "\nAt 4:39\nlog\u00e2\u0082\u0083(27x) as log\u00e2\u0082\u0083(27)+log_3(x), which is simplified as 3+log\u00e2\u0082\u0083(x)\nI mean why is that? cause log\u00e2\u0082\u0083(27) suppose to be log\u00e2\u0082\u0083(3^3), then =3 log\u00e2\u0082\u00833",
        "A": "Your work is ok so far, but is incomplere as 3 log\u00e2\u0082\u00833 can be simplified: `3 log\u00e2\u0082\u00833 = 3*1 = 3. Remember, 3^1 = 3. So, log\u00e2\u0082\u00833 = 1. Hope this helps.",
        "video_name": "pkGrXzakRFs",
        "timestamps": [
            279
        ],
        "3min_transcript": "I'm writing it as an exponential function or exponential equation, instead of a logarithmic equation. So b to the zth power is equal to c. This is the same statement or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, and c is equal to bz, then we can write b to the x power is equal to b to the y power-- that's what a is. We know that already-- times b to the z power. And we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the-- I'll do it in a neutral color-- b to the y plus z power. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing, is that you know how to apply it. And then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents. I know when people first would tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base 3 of 27 times x-- I'll write it that way-- is equal to log base 3 of 27 plus log base 3 of x. This tells us, what power do I have to raise 3 to to get to 27? You could view it as this way. 3 to the ? is equal to 27. Well, 3 to the third power is equal to 27. 3 times 3 is 9 times 3 is 27. So this right over here evaluates to 3. So if we were to simplify-- or I guess I wouldn't even call it simplifying it. I would just call it expanding it out or using this property, because we now have two terms where we started off Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes 3. And then we're left with plus log base 3 of x. So this is just an alternate way of writing this original statement, log base 3 of 27x. So once again, not clear that this is simpler than this right over here. It's just another way of writing it using logarithm properties."
    },
    {
        "Q": "\nAt 4:11, how does one know that the arc that subtends 'theta' is equal to r?",
        "A": "Sal is just defining what a radian is. It is simply the measure of a central angle that subtends an arc equal in length to the radius.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            251
        ],
        "3min_transcript": "What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta. also the same length as the radius of the circle. So this arc is also of length r. So given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? Well, the most obvious one, if you kind of view a radian as another way of saying radiuses, or I guess radii. Well, you say, look, this is subtended by an arc of one radius. So why don't we call this right over here one radian, which is exactly how a radian is defined. When you have a circle, and you have an angle of one radian, the arc that subtends it is exactly one radius long. Which you can imagine might be a little bit useful as we start to interpret more and more types of circles."
    },
    {
        "Q": "\nAt 3:13, what does he mean by \"theta\"? I can't seem to understand what it means.",
        "A": "Theta is the eighth letter of the greek alphabet. Greek letters are commonly used as variables in higher math.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            193
        ],
        "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta."
    },
    {
        "Q": "\nDoes theta at 3:14 mean anything it stands out in a weird way",
        "A": "Not really, it s just used a lot in geometry and trigonometry to indicate angles for whatever reason. It s no different than if you d call it angle x or angle z.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            194
        ],
        "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta."
    },
    {
        "Q": "\nAt 01:14 sal asks what are the theories. What I guessed was I think that since our earth almost takes 360 days around the sun to complete one revolution thats why we said one complete revolution or rotation is 360 degress. Am iCorrect?",
        "A": "Woohoo I placed my guess and it was correct... It was just a guess!",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            74
        ],
        "3min_transcript": "You are by now probably used to the idea of measuring angles in degrees. We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle. And we'd often use this symbol, just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you have essentially done one full rotation. And if you watch figure skating on the Olympics, and someone does a rotation, they'll say, oh, they did a 360. Or especially in some skateboarding competitions, and things like that. But the one thing to realize-- and it might not be obvious right from the get-go-- but this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. And if you think about it, you'll And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already"
    },
    {
        "Q": "\nat 3:22 he calls the angle theta... what does that mean? does it represent a certain angle measure, position, etc?",
        "A": "That is a Greek letter named Theta. In mathematics and science disciplines, Theta is often used as the variable representing an angle measurement. You will see it very often used this way.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            202
        ],
        "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta."
    },
    {
        "Q": "\nAt 3:17 what does theta mean? Is it just a name for any angle, or does it have to be a specific angle to be \"theta\"?",
        "A": "Theta is just like an angle labeled as x. Mathematicians commonly use letters of the greek alphabet to label angles.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            197
        ],
        "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta."
    },
    {
        "Q": "\nSo, basically every degree amount will have the same amount of radians no matter the circle? So every circle is 2pi(r) radians? Is that what this is about? 7:00",
        "A": "Yes. Since radians are just a way of measuring angles, like degrees, they stay the same no matter the size of the circle. As a circle always has 360 degrees around the point, it always has 2pi radians.",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            420
        ],
        "3min_transcript": "have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly the arc length that is subtending the angles. So let's do a couple of thought experiments here. So given that, what would be the angle in radians if we were to go-- so let me draw another circle here. So that's the center, and we'll start right over there. So what would happen if I had an angle-- if I wanted to measure in radians, what angle would this be in radians? And you could almost think of it as radiuses. So what would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? The arc that subtends this angle is the entire circumference of this circle. Well, what's the circumference of a circle in terms of radiuses? So if this has length r, if the radius is length r, what's the circumference of the circle in terms of r? That's going to be 2 pi r. So going back to this angle, the length of the arc that subtends this angle is how many radiuses? Well, it's 2 pi radiuses. It's 2 pi times r. So this angle right over here, I'll call this a different-- well, let's call this angle x. x in this case is going to be 2 pi radians. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. So given that, let's start to think about how we can convert between radians and degrees, and vice versa. If I were to have-- and we can just follow up over here. If we do one full revolution-- that is, 2 pi radians-- how many degrees is this going to be equal to? Well, we already know this. A full revolution in degrees is 360 degrees. Well, I could either write out the word degrees, or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're kind of using units in both cases. Now, if we wanted to simplify this a little bit, we could divide both sides by 2. In which case, on the left-hand side,"
    },
    {
        "Q": "At 4:23, what does Sal mean when he says that the ratio can be represented as a function of n?\n",
        "A": "What Sal means is that your attention should be drawn to the fact that the ratio ((n+1) term over (n) term) dos not simplify to a constant ratio because it has a variable n in it. What it simplifies to is a ratio which is itself a function of the terms in question (the a sub n in question). Hope this helps!",
        "video_name": "av947KCWf2U",
        "timestamps": [
            263
        ],
        "3min_transcript": "we want to figure out whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the N plus oneth term, which is going to be N plus one to the tenth power over N plus one factorial and divide that by the Nth term, Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do is exactly right over here. See if there is some type of a common ratio. Well, if we do a little algebra here N plus one factorial, and factorial algebra is always fun. This is the same thing as N plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies to this N right over here, and why is that useful? Because this N factorial cancels with this N factorial, and we're left with N plus one to the tenth power over N plus one times N to the tenth. So I know what you're thinking. \"Hey, wait, look, this isn't a fixed common ratio here.\" The ratio between consecutive terms here, when I took the N plus oneth term divided by the Nth term, it's changing depending on what my N is. This ratio, I guess you could say, is a function of N so this doesn't seem too useful, but what if I were to say, \"O.K., well look, with any of these series \"we really care about the behavior as our N's \"get really, really, really large as the limit, as our N's, go to infinity.\" So what if we were to look at the behavior here, and if this is approaching some actual values as N approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. Let's take the limit as N approaches infinity of this thing. So remember what we're doing here."
    },
    {
        "Q": "\nYou can't prove that n^10/n! doesn't diverge with the divergence test as stated in 2:40 because lim an = 0 so it can still diverge ou converge. Or am I wrong?",
        "A": "Absolutely correct, a series such as 1/n (harmonic series) diverges, even though the limit as n goes to infinity leads to 0 for an, so Sal made a mistake.",
        "video_name": "av947KCWf2U",
        "timestamps": [
            160
        ],
        "3min_transcript": "If it is not review, I encourage you to watch the videos on geometric series. And what's interesting about this is we've proven to ourselves, in the videos about geometric series, that if the common ratio, if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of R is greater than or equal to one, then the series diverges. And that makes sense, we've proven it as well, but it also makes logical sense that, look, if that absolute value of R is less than one then each term here is going to go down by that common, or it's going to be multiplied by that common ratio, and it's going to decrease on and on and on and on until it makes sense that, even though this is an infinite sum, it will converge to a finite value. Now, with that out of the way for review, let's tackle something a little bit more interesting. we want to figure out whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the N plus oneth term, which is going to be N plus one to the tenth power over N plus one factorial and divide that by the Nth term, Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do is exactly right over here. See if there is some type of a common ratio. Well, if we do a little algebra here N plus one factorial, and factorial algebra is always fun. This is the same thing as N plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies to this N right over here, and why is that useful? Because this N factorial cancels with this N factorial, and we're left with N plus one to the tenth power over N plus one times N to the tenth."
    },
    {
        "Q": "At 2:20 what did Sal mean by you can rewrite this.\n",
        "A": "What he actually means in 2:20 is that -15.08+526.90 is the same thing as saying 526.90-15.08. If you are careful with the numbers, you can rearrange the numbers in an equation for simplification!",
        "video_name": "fFdOr8U4mnI",
        "timestamps": [
            140
        ],
        "3min_transcript": "At the beginning of the week, Stewart's checking account had a balance of negative $15.08. On Monday morning, he deposited a check for $426.90. On Tuesday morning, he deposited another check for $100. How much was in Stewart's checking account after the second deposit, so after both of these deposits right over here. So he starts off with a negative balance. So a negative balance means that he's overdrawn his checking account. He actually owes the bank money now. Luckily, he's now going to put some money in his bank account. So he'll actually have a positive balance in his checking account. So he's starts off with the negative $15.08. And then to that, he adds $426.90. And then he adds another $100. So he started off with negative $15.08. And then to that, he adds $426.90 and $100. And so how much is going to have in his bank account? He started owing $15.08, and then he's going to add $526.90. So one way to visualize it is, if you think about it on a number line, if this is 0 right over here, he's going to start off at negative $15.08. But then he's going to add $526. So this right over here, this is $15.08 to the left. That's how much he owes. And to that, he's going to add $526. So I'm not drawing this to scale. But to that, he is going to add $526.90. So the amount that he's going to be in the positive is going to be $526.90 minus the $15.08. That's how much he's going to be in the positive. And that's going to be $526.90 minus $15.08. So that's going to be, and we can even just rewrite this so it actually looks exactly like that. That's exactly the same thing as $526.90 minus-- adding a negative is the same thing as subtracting a positive-- minus $15.08. And this is-- I will do this in another color-- $526.90 minus $15.08. Let's see, 0 is less than 8. Let's make that a 10 and borrow from this 9. So that becomes an 8, or I guess you could say we're regrouping. Now, everything up here is larger than everything there. So 10 minus 8 is 2. 8 minus 0 is 8. We have our decimal. 6 minus 5 is 1."
    },
    {
        "Q": "0:07, Sal says that Stewart has a checking account with a balance of -$15. Does this mean that Stewart owes money?\n",
        "A": "Yes, he does.",
        "video_name": "fFdOr8U4mnI",
        "timestamps": [
            7
        ],
        "3min_transcript": "At the beginning of the week, Stewart's checking account had a balance of negative $15.08. On Monday morning, he deposited a check for $426.90. On Tuesday morning, he deposited another check for $100. How much was in Stewart's checking account after the second deposit, so after both of these deposits right over here. So he starts off with a negative balance. So a negative balance means that he's overdrawn his checking account. He actually owes the bank money now. Luckily, he's now going to put some money in his bank account. So he'll actually have a positive balance in his checking account. So he's starts off with the negative $15.08. And then to that, he adds $426.90. And then he adds another $100. So he started off with negative $15.08. And then to that, he adds $426.90 and $100. And so how much is going to have in his bank account? He started owing $15.08, and then he's going to add $526.90. So one way to visualize it is, if you think about it on a number line, if this is 0 right over here, he's going to start off at negative $15.08. But then he's going to add $526. So this right over here, this is $15.08 to the left. That's how much he owes. And to that, he's going to add $526. So I'm not drawing this to scale. But to that, he is going to add $526.90. So the amount that he's going to be in the positive is going to be $526.90 minus the $15.08. That's how much he's going to be in the positive. And that's going to be $526.90 minus $15.08. So that's going to be, and we can even just rewrite this so it actually looks exactly like that. That's exactly the same thing as $526.90 minus-- adding a negative is the same thing as subtracting a positive-- minus $15.08. And this is-- I will do this in another color-- $526.90 minus $15.08. Let's see, 0 is less than 8. Let's make that a 10 and borrow from this 9. So that becomes an 8, or I guess you could say we're regrouping. Now, everything up here is larger than everything there. So 10 minus 8 is 2. 8 minus 0 is 8. We have our decimal. 6 minus 5 is 1."
    },
    {
        "Q": "\nAt 3:38, I don't understand why you wouldn't just divide 9 into 2 and have the quotient of 0.22 there instead of \"borrowing\" the 10.\nI calculated 0.22*10^14 which would simplify to 2.2*10^13\nSo i would have gotten it wrong unless i \"borrowed\" the 10 during my calculation.",
        "A": "0.22 x 10^14 is numerically correct, but it isn t valid scientific notation. A number which is written in scientific notation must have the mantissa (the bit before the power of ten) greater than or equal to 1, and less than 10. Here, 0.22 is not greater than or equal to 1, so it fails that test. So although 0.22 x 10^14 = 2.2 x 10^13, only the latter is correct scientific notation.",
        "video_name": "0lOpqmTdtzk",
        "timestamps": [
            218
        ],
        "3min_transcript": "It's clearly not less than 10. But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10, or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times-- let's add these two exponents-- 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit. This is going to be the same thing as 2/9 times 1 to t over 10 to the negative 14. Well what's 1 over 10 to the negative 14? Well that's just 10 to the 14. Now you might say, OK, we just have to figure out what 2/9 is We've written this in scientific notation. But you might have already realized, look, 2/9 is not greater than or equal to 1. How can we make this greater than or equal to 1? Well we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that. So I'm going to multiply this by 10, and I'm going to divide this by 10. So I haven't changed. I've multiplied and divided by 10. So this is equal to 20/9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20/9? This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And I think they said round our answer to two decimal places. So 20 divided by 9-- 9 doesn't go into 2. It does go into 20 two times. 2 times 9 is 18. Subtract. Get a remainder of 2. I think you see where this show is going to go. 9 goes into 20 two times. 2 times 9 is 18. We're just going to keep getting 2's. So we get another 2, bring down a 0. Nine goes into 20 two times. So this thing right over here is really 2.2 repeating. But they said round to two decimal places, so this is going to be equal to 2.22 times 10 to the 13th power."
    },
    {
        "Q": "\nAt 1:28, Sal talks about tuples. This makes me think of \"quintuples,\" except the \"tu\" is pronounced differently. Is that where \"tuples\" comes from? Where does the term \"tuples\" come from?",
        "A": "From Wikipedia: The term originated as an abstraction of the sequence: single, double, triple, quadruple, quintuple, sextuple, septuple, octuple, ..., n\u00e2\u0080\u0091tuple, ..., where the prefixes are taken from the Latin names of the numerals. Although this treats \u00e2\u0080\u0091tuple as the suffix, the original suffix was \u00e2\u0080\u0091ple as in triple (three-fold) or decuple (ten\u00e2\u0080\u0091fold). This originates from a medieval Latin suffix \u00e2\u0080\u0091plus (meaning more ).",
        "video_name": "lCsjJbZHhHU",
        "timestamps": [
            88
        ],
        "3min_transcript": "When you get into higher mathematics, you might see a professor write something like this on a board, where it's just this R with this extra backbone right over here. And maybe they write R2. Or if you're looking at it in a book, it might just be a bolded capital R with a 2 superscript like this. And if you see this, they're referring to the two-dimensional real coordinate space, which sounds very fancy. But one way to think about it, it's really just the two-dimensional space that you're used to dealing with in your coordinate plane. To go a little bit more abstract, this isn't necessarily this visual representation. This visual representation is one way to think about this real coordinate space. If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space-- let me write this down-- and the two-dimensional real coordinate space. the 2 tells us how many dimensions we're dealing with, and then the R tells us this is a real coordinate space. The two-dimensional real coordinate space is all the possible real-valued 2-tuples. Let me write that down. This is all possible real-valued 2-tuples. So what is a 2-tuple? Well, a tuple is an ordered list of numbers. And since we're talking about real values, it's going to be ordered list of real numbers, and a 2-tuple just says it's an ordered list of 2 numbers. So this is an ordered list of 2 real-valued numbers. Well, that's exactly what we did here when we thought about a two-dimensional vector. This right over here is a 2-tuple, and this is a real-valued 2-tuple. Neither of these have any imaginary parts. So you have a 3 and a 4. Order matters. And even if we were trying to represent them in our axes right over here, this vector, 4, 3, would be 4 along the horizontal axis, and then 3 along the vertical axis. And so it would look something like this. And remember, we don't have to draw it just over there. We just care about its magnitude and direction. We could draw it right over here. This would also be 4, 3, the column vector, 4 3. So when we're talking about R2, we're talking about all of the possible real-valued 2-tuples. So all the possible vectors that you can have, where each of its components-- and the components are these numbers right over here-- where each of its components are a real number. So you might have 3, 4. You could have negative 3, negative 4. So that would be 1, 2, 3, 1, 2, 3, 4,"
    },
    {
        "Q": "\nAt 2:20 how come you do not get rid of the exponent after you square 2x^2? Is it because it still contains a variable? because after you squared (8)^2 = 64 ...no more exponent, but when you square (2x)^2, it turns into 4x^2?",
        "A": "In (2x)^2, both the 2 and the X have to be squared. We can calculate 2^2 = 4. But, we don t know the numeric value for X. So, we can t calculate X*X. We just write it in exponent form: x^2. (2x)^2 = (2x) (2x) = (2*2) (x*x) = 4x^2",
        "video_name": "h6HmHjkA034",
        "timestamps": [
            140
        ],
        "3min_transcript": "Find the product 2x plus 8 times 2x minus 8. So we're multiplying two binomials. So you could use FOIL, you could just straight up use the distributive property here. But the whole point of this problem, I'm guessing, is to see whether you recognize a pattern here. This is of the form a plus b times a minus b, where here a is 2x and b is 8. We have 2x plus 8 and then 2x minus 8. a plus b, a minus b. What I want to do is I'm just going to multiply this out for us. And then just see what happens. Whenever you have this pattern, what the product actually looks So if you were to multiply this out, we can distribute the a plus b. We could distribute this whole thing. Distribute the whole a plus b on the a and then distribute it on the b. And I could have done this with this problem right here, and it would have taken us less time to just solve it. But I want to find out the general pattern here. So a plus b times a. So we have a times a plus b, that's this times this. that's negative b times a plus b. So I've done distributive property once, now I could do it again. I can distribute the a onto the a and this b and it gives me a squared. a times a is a squared, plus a times b, which is ab. And now I can do it with the negative b. Negative b times a is negative ab or negative ba, same thing. And negative b times b is negative b squared. Now, what does this simplify to? Well, I have an ab, and then I'm subtracting an ab. So these two guys cancel out and I am just left with a squared minus b squared. So the general pattern, and this is a good one to just kind of know super fast, is that a plus b times a minus b is always going to be a squared minus b squared. So we have an a plus b times an a minus b. So it's going to be 2x squared minus b squared minus 8 squared. 2x squared, that's the same thing as 2 squared times x squared, or 4x squared. And from that, we're subtracting 8 squared. So it's going to be 4x squared minus 64."
    },
    {
        "Q": "at 1:27 Sal said milli ment 1,000 shouldn`t it mean 1,000,000\n",
        "A": "milli means one thousandth, micro means one millionth. :)",
        "video_name": "mI84WDfhuYA",
        "timestamps": [
            87
        ],
        "3min_transcript": "We're asked to arrange the following measurements in order from smallest to largest. And we have the measurements, and they're different units. This is in decameters, then we have meters, then we have millimeters, then we have centimeters. So the way I would tackle this is I would try to convert them all to the same units, maybe meters, and then compare them all in terms of meters. So let's do that. So this is just a screenshot of that exact same screen we just saw there. And so let's just convert each of these into meters. So the first one we have right over here is one decameter. So we have to remind ourselves what deca even means. Well, deca is equal to 10 meters. And actually, let me add more of these, just so that we know all the different prefixes we could have. So you have deca. You have hecto, which would be 100. You have kilo, which would be 1,000. And then, of course, you just have meters, That would just be equal 1 if you have no prefix in front of the meters. And then if you have a tenth of a meter, that is decimeter, so this is 1/10. Then you have centi, which is 1/100. And then you have milli, which is equal to 1/1,000. So let's use this information right over here to figure out how many meters each of these are. So one decameter-- we just saw deca means 10. That's 1 times 10 meters. So this is going to be equal to 10 meters. So this right over here is 10 meters. This is already written in meters. This is 13 meters. Then we have 15,000 millimeters, but milli means 1/1,000, So each millimeter is 1/1,000 of a meter. You could view this as instead of writing a milli here, I wrote 1/1,000. So 15,000 times 1/1,000, that's just going to give me 15. So this is going to be 15 meters. So this is 15 meters. Another way of thinking about it is, look, 1,000 millimeters is equal to a meter. So let's divide this into groups of 1,000. Well, this is literally 15 groups of 1,000. This is 15 groups of 1,000 millimeters, so that's going to be 15 meters. And then, finally, we have 1,900 centimeters. So 1,900, instead of writing centi, I'm going to write 1/100 of a meter."
    },
    {
        "Q": "At 11:40, can (0,0) on a graph be both a x-intercept and a y-intercept?\n",
        "A": "Yes. (0,0) is an x-intercept and a y-intercept. We call this point the origin.",
        "video_name": "_npwsLh0vws",
        "timestamps": [
            700
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n1:20 WHAT did it mean infantine set of eyes",
        "A": "There are infinite y values.",
        "video_name": "_npwsLh0vws",
        "timestamps": [
            80
        ],
        "3min_transcript": "In the following graph, is y a function of x? So in order for y to be a function of x, for any x that you input into the function, any x for which the function is defined. So let's say we have y is equal to f of x. So we have our little function machine. It should spit out exactly one value of y. If it spits out multiple values of y, we don't know what f of x is going to be equal to. It could be equal to any of those possible values for y. So let's see if, for this graph, whether for a given x it spits out exactly one y. Well, the function seems to be only defined so the domain of this function is x is equal to negative 2. That's the only place where we have a definition for it. And if we try to input negative 2 into this little black box, what do we get? Do we get exactly one thing? If we put in negative 2 here, we could get anything. The point negative 2, 9 is on this relation. Negative 2, 7; negative 2, 7.5; negative 2, 3.14159-- they're all on these. So if you put a negative 2 into this relation, essentially, you actually get an infinite set of values. It could be 9. It could be 3.14. It could be 8. It could be negative 8. You get an infinite number of results. So since it does not map to exactly one output of this function, in the following graph, y is not a function of x."
    },
    {
        "Q": "4:54 i think he showed wrong direction of resultant vector c it should be from tail of a to head of b which is shown wrong i think\n",
        "A": "You might want to listen more closely to what Sal was saying. In the second example he was working, the vector C he was using was NOT the sum of vectors A and B. Once he switched directions as you suggest to create the vector -C he then set up the equation A + B = -C.",
        "video_name": "BsBH8nAv5l4",
        "timestamps": [
            294
        ],
        "3min_transcript": "completely perfect. So vector a, just like that. And one way of thinking about subtracting vector b is instead of adding vector b the way we did here, we could add negative b. So negative b would have the same magnitude but just the opposite direction. So that's vector a. Vector negative b will still start right over here, but will go in the opposite direction. So let's do that. So negative b is going to look like this, is going to look something like this. So that is negative b. Notice, same magnitude exactly opposite direction. We've flipped it around 180 degrees, and now the resulting vector is going to be d. So vector d is going to look like that, vector d. So c is a plus b, d is a minus b. Or you can even call this a plus, a plus negative b. and go the other way. See if we could go from the diagrams to the actual equations. So let's start with, let me draw an interesting one. So let's say that this... Let's say that's vector a. Vector a. I'll use green. Let's say that, that is vector b, and I will now use magenta. And let's say that this is vector c. Vector c. So I encourage you to pause the video and see if you can write an equation that defines this relationship. Well, this is interesting because they're all going in a circle right over here. Let's say that you started at... This is your initial point. well, if you're trying to figure out what a plus b is going to be, the resulting vector would start here and end here. But vector c is going in the opposite direction. But we could, instead of thinking about vector c like this, we could think about the opposite of vector c, which would do, so instead of calling this c, I could flip this around by calling it this negative c. So I could flip this around, and now, let me do the same color, this would be equal to negative c. Notice, before I just had vector c here and it started at this point and ended at this point. Now I just flipped it around, it has the exact opposite direction, same magnitude. Now it is negative c. And this makes it easier for us to construct an equation because negative c starts at the tip, at the initial point or the tail of vector a, and it goes to the head of vector b or the terminal point of vector b. So we can now write an equation."
    },
    {
        "Q": "\nIf Vector A - Vector B =Vector D in 4:08 then Vector B - Vector A= -Vector D ??",
        "A": "Yes it is. Just , you are multiplying -1 to the whole equation.",
        "video_name": "BsBH8nAv5l4",
        "timestamps": [
            248
        ],
        "3min_transcript": "and then go to (mumbles) just draw vector a. So vector a. So once again, a vector, I can shift them around as long as I'm not changing the direction or their magnitude. So vector a looks like that. And notice, if you now go and start at the initial point of b and go to the terminal point of a, you still get vector c. So that's why a plus b and b plus a are going to give you the same thing. Now what if instead of saying a plus b I wanted to think about what a minus b is going to be? So let me write that down. Vector a minus vector b, minus vector b. And let's call that vector d. That is equal to vector d. So once again, I could start with vector a and here, order matters. So vector a looks something like this. completely perfect. So vector a, just like that. And one way of thinking about subtracting vector b is instead of adding vector b the way we did here, we could add negative b. So negative b would have the same magnitude but just the opposite direction. So that's vector a. Vector negative b will still start right over here, but will go in the opposite direction. So let's do that. So negative b is going to look like this, is going to look something like this. So that is negative b. Notice, same magnitude exactly opposite direction. We've flipped it around 180 degrees, and now the resulting vector is going to be d. So vector d is going to look like that, vector d. So c is a plus b, d is a minus b. Or you can even call this a plus, a plus negative b. and go the other way. See if we could go from the diagrams to the actual equations. So let's start with, let me draw an interesting one. So let's say that this... Let's say that's vector a. Vector a. I'll use green. Let's say that, that is vector b, and I will now use magenta. And let's say that this is vector c. Vector c. So I encourage you to pause the video and see if you can write an equation that defines this relationship. Well, this is interesting because they're all going in a circle right over here. Let's say that you started at... This is your initial point."
    },
    {
        "Q": "\nwhere did u get 144 on 03:54????",
        "A": "it is given that the side of the square is 12 and hence the area of the square is 12*12=144.....",
        "video_name": "vaOXkt7uuac",
        "timestamps": [
            234
        ],
        "3min_transcript": "the area of all of the parallelograms because they are congruent. So let's see if we can figure out the area of one of the parallelograms. So there is actually a formula for the area of a parallelogram, it's actually just the base times the height. And they actually give us that. But let me show you that they give us that because it might not be obvious to you. Let me try to draw it. I'll use my line tool. Nope, that's not the line tool. One side, then go straight like that, come down like that, good enough. OK, now if I look at just this parallelgoram, they tell us that the height here is 3. And I know it's the height because they told me it's a 90 degree angle. And they tell us at the base is 5. And I'm telling you that the area of a parallelogram is But you shouldn't just take my word for it. That should make intuitive sense to you. And the way to think about it intuitively is imagine if we were to take this part of the parallelogram, and if we were to move it over here. If we were to cut that off and move it over here. Then the parallelogram would look something like this. You'd have the part that we didn't cut off. And then you move the cut-off part over here. And now the dimensions, this base would be 5, and this height would be 3. And the area of this rectangle is 15. And there's no reason why the area of this should be any different than that. We just rearranged its parts. So that's why the area of a parallelogram is just the base times the height. So the area of each of these parallelograms is 15. So the area of all of them combined is 15 times 4, which is 60. So 144 minus 60 is 84. Problem 38. What is the area in square meters of the trapezoid shown below. So to figure out the area and we could break it up into these rectangles and triangles. To figure out the area of this rectangle we need to know its height. And actually we'll need that to figure out the area of the triangles as well. So what's this height right there? Let's see, we know that this distance is going to be 6. It's a rectangle. If that distance is 6 and both of these are 5, both of these triangles here are going to be congruent. Because this length is equal to this length. This length is equal to this length. And we also we make this angle is equal to that angle. But anyway, let me do it in another color."
    },
    {
        "Q": "At 8:15, How does Sal get 25 roots of 3, divided by 2? I don't get where 2 came from.\n",
        "A": "The 2 comes from the equation for the area of a triangle, 1/2 times the base times the height, or [1/2bh]. He divides it by 2 because that is the 1/2 in action. The base is 5 and the height is 5\u00e2\u0088\u009a3. So, the area is (25\u00e2\u0088\u009a3)/2, which could also be simplified to 12.5\u00e2\u0088\u009a3.",
        "video_name": "vaOXkt7uuac",
        "timestamps": [
            495
        ],
        "3min_transcript": "a squared is equal to 16. a is equal to 4. a is equal to 4. And now we're ready to figure out the area. What's the area of the rectangle? 6 times 6, it's 24. What's the area of each of these triangles? 3 times 4 times 1/2. 3 times 4 is 12 times 1/2 is 6. So the area of that triangle is 6. The area of this triangle is 6. So 6 plus 24 plus 6 is 36. B. Problem 39. What is the area in square inches of the triangle below. Interesting. OK, so this is an equilateral triangle, all the sides are equal. And so we could actually say that since these two triangles are symmetric. That's equal to that. And this comes to a general formula for the area of an equilateral triangle. But let's just figure it all out. So this side is going to be 5. If this is 5 and that's 10, what is this side right here? Let's call it x. Pythagorean theorem. This is the hypotenuse. So x squared plus 5 squared plus 25 is going to be equal to the hypotenuse squared, it's equal to 100. x squared is equal to 100 minus 25, 75. x is equal to the square root of 75. 75 is 25 times 3. So that's equal to the square root of 25 times 3. Which is equal to the square root of 25 times the square root of 3. Which is equal to 5 roots of 3. And now, what's the area of just this right triangle right here? This one on the right side. Well its base is 5, its height is 5 roots of 3. So it's going to be 1/2 times the base, 5, times the height, And that's what? 1/2 times 5 times 5. So it's 25 root 3 over 2 and that's just this triangle right there. Well this triangle's going to have the the exact same area. They are congruent triangles. So the area of the figure is this times 2. So 2 times that is equal to just the 25 root 3. And that's choice B. Next problem, problem 40. The perimeter of two squares are in a ratio of 4:9. What is the ratio between the areas of the two squares? Let me draw two squares. That's one square. Let me draw another square. That's another square. Let's say that the sides of this are x and the sides of this one are y."
    },
    {
        "Q": "\nAt 0:42, Sal says that the two angles are supplementary. In a real proof, wouldn't we not be able to assume that unless it was given?? Or do the opposite arrows show that it is a horizontal line?",
        "A": "The arrows do indicate that it is a line. So the angles are supplementary.",
        "video_name": "eTwnt4G5xE4",
        "timestamps": [
            42
        ],
        "3min_transcript": "We're given a bunch of lines here that intersect in all different ways and form triangles. And what I want to do in this video, we've been given the measures of some of the angles, this angle, that angle, and that angle. And what we want to do in this video is figure out what the measure of this angle is. And we're going to call that measure x. And so I encourage you to pause the video right now and try it yourself. And then I'm going to give you the solution. So I'm assuming you've unpaused it. And you've solved it or you've given it at least a good shot of it. And what's fun about these is there's multiple ways to solve these. And you kind of just have to keep figuring out what you can figure out. So let's say you start on the left-hand side right over here. If this is 121 degrees, then you'd say, well look, this angle right over here is supplementary to this angle right over there. So this is 121 degrees plus this green angle, that has to be equal to 180 degrees. So this is going to be 180 minus 121. 80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also"
    },
    {
        "Q": "At 3:44 Sal mentions the \"Exterior angle of 121 to be equal to the sum of the remote interior angles.\" This somehow confuses me as I don't remember him mentioning it before in previous videos. Is this true? If so does any exterior angle of a triangle mean that its equal to the triangles interior angles? Thanks in advance!\n",
        "A": "Yep!! I can t remember the name of the postulate/theorem that says this, but it is definitely true. Any exterior angle of a triangle is equal to the sum of its remote interior angles (the two interior angles that are across from the exterior angle in question). Hope this helps!! :)",
        "video_name": "eTwnt4G5xE4",
        "timestamps": [
            224
        ],
        "3min_transcript": "This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also So over here, they'll also add up to 180. So that will also get us to 59 degrees. We could just get that by taking 180, subtracting 29, subtracting 92. And then if this is 59 degrees, then this angle is also going to be 59 degrees, because they are vertical angles. So we're done. x is equal to 59 degrees. Now there's multiple ways that you could have reasoned through this problem. You could have immediately said-- so let me start over, actually. Maybe a faster way, but you wouldn't have been able to do kind of this basic steps there, is you said, look, this is an exterior angle right over here. It is equal to the sum of the remote interior angles. So 121 is going to be 29 plus this thing right over here. And we ended up doing that when I did it step-by-step before. But here, we're just using kind of a few things that we know about triangles ahead of time Although I like to do it the other way just so we make sure we don't do anything weird. So anyway, this is going to be 129 minus 29, which is going to be 92. And if this is 92, then this is also going to be 92. And then, if this is x, then this is also going to be x. And you could say x plus 92 plus 29 is equal to 180 degrees. And then you'd say x plus 92 plus 29 is going to be 121 degrees. We already knew that before. And so that is going to equal 180 degrees. And so x is equal to 59 degrees. So there's a ton of ways that you could have thought about this problem."
    },
    {
        "Q": "\nat 2:46, i notised that the angel he was trying to solve enisally is about the the same as the one he is going to solve.",
        "A": "Well ya, but there is an even easier way to solve for x. 180-121 is 59. So now we know the angle next to 121. There is another line with the same slope on the other side, so we know the corresponding angles are the same. Therefore we know that x is 59. So Simple.",
        "video_name": "eTwnt4G5xE4",
        "timestamps": [
            166
        ],
        "3min_transcript": "80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also So over here, they'll also add up to 180. So that will also get us to 59 degrees. We could just get that by taking 180, subtracting 29, subtracting 92. And then if this is 59 degrees, then this angle is also going to be 59 degrees, because they are vertical angles. So we're done. x is equal to 59 degrees. Now there's multiple ways that you could have reasoned through this problem. You could have immediately said-- so let me start over, actually. Maybe a faster way, but you wouldn't have been able to do kind of this basic steps there, is you said, look, this is an exterior angle right over here. It is equal to the sum of the remote interior angles. So 121 is going to be 29 plus this thing right over here. And we ended up doing that when I did it step-by-step before. But here, we're just using kind of a few things that we know about triangles ahead of time"
    },
    {
        "Q": "\nWhy in the first place, would you need to find for example an inverse of a matrix like Sal said at 3:40, or a determinant, or add or subtract matrixes?",
        "A": "The why is the means to find solution(s) to matrix operations.",
        "video_name": "0oGJTQCy4cQ",
        "timestamps": [
            220
        ],
        "3min_transcript": "And it has three columns. This is a 1 by 3 matrix. I could have a matrix-- and I think you see where all of this Figuring out the dimensions of a matrix are not too difficult. I could have a matrix that looks like this, where it's 3, 5, 0, 0, negative 1, negative 7. This right over here has three rows. So it's three rows, and it has two columns. So we would call this a 3 by 2. Let me do that in that same color. We would call it a 3 by 2 matrix, three rows and two columns. You know that a matrix is just a rectangular array of numbers. You can say what its dimensions are. You know that each of these numbers that take one of these positions-- we just call those entries. But what are matrices good for? I still might not be clear what the connection is between this and this right over here. this is just a compact representation of a bunch of numbers. It's a way of representing information. They become very valuable in computer graphics because these numbers could represent the color intensity at a certain point. They could represent whether an object is there at a certain point. And as we develop an algebra around matrices, and when we talk about developing an algebra around matrices, we're going to talk about operations that we're going to perform on matrices that we would normally perform with numbers. So we're going to essentially define how to multiply matrices, how to add matrices. We'll learn about taking an inverse of a matrix. And by coming up with an algebra of how we manipulate these things, it'll become very useful in the future when you're trying to write a computer graphics program or you're trying to do an economic simulation or a probability simulation, to say, oh, I have this matrix that represents where different particles are in space. Or I have this matrix that represents the state of some type of a game. And I know ways of doing it very efficiently so that I can multiply a bunch of them. Or I could come run a simulation, and I can actually come up with useful results. So that's all matrices are. But as you'll see through this, we can define operations on them. And then later on, when you take a linear algebra course in college, you'll learn a lot more of the depth of how they can be applied and what you can use them to represent."
    },
    {
        "Q": "At 1:55 in the video, why is it 72 * 7? I understand that 72 is the (9 * 8 * 7) part, but why is it multiplied by seven? Is it because there is only 7 people left after the 3 positions are filled?\n",
        "A": "72 is just the 9*8 part. He multiplied it by 7 because he wasn t done with the whole multiplication of 9*8*7.",
        "video_name": "l9ft9jpriNA",
        "timestamps": [
            115
        ],
        "3min_transcript": "A club of nine people wants to choose a board of three officers, President, Vice President, and Secretary. Assuming the officers are chosen at random, what is the probability that the officers are Marcia for president, Sabita for Vice President, and Robert for Secretary? So to think about the probability of Marcia-- so let me write this-- President is equal to Marcia, or Vice President is equal to Sabita, and Secretary is equal to Robert. This, right here, is one possible outcome, one specific outcome. So it's one outcome out of the total number of outcomes, over the total number of possibilities. Now what is the total number of possibilities? Well to think about that, let's just think about the three positions. and you have Secretary. Now let's just assume that we're going to fill the slot of President first. We don't have to do President first, but we're just going to pick here. So if we're just picking President first, we haven't assigned anyone to any officers just yet, so we have nine people to choose from. So there are nine possibilities here. Now, when we go to selecting our Vice President, we would have already assigned one person to the President. So we only have eight people to pick from. And when we assign our Secretary, we would've already assigned our President and Vice President, so we're only going to have seven people to pick from. So the total permutations here or the total number of possibilities, or the total number of ways, to pick President, Vice President, and Secretary from nine people, is going to be 9 times 8 times 7. Which is, let's see, 9 times 8 is 72. 72 times 7, 2 times 7 is 14, 7 times 7 is 49 plus 1 is 50. So to answer the question, the probability of Marcia being President, Sabita being Vice President, and Robert being Secretary is 1 over the total number of possibilities, which is 1 over 504. That's the probability."
    },
    {
        "Q": "How do you tell the quadrants apart at 0:40\n",
        "A": "the quadrants are usually labeled",
        "video_name": "Jeh5vudjmLI",
        "timestamps": [
            40
        ],
        "3min_transcript": "Plot 4 comma negative 1, and select the quadrant in which the point lies. So 4, the first number in our ordered pair, that's our x-coordinate. That says how far to move in the horizontal or the x-direction. It's a positive 4. So I'm going to go 4 to the right. And then the second coordinate says, what do we do in the vertical direction or in the y-direction? It's a negative 1. Since it's negative, we're going to go down. And it's a negative 1, so we're going to go down 1. So that right over there is the point 4 comma negative 1. So I've plotted it, but now I have to select which quadrant the point lies in. And this is just a naming convention. This is the first quadrant. This is the second quadrant. This is the third quadrant. And this is the fourth quadrant. So the point lies in the fourth quadrant, quadrant IV. And I guess you have to know your Roman numerals a little bit to know that's representing quadrant IV. Let's do a couple more of these. Plot 8 comma negative 4, and select Well, my x-coordinate is 8 so I go 8 in the positive x-direction. And then my y-coordinate is negative 4, so I go 4 down. And this is sitting, again, in not the first, not the second, not the third, but the fourth quadrant, in quadrant IV. Let's do one more of these. Hopefully we get a different quadrant. So we want to plot the point negative 5 comma 5. So now my x-coordinate is negative. It's negative 5. So I'm going to move to the left in the x-direction. So I go to negative 5. And my y-coordinate is positive so I go up 5, so negative 5 comma 5. And this is sitting not in the first quadrant, but the second quadrant. And of course, this is the third and the fourth. So this is sitting in the second quadrant. Check answer, and we got it right."
    },
    {
        "Q": "At 2:55, I cannot understand how does ln(67)=4.205 approximately matches between 2 and 3. plz help:(\n",
        "A": "If we let ln67 = a then e^a = 67 by definition of a logarithm e by definition is about 2.71 I think If we round the 4.2 down for approx. 2^4 = 16 and 3^4 = 81 Therefore since e = 2.71 which is in between 2 and 3 we expect e^4 to be in between 16 and 81, which it is.",
        "video_name": "Dpo_-GrMpNE",
        "timestamps": [
            175
        ],
        "3min_transcript": "this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that-- or seeing that, and the more typical way of seeing that is the natural log. And I think this is ln, so I think it's maybe from French or something, log natural, of 67. So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this ln, it literally means log base e. Now, they let us use a calculator, and that's good because I don't know off the top of my head what power I have to raise 2.71 so on and so forth-- what power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you literally can literally type in the statement natural log So here this is the button for ln, means natural log, log natural, maybe. ln of 67, and then you press Enter, and it'll give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer, but a graphing calculator can literally type it in the way that you would write it out, and then you would press Enter. So 4.20469 and we want to round to the nearest thousandth. So this is the thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. So this is approximately equal to 4.205. And it actually makes a lot of sense, because we know that e is greater than 2, and it is less than 3. And if you think about what 2 to the fourth power And 3 to the fourth power gets you to 81. 67 is between 16 and 81 and e is between 2 and 3. So at least it feels right that's something that's like 2.71 to the little over the fourth power should get you to a number that's pretty close to 3 to the fourth power. Actually that makes sense because it's actually closer to 3. 2.71 is closer to 3 than it is to 2. So this feels right, that you take this to the fourth, little over the fourth power, you get to 67."
    },
    {
        "Q": "At 5:00 in the video, why does Mr. Khan place positive 2 in the x place, if the x value was -2? Was that a mistake or something else?\n",
        "A": "The two aren t related, the positive two is the midpoint of the two x-intercepts and the negative two is one of the intercepts of the parabola",
        "video_name": "EV57jv7JKCs",
        "timestamps": [
            300
        ],
        "3min_transcript": "the x-axis at x equals negative two right over there, and x is equal to six. These are our x-intercepts. So given this, how do we figure out the vertex? Well the key idea here is to recognize that your axis of symmetry for your parabola is going to sit right between your two x-intercepts. So what is the midpoint between, or what is the average of six and negative two? Well, you could do that in your head. Six plus negative two is four divided by two is two. Let me do that. So I'm just trying to find the midpoint between the point, let's use a new color. So I'm trying to find the midpoint between the point negative two comma zero and six comma zero. Well the midpoint, those are just the average of the coordinates. The average of zero and zero is just going to be zero, it's going to sit on the x-axis. or the average negative two plus six over two. Well let's see, that's four over two, that's just going to be two, so two comma zero. And you see that there. You could have done that without even doing the math. You say okay, if I want to go right in between the two, I want to be two away from each of them. And so just like that, I could draw an axis of symmetry for my parabola. So my vertex is going to sit on that axis of symmetry. And so how do I know what the y value is? Well I can figure out, I can substitute back in my original equation, and say well what is y equal when x is equal to two? Because remember the vertex has a coordinate x equals two. It's going to be two comma something. So let's go back, let's see what y equals. So y will equal to one-half times, we're going to see when x equals two, so two minus six, times two plus two. Let's see, this is negative four, Negative four times four is negative 16. So it's equal to one-half times negative 16, which is equal to negative eight. So our vertex when x is equal to two, y is equal to negative eight. And so our vertex is going to be right over here two common negative eight. And now we can draw the general shape of our actual parabola. It's going to look something like, once again this is a hand-drawn sketch, so take it with a little bit of a grain of salt, but it's going to look something like this. And it's going to be symmetric around our axis of symmetry. That's why it's called the axis of symmetry. This art program I have, there's a symmetry tool, but I'll just use this and there you go. That's a pretty good sketch of what this parabola is or what this graph is going to look like which it is, an upward-opening parabola."
    },
    {
        "Q": "at 1:57, how is it equal to each other\n",
        "A": "it means the answer could be that number or less",
        "video_name": "ilWDSYnTEFs",
        "timestamps": [
            117
        ],
        "3min_transcript": "I'm starting to take a little bit more care of my health, and I start counting my actual calories. And let's say C is equal to the number of calories I eat in a given day. And I want to lose some weight. So, in particular, I want to eat less than 1,500 calories in a day. So how can I express that as an inequality? Well, I want the number of calories in a day to be less than-- and remember, the less than symbol, I make it point to the smaller thing. So I want the calories to be less than 1,500. So this is one way of expressing it. I say, look, the number of calories that I consume in a day need to be less than 1,500. Now, one thing to keep in mind when I write that is obviously if I eat no calories in a day, or if I eat 100 calories, or if I eat 1,400 calories, Those are all less than 1,500. But what about 1,500 calories? Is it true that 1,500 is less than 1,500? 1,500 is equal to 1,500. So this is not a true statement. But what if I want to eat up to and including 1,500 calories? I want to make sure that I get every calorie in there. How can I express that? How can express that I can eat less than or equal to 1,500 calories, so I can eat up to and including 1,500 calories? Right now, this is only up to but not including 1,500. How could I express that? Well, the way I would do that is to throw this little line under the less than sign. Now, this is not just less than. This is less than or equal to. So this symbol right over here, this So now 1,500 would be a completely legitimate C, a completely legitimate number of calories to have in a day. And if we wanted to visualize this on a number line, the way we would think about it, let's say that this right over here is our number line. I'm not going to count all the way from 0 to 1,500, but let's imagine that this right over here is 0. Let's say this over here is 1,500. How would we display less than or equal to 1,500 a number line? Well, we would say, look, we could be 1,500, so we'll put a little solid circle right over there. And then we can be less than it, so then we would color in everything less than 1,500,"
    },
    {
        "Q": "At 2:00, it is a little error in video when he writes M_y=cosx+1xe^y ( it should be M_y=cosx+2xe^y).\n",
        "A": "No, he wrote it correctly. It s just a low resolution video, like many of the old KA videos are.",
        "video_name": "Pb04ntcDJcQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "OK, I filled your brain with a bunch of partial derivatives and psi's, with respect to x's and y's. I think now it's time to actually do it with a real differential equation, and make things a little bit more concrete. So let's say I have the differential, y, the differential equation, y cosine of x, plus 2xe to the y, plus sine of x, plus-- I'm already running out of space-- x squared, e to the y, minus 1, times y prime, is equal to 0. Well, your brain is already, hopefully, in exact differential equations mode. But if you were to see this pattern in general, where you see a function of x and y, here-- this is just some function of x and y-- and then you have another function of x and y, times y prime, or times dy, d of x, your brain should immediately say if this is inseparable. because that'll take a lot of time. But if it's not separable, your brain said, oh, maybe this is an exact equation. And, you say, let me test whether this is an exact equation. So if this is an exact equation, this is our function M, which is a function of x and y. And this is our function N, which is a function of x and y. Now, the test is to see if the partial of this, with respect to y, is equal to the partial of this, with respect to x. So let's see. The partial of M, with respect to y, is equal to-- let's see, y is-- so this cosine of x is just a constant, so it's just cosine of x. Cosine of x plus-- now, what's the derivative? Well, 2x is just a constant, what's the derivative of e to the y, with respect to y? Well, it's just e to the y, right? So we have the constant on the outside, 2x times the derivative, with respect to y, so it's 2xe to the y. Fair enough. Now, what is the partial derivative of this, with So N sub x, or the partial of N, with respect to x-- so what's the derivative of sine of x, with respect to x? Well, that's easy, that's cosine of x, plus 2x times e to the y, right? e and y is just a constant, because y is constant when we're taking the partial, with respect to x. So plus 2xe to the y. And then minus 1, the derivative of a constant, with respect to anything is going to be 0. So the derivative of N-- the partial of N, with respect to x, is cosine of x, plus 2xe to the y, which, lo and behold, is the same thing as the derivative, the partial of M, with respect to y. So there we have it. We've shown that M of y is equal to-- or the partial of M, with respect to y-- is equal to the partial of N, with respect to x, which tells us that this is an exact equation. Now, given that this is an exact equation-- oh, my wife"
    },
    {
        "Q": "At around 2:30 Sal tell us that -4 / - 1/2 is equal to -4/1 * - 2/1. I am wondering why the 2 becomes a negative when it's going to be multiplied. I would like to know if -4/1 is the same as - 4/1? I hope I'm clear.\n\nTo be clearer: why doesn't Sal multiply -4/1 by 2/-1, and instead multiplies by -2/1\n",
        "A": "He could multiply by any of those because - 2/1 = -2/1 = 2/(-1).",
        "video_name": "H0q9Fqb8YT4",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2."
    },
    {
        "Q": "at 1:47 it say greatest common divisor what dose that mean\n",
        "A": "It is the largest number that can be divided into both 18 and 20. 18 can be divided by 2, 3, 6, and 9, while twenty can be divided by 2, 4, and 5. The only number that they have in common is 2, so two is the greatest common divisor. Notice that they can each also be divided by 1, but generally we ignore that when simplifying fractions because dividing by 1 gets us nowhere.",
        "video_name": "H0q9Fqb8YT4",
        "timestamps": [
            107
        ],
        "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2."
    },
    {
        "Q": "\nWhy switch the numbers at 0:34",
        "A": "When dividing, you need to make the second fraction a reciprocal, or flip the fraction so that the numerator ends up as the denominator and the denominator as the numerator.",
        "video_name": "H0q9Fqb8YT4",
        "timestamps": [
            34
        ],
        "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2."
    },
    {
        "Q": "at 5:11 Mr Khan says that a 90 degree angle is a right angle so would an 180 degree angle be a straight angle?\n",
        "A": "yes, it could be called a straight angle or just a straight line",
        "video_name": "92aLiyeQj0w",
        "timestamps": [
            311
        ],
        "3min_transcript": "And then I'll point AB in the-- well, assuming that I'm drawing it exactly the way that it's Normally, instead of moving the angle, you could actually move the protractor to the angle. So it looks something like that, and you could see that it's pointing to right about the 30 degree mark. So we could say that the measure of angle BAC is equal to 30 degrees. And so you can look just straight up from evaluating these numbers that 77 degrees is clearly larger than 30 degrees, and so it is a larger angle, which makes sense because it is a more open angle. And in general, there's a couple of interesting angles to think about. If you have a 0 degree angle, you actually have something that's just a closed angled. It really is just a ray at that point. As you get larger and larger or as you get more and more open, is completely straight up and down while the other one is left to right. So you could imagine an angle that looks like this where one ray goes straight up down like that and the other ray goes straight right and left. Or you could imagine something like an angle that looks like this where, at least, the way you're looking at it, one doesn't look straight up down or one does it look straight right left. But if you rotate it, it would look just like this thing right over here where one is going straight up and down and one is going straight right and left. And you can see from our measure right over here that that gives us a 90 degree angle. It's a very interesting angle. It shows up many, many times in geometry and trigonometry, and there's a special word for a 90 degree angle. It is called a \"right angle.\" So this right over here, assuming if you rotate it around, would look just like this. We would call this a \"right angle.\" You draw a little part of a box right over there, and that tells us that this is, if you were to rotate it, exactly up and down while this is going exactly right and left, if you were to rotate it properly, or vice versa. And then, as you go even wider, you get wider and wider and wider and wider until you get all the way to an angle that looks like this. So you could imagine an angle where the two rays in that angle form a line. So let's say this is point X. This is point Y, and this is point Z. You could call this angle ZXY, but it's really so open that it's formed an actual line here. Z, X, and Y are collinear. This is a 180 degree angle where we see the measure of angle ZXY"
    },
    {
        "Q": "\nWhat if your angle is 0 Degrees? At the end 8:18-8:21, Sal says that if your angle is all the way 180 degrees... it forms a line. But, what if the angle is 0 degrees? What does that form? Is it still an acute angle? Or is there a special vocabulary word to describe this type of angle? Appreciate the help! :)",
        "A": "An angle of 0 degrees between two rays forms one ray. Think of the two rays being perfectly on top of each other, going in the same direction.",
        "video_name": "92aLiyeQj0w",
        "timestamps": [
            498,
            501
        ],
        "3min_transcript": "And you can actually go beyond that. So if you were to go all the way around the circle so that you would get back to 360 degrees and then you could keep going around and around and around, and you'll start to see a lot more of that when you enter a trigonometry class. Now, there's two last things that I want to introduce in this video. There are special words, and I'll talk about more types of angles in the next video. But if an angle is less than 90 degrees, so, for example, both of these angles that we started our discussion with are less than 90 degrees, we call them \"acute angles.\" So this is acute. So that is an acute angle, and that is an acute angle right over here. They are less than 90 degrees. What does a non-acute angle look like? And there's a word for it other than non-acute. So, for example-- let me do this in a color I haven't used-- an angle that looks like this, and let me draw it a little bit better than that. An angle it looks like this. So that's one side of the angle or one of the rays and then I'll put the other one on the baseline right Clearly, this is larger than 90 degrees. If I were to approximate, let's see, that's 100, 110, 120, almost 130. So let's call that maybe a 128-degree angle. We call this an \"obtuse angle.\" The way I remember it as acute, it's kind of \"a cute\" angle. It's nice and small. I believe acute in either Latin or Greek or maybe both means something like \"pin\" or \"sharp.\" So that's one way to think about it. An acute angle seems much sharper. Obtuse, I kind of imagine something that's kind of lumbering and large. Or you could think it's not acute. It's not nice and small and pointy. but this is just general terminology for different types of angles. Less than 90 degrees, you have an acute angle. At 90 degrees, you have a right angle. Larger than 90 degrees, you have an obtuse angle. And then, if you get all the way to 180 degrees, your angle actually forms a line."
    },
    {
        "Q": "at 6:00 he is talking about < zxy\nwouldn't it be the same if he called it:\n<xyz\nor\n<yxz ?\n",
        "A": "XYZ? That s a different angle. YXZ? That s the same angle. You might wonder how this is so. Well, when naming angles, you always put the vertex in the middle. In the first angle, there is a different letter in the middle, so there is a different vertex, and thus a different angle. YXZ is the same angle because there is the same vertex and the rays Y and Z are still used in the angle. YXZ is simply a different way to name ZXY but XYZ is a different angle. I hope this helps!",
        "video_name": "92aLiyeQj0w",
        "timestamps": [
            360
        ],
        "3min_transcript": "is completely straight up and down while the other one is left to right. So you could imagine an angle that looks like this where one ray goes straight up down like that and the other ray goes straight right and left. Or you could imagine something like an angle that looks like this where, at least, the way you're looking at it, one doesn't look straight up down or one does it look straight right left. But if you rotate it, it would look just like this thing right over here where one is going straight up and down and one is going straight right and left. And you can see from our measure right over here that that gives us a 90 degree angle. It's a very interesting angle. It shows up many, many times in geometry and trigonometry, and there's a special word for a 90 degree angle. It is called a \"right angle.\" So this right over here, assuming if you rotate it around, would look just like this. We would call this a \"right angle.\" You draw a little part of a box right over there, and that tells us that this is, if you were to rotate it, exactly up and down while this is going exactly right and left, if you were to rotate it properly, or vice versa. And then, as you go even wider, you get wider and wider and wider and wider until you get all the way to an angle that looks like this. So you could imagine an angle where the two rays in that angle form a line. So let's say this is point X. This is point Y, and this is point Z. You could call this angle ZXY, but it's really so open that it's formed an actual line here. Z, X, and Y are collinear. This is a 180 degree angle where we see the measure of angle ZXY And you can actually go beyond that. So if you were to go all the way around the circle so that you would get back to 360 degrees and then you could keep going around and around and around, and you'll start to see a lot more of that when you enter a trigonometry class. Now, there's two last things that I want to introduce in this video. There are special words, and I'll talk about more types of angles in the next video. But if an angle is less than 90 degrees, so, for example, both of these angles that we started our discussion with are less than 90 degrees, we call them \"acute angles.\" So this is acute. So that is an acute angle, and that is an acute angle right over here. They are less than 90 degrees. What does a non-acute angle look like? And there's a word for it other than non-acute."
    },
    {
        "Q": "at 5:37 how can b>1 when at the top of the screen it says b<0\n",
        "A": "It can t. And in the 30 seconds after 5:37, Sal explains that b>1 cannot be part of the solution because b<0 is a restriction.",
        "video_name": "0_VaUYoNV7Y",
        "timestamps": [
            337
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhen you divide A out at 4:20 do you always replace the empty side with 1?",
        "A": "When you divide any variable out of a equation, you always replace the number or variable with 1. 1 is like the benchmark in math, it always takes the place of the number if the initial number is divided out.",
        "video_name": "0_VaUYoNV7Y",
        "timestamps": [
            260
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at the end ( 5:15 - 5:20 ) when sal says b <-1 and b > 1 how does that work b cannot be no number can be b if b was 2 that would violate the first inequality if b was -2 it would violate the second someone please explain.\n",
        "A": "You are trying to use b <-1 AND b > 1 . Sal is using b <-1 OR b > 1. You are correct, AND will not work as it means both conditions need to be true. But, OR is more flexible, only one condition need to be true. Hope this helps.",
        "video_name": "0_VaUYoNV7Y",
        "timestamps": [
            315,
            320
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:49. Don't you mean 19 + 18. You said 19 over 18...",
        "A": "Yeah he meant 19 + 18.. It said it in the bottom right corner..",
        "video_name": "8Eb5MWwcMMY",
        "timestamps": [
            49
        ],
        "3min_transcript": "Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done."
    },
    {
        "Q": "At 1:10, where did the 6 come in?\n",
        "A": "The 18 is bigger than the 3.... so you have to ask yourself what you have to multiply the 3 by to get to 18.... 3x6=18..... Whatever you do to the denominator you have to do to the numerator to arrive at an equivalent fraction... so now you also multiply the numerator by 6 to get the numerator for the equivalent fraction",
        "video_name": "8Eb5MWwcMMY",
        "timestamps": [
            70
        ],
        "3min_transcript": "Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done."
    },
    {
        "Q": "What the heck is a caveat? Sal says it at 3:14. I'm guessing that it is the backwards C like object.\n",
        "A": "It s the RESTRICTION or the CONDITION. You know how Sal writes p \u00e2\u0089\u00a0 -5 , that s the caveat or condition.",
        "video_name": "gcnk8TnzsLc",
        "timestamps": [
            194
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:35, when you cancel out (p - 5), you cross out (p - 6)",
        "A": "It looks like a 6, but it is a 5.",
        "video_name": "gcnk8TnzsLc",
        "timestamps": [
            215
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 4:02 when Sal reached the answer, 4(p+3) over 5, providing p does not equal -5, he did not further simplify the equation. Can you not simplify it into 4p+12 over 5, providing that p does not equal -5?",
        "A": "In more advanced math, you often have to take your result and do other steps. That is usually easier if the result is in factored form. Sometimes in Calculus you will want to multiply the result all the way out. Practice will help you choose...and also little hints like the instructions you receive with the assignment and choices you are given as possible answers.",
        "video_name": "gcnk8TnzsLc",
        "timestamps": [
            242
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 1:02, how did you get 0?",
        "A": "He is talking about the denominator: x cannot make the denominator equal to 0 or that would make the problem undefined. No matter what other solutions we come up with, we always have to exclude that value, which in this case is x.\u00e2\u0089\u00a0 -4",
        "video_name": "2RnS3fSHVV8",
        "timestamps": [
            62
        ],
        "3min_transcript": "Let's tackle a slightly harder problem than what we saw in the last video. I have here x minus 3 over x plus 4 is greater than or equal to 2. So the reason why this is slightly harder is I now have a greater than or equal to. And the other thing that makes it slightly harder is I don't just have a simple 0 here, I actually have a 2 here. So I'm going to do this problem the same way. I'm going to do it two ways. I'm going to do the same two ways we did last time, but I'm going to reverse the order of the two ways that I do it. So the first thing I can do, I might be tempted-- let me multiply both sides of this equation times x plus 4. And like we saw in the last video we had to be very careful because we have an inequality here. If x plus 4 is greater than 0 then we're going to keep the inequality the same. If it's less than 0 we're going to swap it. So let's look at those two situations. So one situation is-- let me make it right here. One situation is x plus 4 is greater then 0. And remember, it can ever be equal to 0 because then this So let's explore x plus 4 is greater than 0. If we subtract 4 from both sides of this equation this is equivalent to saying-- these are equivalent statements. If we subtract 4 from both sides that x is greater than minus 4. So if we assume that x is greater than minus 4 then x plus 4 is going to be greater than 0. And then when we multiply both sides of this times x plus 4-- let's do that. x minus 3 over x plus 4. You have your 2. We're going to multiply both sides of this times x plus 4. Since we're assuming x is greater than minus 4 or that x plus 4 is greater than 0, we don't have to switch the inequality sign. We're multiplying both sides by a positive. So the inequality stays the same as in our original problem. And the whole reason why we did that is because this and this will cancel out. And then we have 2 times x plus 4. Let's see what we get. x minus 3 is greater than or equal to 2 times x plus And now what can we do? We can subtract x from both sides. We get minus 3 is greater than or equal to x plus 8. I just took x from 2x. And then we can subtract 8 from both sides. So you subtract 8 from both sides. You get minus 3 minus 8 is minus 11 is greater than or equal to x. And of course, then we subtract an 8 so this guy disappears. So if we assume that x is greater than minus 4, then x has to be less than or equal to minus 11. Now that seems a little bit nonsensical. In order for this statement to be true, x has to be both greater than minus 4 and it has to be less than minus 11. Anything greater than minus 4 is going to be greater than minus 11. So there's no x that satisfies this equation."
    },
    {
        "Q": "In 3:26, Sal said, \"So all I did, got rid of the exponent...\", but I think he meant the negative sign\n",
        "A": "Yes, he could have said it that way, but what he means is that he got rid of the negative exponent and replaced it with a positive.",
        "video_name": "S34NM0Po0eA",
        "timestamps": [
            206
        ],
        "3min_transcript": "to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the fourth power and then the third power. All this is saying is, if I raise something to a power and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2/3 is the same thing as 1/3 times 2. So we could go in the other direction. We could say, hey look, well this is going to be the same thing as 64 to the 1/3 power and then that thing squared. Notice, I'm raising something to a power and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2/3 power. Now, why did I do this? Well, we already know what 64 to the 1/3 power is. We just calculated it. That's equal to 4. So we could say that-- and I'll write it in that same yellow color-- this is equal to 4 squared, which is equal to 16. The way I think of it, let me find the cube root of 64, which is 4. And then let me square it. And that is going to get me to 16. Now I'll give you in even hairier problem. And I encourage you to try this one on your own before I work through it. So we're going to work with 8/27. And we're going to raise this thing to the-- and I'll try to color code it-- negative 2 over 3 power, to the negative 2/3 power. I encourage you to pause and try this on your own. Well the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says, take the reciprocal of this to the positive exponent. I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power."
    },
    {
        "Q": "At 5:47, why is the answer just 9/4? Can't we just simplify it to 2/1/4?\n",
        "A": "You could, but Sal must consider 9/4 a simplified fraction as well, even if it is improper, because you can t simplify it any more (although, you can change it to a mixed number). Sometimes, whatever is considered simplified depends on the standards of your teacher or professor.",
        "video_name": "S34NM0Po0eA",
        "timestamps": [
            347
        ],
        "3min_transcript": "I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator to that power over the denominator raised to that power. Now, let's think about what this is. Well just like we saw before, this is going to be the same thing. This is going to be the same thing as 27 to the 1/3 power and then that squared because 1/3 times 2 is 2/3. So I'm going to raise 27 to the 1/3 power and then square whatever that is. All this color coding is making this have to switch a lot of colors. This is going to be over 8 to the 1/3 power. And then that's going to be raised to the second power. Same thing we were doing in the denominator-- we raise 8 to the 1/3 and then square that. So what's this going to be? Well, 27 to the 1/3 power is the cube root of 27. times that same number is going to be equal to 27. Well, it might jump out at you already that 3 to the third is equal to 27 or that 27 to the 1/3 is equal to 3. So the numerator, we're going to end up with 3 squared. And then in the denominator, we are going to end up with-- well, what's 8 to the 1/3 power? Well, 2 times 2 times 2 is 8. So this is 8 to the 1/3 third is 2. Let me do that same orange color. 8 to the 1/3 is 2, and then we're going to square that. So this is going to simplify to 3 squared over 2 squared, which is just going to be equal to 9/4. So if you just break it down step by step, it actually is not too daunting."
    },
    {
        "Q": "\ni still don't get it at here the time 0:35\nAnd like M21 how did he got 100/100\nhere the time 0:32\nAnd again help on the last second 1:48",
        "A": "because percent is out of a 100,(for EG,80/100) like an exam, it is out of a 100 percent, if you get 1 wrong you get like 98% or something, and for the last one, he is just multiplying 1.501 times 100%, so you move the decimal twice because 100 has two zero. but let say it was out of 1000, so you move the zero....three times.10, that will be one time.",
        "video_name": "3_caioiRu5I",
        "timestamps": [
            35,
            32,
            108
        ],
        "3min_transcript": "Let's see if we can write 1.501 as a percentage. Now we really just want to write this quantity as something over 100. So what we can do is we can say that this is the same thing as 1.501 over 1. We haven't changed its value. And let's multiply it times 100/100. When we do this product-- and I haven't changed the value. This is just multiplying something times 1. But when we do it, we're going to change the way that we're representing it. The denominator now is going to be 1 times 100. So that's pretty straightforward. That's just going to be 100. And then the numerator, we're going to want to multiply 1.501 times 100. And so if we multiply this times 10, we would move the decimal one over. We want to multiply it by 100, so we want to multiply by 10 again. So we're going to move the decimal over to the right twice. We're multiplying it by 10 twice, or you're multiplying it by 100. This is multiplying by 10. This is multiplying by 100. So this is where the decimal's going to sit now. And so we've rewritten the 1.501 as 150.1 over 100, or 150.1 per 100. Let me write that. 150.1 per 100, which is the same thing as 150.1 per cent, which is the same thing as 150.1%. And we're done. And, essentially, all you're doing-- if you really want to simplify the process-- we're multiplying this by 100 to get 150.1. And then we're saying, that is the percentage. So you want to make sure that you write the percent there or the percent symbol."
    },
    {
        "Q": "At 1:36, it says the answer is 150.1%. Doesn't that equal 150.1/ 100? Is that even possible?\n",
        "A": "Yes, it s possible. The concept is similar to improper fractions to mixed numbers.",
        "video_name": "3_caioiRu5I",
        "timestamps": [
            96
        ],
        "3min_transcript": "Let's see if we can write 1.501 as a percentage. Now we really just want to write this quantity as something over 100. So what we can do is we can say that this is the same thing as 1.501 over 1. We haven't changed its value. And let's multiply it times 100/100. When we do this product-- and I haven't changed the value. This is just multiplying something times 1. But when we do it, we're going to change the way that we're representing it. The denominator now is going to be 1 times 100. So that's pretty straightforward. That's just going to be 100. And then the numerator, we're going to want to multiply 1.501 times 100. And so if we multiply this times 10, we would move the decimal one over. We want to multiply it by 100, so we want to multiply by 10 again. So we're going to move the decimal over to the right twice. We're multiplying it by 10 twice, or you're multiplying it by 100. This is multiplying by 10. This is multiplying by 100. So this is where the decimal's going to sit now. And so we've rewritten the 1.501 as 150.1 over 100, or 150.1 per 100. Let me write that. 150.1 per 100, which is the same thing as 150.1 per cent, which is the same thing as 150.1%. And we're done. And, essentially, all you're doing-- if you really want to simplify the process-- we're multiplying this by 100 to get 150.1. And then we're saying, that is the percentage. So you want to make sure that you write the percent there or the percent symbol."
    },
    {
        "Q": "\nAt 1:03 , can I write -15+j?",
        "A": "Yes, you could write it as -15+j",
        "video_name": "640-86yn2wM",
        "timestamps": [
            63
        ],
        "3min_transcript": "- [Voiceover] Let's do some examples of the writing expressions with variables exercise. So it says \"Write an expression to represent 11 more than a.\" Well you could just have a but if you want 11 more than a, you would wanna add 11 so you could write that as a plus 11. You could also write that as 11 plus a. Both of them would be 11 more than a. So let's check our answer here. We got it right. Let's do a few more of these. \"Write an expression to represent the sum of d and 9.\" So the sum of d and 9, that means you're gonna add d and 9. So I could write that as d plus 9 or I could write that as 9 plus d. And check our answer. Got that right. Let's do a few more of these. \"Write an expression to represent j minus 15.\" Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun. \"Write an expression to represent 7 times r.\" There's a couple ways I could do it. I could use this little dot right over here, do 7 times r like that. That would be correct. I could literally just write 7r. If I just wrote 7r that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could've just done 10 and this is not a decimal, it sits a little bit higher than a decimal. It's multiplication and the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that x-looking thing gets confused with x when you're using x as a variable so that's why this is a lot more useful. So we wanna write 10 times u, 10 times u, let's check our answer. We got it right. Let's do one more. So we could write it as 8 and then I could write a slash like that, 8 divided by d. And there you go. This is 8 divided by d. Let me check, let me check the answer. I'll do one more of these. Oh, it's 6 divided by b. Alright, same thing. So 6, I could use this tool right over here. It does the same thing as if I were to press the backslash. So 6 divided by b. Check my answer. We got it right."
    },
    {
        "Q": "Hi can I please get help Asap Please ?! I understood what sal was saying until he got to the slope part. To sum it up , I don't understand from 1:38 to the end\n",
        "A": "If x = 1 what is y? If x = 2 what is y? How much did x change? y? What s (change in y)/(change in x)?",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            98
        ],
        "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done."
    },
    {
        "Q": "At 0:27, how do you actually tell if -2 equals y? How is x=0 when there is 1/3 next to it?\n",
        "A": "A little after your time stamp, Sal states that we know that the y intercept is -2 because the y intercept is where x=0. The idea of putting an equation in slope intercept form is so that we can quickly recognize the slope (m) and the y intercept (b). For the y intercept (where a function crosses the y axis), x will always be zero.",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            27
        ],
        "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done."
    },
    {
        "Q": "\nAt 0:38, what does he mean by \"the y-intercept occurs when x equals to zero\"?",
        "A": "The Y-intercept is the point where the graph crosses the Y-Axis and a line crosses the Y-axis when x=0, since the Y-axis can be thought of as the line X=0.",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            38
        ],
        "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done."
    },
    {
        "Q": "At 6:22, Sal says that we are going to find the volume of the coin shaped object by multiplying pi times f(x)^2 times dx, but how is this possible because the side of the coin also slightly slopes like the rest of the cone?\n",
        "A": "The point of the integral is to take the sum of an infinite number of coins, so in theory the slope is 0.",
        "video_name": "btGaOTXxXs8",
        "timestamps": [
            382
        ],
        "3min_transcript": "you rotate it. What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. And let me draw it out here. So that same disk out here would look something like this. And it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt at the x-axis sits it right over there. It comes out of the center. And then this is the surface of my disk. And then this right over here is my depth dx. So that looks pretty good. to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this base? Well, we know that the area of a circle is equal to pi r squared. So if we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle. And for any x, the height over here is going to be equal to f of x. And in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Well, our volume is going to be our area times the depth here. It's going to be that times the depth, times dx. So the volume of this thing right over here-- so the volume just of this coin, I guess you could call it, is going to be equal to. So my volume is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi-- x squared squared is x to the fourth-- pi x to the fourth dx. Now, this expression right over here, this gave us the volume just of one of those disks. But what we want is the volume of this entire hat, or this entire bugle or cone-looking, or I guess you could say the front-of-a-trumpet-looking"
    },
    {
        "Q": "\nThis seems too simple for a 10-minute video, so someone please confirm that I understand. If I fully get the concepts of span, subspace, and a matrix being defined as a set of column vectors, then it's dead simple to grasp the concept of the column space of a matrix, right? So, anyone who gets the videos in this series up to this point can watch up to 1:28, then skip to the next video, right?",
        "A": "I believe the point to drive home is that any vector multiplied with the matrix must result in a vector that exists in the column space of that matrix, which in turn is the span of those column vectors. This implies that the column space is a subspace.",
        "video_name": "st6D5OdFV9M",
        "timestamps": [
            88
        ],
        "3min_transcript": "We spent a good deal of time on the idea of a null space. What I'm going to do in this video is introduce you to a new type of space that can be defined around a matrix, it's called a column space. And you could probably guess what it means just based on what it's called. But let's say I have some matrix A. Let's say it's an m by n matrix. So I can write my matrix A and we've seen this multiple times, I can write it as a collection of columns vectors. So this first one, second one, and I'll have n of them. How do I know that I have n of them? Because I have n columns. And each of these column vectors, we're going to have how many components? So v1, v2, all the way to vn. This matrix has m rows. So each of these guys are going to have m components. So they're all members of Rm. So the column space is defined as all of the possible linear So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors. What's all of the linear combinations of a set of vectors? It's the span of those vectors. So it's the span of vector 1, vector 2, all the way to vector n. And we've done it before when we first talked about span and subspaces. But it's pretty easy to show that the span of any set of vectors is a legitimate subspace. It definitely contains the 0 vector. If you multiply all of these guys by 0, which is a valid linear combination added up, you'll see that it contains the 0 vector. If, let's say that I have some vector a that is a member of the column space of a. That means it can be represented as some linear combination. So a is equal to c1 times vector 1, plus c2 times vector Now, the question is, is this closed under multiplication? If I multiply a times some new-- let me say I multiply it times some scale or s, I'm just picking a random letter-- so s times a, is this in my span? Well s times a would be equal to s c1 v1 plus s c2 v2, all the way to s Cn Vn Which is once again just a linear combination of these column vectors. So this Sa, would clearly be a member of the column space of a. And then finally, to make sure it's a valid subspace-- and this actually doesn't apply just to column space, so this applies to any span. This is actually a review of what we've done the past. We just have to make sure it's closed under addition. So let's say a is a member of our column space."
    },
    {
        "Q": "\nAt 9:48 to the end of the video the lines get confusing",
        "A": "Yes, They Do. Don t Worry You ll get the hang of it Good Luck ! And Happy Learning .",
        "video_name": "K759mIqpvOU",
        "timestamps": [
            588
        ],
        "3min_transcript": "it's gonna be negative six, and L is this X coordinate, it's going to be, it's going to be positive six. So the absolute value of J minus L is going to be our, is going to the the difference in the horizontal axis, or it's going to be the distance on the horizontal axis between this point and that point. Or you say the horizontal distance between those two points. So it would be the length of this line segment. So that is the absolute value of J minus L. Once again, the X coordinate here is negative six, the X coordinate here is positive six, and you can even figure it out. It's going to be negative six minus six, which would be negative 12, and then you take the absolute value of that, this is going to be 12. And you don't even have to figure that out here, we just know that the length of this line is the absolute value of J minus L. So that's that. And then they have the absolute value of M minus Q. the absolute value of M minus Q. So they have M over here, that's the Y coordinate here, and Q is the Y coordinate down here. So the absolute value of M minus Q is going to be the distance, the vertical distance between these two points, which is really just, 'cause the X value isn't changing, this is actually going to be the length. This is going to be the length of that side. That's going to be the absolute value of M minus Q. So yeah, if you multiply this length times this length, you're going to get the area of the rectangle. So I didn't even have to look at the other choices, I would definitely go with this one. But let's see where the other ones probably aren't correct. So this is the absolute value of J minus M. So here you're taking the difference of the X coordinate here and the Y coordinate over there. So that's kind of bizarre. This this already looks suspicious. Here you're saying the absolute value of J minus N. Absolute value of J, Well, their X coordinates are the same, so this is actually going to be, this we actually know is going to be zero. J is equal to N, they're both equal to negative six. That's not gonna give you the length of this line, because we have no change along X here. All the change is along Y. If we wanted to figure out the length of this line right over here we would have to find the absolute value of K, of the change in our Y coordinates. So absolute value of K minus O would give you the length of this line. And if you wanted the length of this right over here, you'd want your change in X, so that would be the absolute value of N minus P, or you could say the absolute value of P minus N. But they didn't use those choices. So yeah, we feel good about that."
    },
    {
        "Q": "\nat 5:21 How does he know it will be a right triangle?",
        "A": "He knows that triangle BDM is a right triangle because point D is on a line perpendicular to line AB. As he says at 5:30, All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. Therefore, line DM intersects line BM at a right angle, making BDM a right triangle.",
        "video_name": "smtrrefmC40",
        "timestamps": [
            321
        ],
        "3min_transcript": "the average of 24 and 0, is 12. That's fair enough. Now let's see what we can figure out about the sides. So we know this is a right triangle. So our gut reaction is always to use the Pythagorean theorem. We know this side and that side. So if we wanted to figure out AB, we could just say that-- we just know that 24 squared plus 7 squared is equal to AB squared. And 24 squared is 576, plus 49 is equal to AB squared. And let's see-- 576 plus 49. If it was plus 50, it would get us to 626, but it's 1 less than that. So it's 625. So 625 is equal to AB squared. So this distance, the distance of this big hypotenuse here, is 25. Or half of the distance, this is going to be 25/2, from B to M. From M to A is also going to be 25/2. Now, the other thing we know is that M right over here, the triangle CMA is an isosceles triangle. How do we know that? Well, M, if you look at its x-coordinate-- its x-coordinate is directly in between the x-coordinates for C and A. 7/2-- it's the average. This is 7. This is 0. Its M-coordinate is right over there. It's directly above the midpoint of this base over here. So this is going to be an isosceles triangle. It's symmetric. You could flip the triangle over. So this length-- and this seems useful because this is kind of the base of the triangle we care about. this is also going to be 25/2. It's an isosceles triangle. This is going to be the same as that because we're symmetric around this right over here. So let's see. We know one side of this triangle. Let's see if we could figure out this side over here. It seems pretty straightforward. This is going to be a right triangle right over here, because this line from D to M is going to be perpendicular to AB. All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. So this is going to be a right triangle. So we can figure out DM using the Pythagorean theorem again. We get 25/2 squared plus DM squared is going to be equal to 15 squared."
    },
    {
        "Q": "I dont understand, how at 10:37 he got cosx=sin90-x?\n",
        "A": "Its just a property of sin and cos. If you look at the unit circle you can see that the sin of say 30 is the same as the cos of 60. The sin of an angle is the same as the cos of the complementary angle.",
        "video_name": "smtrrefmC40",
        "timestamps": [
            637
        ],
        "3min_transcript": "And this over here is 24. This is 25/2, and this is 25/2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25/2 squared, plus 25/2 squared, minus 2 times 25/2 times 25/2 times the cosine of this angle, times-- let me scroll over-- the cosine of theta plus 90 degrees. this has it in terms of cosine of theta plus 90 degrees. How can we figure out the sine of theta? That's what we actually care about, to figure out the area of this triangle, or actually to figure out the height of this triangle. And to do that, you just have to make the realization. We know the trig identity that the cosine of theta is equal-- I won't use theta because I don't want to overload theta-- cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of-- let me put it in parentheses-- 90 minus whatever is here. 90 minus theta minus 90, which is equal to-- the 90's cancel out-- sine of negative theta. And we know sine of negative theta is equal to the negative sine of theta. is the negative sine of theta. So we could write the sine of theta here and then put the negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared, which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared plus 25 squared. This is 2 times 25/2 squared plus 2 times-- this is 25/2 squared again-- times sine of theta. Now we just have to solve for sine of theta. So this is going to be equal to-- well, this is 576 is equal to 2 times 25/2 squared."
    },
    {
        "Q": "I dont understand, how at 10:37 he got cosx=sin90-x? Also at 14:39 it is really confusing?\n",
        "A": "If you take a right triangle where there are the points A,B,and C(with C being the right angle), you can see that sin(x)=cos(90-x) by looking at the sine of angle A and the cosine of angle B. (90-angle A= angle B)",
        "video_name": "smtrrefmC40",
        "timestamps": [
            637,
            879
        ],
        "3min_transcript": "And this over here is 24. This is 25/2, and this is 25/2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25/2 squared, plus 25/2 squared, minus 2 times 25/2 times 25/2 times the cosine of this angle, times-- let me scroll over-- the cosine of theta plus 90 degrees. this has it in terms of cosine of theta plus 90 degrees. How can we figure out the sine of theta? That's what we actually care about, to figure out the area of this triangle, or actually to figure out the height of this triangle. And to do that, you just have to make the realization. We know the trig identity that the cosine of theta is equal-- I won't use theta because I don't want to overload theta-- cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of-- let me put it in parentheses-- 90 minus whatever is here. 90 minus theta minus 90, which is equal to-- the 90's cancel out-- sine of negative theta. And we know sine of negative theta is equal to the negative sine of theta. is the negative sine of theta. So we could write the sine of theta here and then put the negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared, which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared plus 25 squared. This is 2 times 25/2 squared plus 2 times-- this is 25/2 squared again-- times sine of theta. Now we just have to solve for sine of theta. So this is going to be equal to-- well, this is 576 is equal to 2 times 25/2 squared."
    },
    {
        "Q": "\nIn 2:15 did he make a mistake?\nYou are supposed to divide with the biggest number inside the box, but he did the opposite. Is that a mistake?\nM&M",
        "A": "No, it is not a mistake. The problem is 4 / 16 = 4 \u00c3\u00b7 16. 16 is the divisor, it goes outside. Maybe you need to review decimal division. This usually comes up at that time.",
        "video_name": "FaDtge_vkbg",
        "timestamps": [
            135
        ],
        "3min_transcript": "Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. The word \"cent\" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80."
    },
    {
        "Q": "\nisn't it 1/2 on 0:47?",
        "A": "No, it s 1/4. Simplify 4/16 by going like this: 4 divided by four equals 1 16 divided by four equals 4",
        "video_name": "FaDtge_vkbg",
        "timestamps": [
            47
        ],
        "3min_transcript": "Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. The word \"cent\" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80."
    },
    {
        "Q": "At 2:37, how do you know the slope of the tangent is one?\n",
        "A": "It is given that tan \u00ce\u00b8 = 1 in the problem.",
        "video_name": "MABWdzmZFIQ",
        "timestamps": [
            157
        ],
        "3min_transcript": "is going to be the inverse tangent of one. So it might be tempting to just pick this one right over here. Type inverse tangent of one into his calculator. So maybe this looks like the best choice. But remember, I said if we restrict the domain right over here. if we restrict the possible values of tangent, of theta here appropriately, then this is going to simplify to this. But there is a scenario where this does not happen. And that's if we pick thetas that are outside of the range of the inverse tangent function. What do I mean by that? Well it's really just based on the idea that there are multiple angles that have... or multiple angles whose tangent is one. And let me draw that here with a unit circle here. So we draw a unit circle, so that's my x axis, that's my y axis, let me draw my unit circle here. have to draw the unit circle, because the tangent is really much more about the slope of the ray created by the angle, than where it intersects the unit circle as would be the case with sine and cosine. So if you have.. so you could have this angle right over here. So let's say this is a candidate theta, where the tangent of this theta is the slope of this line, and this terminal angle, the terminal ray, you could say of the angle. The other side, the initial ray, is along the positive x axis. And so you could say, okay the tangent of this theta, the tangent of this theta is one. Because the slope of this line is one. Let me scroll over a little bit. Well, so let me write it this way. So tangent theta is equal to one. But I can construct another theta whose tangent is equal to one by going all the way over here and essentially going in the opposite direction but the slope of this line, so let's call this theta two, tangent of theta two is also going to be equal to one. And of course you could go another but that's functionally the same angle in terms of where it is relative to the positive x axis, or what direction it points into, but this one is fundamentally a different angle. So we do not know, we do not have enough information just given what we've been told to know exactly which theta we're talking about, whether we're talking about this orange theta or this mauve theta. So I would say the get more information, there are multiple angles which fit this description."
    },
    {
        "Q": "\nAt 7:43, I would have separated it into the span of [.5,0] and [0,1]. Why is that wrong?",
        "A": "The span of (1/2, 1) is a line parallel to (1/2, 1). The span of {(1/2, 0), (0, 1)} is the span of {(1, 0), (0, 1)} - the xy plane.",
        "video_name": "3-xfmbdzkqc",
        "timestamps": [
            463
        ],
        "3min_transcript": "corresponds to this pivot column, plus or minus 1/2 times my second entry has got to be equal to that 0 right there. Or, v1 is equal to 1/2 v2. And so if I wanted to write all of the eigenvectors that satisfy this, I could write it this way. My eigenspace that corresponds to lambda equals 5. That corresponds to the eigenvalue 5 is equal to the set of all of the vectors, v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1. or 1/2 times t. Just like that. For any t is a member of the real numbers. If we wanted to, we could scale this up. We could say any real number times 1, 2. That would also be the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1/2 and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy-- that work for the equation where the eigenvalue is equal to 5. Now what about when the eigenvalue is equal to minus 1? So let's do that case. When lambda is equal to minus 1, then we have-- it's going So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row."
    },
    {
        "Q": "at 9:40, i got 1 and 1 at the bottom row instead because I kept the bottom row the same rather than the top row. Is that also correct?\n",
        "A": "Yes, that has the same meaning. You will still end up with v1 + v2 = 0.",
        "video_name": "3-xfmbdzkqc",
        "timestamps": [
            580
        ],
        "3min_transcript": "So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row. Minus 2, minus 2. And then my second row, I'll replace it with two times-- I'll replace it with it plus 2 times the first. Or even better, I'm going to replace it with it plus minus 2 times the first. So minus 4 plus 4 is 0. And then if I divide the top row by minus 2, the reduced row echelon form of this matrix right here or this matrix right here is going to be 1, 1, 0. So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here It's the set of vectors that satisfy this equation: 1, 1, 0, 0. And then you have v1, v2 is equal to 0. Or you get v1 plus-- these aren't vectors, v1 plus v2 is equal to 0. Because 0 is just equal to that thing right there. So 1 times v1 plus 1 times v2 is going to be equal to that 0 right there. Or I could write v1 is equal to minus v2. Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors, v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1. So let's just graph this a little bit just to understand what we just did. We were able to find two eigenvalues for"
    },
    {
        "Q": "why couldn't the five at 0:47 be plus or minus five?\n",
        "A": "mj, You re correct. 25x\u00e2\u0081\u00b4-30x\u00c2\u00b2+9 can be factored to either (5x\u00c2\u00b2-3)(5x\u00c2\u00b2-3) or (-5x\u00c2\u00b2+3)(-5x\u00c2\u00b2+3)",
        "video_name": "o-ZbdYVGehI",
        "timestamps": [
            47
        ],
        "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9."
    },
    {
        "Q": "Can anyone explain to me where the 2 came from at 1:26? Thanks!\n",
        "A": "He s just checking to make sure that it is a perfect square",
        "video_name": "o-ZbdYVGehI",
        "timestamps": [
            86
        ],
        "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9."
    },
    {
        "Q": "At 4:34, How did you turn the 370, to a 371?? Why add the 1??\n",
        "A": "The solution to the problem was that she should sell 370.3 computers in order to make a profit. However, in real life you can t sell a fraction of a computer. If she sold 370 computers she would make too little money to make a profit, so you have to round up and say that she should sell 371.",
        "video_name": "roHvNNFXr4k",
        "timestamps": [
            274
        ],
        "3min_transcript": "And then to solve for x, we just have to divide both sides by 27. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times. 6 times 7 is 42. 6 times 2 is 12. Plus 4. 16. Let's see, 90 minus 62 is actually 28. Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14. Plus 4 is 18. There you go. So 190 minus 189. We get 1. Let's bring down another 0. We bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we'll put a 0 right there. 0 times 27 is 0. Then we subtract. And then, we get 10 again. And now we're in the decimal range. Or we're going to start getting decimal values. We can bring down another 0. We could get 27 goes into 100 three times. And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third of a computer. She could either sell 370 computers. If she sells 370 computers, then she's not going to get to break even because that's less than the quantity she needs for break even. So she needs to sell 371. She needs to sell 371 computers in a month to make a profit."
    },
    {
        "Q": "\nWhy did Sal switch sides 2:28 (the variable and the number)",
        "A": "He switched the sides simply because it s customary to write equations with the variable on the left and the number on the right. Switching the sides doesn t change the validity of the equation. For example, if 2 + 7 = 13 - 4, then 13 -4 = 2 + 7.",
        "video_name": "roHvNNFXr4k",
        "timestamps": [
            148
        ],
        "3min_transcript": "Marcia has just opened her new computer store. She makes $27 on every computer she sells and her monthly expenses are $10,000. What is the minimum number of computer she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now, let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit? So I'll write her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computer she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0. And then, maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even-- that's 0 profit. Neither positive or negative-- is equal to 27 times-- and I'll do it all in one color now. 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides. Add 10,000 to both sides, so it's no longer on the right-hand side. And we are left with 10,000. And then to solve for x, we just have to divide both sides by 27. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times."
    },
    {
        "Q": "\nAt 2:03, why does Khan leave a blank space?\n\nshouldn't it be; (assume \"///\" is just a space for formatting purpose)\n///////////////////// (4/9)^1 + (4/9)^2 + (4/9)^3 + . . .\n///////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + . . .\n//////////////////// ----------------------------------------- -\nnot\n/////////////////////////// (4/9) + (4/9)^2 + (4/9)^3 + ...\n///////////////////////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + ...\n///////////////////// ----------------------------------------- -",
        "A": "Use code blocking for monospace formatting. (Click the Formatting tips hyperreference below the entry block) (4/9)^1 + (4/9)^2 + (4/9)^3 + ... (4/9)^2 + (4/9)^3 + (4/9)^4 + ... You want the addition to be applied to equivalent exponents: (4/9)^1 + (4/9)^2 + (4/9)^3 + ... (4/9)^2 + (4/9)^3 + (4/9)^4 + ... (4/9)^1 + 2*(4/9)^2 + 2*(4/9)^3 + 2*...",
        "video_name": "vBlR2xNAGmo",
        "timestamps": [
            123
        ],
        "3min_transcript": "In the last video we got as far as figuring out that the area of this Koch snowflake This thing that has an infinite perimeter, can be expressed as this infinite sum over here So our job in this video is to try to simplify this, and hopefully get a finite value Let's do our best to actually simplify this thing right over here So the easiest part of this thing to simplify is this right over here So let's just focus on that Then if we can get a value for this part that I am bracketing off Then we can just place that value here and simplify the rest of it So what I've just bracketed off can be re-written as three times four ninths plus four ninths squared plus four ninths to the third power And you can go on and on and on Plus four ninths to every other power, all the way through infinity! Lucky for us, there is a way to figure out this infinite (geometric) series But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas So let's say that we define some sum, this one over here (let's call it S) Let's say that S is equal to what we have in parentheses over here It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third all the way to infinity Now let's also say that we multiply S by four ninths What's four ninths S going to look like? So then, I'm just essentially multiplying every term here by four ninths So if I take this first term and multiply it by four ninths, what am I going to get? If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power And we are going to go all the way to infinity So this is interesting When I multiply four ninths times this I get all of the terms here except for this first four ninths Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series We can subtract this term right over here (this pink line) from this green line If we do that, clearly this is equal to that and this is equal to that So if we subtract this from that its equivalent to subtracting the pink from the green So we get S minus four ninths S is equal to... Well, every other term, this guy minus this guy is going to cancel out"
    },
    {
        "Q": "Why does Sal use shorthand at 5:28?\n",
        "A": "conserving space.",
        "video_name": "aTjNDKlz8G4",
        "timestamps": [
            328
        ],
        "3min_transcript": "Then I'll just cancel this out and I would have gotten inches per minute. So anyway, I don't want to confuse you too much with all of that unit cancellation stuff. The bottom line is you just remember, well if I'm going 28 feet per minute, I'm going to go 12 times that many inches per minute, right, because there are 12 inches per foot. So I'm going 336 inches per minute. So now I have the question, but we're not done, because the question is how many inches am I going to be traveling in 1 second. So let me erase some of the stuff here at the bottom. So 336 inches -- let's write it like that -- inches per minute, Well what do we know? We know that 1 minute -- and notice, I write it in the numerator here because I want to cancel it out with this minute here. 1 minute is equal to how many seconds? It equals 60 seconds. And this part can be confusing, but it's always good to just take a step back and think about what I'm doing. If I'm going to be going 336 inches per minute, how many inches am I going to travel in 1 second? Am I going to travel more than 336 or am I going to travel less than 336 inches per second. Well obviously less, because a second is a much shorter period of time. So if I'm in a much shorter period of time, I'm going to be traveling a much shorter distance, if I'm So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we"
    },
    {
        "Q": "\n6:39 how did you get 5.6 shouldn't it be 5.3?\nSorry, I'm confused. :/",
        "A": "Its basically 336 divided by 60. Your answer is 5.6. Hope that helps :)",
        "video_name": "aTjNDKlz8G4",
        "timestamps": [
            399
        ],
        "3min_transcript": "So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we 6 goes into 336, what, 56 times? 56 over 10, and then we can divide that again by 2. So then that gets us 28 over 5. And 28 over 5 -- let's see, 5 goes into 28 five times, 25. 3, 5.6. So this equals 5.6. So I think we now just solved the problem. If Zack is going 28 feet in every minute, that's his speed, he's actually going 5.6 inches per second. Hopefully that kind of made sense. Let's try to see if we could do another one. per hour is that? Well, 91 feet per second. If we want to say how many miles that is, should we be dividing or should we be multiplying? We should be dividing because it's going to be a smaller number of miles. We know that 1 mile is equal to -- and you might want to just memorize this -- 5,280 feet. It's actually a pretty useful number to know. And then that will actually cancel out the feet. Then we want to go from seconds to hours, right? So, if we go from seconds to hours, if I can travel 91 feet"
    },
    {
        "Q": "You lost me at 6:39 >.<\n",
        "A": "Its basically 336 divided by 60. Your answer is 5.6. Hope that helps :)",
        "video_name": "aTjNDKlz8G4",
        "timestamps": [
            399
        ],
        "3min_transcript": "So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we 6 goes into 336, what, 56 times? 56 over 10, and then we can divide that again by 2. So then that gets us 28 over 5. And 28 over 5 -- let's see, 5 goes into 28 five times, 25. 3, 5.6. So this equals 5.6. So I think we now just solved the problem. If Zack is going 28 feet in every minute, that's his speed, he's actually going 5.6 inches per second. Hopefully that kind of made sense. Let's try to see if we could do another one. per hour is that? Well, 91 feet per second. If we want to say how many miles that is, should we be dividing or should we be multiplying? We should be dividing because it's going to be a smaller number of miles. We know that 1 mile is equal to -- and you might want to just memorize this -- 5,280 feet. It's actually a pretty useful number to know. And then that will actually cancel out the feet. Then we want to go from seconds to hours, right? So, if we go from seconds to hours, if I can travel 91 feet"
    },
    {
        "Q": "At 1:01, why is it called the Foil method? Does 'Foil' stand for something or is it just called that?\n",
        "A": "FOIL stands for First, Outside, Inside, Last . You multiply together the first term in each binomial, the outside (leftmost and rightmost) terms, the inside terms, and the last term in each binomial. Take those four products, add them up, and you have the expanded expression.",
        "video_name": "HLNSouzygw0",
        "timestamps": [
            61
        ],
        "3min_transcript": "- [Instructor] We're now going to explore factoring a type of expression called a difference of squares and the reason why it's called a difference of squares is 'cause it's expressions like x squared minus nine. This is a difference. We're subtracting between two quantities that are each squares. This is literally x squared. Let me do that in a different color. This is x squared minus three squared. It's the difference between two quantities that have been squared and it turns out that this is pretty straightforward to factor. And to see how it can be factored, let me pause there for a second and get a little bit of review of multiplying binomials. So put this on the back burner a little bit. Before I give you the answer of how you factor this, let's do a little bit of an exercise. Let's multiply x plus a times x minus a where a is some number. Now, we can use that, but I like just thinking of this as a distributive property twice. We could take x plus a and distribute it onto the x and onto the a. So when we multiply it by x, we would get x times x is x squared, a times x is plus ax and then when we multiply it by the negative a, well, it'll become negative a times x minus a squared. So these middle two terms cancel out and you are left with x squared minus a squared. You're left with a difference of squares. x squared minus a squared. So we have an interesting result right over here that x squared minus a squared is equal to, is equal to x plus a, x plus a times x minus a. And so we can use, and this is for any a. Here, what is our a? Our a is three. This is x squared minus three squared or we could say minus our a squared if we say three is a and so to factor it, this is just going to be equal to x plus our a which is three times x minus our a which is three. So x plus three times x minus three. Now, let's do some examples to really reinforce this idea of factoring differences of squares. So let's say we want to factor, let me say y squared minus 25 and it has to be a difference of squares. It doesn't work with a sum of squares. Well, in this case, this is going to be y and you have to confirm, okay, yeah, 25 is five squared and y squared is well, y squared. So this gonna be y plus something times y minus something"
    },
    {
        "Q": "\nWhy add -11 to the other side when you could add 4 and equal them out 0:31",
        "A": "Lauri, The problem was to Use the drop down menu to form a linear equation with no solution. If you chose the 4, you would have a linear equation which is valid for any x and so it has an infinite number of solutions for x.",
        "video_name": "uQs100shv-A",
        "timestamps": [
            31
        ],
        "3min_transcript": "We're asked to use the drop-downs to form a linear equation with no solutions. So a linear equation with no solutions is going to be one where I don't care how you manipulate it, the thing on the left can never be equal to the thing on the right. And so let's see what options they give us. One, they want us to-- we can pick the coefficient on the x term and then we can pick the constant. So if we made this negative 11x, so now we have a negative 11x on both sides. Here on the left hand side, we have negative 11x plus 4. If we do something other than 4 here, so if we did say negative 11x minus 11, then here we're not going to have any solutions. And you say, hey, Sal how did you come up with that? Well think about it right over here. We have a negative 11x here, we have a negative 11x there. If you wanted to solve it algebraically you could add 11x to both sides and both of these terms will cancel out with each other and all you would be left with is a 4 is equal to a negative 11, which is not possible for any x that you pick. Another way that you think about it is here we have negative 11 times some number taking negative 11 times that same number and we're subtracting 11 from it. So if you take a negative 11 times some number and on one side you add four, and on the other side you subtract 11, there's no way, it doesn't matter what x you pick. There's no x for which that is going to be true. But let's check our answer right over here."
    },
    {
        "Q": "At 1:20 Sal says that there is only one triangle, but couldn't he have pulled it to the right?\n",
        "A": "If you mean, could he have pulled the bottom right point to the right more , then no. If he pulled it to the right more, then the triangle wouldn t have two sides of length 3, which is what the directions are asking for.",
        "video_name": "lohMwoq3WFA",
        "timestamps": [
            80
        ],
        "3min_transcript": "They're asking us to draw a right triangle. So that means it has to have a 90-degree angle. But it's also an isosceles triangle, so that means it has to have at least two sides equal and has two sides of length 3. So those two sides that are going to be equal are going to be of length 3, and it's got to be a right triangle. So let's see if we can do that. So let's try to make this right over here the right angle. And let's make this side and this side have length 3, so 3 and then 3 right over there. Let me make sure I get that right angle right. OK, there you go. So it's a right angle. It's isosceles. At least two sides are equal. And the two sides have length 3. So it seems like we've met all of our constraints. Now they say, is there a unique triangle that satisfies this condition? So another way of rephrasing that, is this the only triangle that I could have drawn that meets these conditions? I can't change this angle if I want to meet these conditions. I can't change these two lengths. And if you keep this angle constant and you keep these two lengths constant, then this point and this point are going to be So this is the only side that can connect those two points. So this is the only triangle that meets those conditions. You can't have different side lengths, or you couldn't have different angles right over here and also meet those conditions. So is there a unique triangle that satisfies the given conditions? Yes, there's only one unique triangle."
    },
    {
        "Q": "at 2:37.... someone tell me that was a typo on the transcript.\n",
        "A": "Of course it is, it should say when is f of x increasing . A very unfortunate typo.",
        "video_name": "KxOp3s9ottg",
        "timestamps": [
            157
        ],
        "3min_transcript": "between a and b. At x equals a or at x equals b the value of our function is zero but it's positive when x is between a and b, a and b or if x is greater than c. X is, we could write it there, c is less than x or we could write that x is greater than c. These are the intervals when our function is positive. Let me write this, f of x, f of x positive when x is in this interval or this interval or that interval. So when is f of x negative? Let me do this in another color. F of x is going to be negative. Well, it's gonna be negative if x is less than a. So this is if x is less than a or if x is between b and c F of x is down here so this is where it's negative. So here or, or x is between b or c, x is between b and c. And I'm not saying less than or equal to because at b or c the value of the function f of b is zero, f of c is zero. That's where we are actually intersecting the x-axis. So that was reasonably straightforward. Now let's ask ourselves a different question. When is the function increasing or decreasing? So when is f of x, f of x increasing? Well increasing, one way to think about it is every time that x is increasing then y should be increasing or another way to think about it, you have a, you have a positive rate of change of y with respect to x. We could even think about it as imagine if you had a tangent line at any of these points. of that tangent line is going to be positive. But the easiest way for me to think about it is as you increase x you're going to be increasing y. So where is the function increasing? Well I'm doing it in blue. So it's increasing right until we get to this point right over here, right until we get to that point over there then it starts decreasing until we get to this point right over here and then it starts increasing again. It starts, it starts increasing again. So let me make some more labels here. So let's say that this, this is x equals d and that this right over here, actually let me do that in green color, so let's say this is x equals d. Now it's not a, d, b but you get the picture and let's say that this is x is equal to, x is equal to,"
    },
    {
        "Q": "at about 6:45 mins into the video, sal differentiates 6tan(theta). why is 6 not differentiated but tan is?\n",
        "A": "hmm.... You know what... a constant can never be differentiated...... !!!",
        "video_name": "fD7MbnXbTls",
        "timestamps": [
            405
        ],
        "3min_transcript": "Let me scroll over more. So that's the derivative of the outside. If the cosine theta was just an x, you would say x to the minus 1 derivative is minus 1 x to the minus 2. Now times the derivative of the inside. Of cosine of theta with respect to theta. So that's times minus sine of theta. I'm going to multiply all of that times sine of theta. The derivative of this thing, which is the stuff in green, times the first expression. So what does this equal? These cosine of theta divided by cosine of theta, that is equal to 1. And then I have a minus 1 and I have a minus sine of theta. That's plus plus. What do I have? I have sine squared, sine of theta time sine of theta over cosine squared. So plus sine squares of theta over cosine squared of theta. Which is equal to 1 plus tangent squared of theta. What's 1 plus tangent squared of theta? That's equal to secant squared of theta. secant squared of theta. All that work to get us fairly something-- it's nice when it comes out simple. So d x d theta, this is just equal to secant squared of theta. If we want to figure out what d x is equal to, d x is equal to just both sides times d theta. So it's 6 times secant squared theta d theta. That's our d x. Of course, in the future we're going to have to back substitute, so we want to solve for theta. That's fairly straightforward. Just take the arctangent of both sides of this equation. You get that the arctangent of x over 6 is equal to the theta. We'll save this for later. So what is our integral reduced to? Our integral now becomes the integral of d x? What's d x? It is 6 secant squared theta d theta. times 1 plus tangent squared of theta. We know that this right there is secant squared of theta. I've shown you that multiple times. So this is secant squared of theta in the denominator. We have a secant squared on the numerator, they cancel out. So those cancel out. So are integral reduces to, lucky for us, 6/36 which is just 1/6 d theta. Which is equal to 1/6 theta plus c. Now we back substitute using this result. Theta is equal to arctangent x over 6. The anti-derivative 1 over 36 plus x squared is equal to 1/6 times theta. Theta's just equal to the arctangent x over 6 plus c."
    },
    {
        "Q": "\nAt 1:05 for the slope formula of y=6 i thought the slope is 6 and the y intercept is 0? Is that right? If not, why isn't it?",
        "A": "If the slope was 6, there would be a x after the 6.",
        "video_name": "y5yNi08cr6I",
        "timestamps": [
            65
        ],
        "3min_transcript": "We're asked which of these lines are parallel. So they give us three equations of three different lines and if they're parallel, then they have to have the So all we have to do over here is figure out the slopes of each of these lines, and if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form, we can just divide both sides of this equation by 2. We get y is equal to 6x-- right, 12 divided by 2 -- 6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to six. You might say this hey, this is a bizarre character, how do I get this into slope-intercept form, where's the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. slope here is 0. y is going to be equal to six no matter how much you change x. Change in y is always going to be 0, it's always going to be 6. So here, our slope is 0, so these two lines are definitely not parallel, they have different slopes. So let's try line C. Line C-- I'll do it down here. Line C, so it's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So the point negative 2, 2, is being represented here because you're subtracting the points. And the slope is 6, so we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form, so let's put it in slope-intercept form will still be equal to 6. So if we distribute the 6, we get y minus 2 is equal to 6 times x, 6x, plus 6 times 2 is 12. And if you add this 2 -- if you add 2 to both sides of the equation, you get y-- because these guys cancel out-- is equal to 6x plus 14. So you see, once again, the slope is 6. So line A and line C have the same the slope, so line A and line C are parallel. And they're different lines. If they had the same y-intercept, then they would just be the same line."
    },
    {
        "Q": "At 0:25, doesn't Sal mean \"letters,\" and not \"alphabets\"?\n",
        "A": "In the general idea... letters... some people say different things.",
        "video_name": "VYbqG2NuOo8",
        "timestamps": [
            25
        ],
        "3min_transcript": "- [Voiceover] So let's ask ourselves some interesting questions about alphabets in the English language. And in case you don't remember and are in the mood to count, there are 26 alphabets. So if you go, \"A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, and Z,\" you'll get, you'll get 26, 26 alphabets. Now let's ask some interesting questions. So given that there are 26 alphabets in the English language, how many possible three letter words are there? And we're not going to be thinking about phonetics or how hard it is to pronounce it. So, for example, the word, the word ZGT would be a legitimate word in this example. Or the word, the word, the word SKJ would be a legitimate word in this example. So how many possible three letter words are there in the English language? I encourage you to pause the video and try to think about it. Alright, I assume you've had a go at it. So let's just think about it, for three letter words possibilities are there for the first one? Well, there's 26 possible letters for the first one. Anything from a to z would be completely fine. Now how many possibilities for the second one? And I intentionally ask this to you to be a distractor because we've seen a lot of examples. We're saying, \"Oh, there's 26 possibilities \"for the first one and maybe there's 25 for the second one, \"and then 24 for the third,\" but that's not the case right over here because we can repeat letters. I didn't say that all of the letters had to be different. So, for example, the word, the word HHH would also be a legitimate word in our example right over here. So we have 26 possibilities for the second letter and we have 26 possibilities for the third letter. So we're going to have, and I don't know what this is, 26 to the third power possibilities, or 26 times 26 time 26 and you can figure out what that is. That is how many possible three letter words we can have for the English language they are, if they meant anything and if we repeated letters. Now let's ask a different question. What if we said, \"How many possible three letter words \"are there if we want all different letters?\" So we want all different letters. So these all have to be different letters. Different, different letters and once again, pause the video and see if you can think it through. Alright, so this is where permutations start to be useful. Although, I think a lot of things like this, it's always best to reason through than try to figure out if some formula applies to it. So in this situation, well, if we went in order, we could have 26 different letters for the first one, 26 different possibilities for the first one. You know, I'm always starting with that one, but there's nothing special about the one on the left. We could say that the one on the right, there's 26 possibilities, well for each of those possibilities, for each of those 26 possibilities, there might be 25 possibilities"
    },
    {
        "Q": "At 4:11 when You got the 3rd row did you not make a mistake when you are converting the row, You said it was the 3rd row- 2X the second row. so for the 3rd row shouldn't it be 0 3 3 2 ? Because in the video you did the 3rd- 2x the first row.....\n",
        "A": "I guess that was just misspoken, since the first row was the only row kept original. Sal meant to do the 3rd row minus 2 times the first row.",
        "video_name": "QV0jsTiobU4",
        "timestamps": [
            251
        ],
        "3min_transcript": "Everything above the main diagonal is 0, so this is the main diagonal right here, all the way down like that. All of these guys are going to be non-zero. All of that's going to be non-zero, and then the 0's are going to be above the diagonal, like that. We saw in the last video that the determinant of this guy is just equal to the product of the diagonal entries, which is a very simple way of finding a determinant. And you could use the same argument we made in the last video to say that the same is true of the lower triangular matrix, that its determinant is also just the product of those entries. I won't prove it here, but you can use the exact same argument you used in the video that I just did on the upper triangular. So given this, that the determinant of this is just the product of those guys, and that I can perform row operations on this guy and not change the determinant, maybe a simpler way to calculate this determinant is to get multiply the entries down the diagonal. So let's do that. So we want to find the determinant of A. It's 1, 2, 2, 1, 1, 1, 2, 4, 2, 2, 7, 5, 2, minus 1, 4, minus 6, 3. Now let's try to get this into upper triangular form. So let's replace the second row with the-- so I'm just going to keep the first row the same. 1, 2, 2, 1. And let's replace the second row with the second row minus the first row. The second row minus the first row is going to be equal to 1 minus 1 is 0. So in this case the constant is just 1. 2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. Let me get a good color here. I'll do pink. Let's replace the last row with the last row, essentially, plus the first row. You could say minus minus 1 times the first row is the same thing as the last row plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6. Minus 6 plus 2 is minus 4."
    },
    {
        "Q": "\nat 2:32 why does Sal get -xe^-x^2 instead of xe^x^2?",
        "A": "Those negatives do not cancel.",
        "video_name": "DL-ozRGDlkY",
        "timestamps": [
            152
        ],
        "3min_transcript": "If you can on one side of this equation through algebra separate out the Ys and the DYs and on the other side have all the Xs and DXs, and then integrate. Perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it don't worry because we're about to work through it. So like I said, let's use a little bit of algebra to get all the Ys and DYs on one side and all the Xs and DXs on the other side. So one way, let's say I want to get all the Ys and DYs on the left hand side, and all the Xs and DXs on the right hand side. Well, I can multiply both sides times Y. So I can multiply both sides times Y that has the effect of putting the Ys on the left hand side and then I can multiple both sides times DX. I can multiple both sides times DX and we kind of treat, you can treat these differentials when you're manipulating it to essentially separate out the variables. And so, this will cancel with that. And so, we are left with Y, DY. Y, DY is equal to negative X. And actually let me write it this way. Let me write it as negative X, E. Actually, I might need a little more space. So negative X E to the negative X squared DX. DX. Now why is this interesting? Because we could integrate both sides. And now this also highlights why we call it the separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the Ys and DYs on one side and the Xs and DXs on the other side. But this one we were able to. a separable differential equation. Differential equation. And it's usually the first technique that you should try. Hey, can I separate the Ys and the Xs and as I said, this is not going to be true of many, if not most differential equations. But now that we did this we can integrate both sides. So let's do that. So, I'll find a nice color to integrate with. So, I'm going to integrate both sides. Now if you integrate the left hand side what do you get? You get and remember, we're integrating with respect to Y here. So this is going to be Y squared over two and we could put some constant there. I could call that plus C one. And if you're integrating that that's going to be equal to. Now the right hand side we're integrating with respect to X. And let's see, you could do U substitution or you could recognize that look, the derivative of negative X squared"
    },
    {
        "Q": "In 0:28 sal says 8 goes into 37 three times but can you add the remainder??\n",
        "A": "8 goes into 37 4 times, ( because 8 x 4 = 32 ) and that is why he writes 4 on top of the line.. The reminder then is 5, ( because 37- 32 = 5 ) and he then uses that when he brings down another place value - 7.. So the next thing to figure out then is how many times 8 goes into 57, and so on.... Did that make any sense ?",
        "video_name": "MbpmP1esh-Q",
        "timestamps": [
            28
        ],
        "3min_transcript": "Let's divide 3,771 by 8. So we're going to do a little bit of long division is the way we set it up. So the first thing we want to do and think about is, well, does 8 go into 3? Well, no, 8 does not go into 3. So we'll keep moving. Does 8 go into 37? Sure. 8 times 4 is 37. 8 times 5 is 40. So that's too big. So 8 goes into 37 four times. 4 times 8 is 32. And we subtract, and we get 37 minus 32 is 5. And just so you have a little glimpse of what's actually going on here, we're really saying how many times does 8 go into 3,700? And we're saying that 8 goes into 3,700 400 times. There's 4 sitting in the hundreds place, and that 400 times 8 is 3,200. And then when you subtract 3,200 from 3,700, you're getting 500. But we'll revert back to just the traditional long division But that's what's really going on here. we're really saying 8 goes into 3,700 400 times. But let's keep going. So now, we have this 5 to deal with, which is, as we already talked about, it's in the hundreds place. It's really a 500. But now, let's bring down another place value. So let's bring down the 7. And let's think about how many times does 8 go into 57. And we're really thinking about how much 8 goes into 570. But 8 goes into 57-- let's see, 8 times 5 is 40. 8 times 6 is 48. 8 times 7 is 54. So let's go 7 times. 7 times 8. Oh, actually, I did that right. 8 times 7-- my multiplication tables are weak. It is 8 times-- so it does go in seven times. But 7 times 8 isn't 54; it's 56. I always have trouble with that one. It is 56. And so now, we subtract again. And now we have one more place value to bring down. We have-- let me do it in in a color I haven't used yet. I'll do blue. We have a 1 to bring down. And now we're literally saying, how many times does 8 go into 11? Well, 8 goes into 11 one time. 1 times 8 is 8. And we are left with-- because we have no more places to bring down-- 11 minus 8 is 3. So 8 goes into 3,771 471 times, and the remainder here is 3."
    },
    {
        "Q": "\nCan someone explain to me why the cube root and 3 exponent got erased? at 3:50ish",
        "A": "Something raised to the power of 3 is a perfect cube, like: 2^3 = 8 You can simplify cube roots of perfect cubes: cuberroot (8) = 2. The video is just a more complicated version of the example I just did. Sal has cuberoot [(x+1)^3]. The (x+1)^3 is a perfect cube. Take the cube root, and you end up with just (x+1). Hope this helps.",
        "video_name": "8GEGnSEJA2s",
        "timestamps": [
            230
        ],
        "3min_transcript": "this part just simplifies to x plus one, and then I subtract one, so it all simplified out to just being equal to x. So we're just left with an x. So, f of g of x is just x. So now, let's try what g of f of x is. So, g of f of x is going to be equal to, I'll do it right over here, this is going to be equal to the cube root of actually, let me write it out, wherever I see an x, I can write f of x instead, I didn't do it that last time, I went directly and replaced with the definition of f of x but just to make it clear what I'm doing everywhere I'm seeing an x, I replace it with an f of x. So, the cube root of f of x plus one, minus seven. Well, that's going to be equal to the cube root of cube root of f of x, which is all of this business over here minus one, and then we add one and we add one, and then we subtract the seven lucky for us, subtracting one and adding one, those cancel out. Next, we're gonna take the cube root of x plus seven to the third power. Well, the cube root of x plus seven to the third power is just going to be x plus seven so, this is going to be x plus seven, for all of this business simplifies to x plus seven, and then we do subtract seven and these two cancel out, or they negate each other and we are just left with x. So, we see something very interesting. f of g of x is just x and g of f of x is x. So, in this case, if we start with an x if we start with an x, we input it into the function g and we get g of x and then we input that into the function f, then we input that into the function f, f of g of x gets us back to x. It gets us back to x. So we kind of did a round-trip. And the same thing is happening over here. If I put x into f of x... I'm sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g into the function g, once again I do this round-trip and I get back to x. Another way to think about it, these are both composite functions, one way to think about it is, if these are the set of all possible inputs into either of these composite functions, and then, these are the outputs, so you're starting with an x, I'll do this case first,"
    },
    {
        "Q": "\nat 1:09 he says 100 times the light goes on and off. How long would that take?!?!?!",
        "A": "ok... you re weird -Midnite Blaze",
        "video_name": "-xYkTJFbuM0",
        "timestamps": [
            69
        ],
        "3min_transcript": "So we had the hundred logicians. All of their foreheads were painted blue. And before they entered the room, they were told that at least one of you hundred logicians has your forehead painted blue. And then every time that they turned on the lights, so that they could see each other, they said OK, once you've determined that you have a blue forehead, when the lights get turned off again, we want you to leave the room. And then once that's kind of settled down, they'll turn the lights on again. And people will look at each other again. And then they'll turn them off again. And maybe people will leave the room. And so forth and so on. And they're also all told that everyone in the room is a perfect logician. They have infallible logic. So the question was, what happens? And actually maybe an even more interesting question is why does it happen? So I'll answer the first, what happens? And if just take the answer, and you don't know why, it almost seems mystical. That essentially the light gets turned on and off 100 gets turned on, and the lights get turned off again, all of them leave. They all leave. So I mean, it's kind of weird, right? Let's say I'm one of them. Or you're one of them. I go into this room. The lights get turned on. And I see 99 people with blue foreheads. And I can't see my own forehead. They see my forehead, of course. But to any other person, I'm one of the 99, right? But I see 99 blue foreheads. So essentially what happens if we were to watch the show is, the lights get turned on. You see 99 blue foreheads. Then the lights get turned off again. And then the lights get turned on again. And everyone's still sitting there. And I still see 99 blue foreheads. And that happens 100 times. everyone leaves the room. And at first glance, that seems crazy, because nothing changes. Nothing changes between every time we turn on the light. But the way you need to think about this-- and this is what makes it interesting-- is what happens instead of 100, let's say there was one person in the room. So before the show starts-- they never told me that there were going to be 100 people in the room. They just said, at least one of you, at least one of the people in the room, has your forehead painted blue. And as soon as you know that your forehead is painted blue, you leave the room. And that everyone's a perfect logician. So imagine the situation where instead of 100 there's only one perfect logician. Let's say it's me. So that's the room. I walk in. And I sit down. And maybe I should do it with blue."
    },
    {
        "Q": "\nso would the ratio 9.999... : .999... would be equal to 10:1",
        "A": "Yup! just like 1.111\u00e2\u0080\u00a6 to 0.111\u00e2\u0080\u00a6 :) (10/9 to 1/9)",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            601
        ],
        "3min_transcript": "Anyway, it's 1/2 plus 1/4 plus 1/8, dot, dot, dot, dot, dot, to get 1. Each time, you fall short of 1. So how can you ever do anything? Luckily, infinity has got our backs. I mean, that's like the definition of infinity, a numbers so large, you can never get there, no matter how many steps you do, no matter how high you count. This way of writing numbers with this dot, dot, dot business, or with a bar over the repeating part, is a shorthand for an infinite series, whether it be 9/10 plus 9/100, and so on to get 1. Or 3/10 plus 3/100, and so on, to get 1/3. No matter how many 3s you write down, it will always be less than 1/3, but it will also always be less than infinity 3s. Infinity is what gets us there when no real number can. The binary equivalent of 0.9 repeating is 0.1 repeating. That's exactly 1/2, plus 1/4, plus 1/8, and so on. That's how we know a dotted, dotted, dot, dot, dot The ultimate reason that 0.9 repeating equals 1 is because it works. It's consistent, just like 1 plus 1 equals 2 is consistent, and just like 1 divided by 0 equals infinity isn't. Mathematics is about making up rules and seeing what happens. And it takes great creativity to come up with good rules. The only difference between mathematics and art is that if you don't follow your invented rules precisely in mathematics, people have a tendency to tell you you're wrong. Some rules give you elementary algebra and real numbers, and these rules can't tell the difference between 0.9 repeating and 1, just like they can't tell the difference between 0.5 and 1/2, or between 0 and negative 0. I hope you see now that the view that 9.9 repeating does not equal 10 is simply un-- 9.9 repeating --able. If you started this video thinking, I h-- 7.9 repeating that 7.9 repeating is 8, I hope now, you're thinking, oh, sweet, 4.9 repeating is 5? High 4.9 repeating! I got 98.9 repeating problems, but 0.9 repeating is 1. Here's the moral of the story. The idea of a number infinitely close to but less than 1 is not stupid or wrong, but wonderful, and beautiful, and interesting. The true mathematician takes \"you can't do that\" as a challenge. If someone tells you can't subtract a bigger number from a smaller number, just invent negative numbers. If someone tells you can't multiply a number by itself to get a negative number, then invent imaginary numbers. If someone tells you can't multiply two non-zero numbers together to get 0, or raise one non-zero number to the power of another and get 0, you should probably say, I'll do both at once, and in 8 dimensions. And if you ignore them telling you that numbers aren't 8-dimensional, and that inventing fake numbers is a useless waste of time, and then actually try to figure it out, the next thing you know, you've got split octonions, which besides being super awesome, just happen to be the perfect way to describe the wave equation of electrons and stuff."
    },
    {
        "Q": "\ni just still don't get why .9999..... equals 1 it doesn't make any sense.\nalso starting at 01:28 it starts to get confusing... can someone explain it to me???",
        "A": "It s confusing because it deals with infinities, something that algebra and arithmetic are quite incapable of adequately describing. Our brains too, because infinities are actually quite hard for us to accurately imagine. calculus, on the other hand, is full of wonderful things like numbers infinitely close but not equal to other numbers and dividing by zero and such.",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            88
        ],
        "3min_transcript": "99.9 repeating percent of mathematicians agree, 0.9 repeating equals 1. If because I said so works for you, you can go ahead and do something else now. Maybe you're like, 0.9 repeating equals one, that's this 0.9 repeating-derful! Otherwise, on to reason number 2, or reason 1.9 repeating. See, it's weird, because when we think of the number 1 or 2, in most contexts we mean it as a natural number, like 1, 2, 3, 4, 5. In the sense then, the next number after 1 is 2. 9.9 repeating may equal 10, but you wouldn't say you have 9.9 repeating lords a leaping, in the same way you wouldn't say you had 9.75 lords a leaping plus 1/4 lord a leaping. Lords, leaping or otherwise, come in natural numbers. So what does this statement mean? 0.9 repeating is the same as 1? It looks pretty different, but it equals 1 in the same way that a 1/2 equals 0.5. They have the same value, You can philosophize over whether, if 1 is the loneliest number, 0.78 plus 0.22 is just as lonely, but there's no mathematical doubt that they have the same value, just as 100 years of solitude or 99.9 repeating years of solitude. So reason 2 is not a proof, but a reason to stay open minded. Numbers that look different can have the same value. Another example of this is that, in algebra, 0 equals negative 0. 0.9 repeating is a decimal number, a real number. See, if you want 0.9 repeating to be that number infinitesimally close to 1, but not 1-- and let's face it, some of you do-- then you're writing down the wrong number when you write 0.9 repeating. That number infinitely close to 1, but less than 1, is a number, but it's not 0.9 repeating or any real number. OK, let's do more 3.9 repeating-mal proof for reason 3.9 repeating. According to this 3.9 repeating-mula, 3.9 repeating is 4. First step, say 0.9 repeating equals x. Then multiply each side by 10. Third, subtract 0.9 repeating from this side, which equals x, which we subtract from the other side. And 10x minus x is 9x. Divide by 9, and you get 1 equals x, which you might notice also equals 0.9 repeating. There's no tricks here. It's simple multiplication, subtraction and division by 9, which are all allowed because they are consistent. When something is inconsisten-- 9.9 repeating --t, we just throw it out of algebra altogether. For example, in algebra if you try to divide by 0, you get this problem where anything can equal anything. I mean, if you want to say everything is equal, fine, but your algebra sucks. Normal, everyday elementary algebra, the one they shove down students throats as if it were the only algebra, doesn't allow dividing by 0. So it stays consistent and suspiciously practical. We also could have shifted the decimal point twice, multiplying by 100 to prove that if you have 99.9 repeating bottles of beer on the wall, 99.9 repeating bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall."
    },
    {
        "Q": "At around 05:58, what did she write on the left hand side in green?\n",
        "A": "Infinity times magnified",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            358
        ],
        "3min_transcript": "a long time ago. But elementary algebra can't deal with infinity. If you allow infinity in your algebrizations, once again, you get contradictions. Infinity may not be a real number, but it is a number, a hyperreal number. Hyperreals, like infinity and the infinitesimal, follow different rules. And while algebra can't handle them, some people thought they should be numbers, and you should be able to use them. And so they figured out how, and bam! you get something like calculus. Reason 6. Take the number 1, and subtract 0.9 repeating. It's pretty clear that it's infinite 0s, but you might be tempted to think there's some sort of final 1 beyond infinity. Let's write that down as 0.0 repeating 1. Of course, if the 0 repeats infinitely, then you never get to the 1, so you might as well leave off the number. Thus the difference between 0.9 repeating and 1 is 0. There is no difference. Here's another \"there's no difference\" proof. What's the next higher real number? The game, is for any number you claim is the next real number, I can find one that's even closer to 1. Of the many delightful things about real numbers is that for any two numbers, no matter how close, there's still an infinite amount of numbers between them, and an infinite amount of numbers between those, and so on. There is no next higher real number after 1. Likewise, there's no next lower number. If 0.9 repeating and 1 were different real numbers, there would have to be infinite other real numbers between them. If you can't name a number higher than 0.9 repeating, but lower than 1, it can only be because 0.9 repeating is 1. If you don't like it, well, go to college and learn about hyperreals, or better yet, surreals. That's a system where you can have a number that's infinitely close to 1, but not 1. But even weirder, there's infinity more numbers that are infinitely even closer. Take 0.3 repeating, a repeating decimal equal to 1/3. Multiply it by 3. Obviously, by definition, 3/3 is 1, and 0.3 repeating times 3 is 0.9 repeating, which you might have noticed is also 1. The only assumption here is that 0.3 repeating equals 1/3. Maybe you don't like decimal notation in general, which brings us to reason number 9, this sum of an infinite series thing. 9/10 plus 9/100 plus 9/1000. And we can sum this series and get 1. But I can see why you might be unhappy with this. It recalls Zeno's paradoxes. How can you get across a room, when first you have to walk halfway, and then half of that, and so on. Or, how can you shoot an arrow into a target, when first it needs to go halfway, but before it can get halfway, it needs to go half of halfway, and before that, half of half of halfway, and half of half of half of halfway, and so on."
    },
    {
        "Q": "my qs is two no. are in the ratio 5:6. If 8 is subtracted from each of the no. the ratio becomes 4:5. find the no.s\n",
        "A": "the numbers are 40 and 48",
        "video_name": "DopnmxeMt-s",
        "timestamps": [
            306,
            245
        ],
        "3min_transcript": "if you want to try working this out yourself, you might want to pause the video and then play it once you're ready to see how I do it. Anyway, let me move forward assuming you haven't paused it. If we want to get rid of this 3/4, all we do is we subtract 3/4 from both sides of this equation. Minus 3/4. Well, the left hand side, the two 3/4 will just cancel. We get minus 1/2x equals, and then on the right hand side, we just have to do this fraction addition or fraction So the least common multiple of 6 and 4 is 12. So this becomes 5/6 6 is 10/12 minus 3/4 is 9/12, so we get minus 1/2x is equal to 1/12. And if that step confused you, I went a little fast, you might just want to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do is, well, the coefficient on the x term is minus 1/2, and this is now a level one problem. So to solve for x, we just multiply both sides by the reciprocal of this minus 1/2x, and that's minus 2/1 times minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and you're used to this by now, simplifies to x. The right hand side becomes minus 2/12, and we could simplify that further to minus 1/6. Well, let's check that just to make sure we got it right. So let's try to remember that minus 1/6. So the original problem was minus 1/2x, I just wrote only the left hand side of the original problem. So minus 1/2 times minus 1/6, well, that's positive 1/12 plus 3/4. Well, that's the same thing as 12, the 1 stays the same, plus 9. 1 plus 9 is 10 over 12. And that is equal to 5/6, which is what our original problem was. This stuff I wrote later. So it's 5/6, so the problem checks out. So hopefully, you're now ready to try some level two problems on your own. I might add some other example problems. But the only extra step here relative to level one problems is you'll have this constant term that you need to add or subtract from both sides of this equation, and you'll essentially turn it into a level one problem."
    },
    {
        "Q": "\nAt 4:34, I know that Sal means the twos cancel out and then it turns to one over one, which equals one, etc. But WHY do the twos cancel out?",
        "A": "2/2 means 2 divided by 2, any number divided by itself = 1. To rephrase it: How many 2 s are in 2? One. Works for any number.",
        "video_name": "DopnmxeMt-s",
        "timestamps": [
            274
        ],
        "3min_transcript": "if you want to try working this out yourself, you might want to pause the video and then play it once you're ready to see how I do it. Anyway, let me move forward assuming you haven't paused it. If we want to get rid of this 3/4, all we do is we subtract 3/4 from both sides of this equation. Minus 3/4. Well, the left hand side, the two 3/4 will just cancel. We get minus 1/2x equals, and then on the right hand side, we just have to do this fraction addition or fraction So the least common multiple of 6 and 4 is 12. So this becomes 5/6 6 is 10/12 minus 3/4 is 9/12, so we get minus 1/2x is equal to 1/12. And if that step confused you, I went a little fast, you might just want to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do is, well, the coefficient on the x term is minus 1/2, and this is now a level one problem. So to solve for x, we just multiply both sides by the reciprocal of this minus 1/2x, and that's minus 2/1 times minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and you're used to this by now, simplifies to x. The right hand side becomes minus 2/12, and we could simplify that further to minus 1/6. Well, let's check that just to make sure we got it right. So let's try to remember that minus 1/6. So the original problem was minus 1/2x, I just wrote only the left hand side of the original problem. So minus 1/2 times minus 1/6, well, that's positive 1/12 plus 3/4. Well, that's the same thing as 12, the 1 stays the same, plus 9. 1 plus 9 is 10 over 12. And that is equal to 5/6, which is what our original problem was. This stuff I wrote later. So it's 5/6, so the problem checks out. So hopefully, you're now ready to try some level two problems on your own. I might add some other example problems. But the only extra step here relative to level one problems is you'll have this constant term that you need to add or subtract from both sides of this equation, and you'll essentially turn it into a level one problem."
    },
    {
        "Q": "\nAt 2:50, not enough to solve but we can simply i think,\n15a+15b = 3a+3a+3a+3a+3a+5b+5b+5b\ni guess you see know ...\n\n3a+5b = 2\n3a+5b = 2\n3a+5b = 2 THEN\n\n2+2+2+3a+3a = ?\n6+6a= 6 . ( 1+a)\n\nis it correct ? or am i too wrong in there =)",
        "A": "You are right",
        "video_name": "CLQRZ2UbQ4Q",
        "timestamps": [
            170
        ],
        "3min_transcript": "Let's do a few more examples where we're evaluating expressions with unknown variables. So this first one we're told 3x plus 3y plus 3z is equal to 1, and then we're asked what's 12x plus 12y plus 12z equal to? And I'll give you a few moments to think about that. Well let's rewrite this second expression by factoring out the 12, so we get 12 times x plus y plus z. That's this second expression here, and you can verify that by distributing the 12. You'll get exactly this right up here. Now, what is 12 times x plus y plus z? Well, we don't know yet exactly what x plus y plus z is equal to, but this first equation might help us. This first equation, we can rewrite this left-hand side by factoring out the 3, so we could rewrite this as 3 times x plus y plus z is equal to 1. All I did is I factored the 3 out on the left-hand side. I just divide both sides of this equation by 3, and I'm left with x plus y plus z is equal to 1/3, and so here, instead of x plus y plus z, I can write 1/3. So this whole thing simplified to 12 times 1/3. 12 times 1/3 is the same thing as 12 divided by 3, which is equal to 4. Let's try one more. So here we are told that 3a plus 5b is equal to 2, and then we're asked what's 15a plus 15b going to be equal to? So we might-- let's see. I'll give you a few moments to try to tackle this on your own. Let's see how we might do it. We could approach it the way we've approached the last few problems, trying to rewrite the second expression. We could rewrite it as 15 times a plus b, and so we just have to figure out what a plus b is, And so, it's tempting to look up here, and say maybe we can solve for a plus b somehow, but we really can't. If we divide-- if we try to factor out a 3, we'll get 3 times a plus 5/3b, so this doesn't really simplify things in terms of a plus b. If we try to factor out a 5, we'd get 5 times 3/5a plus b is equal to 2, but neither of these gets us in a form where we can then solve for a plus b. So in this situation, we actually do not have enough information to solve this problem. So it's a little bit of a trick. Not enough info to solve. Anyway, hopefully you enjoyed that."
    },
    {
        "Q": "at 4:27 it shows reflection 0,1 to 1,1 and I see this as Y=1 but for the question y=x\u00e2\u0088\u00922 answered 0,2 to 2,4 and not sure why its wrong.\n",
        "A": "You weren t plugging in the numbers correctly. y=x-2 for (0,2) would give you 2=0-2 which is not correct since 2 does not equal -2. Plugging in (2,4) for the second part would give you 4=2-2 which is a false statement as well.",
        "video_name": "vO1Ur38PGCY",
        "timestamps": [
            267
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nwait at 6:39 why did he simplify",
        "A": "The sooner you simplify the more extra work you avoid.",
        "video_name": "bAerID24QJ0",
        "timestamps": [
            399
        ],
        "3min_transcript": "This 4 becomes a 1 and this becomes a 10. If you remember when you're multiplying fractions, you can simplify it like that. So it actually becomes minus-- actually, plus 30, because we have a minus times a minus and 3 times 10, over, the 4 is now 1, so all we have left is 39. And 30/39, if we divide the top and the bottom by 3, we get 10 over 13, which is the same thing as what the equation said we would get, so we know that we've got the right answer. Let's do one more problem. Minus 5/6x is equal to 7/8. yourself, now's a good time to pause, and I'm going to start doing the problem right now. So same thing. What's the reciprocal of minus 5/6? Well, it's minus 6/5. We multiply that. If you do that on the left hand side, we have to do it on the right hand side as well. Minus 6/5. The left hand side, the minus 6/5 and the minus 5/6 cancel out. We're just left with x. And the right hand side, we have, well, we can divide both the 6 and the 8 by 2, so this 6 becomes negative 3. This becomes 4. 7 times negative 3 is minus 21/20. And assuming I haven't made any careless mistakes, that should be right. Actually, let's just check that real quick. Minus 5/6 times minus 21/20. Turn this into a 4. Make this into a 2. Make this into a 7. Negative times negative is positive. So you have 7. 2 times 4 is 8. And that's what we said we would get. So we got it right. I think you're ready at this point to try some level one equations. Have fun."
    },
    {
        "Q": "\nat 2:10 what do u mean by recipical",
        "A": "a reciprocal is the inverted number. if you have 3/4 the reciprocal is 4/3. if the number is a whole we know the denominator is 1 therefore any whole number a has the reciprocal 1/a. A number like 3 has the reciprocal of 1/3.",
        "video_name": "bAerID24QJ0",
        "timestamps": [
            130
        ],
        "3min_transcript": "Welcome to level one linear equations. So let's start doing some problems. So let's say I had the equation 5-- a big fat 5, 5x equals 20. So at first this might look a little unfamiliar for you, but if I were to rephrase this, I think you'll realize this is a pretty easy problem. This is the same thing as saying 5 times question mark equals 20. And the reason we do the notation a little bit-- we write the 5 next to the x, because when you write a number right next to a variable, you assume that you're multiplying them. So this is just saying 5 times x, so instead of a question mark, we're writing an x. So 5 times x is equal to 20. Now, most of you all could do that in your head. You could say, well, what number times 5 is equal to 20? Well, it equals 4. But I'll show you a way to do it systematically just in case that 5 was a more complicated number. So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4."
    },
    {
        "Q": "what are conventions? 3:15\n",
        "A": "A convention is an unspoken agreement to do something a certain way. Grammar, spelling, the order of operations, etc. are all systems of conventions.",
        "video_name": "bAerID24QJ0",
        "timestamps": [
            195
        ],
        "3min_transcript": "So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4. to use times, and you're probably wondering why in algebra, there are all these other conventions for doing times as opposed to just the traditional multiplication sign. And the main reason is, I think, just a regular multiplication sign gets confused with the variable x, so they thought of either using a dot if you're multiplying two constants, or just writing it next to a variable to imply you're multiplying a variable. So if we multiply the left hand side by negative 4/3, we also have to do the same thing to the right hand side, minus 4/3. The left hand side, the minus 4/3 and the 3/4, they cancel out. You could work it out on your own to see that they do. They equal 1, so we're just left with x is equal to 10 times minus 4 is minus 40, 13 times 3, well, that's equal to 39."
    },
    {
        "Q": "I always learned the equation was A(n)=a+(n-1)*(d) where a equals the first term, n is the term number, and d is the common difference. So the equation for 3:14 is A(100)=15+(100-1)*(-6)=-579. Is this correct?\n",
        "A": "Yes, that is correct (5:53). Good job!",
        "video_name": "JtsyP0tnVRY",
        "timestamps": [
            194
        ],
        "3min_transcript": "Then to go from 9 to 3, well, we subtracted 6 again. And then to go from 3 to negative 3, well, we subtracted 6 again. So it looks like, every term, you subtract 6. So the second term is going to be 6 less than the first term. The third term is going to be 12 from the first term, or negative 6 subtracted twice. So in the third term, you subtract negative 6 twice. In the fourth term, you subtract negative 6 three times. So whatever term you're looking at, you subtract negative 6 one less than that many times. Let me write this down just so-- Notice when your first term, you have 15, and you don't subtract negative 6 at all. Or you could say you subtract negative 6 0 times. let me write it better this way --minus 0 times negative 6. That's what that first term is right there. What's the second term? This is 15. We just subtracted negative 6 once, or you could say, minus 1 times 6. Or you could say plus 1 times negative 6. Either way, we're subtracting the 6 once. Now what's happening here? This is 15 minus 2 times negative 6-- or, sorry --minus 2 times 6. We're subtracting a 6 twice. What's the fourth term? This is 15 minus-- We're subtracting the 6 three times from the 15, so minus 3 times 6. So, if you see the pattern here, when we have our fourth The fourth term, we have a 3. The third term, we have a 2. The second term, we have a 1. So if we had the nth term, if we just had the nth term here, what's this going to be? It's going to be 15 minus-- You see it's going to be n minus 1 right here. When n is 4, n minus 1 is 3. When n is 3, n minus 1 is 2. When n is 2, n minus 1 is 1. When n is 1, n minus 1 is 0. So we're going to have this term right here is n minus 1. So minus n minus 1 times 6. So if you want to figure out the 100th term of this sequence, I didn't even have to write it in this general term, you can just look at this pattern. It's going to be-- and I'll do it in pink --the 100th term in our sequence-- I'll continue our table down --is going to be what? It's going to be 15 minus 100 minus 1, which is 99, times 6."
    },
    {
        "Q": "\nWhy did he put the 0 at 2:16?",
        "A": "The 0 represents how many times we want to subtract 6 (or add -6, that s the same thing) from our starting value, which is 15. More generally, the value x represents how many times we want to subtract 6 from our starting value of 15. Because the very first value is 15, we don t want to subtract 6 from it, because that would give us 9, which is the second term, so we write 15 - (0)*6. Anything times 0 is 0, so that becomes 15 - 0, but subtracting or adding nothing to an equation doesn t change it s value.",
        "video_name": "JtsyP0tnVRY",
        "timestamps": [
            136
        ],
        "3min_transcript": "We are asked, what is the value of the 100th term in this sequence? And the first term is 15, then 9, then 3, then negative 3. So let's write it like this, in a table. So if we have the term, just so we have things straight, and then we have the value. and then we have the value of the term. I'll do a nice little table here. So our first term we saw is 15. Our second term is 9. Our third term is 3. I'm just really copying this down, but I'm making sure we associate it with the right term. And then our fourth term is negative 3. And they want us to figure out what the 100th term of this sequence is going to be. So let's see what's happening here, if we can discern some type of pattern. So when we went from the first term to the second term, what happened? 15 to 9. Looks like we went down by 6. It's always good to think about just how much the numbers changed by. That's always the simplest type of pattern. Then to go from 9 to 3, well, we subtracted 6 again. And then to go from 3 to negative 3, well, we subtracted 6 again. So it looks like, every term, you subtract 6. So the second term is going to be 6 less than the first term. The third term is going to be 12 from the first term, or negative 6 subtracted twice. So in the third term, you subtract negative 6 twice. In the fourth term, you subtract negative 6 three times. So whatever term you're looking at, you subtract negative 6 one less than that many times. Let me write this down just so-- Notice when your first term, you have 15, and you don't subtract negative 6 at all. Or you could say you subtract negative 6 0 times. let me write it better this way --minus 0 times negative 6. That's what that first term is right there. What's the second term? This is 15. We just subtracted negative 6 once, or you could say, minus 1 times 6. Or you could say plus 1 times negative 6. Either way, we're subtracting the 6 once. Now what's happening here? This is 15 minus 2 times negative 6-- or, sorry --minus 2 times 6. We're subtracting a 6 twice. What's the fourth term? This is 15 minus-- We're subtracting the 6 three times from the 15, so minus 3 times 6. So, if you see the pattern here, when we have our fourth"
    },
    {
        "Q": "At 5:14, Why would the sqrt of 1/2 be 1/sqrt2 and not sqrt 1/2?\n",
        "A": "This is because the sqrt(1/2) is the same thing as saying sqrt(1)/sqrt(2) and since the square root of 1 is just 1 we can simplify the numerator and rewrite it as 1/sqrt(2).",
        "video_name": "KoYZErFpZ5Q",
        "timestamps": [
            314
        ],
        "3min_transcript": "So we know the length of segment AB, which is a radius of the circle or is the radius of the circle is length one, what else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, maybe not completely fit it over but if we were to make it look like this. So the triangle, we could make it look like, a little bit more like this. Actually I want to make it look like a right angle though. So my triangle, let me make it look like, there you go. So if this is D, this is B, this is A, this is our right angle. this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle. Because they're not all the same, it's not equilateral. If all the angles were the same this would be equilateral. This is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same, this is an isosceles triangle. So how does that help us figure out the lengths of the sides? Well if you say that this side has length X, that means that this side has length X. If this side has length X, then this side has length X. and now we can use the Pythagorean Theorem. We can say that this X squared plus this X squared is equal to the hypotenuse square is equal to one square. Or we could write that two X squared is equal two or taking the principal root of both sides, we get X is equal to one over the square root of two. And a lot of folks don't like having a radical denominator, they don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, let's see the numerator will have the square root of two, and the denominator we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things We've been able to figure the angle of ABD in radians. We're able to figure out lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work we have done? So let's first think about, what is the sine? We'll do this in the color orange."
    },
    {
        "Q": "\nat 5:17 shouldn't it be sqrt 1/2 ?",
        "A": "Yes, but you need to go further and simplify the radical. When your radical is simplified, 1) it will not have a fraction inside the radical. sqrt(1/2) = sqrt(1) / sqrt(2) 2) all perfect squares inside the radical will be simplified: sqrt(1) / sqrt(2) = 1 / sqrt(2) 3) there will be no radical in the denominator - rationalize the denominator: 1 / sqrt(2) = 1 / sqrt(2) * sqrt(2) / sqrt(2) = sqrt(2) / 2 hope this helps.",
        "video_name": "KoYZErFpZ5Q",
        "timestamps": [
            317
        ],
        "3min_transcript": "So we know the length of segment AB, which is a radius of the circle or is the radius of the circle is length one, what else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, maybe not completely fit it over but if we were to make it look like this. So the triangle, we could make it look like, a little bit more like this. Actually I want to make it look like a right angle though. So my triangle, let me make it look like, there you go. So if this is D, this is B, this is A, this is our right angle. this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle. Because they're not all the same, it's not equilateral. If all the angles were the same this would be equilateral. This is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same, this is an isosceles triangle. So how does that help us figure out the lengths of the sides? Well if you say that this side has length X, that means that this side has length X. If this side has length X, then this side has length X. and now we can use the Pythagorean Theorem. We can say that this X squared plus this X squared is equal to the hypotenuse square is equal to one square. Or we could write that two X squared is equal two or taking the principal root of both sides, we get X is equal to one over the square root of two. And a lot of folks don't like having a radical denominator, they don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, let's see the numerator will have the square root of two, and the denominator we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things We've been able to figure the angle of ABD in radians. We're able to figure out lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work we have done? So let's first think about, what is the sine? We'll do this in the color orange."
    },
    {
        "Q": "\nAt around 2:06, how come Sal doesn't put (square root of y)^2 in the denominator? I thought when using the product rule, you put (g(x))^2 in the denominator.",
        "A": "I suspect you re confusing the product rule with the quotient rule. Product rule: derivative of uv = u v + uv Quotient rule: derivative of u/v = (u v - uv )/v\u00c2\u00b2",
        "video_name": "2CsQ_l1S2_Y",
        "timestamps": [
            126
        ],
        "3min_transcript": "What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we can say, explicit differentiation when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1. This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1/x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. So we get dy dx is equal to negative 2 x to the negative 2 minus 1-- x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. to both sides of this equation. And so let me make it clear what we're doing-- x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, well, actually, we're going to apply both the product rule and the chain rule. The product rule tells us-- so we have the product of two functions of x. You could view it that way. So this, the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, times the derivative with respect to x of the square root of y. Let me make it clear, this bracket. the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. This simplifies to 1, so we're just going to be left with the square root of y right over here. So this is going to simplify to a square root of y. And what does this over here simplify to? Well the derivative with respect to x of the square root of y, here we want to apply the chain rule. So let me make it clear. So we have plus this x plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something,"
    },
    {
        "Q": "His algebra is wrong at 2:37 3*1/4 is not 13/4 and 2*1/2 is not 5/2\n",
        "A": "His math is not wrong. You are confusing mixed numbers: 3 1/4 with multiplication of 3 * 1/4. In a mixed number like 3 1/4 of 2 1/2 there is no multiplication symbol between the numbers. 3 1/4 is 3 whole units + 1/4 unit, or 3.25 in decimal. 2 1/2 is 2 whole units + 1/2, or 2.5 in decimal form. Seems like you need to review basic concepts of dealing with fractions and mixed numbers.",
        "video_name": "x5EJG_rAtkY",
        "timestamps": [
            157
        ],
        "3min_transcript": "And that's exactly what these two statements are saying. x has to be less than 2 and 1/2, and it has to be greater than negative 2 and 1/2. If this absolute value were the other way, that the absolute value of x has to be greater than 2 and 1/2, then it would be the numbers outside of this, and it would be an or. But we're dealing with the less than situation right there, so let's just do what we were able to figure out when it was just an x. The distance from this thing to 0 has to be less than 2 and 1/2, so we can write that 2r minus 3 and 1/4 has to be less than 2 and 1/2 and 2r minus 3 and 1/4 has to be greater than negative 2 and 1/2. Same exact reasoning here. Let me draw a line so we don't get confused. This quantity right here has to be between negative 2 and 1/2. It has to be greater than negative 2 and 1/2 right there. And it has to be less than 2 and 1/2, so that's all I wrote there. So let's solve each of these independently. Well, this first went over here, you've learned before that I don't like improper fractions, and I don't like fractions in general. So let's make all of these fractions. Sorry, I don't like mixed numbers. I want them to be improper fractions. So let's turn all of these into improper fractions. So if I were to rewrite it, we get 2r minus 3 and 1/4 is the same thing as 3 times 4 is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question-- and do the same thing here-- we have 2r minus 13 over 4 has to be a greater All right, now let's solve each of these independently. To get rid of the fractions, the easiest thing to do is to multiply both sides of this equation by 4. That'll eliminate all of the fractions, so let's do that. Let's multiply-- let me scroll to the left a little bit-- let's multiply both sides of this equation by 4. 4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than-- and I multiplied by a positive number so I didn't have to worry about swapping the inequality-- is less than 5/2 times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13 is less than 10. Now we can add 13 to both sides of this equation so that we get rid of it on the left-hand side. Add 13 to both sides and we get 8r-- these guys cancel out-- is less than 23, and then we divide"
    },
    {
        "Q": "\n00:32 when you said 'any m&b on the surface = SE for that line', what does 'that line' exactly mean?",
        "A": "That line is referring to a line drawn through your data. If you have your data plotted on a graph, you can draw a line through it to fit to the data. This line will have a gradient, m, and a y-intercept, b. It will also have a squared error - the square distance of each point of data from the line. The aim is to find the line with the minimum squared error.",
        "video_name": "u1HhUB3NP8g",
        "timestamps": [
            32
        ],
        "3min_transcript": "All right, so where we left off, we had simplified our algebraic expression for the squared error to the line from the n data points. We kind of visualized it. This expression right here would be a surface, I guess you could view it as a surface in three dimensions, where for any m and b is going to be a point on that surface that represents the squared error for that line. Our goal is to find the m and the b, which would define an actual line, that minimize the squared error. The way that we do that, is we find a point where the partial derivative of the squared error with respect to m is 0, and the partial derivative with respect to b is also equal to 0. So it's flat with respect to m. So that means that the slope in this direction is going to be flat. Let me do it in the same color. So the slope in this direction, that's the partial derivative with respect to m, is going to be flat. It's not going to change in that direction. The partial derivative with respect to b is going to be flat. So it will be a flat point right over there. and that is our minimum point. So let's figure out the m and b's that give us this. So if I were to take the partial derivative of this expression with respect to m. Well this first term has no m terms in it. So it's a constant from the point of view of m. Just as a reminder, partial derivatives, it's just like taking a regular derivative. You're just assuming that everything but the variable that you're doing the partial derivative with respect to, you're assuming everything else is a constant. So in this expression, all the x's, the y's, the b's, the n's, those are all constant. The only variable, when we take the partial derivative with respect to m, that matters is the m. So this is a constant. There's no m here. This term right over here, we're taking with respect to m. So the derivative of this with respect to m, it's kind of the coefficients on the m. So negative 2 times n times the mean of the xy's, that's Then this term or right here has no m's in it. So it's constant with respect to m. So its partial derivative with respect to m is 0. Then this term here, you have n times the mean of the x squared times m squared. So this is going to be-- we're talking about a partial derivative with respect to m-- so it's going to be 2 times n times the mean of the x [? squareds ?] times m. The derivative of m squared is 2m, and then you just have this coefficient there as well. Now this term, you also have an m over there. So let's see, everything else is just kind of a coefficient on this m. So the derivative with respect to m is 2bn times the mean of the x's. If I took the derivative of 3m, the derivative is just 3."
    },
    {
        "Q": "at around 1:40 how do we know that the top side (c or 2x) in not equal to the right side (b or x=10) ?\n",
        "A": "It s because triangle ABC is an isosceles triangle. If 2x was equal to x+10, then it wouldn t be an isosceles triangle, it would be an equilateral triangle, because an equilateral triangle s angles are all equal.",
        "video_name": "CVKAro3HUxQ",
        "timestamps": [
            100
        ],
        "3min_transcript": "So what do we have here? We have a triangle, and we know that the length of AC is equal to the length of CB. So this is an isosceles triangle, we have two of its legs are equal to each other. And then they also tell us that this line up here, they didn't put another label there. Let me put another label there just for fun. Let's call this, you could even call this a ray because it's starting at C, that line or ray CD is parallel to this segment AB over here, and that's interesting. Then they give us these two angles right over here, these adjacent angles. They give it to us in terms of x. And what I want to do in this video is try to figure out what x is. And so given that they told us that this line and this line are parallel, and we can turn this into line CD, so it's not just a ray anymore, so it just keeps going on and on in both directions. The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here. And you might recognize that this right over here, You might recognize that line CB is a transversal for those two parallel lines. Let's let me draw both of the parallel lines a little bit more so that you can recognize that as a transversal, and then a few things might jump out. You have this x plus 10 right over here, and its corresponding angle is right down here. This would also be x plus 10. And if this is x plus 10, then you have a vertical angle right over here that would also be x plus 10. Or you could say that you have alternate interior angles that would also be congruent. Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed as functions of-- expressed in terms of x. So when we take their sum, they need to be equal to 180, and then we can actually solve for x. We get 2x plus x plus 10 plus x plus 10 And then we can add up the x's. So we have a 2x there plus an x plus another x, that gives us 4x. 4 x's. And then we have a plus 10 and another plus 10, so that gives us a plus 20, is equal to 180. And we can subtract 20 from both sides of that, and we get 4x is equal to 160. Divide both sides by 4, and we get x is equal to 40. And we're done. We've figured out what x is, and then we could actually figure out what these angles are. If this is x plus 10 then you have 40 plus 10, this right over here is going to be a 50-degree angle. This is 2x, so 2 times 40, this is an 80-degree angle. It doesn't look at it the way I've drawn it, and that's why you should never assume anything based on how a diagram is drawn."
    },
    {
        "Q": "\nAt 2:55 Sal suggests that it is necessary to take the natural log of both sides but I solved it by changing it to e^(ln(x)*x) and then taking the derivative as normal and I got the right answer.",
        "A": "There is always different ways to solve the same problems. Sal only mentioned one of the ways by taking the natural log (mentioned at 4:37), but using e is another way.",
        "video_name": "N5kkwVoAtkc",
        "timestamps": [
            175
        ],
        "3min_transcript": "natural log of x. And now we can take the derivative of both sides of this with respect to x. So the derivative with respect to x of that, and then the derivative with respect to x of that. Now we're going to apply a little bit of the chain rule. So the chain rule. What's the derivative of this with respect to x? What's the derivative of our inner expression with respect to x? It's a little implicit differentiation, so it's dy with respect to x times the derivative of this whole thing with respect to this inner function. So the derivative of natural log of x is 1/x. So the derivative of natural log of y with respect to y is 1/y. So times 1/y. And the derivative of this-- this is just the product rule, and I'll arbitrarily switch colors here-- is the derivative of the first term, which is 1, times the second term, so times which is 1/x times the first term. So times x. And so we get dy/dx times 1/y is equal to natural log of x plus-- this just turns out to be 1-- x divided by x, and then you multiply both sides of this by y. You get dy/dx is equal to y times the natural log of x plus 1. And if you don't like this y sitting here, you could just make the substitution. y is equal to x to the x. So you could say that the derivative of y with respect to x is equal to x to the x times the natural log of x plus 1. And that's a fun problem, and this is often kind of given as a trick problem, or sometimes even a bonus problem if people don't know to take the natural log of both sides of that. that's what we're going to tackle in this. But it's good to see this problem done first because it gives us the basic tools. So the more difficult problem we're going to deal with is this one. Let me write it down. So the problem is y is equal to x to the-- and here's the twist-- x to the x to the x. And we want to find out dy/dx. We want to find out the derivative of y with respect to x. So to solve this problem we essentially use the same tools. We use the natural log to essentially breakdown this exponent and get it into terms that we can deal with. So we can use the product rule. So let's take the natural log of both sides of this equation like we did last time. You get the natural log of y is equal to the natural log of x to the x to the x."
    },
    {
        "Q": "\nAt 4:11, why does Sal ask us how many times 16 goes into 1388 when we can see the answer on the other side of the screen?",
        "A": "Sal is explaining how to do it and he is asking himself he same questions that you will ask yourself.",
        "video_name": "R486L0M5cWk",
        "timestamps": [
            251
        ],
        "3min_transcript": "So that's still too large, so it's going to be 6. But notice, we had to do this little side work on the side right over here to realize it wasn't seven. Now six is the largest how many times you go into 108 without going over it. So 6 times 6 is 36. Carry the 3, or regroup the 3, depending on how you think about it. 6 times 1 is 6, plus 3 is 9. Then you subtract again. 8 minus 6 is 2. And then you can just say 10 minus 9 is 1, or you could even borrow. You could make this a 10. And then that goes away. 10 minus 9 is 1. So then you have 12. And if we're not going into decimals, you're done. Because 16 does not go into 12. So 16 goes into 1,388 86 times with a remainder of 12. That right over there is your remainder. And that's all a decent way of doing it. And that's the way you traditionally But what I want to do is introduce another maybe a little more interesting way So once again, let's do our 16 goes into 1,388. And what we're going to do is give us much more leeway for approximation, or for essentially guessing. And what we want to do is just guess. We're going to make guesses for how many times 16 goes into the numbers without overestimating, without jumping too high. And now we're not just going to think about the 1 or the 13 or the 138. We're going to think about the whole number as a whole. And before we do that, I'm going to get two things out of the way, just because it will help us. I'm just going to remind ourselves what 16 times 2 and 16 times 5 are. And I'm just picking these as random numbers that we can use to approximate. You don't have to use 2 and 5. You can use any numbers. Maybe I'll show other examples there. So 16 times 2, we know, is 32. And 16 times 5 is 50 plus 30, is 80. So let's just keep these two results in mind So the first thing to think about is our best guess for how many times does 16 go into 1,388. Or another way to think about, how many times does 16 go into 1,000? Let's just do something at a very rough approximation. Well, we know it's not going to be 100, because 100 times 16 would be 1,600. You would just throw those two 0's at the end of it. And [? you'd ?] say, how many times does it go into 1,000? Well, we know if 16 times 5 is 80, we know that 16 times 50 would be 800. So let's use that. And I'm using the 5-- I'm multiplying it by another 10 to get to 50-- instead of the 2 because 800 is a lot closer than 320 to our 1,000 that we care about. So what we could say is, well, 16 times 50 will get us to 800."
    },
    {
        "Q": "\n@3:35 Sal Crosses out the \"hours.\" Why can he do this?",
        "A": "Because the hours part can be cutout as it is both at the numerator as well as at the denominator.",
        "video_name": "Uc2Tm4Lr7uI",
        "timestamps": [
            215
        ],
        "3min_transcript": "at just 8 miles per hour.\" So we're given a time. And we are given a speed. We should be able to figure out a distance. So let's just do a little bit of aside here. We should be able to figure out the distance to the-- actually let me write it this way. The distance to the store will be equal to-- now we've got to make sure we have our units right. Here they gave it in minutes. Here they have 8 miles per hour. So let's convert this into hours. So 45 minutes in hours, so it's 45 minutes out of 60 minutes per hour. So that's going to give us 45/60. Divide both by 15. That's the same thing as 3/4. So it's going to be 3/4 hours is the time times an average speed So what is the distance to the store? Well, 3/4 times 8. Or you could view it as 3/4 times 8 times 1, is going to be-- well, it's going to be 24 over 4. Let me just write that. That's going to be 24 over 4 which is equal to-- did I get it? Yeah. 24 over 4, which is equal to 6. And units-wise, we're just left with miles. So the distance to the store is 6 miles. 2 times the distance to the gift store, well, this whole thing is going to be 12 miles. 12 miles is the total distance she traveled. Now, what is the time to the gift store? Well, they already told that to us. They already told us that it's 45 minutes. Now, I want to put everything in hours. I'm assuming that they want our average speed in hours. So the time to the gift store was 3/4 of an hour. And what's the time coming back from the gift store? Well, we know her speed. We know her speed coming back. We already know the distance from the gift store. It's the same as the distance to the gift store. So we can take this distance, we can take 6 miles, that's the distance to the gift store, 6 miles divided by her speed coming back, which is 24 miles per hour, so divided by 24 miles per hour. It gives us-- well, let's see. We're going to have 6 over 24 is the same thing as 1/4. It's going to be 1/4. And then miles divided by miles per hour is the same thing as miles times hours per mile. The miles cancel out. And you'll have 1/4 of an hour. So it takes her 1/4 of an hour to get back. And that fits our intuition."
    },
    {
        "Q": "\nAt 2:34, Sal arrives at -(sqrt of 1/7)... I tried to simplify this more got -sqrt or 1 / sqrt of 7. I squared the top and the bottom. -1^2 and sqrt 7^2 and got 1/7. Is this correct or incorrect and why?",
        "A": "No, you started off ok. -(sqrt of 1/7) = - sqrt(1) / sqrt(7) But, you can not square the fraction. It isn t equivalent to the original fraction. Instead, you should 1) Simplify sqrt(1): - sqrt(1) / sqrt(7) = -1 / sqrt(7) 2) Rationalize the denominator. This means we want to convert the denominator to a rational number. We do this by multiplying top and bottom by sqrt(7) and simplify. -1 / sqrt(7) = -1 / sqrt(7) * sqrt(7) / sqrt(7) = - sqrt(7) / sqrt(49) = - sqrt(7) / 7 Hope this helps.",
        "video_name": "suwJmCrSDI8",
        "timestamps": [
            154
        ],
        "3min_transcript": "Now, are there any perfect squares in 35? 35 is seven times five. No, neither of those are perfect squares. So, I could just leave that as square root of 35. And, let's see, the square root of four? Well, thats going to be two. This is the principal root, so we're thinking about the positive square root. The square root of nine is three. And so, this part right over is going to be four times two, times the square root of five, so it's going to be eight square roots of five. And then, this part over here is going to be minus three, times three, times the square root of five. So, minus nine square roots of five. And all of that is going to be over the square root of 35. Square root of 35. And, so let's see, if I have eight of something, and I subtract nine of that something, I'm gonna have negative one of that something. or I could just say negative square root of five. Negative square root of five over the square root of 35. I actually think I could simplify this even more, because this is the same thing. This is equal to the negative of the square root of five over 35. Both the numerator and the denominator are divisible by five. So, we could divide them both by five, and we would get the square root of divide the numerator by five, you get one. Divide the denominator by five, you get seven. So, we can view this as the square root of 1/7th. Square root of 1/7th. And we are all done. Let's do another one of these. These are strangely, strangely fun. And once again, pause it, and see if you can work it out on your own. Perform the indicated operations. Alright, so let's first multiply. So, this essentially is doing the distributive property twice. And, actually let me just do it that way, plus the square root of six. Let's first multiply it times the square root of five. So, the square root of five times the square root of five is going to be five. Square root of five times the square root of six is the square root of 30. So, five plus the square root of 30. And then, when I take this expression, and I multiply it times the second term, times the negative square root of six. Well, negative square root of six times the square root of five is going to be the negative of the square root of 30. And then the negative of the square root of six times the square root of six is going to be, we're gonna subtract six. The square root of six times the square root of six is six, then we have the negative out there. And so, just like that, we are left with, well, let's see. Square root of 30 minus the square root of 30, well those cancel out, thats zero. And we're left with five minus six, which is going to be equal to negative one."
    },
    {
        "Q": "\nWith reference to 9:15 (appx) ,\n\nIs it right to say that all graphs where there is ONLY a discontinuity at one end point, are graphs with removable discontinuity ? By the looks of it, it seems so. Was just wondering if there are any special cases.",
        "A": "If you have a piecewise function, and your endpoint is at the point where the function changes, then the discontinuity is not removable. It s not a requirement that a discontinuity at an endpoint must be removable or that a discontinuity should be at the endpoints.",
        "video_name": "kdEQGfeC0SE",
        "timestamps": [
            555
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n4:54 why is rationalizing important?",
        "A": "Rationalizing is like simplifying. To always have to simplify your answer unless you re told not to.",
        "video_name": "tSHitjFIjd8",
        "timestamps": [
            294
        ],
        "3min_transcript": "We know from the Pythagorean theorem-- let's say the hypotenuse is equal to C-- the Pythagorean theorem tells us that A squared plus B squared is equal to C squared. Right? Well we know that A equals B, because this is a 45-45-90 triangle. So we could substitute A for B or B for A. But let's just substitute B for A. So we could say B squared plus B squared is equal to C squared. Or 2B squared is equal to C squared. Or B squared is equal to C squared over 2. Or B is equal to the square root of C squared over 2. the numerator and the square root of the denominator-- C over the square root of 2. And actually, even though this is a presentation on triangles, I'm going to give you a little bit of actually information on something called rationalizing denominators. So this is perfectly correct. We just arrived at B-- and we also know that A equals B-- but that B is equal to C divided by the square root of 2. It turns out that in most of mathematics, and I never understood quite exactly why this was the case, people don't like square root of 2s in the denominator. Or in general they don't like irrational numbers in the denominator. Irrational numbers are numbers that have decimal places that never repeat and never end. So the way that they get rid of irrational numbers in the denominator is that you do something called rationalizing the denominator. And the way you rationalize a denominator-- let's take our example right now. If we had C over the square root of 2, we just multiply both the numerator and the denominator by the Because when you multiply the numerator and the denominator by the same number, that's just like multiplying it by 1. The square root of 2 over the square root of 2 is 1. And as you see, the reason we're doing this is because square root of 2 times square root of 2, what's the square root of 2 times square root of 2? Right, it's 2. Right? We just said, something times something is 2, well the square root of 2 times square root of 2, that's going to be 2. And then the numerator is C times the square root of 2. So notice, C times the square root of 2 over 2 is the same thing as C over the square root of 2. And this is important to realize, because sometimes while you're taking a standardized test or you're doing a test in class, you might get an answer that looks like this, has a square root of 2, or maybe even a square root of 3 or whatever, in the denominator. And you might not see your answer if it's a multiple choice question. What you ned to do in that case is rationalize the denominator. So multiply the numerator and the denominator by square root of 2 and you'll get square root of 2 over 2."
    },
    {
        "Q": "\nI am confused. I think that this new formula (B=C2/square root 2) is unnecessary. Sal stated at 1:04 that in these 45-45-90 triangels, both non-hypotenuse sides are equal. Therefore, if one non-hypotenuse side of the triangle is 8, we could automatically assume its other congruent side is 8. We could then just use the Pythagorean Theorem to find the hypotenuse. Could someone please explain why we must use a different formula? I'll give a more detailed example of my point in the questions area.",
        "A": "For example, in a 45-45-90 triangle where one side is labeled 2 and the hypotenuse is unknown, we can label the other non-hypotenuse side 2 also. From this point, we can simply use the Pythagorean Theorem to find the hypotenuse. We don t need the formula which Sal gives at 4:27 in the video.",
        "video_name": "tSHitjFIjd8",
        "timestamps": [
            64
        ],
        "3min_transcript": "Welcome to the presentation on 45-45-90 triangles. Let me write that down. How come the pen-- oh, there you go. 45-45-90 triangles. Or we could say 45-45-90 right triangles, but that might be redundant, because we know any angle that has a 90 degree measure in it is a right triangle. And as you can imagine, the 45-45-90, these are actually the degrees of the angles of the triangle. So why are these triangles special? Well, if you saw the last presentation I gave you a little theorem that told you that if two of the base angles of a triangle are equal-- and it's I guess only a base angle You could draw it like this, in which case it's maybe not so obviously a base angle, but it would still be true. If these two angles are equal, then the sides that they don't share-- so this side and this side in this example, or this are going to be equal. So what's interesting about a 45-45-90 triangle is that it is a right triangle that has this property. And how do we know that it's the only right triangle that has this property? Well, you could imagine a world where I told you that this is a right triangle. This is 90 degrees, so this is the hypotenuse. Right, it's the side opposite the 90 degree angle. And if I were to tell you that these two angles are equal to each other, what do those two angles have to be? Well if we call these two angles x, we know that the angles in a triangle add up to 180. So we'd say x plus x plus-- this is 90-- plus 90 is equal to 180. Or 2x plus 90 is equal to 180. Or 2x is equal to 90. So the only right triangle in which the other two angles are equal is a 45-45-90 triangle. So what's interesting about a 45-45-90 triangle? Well other than what I just told you-- let me redraw it. I'll redraw it like this. So we already know this is 90 degrees, this is 45 degrees, this is 45 degrees. And based on what I just told you, we also know that the sides that the 45 degree angles don't share are equal. So this side is equal to this side. And if we're viewing it from a Pythagorean theorem point of view, this tells us that the two sides that are not the hypotenuse are equal. So this is a hypotenuse."
    },
    {
        "Q": "At 8:30, why is 2 being multiplied by 8, but \u00e2\u0088\u009a2 isn't being multiplied by 8?\n",
        "A": "Think of it as 8/1 * 2/\u00e2\u0088\u009a2. In normal multiplication, you just multiply across for example, 2/3 * 3/4 = (2*3)/(3*4) = 6/12 = 1/2",
        "video_name": "tSHitjFIjd8",
        "timestamps": [
            510
        ],
        "3min_transcript": "So what did we learn? This is equal to B, right? So turns out that B is equal to C times the square root of 2 over 2. So let me write that. So we know that A equals B, right? And that equals the square root of 2 over 2 times C. Now you might want to memorize this, though you can always derive it if you use the Pythagorean theorem and remember that the sides that aren't the hypotenuse in a 45-45-90 triangle are equal to each other. But this is very good to know. Because if, say, you're taking the SAT and you need to solve a problem really fast, and if you have this memorized and someone gives you the hypotenuse, you can figure out what are the sides very fast, or i8f someone gives you one of the sides, you can figure out the hypotenuse very fast. Let's try that out. I'm going to erase everything. So we learned just now that A is equal to B is equal to the So if I were to give you a right triangle, and I were to tell you that this angle is 90 and this angle is 45, and that this side is, let's say this side is 8. I want to figure out what this side is. Well first of all, let's figure out what side is the hypotenuse. Well the hypotenuse is the side opposite the right angle. So we're trying to actually figure out the hypotenuse. Let's call the hypotenuse C. And we also know this is a 45-45-90 triangle, right? Because this angle is 45, so this one also has to be 45, because 45 plus 90 plus 90 is equal to 180. So this is a 45-45-90 triangle, and we know one of the sides-- this side could be A or B-- we know that 8 is equal to the C is what we're trying to figure out. So if we multiply both sides of this equation by 2 times the square root of 2-- I'm just multiplying it by the inverse of the coefficient on C. Because the square root of 2 cancels out that square root of 2, this 2 cancels out with this 2. We get 2 times 8, 16 over the square root of 2 equals C. Which would be correct, but as I just showed you, people don't like having radicals in the denominator. So we can just say C is equal to 16 over the square root of 2 times the square root of 2 over the square root of 2. So this equals 16 square roots of 2 over 2. Which is the same thing as 8 square roots of 2."
    },
    {
        "Q": "At 0:54 , does Sal mean that if a triangle has less then 3 acute angles, it isn't an acute triangle? Is it true? If it is, why?\n",
        "A": "Yes, it s true. You must have 3 acute angles to make an acute triangle. You need one right angle to make a right triangle. You need one obtuse angle to make an obtuse triangle.",
        "video_name": "PiQxA9O7Rd8",
        "timestamps": [
            54
        ],
        "3min_transcript": "Which side is perpendicular to side BC? So BC is this line segment right over here. And for another segment to be perpendicular to it, perpendicular just means that the two segments need to intersect at a right angle, or at a 90-degree angle. And we see that BC intersects AB at a 90-degree angle. This symbol right over here represents a 90-degree, or a right angle. So we just have to find side AB or BA. And that's right over here. Side AB is perpendicular to side BC. Let's do a few more of these. Put the triangles into the correct categories, so this right over here. So let's see. Let's think about our categories. Right triangles-- so that means it has a 90-degree angle in it. Obtuse triangles-- that means it has an angle larger than 90 degrees in it. Acute triangles-- that means all three angles are less than 90 degrees. So this one has a 90-degree angle. It has a right angle right over here. This one right over here, all of these angles are less than 90 degrees, just eyeballing it. So this is going to be an acute-- that's going to be an acute triangle. I'll put it under acute triangles right over there. Then this one over here, this angle up here, this is-- and we can assume that these actually are drawn to scale, this is more open than a 90-degree angle. This is an obtuse angle right over here. It's going to be more than 90 degrees. So this is an obtuse triangle. Now, this one over here, all of them seem acute. None of them even seem to be a right angle. So I would put this again into acute-- acute triangles. This one here clearly has a right angle. It's labeled as such. So we'll throw it right over here. And then this one, this angle right over here is clearly even larger. It has a larger measure than a right angle. So this angle right over here is more than 90 degrees. It's going to be an obtuse angle. So we got two in each of these. And let's check our answer. We got it right."
    },
    {
        "Q": "\nAt 0:30, The selected answer was side AB, but couldn't side CD also be considered a correct answer?",
        "A": "Perpendicular means the two lines are forming a right angle.",
        "video_name": "PiQxA9O7Rd8",
        "timestamps": [
            30
        ],
        "3min_transcript": "Which side is perpendicular to side BC? So BC is this line segment right over here. And for another segment to be perpendicular to it, perpendicular just means that the two segments need to intersect at a right angle, or at a 90-degree angle. And we see that BC intersects AB at a 90-degree angle. This symbol right over here represents a 90-degree, or a right angle. So we just have to find side AB or BA. And that's right over here. Side AB is perpendicular to side BC. Let's do a few more of these. Put the triangles into the correct categories, so this right over here. So let's see. Let's think about our categories. Right triangles-- so that means it has a 90-degree angle in it. Obtuse triangles-- that means it has an angle larger than 90 degrees in it. Acute triangles-- that means all three angles are less than 90 degrees. So this one has a 90-degree angle. It has a right angle right over here. This one right over here, all of these angles are less than 90 degrees, just eyeballing it. So this is going to be an acute-- that's going to be an acute triangle. I'll put it under acute triangles right over there. Then this one over here, this angle up here, this is-- and we can assume that these actually are drawn to scale, this is more open than a 90-degree angle. This is an obtuse angle right over here. It's going to be more than 90 degrees. So this is an obtuse triangle. Now, this one over here, all of them seem acute. None of them even seem to be a right angle. So I would put this again into acute-- acute triangles. This one here clearly has a right angle. It's labeled as such. So we'll throw it right over here. And then this one, this angle right over here is clearly even larger. It has a larger measure than a right angle. So this angle right over here is more than 90 degrees. It's going to be an obtuse angle. So we got two in each of these. And let's check our answer. We got it right."
    },
    {
        "Q": "At 11:28, when Sal multiplies by b^2 shouldn't it be b^2 x^2/a^2 instead of b^2/ a^2 x^2 ?\n",
        "A": "Actually, both are equivalent. First Case: b^2 x^2/a^2 (b^2*x^2)/a^2 Second Case: b^2/a^2 x^2 (b^2/a^2)*x^2 x^2=x^2/1. (b^2/a^2)*(x^2/1) Multiply the numerators and denominators of the fractions. (b^2*x^2)/a^2 We end up with (b^2*x^2)/a^2 in both cases, meaning that both expressions is equivalent. I hope this helps!",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            688
        ],
        "3min_transcript": "never get to x equal to 0. You get to y equal 0, right here and here. But you never get to x equals 0. And actually your teacher might want you to plot these points, and there you just substitute y equals 0. And you can just look at the original equation. Actually, you could even look at this equation right here. Can x ever equal 0? If you look at this equation, if x is equal to 0, this whole term right here would cancel out, and you'd just be left with a minus b squared. Which is, you're taking b squared, and you put a negative sign in front of it. So that's a negative number. And then you're taking a square root of a negative number. So we're not dealing with imaginaries right now. So you can never have x equal to 0. But y could be equal to 0, right? You can set y equal to 0 and then you could solve for it. So in this case, actually let's do that. If y is equal to 0, you get 0 is equal to the square root of b squared over a squared x squared minus b squared. If you square both sides, you get b squared over a I know this is messy. So then you get b squared over a squared x squared is equal to b squared. You could divide both sides by b squared, I guess. You get a 1 and a 1. And then you could multiply both sides by a squared. You get x squared is equal to a squared, and then you get x is equal to the plus or minus square root of a. So this point right here is the point a comma 0, and this point right here is the point minus a comma 0. Now let's go back to the other problem. I have a feeling I might be running out of time. So notice that when the x term was positive, our hyperbola opened to the right and the left. And you could probably get from detective reasoning that when the y term is positive, which is the case in this one, we're probably going to open up and down. So let's solve for y. You get y squared over b squared. We're going to add x squared over a squared to both sides. So you get equals x squared over a squared plus 1. Multiply both sides by b squared. y squared is equal to b squared over a squared x squared plus b squared. You have to distribute the b squared. Now take the square root. I'll switch colors for that. So y is equal to the plus or minus square root of b squared over a squared x squared plus b squared. And once again-- I've run out of space-- we can make that same argument that as x approaches positive or negative infinity, this equation, this b, this little constant term right here isn't going to matter as much. You're just going to take the square root of this term right here."
    },
    {
        "Q": "\nAt 2:46 did Sal mean hyperbolas? It sounded like he said parabolas.",
        "A": "I think you are correct. He said parabolas , but he should have said hyperbolas .",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            166
        ],
        "3min_transcript": "And so this is a circle. And once again, just as review, a circle, all of the points on the circle are equidistant from the center. Or in this case, you can kind of say that the major axis and the minor axis are the same distance, that there isn't any distinction between the two. You're always an equal distance away from the center. So that was a circle. An ellipse was pretty much this, but these two numbers could be different. Because your distance from the center could change. So it's x squared over a squared plus y squared over b squared is equal to 1. That's an ellipse. And now, I'll skip parabola for now, because parabola's kind of an interesting case, and you've already touched on it. So I'll go into more depth in that in a future video. But a hyperbola is very close in formula to this. And so there's two ways that a hyperbola could be written. And I'll do those two ways. minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it. you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared."
    },
    {
        "Q": "at 10:32, why is the a sqrt(a), isn't it supposed to me +- a? Because x^2=a^2 and then root both sides, x = +-a? I can't find the report a problem button though. But I might be missing something.\n",
        "A": "If you keep playing the video, you should notice that in the bottom right hand corner Sal explains his mistake.",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            632
        ],
        "3min_transcript": "never get to x equal to 0. You get to y equal 0, right here and here. But you never get to x equals 0. And actually your teacher might want you to plot these points, and there you just substitute y equals 0. And you can just look at the original equation. Actually, you could even look at this equation right here. Can x ever equal 0? If you look at this equation, if x is equal to 0, this whole term right here would cancel out, and you'd just be left with a minus b squared. Which is, you're taking b squared, and you put a negative sign in front of it. So that's a negative number. And then you're taking a square root of a negative number. So we're not dealing with imaginaries right now. So you can never have x equal to 0. But y could be equal to 0, right? You can set y equal to 0 and then you could solve for it. So in this case, actually let's do that. If y is equal to 0, you get 0 is equal to the square root of b squared over a squared x squared minus b squared. If you square both sides, you get b squared over a I know this is messy. So then you get b squared over a squared x squared is equal to b squared. You could divide both sides by b squared, I guess. You get a 1 and a 1. And then you could multiply both sides by a squared. You get x squared is equal to a squared, and then you get x is equal to the plus or minus square root of a. So this point right here is the point a comma 0, and this point right here is the point minus a comma 0. Now let's go back to the other problem. I have a feeling I might be running out of time. So notice that when the x term was positive, our hyperbola opened to the right and the left. And you could probably get from detective reasoning that when the y term is positive, which is the case in this one, we're probably going to open up and down. So let's solve for y. You get y squared over b squared. We're going to add x squared over a squared to both sides. So you get equals x squared over a squared plus 1. Multiply both sides by b squared. y squared is equal to b squared over a squared x squared plus b squared. You have to distribute the b squared. Now take the square root. I'll switch colors for that. So y is equal to the plus or minus square root of b squared over a squared x squared plus b squared. And once again-- I've run out of space-- we can make that same argument that as x approaches positive or negative infinity, this equation, this b, this little constant term right here isn't going to matter as much. You're just going to take the square root of this term right here."
    },
    {
        "Q": "At 2:35, do you have to distribute the y, could you just divide the whole equation by x?\n",
        "A": "Keep in mind that we want to solve the equation for x. To do that we have to unlock the parentheses and then gather all the x terms on one side of the equation and everything else on the other side. That s why the y must be distributed. Dividing by x would severely complicate things, as you would end up with [ y ( x + 3 ) ] / x = 2 - 1/x and we would still not have only the x terms on one side of the equation.",
        "video_name": "VzWxvDe8TUQ",
        "timestamps": [
            155
        ],
        "3min_transcript": "You start with a member of the range and you go back to a member of the domain. That's what g inverse does. And so, there's a couple of ways to think about it. If we were to set, if we were to say, look, let's just set y is equal to g of x. So we could just call that point y, as well. So this expression here, let's just, okay, if we have an x, this is what we do. We multiply by two times and we subtract one, divided by x plus three, and then that tells us what the associated g of x, what the associated y is. But what if we're given the y, or the g of x. How do we get x? Well, we could just solve for x. So let's do that. If we said that y is equal to two x minus one over x plus three, let's solve for x, so that we know for any y we get, what the corresponding x we'll get. So how do we do that? Well, we can multiply both sides times x plus three. we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to"
    },
    {
        "Q": "\nAt 3:52, I think Sal forgets the negative sign next to the 1. If this is true, does that mean my answer of (3y+1)/(-y+2) is correct? I took a different approach by leaving the 3y and adding the 1, instead of Sal's method, so I was hoping that it worked both ways.",
        "A": "Sal does forget the sign, and a pop-up box in the lower right-hand corner mentions that. Your answer is correct. ( -1 - 3y ) / ( y - 2 ) = -1 ( 1 + 3y ) / [ -1 ( -y + 2 ) = ( 1 + 3y ) / ( -y + 2 )",
        "video_name": "VzWxvDe8TUQ",
        "timestamps": [
            232
        ],
        "3min_transcript": "we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to over y minus two. So one way you could think about it is, you could say, well x is equal to, this expression is equal to the inverse function as a function of y. You give me a y, you give me a y that's sitting in this range, I can input it into this function definition here and I can give you the corresponding x. I can give you the corresponding x. Now, we didn't ask for g inverse of y. We asked for g inverse of x. But the important thing to remember is this variable that we use inside of, the variable that we use inside of functions right over here, they're chosen somewhat arbitrarily. They just say, okay, I'm going to call the input y, and if you're going to call the input y, well, this is how I would give you the output. But we could call the input a, and that would be one minus three a over a minus twp. Or we could even call the input, so let me make this clear."
    },
    {
        "Q": "\nAt 3:47, shouldn't it be -1-3y instead of 1-3y, so that we get (y-2)x = -1-3y ?? In other words, did he forget to keep the 1 negative?",
        "A": "This is a known mistake in the video. There is a box that pops up at about 3:50 that states there is an error and provides the correct value.",
        "video_name": "VzWxvDe8TUQ",
        "timestamps": [
            227
        ],
        "3min_transcript": "we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to over y minus two. So one way you could think about it is, you could say, well x is equal to, this expression is equal to the inverse function as a function of y. You give me a y, you give me a y that's sitting in this range, I can input it into this function definition here and I can give you the corresponding x. I can give you the corresponding x. Now, we didn't ask for g inverse of y. We asked for g inverse of x. But the important thing to remember is this variable that we use inside of, the variable that we use inside of functions right over here, they're chosen somewhat arbitrarily. They just say, okay, I'm going to call the input y, and if you're going to call the input y, well, this is how I would give you the output. But we could call the input a, and that would be one minus three a over a minus twp. Or we could even call the input, so let me make this clear."
    },
    {
        "Q": "I'm not following what Sal did at 3:18 when he got out the calculator and got a number for r. Did he just rearrange the whole logistic equation in his head and solve for r, or what?\n",
        "A": "this is the equation he used: future value / present value = (1+i)^n (growth rate equation google it) i= growth rate n=number of periods. 150/100=(1+i)^20---> i=[(1.5)^(1/20)] - 1",
        "video_name": "-fIsaqN-aaQ",
        "timestamps": [
            198
        ],
        "3min_transcript": "This logistic function. This logistic function is a nonconstant solution, and it's the interesting one we care about if we're going to model population to the logistic differential equation. So now that we've done all that work to come up with this, let's actually apply it. That was the whole goal, was to model population growth. So let's come up with some assumptions. Let's first think about, well let's say that I have an island. So let's say that this is my island, and I start settling it with a 100 people. So I'm essentially saying N naught So I'm saying N naught is equal to 100. Let's say that this environment, given current technology of farming and agriculture, and the availability of water and whatever else, let's say it can only support 1,000 people max. So you get the idea, so we get K is equal to 1000. That's the limit to the population. So now what we have to think about is what is r going to be? So we have to come up with some assumptions. So, let's say in a generation which is about 20 years, well I'll just assume in 20 years, yeah I think it's reasonable that the population grows by, let's say that the population grows by 50%. In 20 years you have 50% growth. 50% increase, increase in the actual population. in order to after 20 years to grow by 50%. Well to think about that I'll get out my calculator. One way to think about it, growing by 50%, that means that you are at 1.5 your original population, and if I take that to the 120th power, and we'll just do 1 divided by 20th. This essentially says how much am I going to grow by or what is going to, this is telling me I'm going to grow by a factor of 1.02 every year, 1.02048. So one way to think about it is if every year I grow by 0.020 I'll just round five then over 20 years"
    },
    {
        "Q": "On 2:32 is scaling a type of measuring? That's all I'm asking thank you!\n",
        "A": "Sort of. When you scale something up or down, you are keeping the same proportions, but making it bigger or smaller. So, if you have a 1 cm cube and scale it up by 10, then each of the measurements would increase x10. So, it keeps the same proportions relative to the other sides, but gets bigger. If one side gets bigger out of proportion, then it would lose its original shape.",
        "video_name": "yUYDhmQsiXY",
        "timestamps": [
            152
        ],
        "3min_transcript": "it's already written as 8/7 times 2/3. And then, this last expression, we could write it as, in the numerator, 5 times 2. And then in the denominator, it's over 5 times 3. 5 times 3, which is of course the same thing as 5/5 times 2/3. So you see, all three of these expressions involve something times 2/3. Now, looking at it this way, does it become easier to pick out which of these are the largest, which of these are the smallest, and which of these are someplace in between? I encourage you to pause it again if you haven't thought about it yet. So let's visualize each of these expressions by first trying to visualize 2/3. So let's say the height of what I am drawing right now, let's say the height of this bar right over here is 2/3. The height here is 2/3. So first, let's think about what this one on the right here represents. This is 5/5 times 2/3. Well, what's 5/5? 5/5 is the same thing as 1. This is literally just 1 times 2/3. This whole expression is the same thing as 1 times 2/3, or really, just 2/3. So this, the height here, 2/3, this is the same thing as this thing over here. This is going to be equal to-- this could also be viewed as 5 times 2 over 3 times 5, which was this first expression right over here. Now, let's think about what these would look like. So this is 7/8 times 2/3. So it's less than 8/8 times 2/3. It's less than 1 times 2/3. So we're going to scale 2/3 down. It's going to be 7/8 of 2/3. So this one right over here would look something like this. Let me see if I can draw it. Yeah, it would look something like this. If the yellow height is 2/3, then this right over here, then this height right over here-- let me make it clear. This height right over here would be 7/8 times 2/3. Likewise, let's look at this one right over here. Let's look at this one in the middle, 8/7 times 2/3. Well, 8/7 is bigger than 7/7. It's more than 1. This is more than 2/3. This is 1 and 1/7 times 2/3. So it's going to be the same height as 2/3 plus another 1/7. So it's going to look something like this. It's going to look something like this. So its height-- now we scaled the 2/3 up because 8/7"
    },
    {
        "Q": "how does sal know that at 2:10 the graph g(x) ill be equal to f(x)-horizontal shift + the vertical shift and not f(x) + horizontal shift + vertical shift?\n",
        "A": "He explains this at 3:00 and on.",
        "video_name": "MDav5OMpCto",
        "timestamps": [
            130
        ],
        "3min_transcript": "- [Voiceover] So we have these two graphs that look pretty similar, y equals f of x and y is equal to g of x. And what they asked us to do is write a formula for the function g in terms of f. So let's think about how to do it and like always pause the video and see if you can work through it on your own. All right, well, what I'd like to do is I'd like to focus on this minimum point because I think that's a very easy thing to look at because of them have that minimum point right over there. And so we can think, but how do we shift f? Especially this minimum point, how do we shift it to get to overlapping with g? Well, the first thing that might jump out of this is that we would want to shift to the left. And we'd wanna shift to the left four. So let me do this in a new color. So I would wanna shift to the left by four. We have shifted to the left by four or you could say we shifted by negative four. Either way, you could think about it. And then we're gonna shift down. from y equals two to y is equal to negative five. So let me do that. So let's shift down. So we shift down by seven or you could say we have a negative seven shift. So, how do you express g of x if it's a version of f of x that shifted to the left by four and shifted down by seven. Or, you could say I have a negative four horizontal shift. I have a negative seven vertical shift. Well, one thing to think about it is g of x, g of x is going to be equal to f of, let me do it in a little darker color, it's going to be equal to f of x minus your horizontal shift, all right, horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical Well, what is our horizontal shift here? Well, we're shifting to the left so it's a negative shift. So our horizontal shift is negative four. Now, what's our vertical shift? Well, we went down so our vertical shift is negative seven. So it's negative seven. So there you have it. We get g of x. Let me do that in the same color. We get g of x is equal to f of x minus negative four or x plus four and then we have plus negative seven or you can just say minus seven. And we're done. And when I look at things like this, the negative seven is somewhat, it's more intuitive to me. As I shifted it down, it made sense that I have a negative seven. But first, when you work on these, you say, \"I shifted to the left. \"Why is it a plus four?\""
    },
    {
        "Q": "\n0:33 so a derivative of a function for critical points can only either equal to zero or undefined? a derivative can't have both places where it equals to zero and and places where it's undefined?\ncould you please do a video or explain how to identify the minima and maxima for a function where the derivative can be undefined instead of equalling to zero?",
        "A": "Some calculus textbooks (perhaps most) refer only to points where the derivative is zero as critical points. Others may include points where the derivative is undefined. Consider this function: y = sqrt(36 - x^2) That s the equation for the top half of a circle. The derivative is zero at x=0, a maximum, and undefined at x=-6 and x=6, which are minima.",
        "video_name": "pInFesXIfg8",
        "timestamps": [
            33
        ],
        "3min_transcript": "We've got the function f of x is equal to x to the third power minus 12x plus 2. And what I want to do in this video is think about at what points does my function f take on minimum or maximum values? And to figure that out I have to first figure out what are the critical points for my function f. And then which of those critical points do we achieve a minimum or maximum value? And to determine the critical points we have to find the derivative of our function because our critical points are just the point at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you can call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x is either undefined or 0. Well this thing is defined for all values of x. So the only places we're going to find critical points So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12."
    },
    {
        "Q": "\nAt 3:08 , why did sal crossed the curve of -12 from 2 and -2 ?\nPlease explain.\nThank you",
        "A": "I dont understand your question, please explain your problem.",
        "video_name": "pInFesXIfg8",
        "timestamps": [
            188
        ],
        "3min_transcript": "So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function"
    },
    {
        "Q": "\n1:00\nSo sin(x)^-1 is not 1/sin(x)? Why is it written this way?",
        "A": "It really isn t a well thought out syntax. If it confuses you, you can use arcsin instead of sin^-1. The difference is subtle. Btw, 1/sin(x) can be written as csc(x) (csc stands for cosecant ).",
        "video_name": "57BiI_iD3-U",
        "timestamps": [
            60
        ],
        "3min_transcript": "Let's see if we can remove the parameter t from a slightly more interesting example. So let's say that x is equal to 3 times the cosine of t. And y is equal to 2 times the sine of t. We can try to remove the parameter the same way we did in the previous video, where we can solve for t in terms of either x or y and then substitute back in. And I'll do that. But I want to do that first, just to show you that it kind of leads to a hairy or an unintuitive answer. So if we solve for-- let's solve for t here. We could do it either one, they're equally complex. So if we solve for t here, we would say divide both sides by 2. You'd get y over 2 is equal to sine of t. And then you would take the arcsine of both sides, or the inverse sine of both sides, and you would get-- I like writing arcsine, because inverse sine, people often confuse it with an Arcsine of y over 2 is equal to t. Actually, let me do that little aside there. I should probably do it at the trigonometry playlist, but it's a good thing to hit home. Because I think people get confused. So arcsine of anything, let's say, y. The other way of writing that is sine minus 1 of y. These two things are equivalent, when they're normally used. But I don't like using this notation most of the time, because it can be ambiguous. This could mean sine of y to the negative 1 power, which equals 1 over sine of y. And arcsine and this are definitely not the same thing. So you want to be very careful there to make sure that you don't get confused when someone writes an inverse sine like this. But they're not actually taking sine of y to On the other hand, if someone were to write sine squared of y, this is unambiguously the same thing as sine of y squared. In fact, I wish this was the more conventional notation because it wouldn't make people think, oh, 2 and minus 1 there, and of course, that's just sine of y squared. So it can be very ambiguous. And of course, if this was a negative, this would be a minus 2, and then this really would be 1 over sine of y squared. That's why, just a long-winded way of explaining why I wrote arcsine, instead of inverse sine right there. Needless to say, let's get back to the problem. So we've solved for t in terms of y. Now we can substitute back here. And we've got an expression for x in terms of y. So we get x is equal to 3 times the cosine of t. But we just solved for t. t is this thing right here. So it's the cosine of arcsine of y over 2."
    },
    {
        "Q": "\nAt 0:23, the equations were x=3cos(t) and y=2sin(t). Why couldn't we have written x=3sin(t) instead of cosine and y=2cos(t) instead of sine? Would it have made a different graph? Would it have been a hyperbola instead of an ellipse?",
        "A": "It would have made the same graph. Try graphing it on your graphing calculator (or a free online one). Be sure that the graph is in parametric mode (or PAR ) and in radians, not degrees. The only difference is in which direction the equation goes. As Sal was mentioning at around 7:45, we don t know which direction the graph was created by simplifying the equation. If sin and cos were switched, then the graph would go in a different direction, but would look the same. Hope this helps!",
        "video_name": "57BiI_iD3-U",
        "timestamps": [
            23
        ],
        "3min_transcript": "Let's see if we can remove the parameter t from a slightly more interesting example. So let's say that x is equal to 3 times the cosine of t. And y is equal to 2 times the sine of t. We can try to remove the parameter the same way we did in the previous video, where we can solve for t in terms of either x or y and then substitute back in. And I'll do that. But I want to do that first, just to show you that it kind of leads to a hairy or an unintuitive answer. So if we solve for-- let's solve for t here. We could do it either one, they're equally complex. So if we solve for t here, we would say divide both sides by 2. You'd get y over 2 is equal to sine of t. And then you would take the arcsine of both sides, or the inverse sine of both sides, and you would get-- I like writing arcsine, because inverse sine, people often confuse it with an Arcsine of y over 2 is equal to t. Actually, let me do that little aside there. I should probably do it at the trigonometry playlist, but it's a good thing to hit home. Because I think people get confused. So arcsine of anything, let's say, y. The other way of writing that is sine minus 1 of y. These two things are equivalent, when they're normally used. But I don't like using this notation most of the time, because it can be ambiguous. This could mean sine of y to the negative 1 power, which equals 1 over sine of y. And arcsine and this are definitely not the same thing. So you want to be very careful there to make sure that you don't get confused when someone writes an inverse sine like this. But they're not actually taking sine of y to On the other hand, if someone were to write sine squared of y, this is unambiguously the same thing as sine of y squared. In fact, I wish this was the more conventional notation because it wouldn't make people think, oh, 2 and minus 1 there, and of course, that's just sine of y squared. So it can be very ambiguous. And of course, if this was a negative, this would be a minus 2, and then this really would be 1 over sine of y squared. That's why, just a long-winded way of explaining why I wrote arcsine, instead of inverse sine right there. Needless to say, let's get back to the problem. So we've solved for t in terms of y. Now we can substitute back here. And we've got an expression for x in terms of y. So we get x is equal to 3 times the cosine of t. But we just solved for t. t is this thing right here. So it's the cosine of arcsine of y over 2."
    },
    {
        "Q": "\nAt 1:57, why does Sal say that you divide x^4 by x^4 but then simply divides x^4 by 4?",
        "A": "just misspoke. just taking the antiderivative of x^3",
        "video_name": "cBi4a1iSaPk",
        "timestamps": [
            117
        ],
        "3min_transcript": "So our goal in this video is to take the antiderivative of this fairly crazy looking expression. Or another way of saying it is to find the indefinite integral of this crazy looking expression. And the key realization right over here is that this expression is made up of a bunch of terms. And the indefinite integral of the entire expression is going to be equal to the indefinite integral of each of the term. So this is going to be equal to, we could look at this term right over here, and just take the indefinite that, 7x to the third dx. And then from that, we can subtract the indefinite integral of this thing. So we could say this is, and then minus the indefinite integral of 5 times the square root of x dx. And then we can look at this one right over here. So then we could say plus the indefinite integral of 18 square root of x. And then finally, I'm running out of colors here, finally I need more colors in my thing. We can take the antiderivative of this. So plus the antiderivative of x to the negative 40th power dx. So I've just rewritten this and color-coded things. So let's take the antiderivative of each of these. And you'll see that we'll be able to do it using our whatever we want to call it. The inverse of the power rule, or the anti-power rule, whatever you might want to call it. So let's look at the first one. So we have-- what I'm going to do is, I'm just going to find the antiderivative without the constant, and just add the constant at the end. For the sake of this one. Just to make sure we get the most general antiderivative. So here the exponent is a 3. So we can increase it by 1. So it's going to be x to the 4th. Let me do that same purple color, or pink color. It's going to be x to the 4th, or we're going to divide by x to the 4th. So it's x to the 4th over 4 is the antiderivative of x to the 3rd. And you just had this scaling quantity, the seven out front. So we get 7x to the 4th over 4. Fair enough. From that, we're going to subtract the antiderivative of this. Now at first this might not be obvious, that you could use our inverse power rule, or anti-power rule here. But then you just need to realize that 5 times the principal root of x is the same thing as 5 times x to the 1/2 power. And so once again, the exponent here is 1/2. We can increment it by 1. So it's going to be x to the 3/2. And then divide by the incremented exponent. So divide by 3/2. And of course we had this 5 out front, so we still want to have the 5 out front. Now this next expression looks even wackier. But once again, we can simplify a little bit. This is the same thing-- let me do it right over here-- this is the same thing as 18 times x to the 1/2 times"
    },
    {
        "Q": "At 5:00,\n5/3/2 = 5* 2/3 = 10/3\nI get 5/3/2 = 5/6\n\nWhere am I going wrong?\n",
        "A": "The fraction Sal is simplifying is 5 divided by 3/2 . To write it horizontally requires parentheses (5)/(3/2). To divide 5 by 3 before dividing 3 by 2 misreads the problem leading to your wrong answer. The large fraction bar acts as a grouping symbol in common mathematical usage. Thus in order of operations you can think of there being parentheses around the numerator and denominator of any fraction written using the fraction bar.",
        "video_name": "cBi4a1iSaPk",
        "timestamps": [
            300
        ],
        "3min_transcript": "x to the 3rd in the denominator is the same thing as x to the negative 3. We have the same base, we could just add the exponents. So this is going to be equal to 18 times x to the 2 and 1/2 power. Or another way of thinking about it, this is the same thing as 18 times x to the 5/2 power. Did I do that right? Negative 3. Oh sorry, this is negative 2 and 1/2. Let me make this very clear. And this is going to be the negative 5/2 power. x to the negative 3 is the same thing as x to the negative 6/2. Negative 6/2 plus 1/2 is negative 5/2. So once again, we just have to increment this exponent. So negative 5/2 plus 1 is going to be negative 3/2. And then you divide by what your exponent is when you increment it. So divide it by negative 3/2. And then you had the 18 out front. And we obviously are going to have to simplify this. And then finally, our exponent in this term. Let me not use that purple anymore. The exponent in this term right over here is negative 40. If we increment it, we get x to the negative 39 power, all of that over negative 39. And now we can add our constant, c. And all we need to do is simplify all of this craziness. So the first one is fairly simplified. We can write it as 7/4 x to the 4th. Now this term right over here is essentially negative 5 divided by 3/2. So 5 over 3/2 is equal to 5 times 2/3, So this thing right over here simplifies to negative 10/3 x to the 3/2. And then we have all of this craziness. Now 18 divided by negative 3/2 is equal to 18 times negative 2/3. Which is equal to, well we can simplify this a little bit, this is the same thing as 6 times negative 2. Which is equal to negative 12. So this expression right over here is negative 12x to the negative 3/2. And then finally, this one right over here, we can just rewrite it as, if we want, we could, well, we could just write negative 1/39 x to the negative 39 plus c. And we're done. We've found the indefinite integral of all this craziness."
    },
    {
        "Q": "\nat 1:17 when you try to graph the slope of f(x)'s assumed parabola,how do you know that slope will decrease linearly",
        "A": "though Sal didn t want to share it with you here .. but, if you want to know .. a standard parabolic function is of type a x^2 + b x + c .. now help yourself and differentiate the above function .. for x^2, we got the function to be 2x .. when you differentiate the above standard parabolic function, you will necessarily get the function 2ax + b .. this is the equation of a line or a linear relation in x irrespective of the values of a, b and c ..",
        "video_name": "eVme7kuGyuo",
        "timestamps": [
            77
        ],
        "3min_transcript": "So I've got this crazy discontinuous function here, which we'll call f of x. And my goal is to try to draw its derivative right over here. So what I'm going to need to think about is the slope of the tangent line, or the slope at each point in this curve, and then try my best to draw that slope. So let's try to tackle it. So right over here at this point, the slope is positive. And actually, it's a good bit positive. And then as we get larger and larger x's, the slope is still positive, but it's less positive-- and all the way up to this point right over here, where it becomes 0. So let's see how I could draw that over here. So over here we know that the slope must be equal to 0-- right over here. Remember over here, I'm going to try to draw y is equal to f prime of x. And I'm going to assume that this is some type of a parabola. But let's say that, so let's see, here the slope is quite positive. So let's say the slope is right over here. And then it gets less and less and less positive. And I'll assume it does it in a linear fashion. That's why I had to assume that it's some type of a parabola. So it gets less and less and less positive. Notice here, for example, the slope is still positive. And so when you look at the derivative, the slope is still a positive value. But as we get larger and larger x's up to this point, the slope is getting less and less positive, all the way to 0. And then the slope is getting more and more negative. And at this point, it seems like the slope is just as negative as it was positive there. So at this point right over here, the slope is just as negative as it was positive right over there. So it seems like this would be a reasonable view of the slope of the tangent line over this interval. Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this."
    },
    {
        "Q": "Why doesn't Sal shade in the yellow circle (which marks the beginning of the red line) at 4:04?\n\nEdit: Isn't that point defined (just a jump discontinuity)?\n",
        "A": "It is not defined, because the definition of derivative uses limits, and since the limit from the left diverges (meaning, f(x) is the red point, and f(x+e) where e<0 has (f(x+e)-f(x))/e going to infinity), we can t define the derivative here. It just stems from the limit definitions is all.",
        "video_name": "eVme7kuGyuo",
        "timestamps": [
            244
        ],
        "3min_transcript": "Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity. But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope. In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope. And once again, it's undefined here So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0. The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative. But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. And by doing so, we have essentially drawn the derivative over that interval."
    },
    {
        "Q": "\nAt 0:41 \"How many times does 4 go into 9 ?\" Why do we need to know how many times does 4 go into 9 ?",
        "A": "Because that would go in the ones place for long dividing 4 / 9.",
        "video_name": "KFzcwWTEDDI",
        "timestamps": [
            41
        ],
        "3min_transcript": "In this video, I'll introduce you to a new way of computing division, especially for larger numbers. And then we'll think a little bit about why it works. So we're going to try to compute what 96 divided by 4 is. And I'm going to write it a little bit differently. I'm going to write 96 divided by-- so I'm going to write this strange-looking symbol right over here, this thing that covers the 96. But you could view this as 96 divided by 4. And I'll show you in a second why we write it this way. This is actually a very useful way of actually computing it. So the first thing we'll do is we'll say, well, how many times does 4 go into 9? Well, we know that 4 times 2 is equal to 8 and that 4 times 3 is equal to 12. So 3 would be too much. We would go above 9. So we want to be below 9 but not have too much left over. We want the largest number that gets us into 9 without going over 9. 4 goes into 9 two times. And then we say, what's 2 times 4? Well, 2 times 4 is 8. 4 times 2 is 8. Or 2 times 4 is 8. And now, we subtract. We subtract the 8 from the 9. And we get 1. And now we bring down the next digit, which is the 6. And then we ask ourselves, well, how many times does 4 go into 16? Well, in this case, we know that 4 goes into 16 exactly four times. 4 times 4 is 16. So we say 4 goes into 16 four times. Then we multiply 4 times 4 is 16. We subtract. And 16 minus 16, we have absolutely nothing left over. And there we have our answer. I know it seems kind of magical at this point. But in a few seconds, we're going to think about why this actually worked. We got that 96 divided by 4 is equal to-- I want to do that in a different color-- 24. and think about why did this actually work. How did we magically get the right answer here? And you can verify this. Multiply 4 times 24, and you will get 96. Well, I'm assuming you gave a go at it. And the important thing always is to keep track of the place value. And it really tells you what's going on when we do this process. When we looked at this 9 right over here, this 9 is in the tens place. This is actually representing 90. It represents 9 tens. So we're saying, well, how many times does 4 go into 90 if we're thinking about multiples of 10? Well, it goes 20 times. 4 times 20 is 80. And so we said, well, 4 times 20 is 80. But we still have 16 left over. You do 96 minus 80. You have 16 left over to divide 4 into. And then 4 goes into 16 four times."
    },
    {
        "Q": "\nAt 4:15 why did he flip 3/7?",
        "A": "A simpler way to divide fractions is to take the first fraction and multiply it with the reciprocal of the second fraction, example: 5/7 divided by 3/2 is equal to 5/7 times 2/3.",
        "video_name": "K2b8iMPY11I",
        "timestamps": [
            255
        ],
        "3min_transcript": "So, this is going to be the same thing as two times, two times the reciprocal of 2/3, which is, we swap the denominator and the numerator, two times three over two. So, two times three halves, if I have three halves twice, well, that's just going to be equal to, that's going to be equal to six halves, and six halves, you can view that as, well, two halves is a whole, so this is gonna be three wholes. Or you can say six divided by two. Well, that's just gonna be equal to three. You can view it either way. And then that is equal to three. Let's do a few more of these. And let's keep making them a little bit more complicated. Just let me get some good practice. And like always, pause the video. You should get excited when you see one of these things. And pause the video and see if you can do it on your own. All right, let's do something really interesting. Let's say negative 16 over nine over, over, I'll do this in a tan color, What is this? Can you simplify this complex fraction? Can you simplify this expression over here? Well, once again, we can view this as negative 16/9 divided by 3/7. So, this could be rewritten as, and I can write it either as negative 16 over 9 or I could rewrite it as negative 16/9. So, I can put the negative in front of the whole, in front of the whole fraction like that, or I could say that it's negative 16 over nine, or I can even write this as negative 16 over negative nine. Those would all be equivalent. But right now I'm writing at negative 16/9 and I am going to now divide, we can interpret this complex fraction as dividing it by 3/7. Dividing by 3/7. And so, this is going to be equal to, this is going to be equal to negative 16/9, Actually, let me just rewrite it as negative 16 over nine. Negative 16 over nine just to show that we can do it. Times the reciprocal of this. So, times 7/3. I just swap the numerator and the denominator. Times, do that same brown color, times, times 7/3, and now what am I going to get? In the numerator, I have negative 16 times seven. Let me think about it. 10 times seven is 70. Six times seven is 42. 70 plus 42 is 112. So, this, so this part right over here, this part right over here would be negative 112. Negative 112 over nine times three. Nine times three is 20, nine times three is 27. And there you have it. You can view this as negative 112 over 27 or you can put the negative up front and you can say, \"Hey, this is hinting"
    },
    {
        "Q": "7:20 After the summation sign brackets are needed around the two terms following, else the summation sign only applies to the first term?\n",
        "A": "You are right. A parenthesis is needed to enclose the two terms in order to include both terms in the summation. Otherwise, it would be ambiguous. Sal sometimes makes mistakes like that. However, judging from this context, even without a parenthesis, it was clear that he intended to include both terms in the sum.",
        "video_name": "qUNGPqCPzMg",
        "timestamps": [
            440
        ],
        "3min_transcript": "this as negative 2, you could view this as negative 2 plus 0n, plus 0 times n. That's not \"on.\" That's 0-- let me write it this way-- 0 times n. So when you look at it this way, it's clear that A plus B is the coefficient on n. That must be equal to 0. A plus B must be equal to 0. And this is kind of bread-and-butter partial fraction expansion. We have other videos on that if you need to review that. And the constant part, 2A plus B, is equal to negative 2. And so now we have two equations in two unknowns. And we could solve it a bunch of different ways. But one interesting way is let's multiply the top equation by negative 1. So then this becomes negative A minus B is equal to-- well, Now we can add these two things together. And we are left with 2A minus A is A, plus B minus B-- well, those cancel out. A is equal to negative 2. And if A is equal to negative 2, A plus B is 0, B must be equal to 2. Negative 2 plus 2 is equal to 0. We solved for A. And then I substituted it back up here. So now we can rewrite all of this right over here. We can rewrite it as the sum-- and actually, let me do a little bit instead. Let me just write it as a finite sum as opposed to an infinite sum. And then we can just take the limit as we go to infinity. So let me rewrite it like this. So this is the sum from n equals 2-- instead to infinity, I'll just say to capital N. And then later, we could take the limit as this goes to infinity of-- well, instead of writing this, I can write this right over here. So it's negative 2 over n plus 1. And then B is 2, plus B over n plus 2. So once again, I've just expressed this as a finite sum. Later, we can take the limit as capital N approaches infinity to figure out what this thing is. Oh, sorry, and B-- let me not write B anymore. We now know that B is 2 over n plus 2. Now, how does this actually go about helping us? Well, let's do what we did up here. Let's actually write out what this is going to be equal to. This is going to be equal to-- when n is 2, this is negative 2/3, so it's negative 2/3, plus 2/4. So that's n equals-- let me do it down here, because I'm"
    },
    {
        "Q": "\nSal loses me at about 6:00. I don't understand what he's doing with the\nA+B=0\n2A+B= -2\n-A-B=0\nA= -2\nB=2\nWhat is the thought process here? I don't understand what he did.",
        "A": "When you have the same variable on both sides of the equal sign, the coefficients of that variable are going to equal each other so in this case (A + B)n = 0(n) so (A + B) = 0. Same with (2A + B). They are both constant terms and you can rewrite them as (2A + B)n^0 = -2n^0. Then set coefficients of the same variable equal to each other and you get (2A + B) = -2. Now you solve the system of equations (Sal does it here by the method of elimination) and you get A = -2 and B = 2.",
        "video_name": "qUNGPqCPzMg",
        "timestamps": [
            360
        ],
        "3min_transcript": "And, of course, all of that is over n plus 1 times n plus 2. So how do we solve for A and B? Well, the realization is that this thing must be equal to negative 2. These two things must be equal to each other. Remember, we're making the claim that this, which is the same thing as this, is equal to this. That's the whole reason why we started doing this. So we're making the claim that these two things are equivalent. We're making this claim. So everything in the numerator must be equal to negative 2. So how do we do that? It looks like we have two unknowns here. To figure out two unknowns, we normally need two equations. Well, the realization here is, look, we have an n term on the left-hand side here. We have no n term here. this as negative 2, you could view this as negative 2 plus 0n, plus 0 times n. That's not \"on.\" That's 0-- let me write it this way-- 0 times n. So when you look at it this way, it's clear that A plus B is the coefficient on n. That must be equal to 0. A plus B must be equal to 0. And this is kind of bread-and-butter partial fraction expansion. We have other videos on that if you need to review that. And the constant part, 2A plus B, is equal to negative 2. And so now we have two equations in two unknowns. And we could solve it a bunch of different ways. But one interesting way is let's multiply the top equation by negative 1. So then this becomes negative A minus B is equal to-- well, Now we can add these two things together. And we are left with 2A minus A is A, plus B minus B-- well, those cancel out. A is equal to negative 2. And if A is equal to negative 2, A plus B is 0, B must be equal to 2. Negative 2 plus 2 is equal to 0. We solved for A. And then I substituted it back up here. So now we can rewrite all of this right over here. We can rewrite it as the sum-- and actually, let me do a little bit instead. Let me just write it as a finite sum as opposed to an infinite sum. And then we can just take the limit as we go to infinity. So let me rewrite it like this. So this is the sum from n equals 2-- instead to infinity, I'll just say to capital N. And then later, we could take the limit as this goes to infinity of-- well, instead of writing this, I can write this right over here."
    },
    {
        "Q": "\nat 1:00 Sal says, \"We also have to subtract 54 from this side\" When he should have said \"We also have to subtract 5 from this side\"",
        "A": "i know. it was just a typo. :D",
        "video_name": "VidnbCEOGdg",
        "timestamps": [
            60
        ],
        "3min_transcript": "Solve for a and check your solution. And we have a plus 5 is equal to 54. Now, all this is saying is that we have some numbers, some variable a. And if I add 5 to it, I will get 54. And you might be able to do this in your head. But we're going to do it a little bit more systematically. Because that'll be helpful for you when we do more complicated problems. So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5, or a positive 5, is to subtract 5. So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Because if you add 5 and you subtract 5, they cancel out. So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. A is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49, 49 plus-- let me do that in that same shade of green. 49 plus 5 is equal to 54. We're trying to check this. 49 plus 5 is 54. And that, indeed, is equal to 54. So it all checks out."
    },
    {
        "Q": "At 0:38, Sal says that we need to get rid of everything on the left-hand side. Why?\n",
        "A": "you do this to isolate x, or the other variable, on one side to where x equals a number. :-) hope this helps!",
        "video_name": "VidnbCEOGdg",
        "timestamps": [
            38
        ],
        "3min_transcript": "Solve for a and check your solution. And we have a plus 5 is equal to 54. Now, all this is saying is that we have some numbers, some variable a. And if I add 5 to it, I will get 54. And you might be able to do this in your head. But we're going to do it a little bit more systematically. Because that'll be helpful for you when we do more complicated problems. So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5, or a positive 5, is to subtract 5. So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Because if you add 5 and you subtract 5, they cancel out. So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. A is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49, 49 plus-- let me do that in that same shade of green. 49 plus 5 is equal to 54. We're trying to check this. 49 plus 5 is 54. And that, indeed, is equal to 54. So it all checks out."
    },
    {
        "Q": "At 0:54, what are \"real numbers\"?\n",
        "A": "real numbers are a value that represents a quantity along a continuous line. Both plus and minus. like 1, 2, 3, 4, 5, blah blah blah",
        "video_name": "GVZUpOm3XUg",
        "timestamps": [
            54
        ],
        "3min_transcript": "What I want to do in this video is introduce the idea of a universal set, or the universe that we care about, and also the idea of a complement, or an absolute complement. If we're for doing it as a Venn diagram, the universe is usually depicted as some type of a rectangle right over here. And it itself is a set. And it usually is denoted with the capital U-- U for universe-- not to be confused with the union set notation. And you could say that the universe is all possible things that could be in a set, including farm animals and kitchen utensils and emotions and types of Italian food or even types of food. But then that just becomes somewhat crazy, because you're thinking of all possible things. Normally when people talk about a universal set, they're talking about a universe of things that they care about. So the set of all people or the set of all real numbers or the set of all countries, whatever the discussion is being focused on. But we'll talk about in abstract terms right now. Now, let's say you have a subset of that universal set, set A. that I have just shaded in. What we're going to talk about now is the idea of a complement, or the absolute complement of A. And the way you could think about this is this is the set of all things in the universe that aren't in A. And we've already looked at ways of expressing this. The set of all things in the universe that aren't in A, we could also write as a universal set minus A. Once again, this is a capital U. This is not the union symbol right over here. Or we could literally write this as U, and then we write that little slash-looking thing, U slash A. So how do we represent that in the Venn diagram? Well, it would be all the stuff in U that is not in A. One way as the relative complement of A that is in U. But when you're taking the relative complement of something that is in the universal set, you're really talking about the absolute complement. Or when people just talk about the complement, that's what they're saying. What's the set of all the things in my universe that are not in A. Now, let's make things a little bit more concrete by talking about sets of numbers. Once again, our sets-- we could have been talking about sets of TV personalities or sets of animals or whatever it might be. But numbers are a nice, simple thing to deal with. And let's say that our universe that we care about right over here is the set of integers. So our universe is the set of integers. So I'll just write U-- capital U-- is equal to the set of integers. And this is a little bit of an aside, but the notation for the set of integers tends to be a bold Z. And it's Z for Zahlen, from German, for apparently integer."
    },
    {
        "Q": "Why is the integral of cos(x) not -sin(x) at 2:47?\n",
        "A": "The fundamental theorem of calculus says that f(x) is a differentiable function, then the integral of f (x) is just f(x). Now if f(x) = sin(x), then f (x) = cos(x). This means that the integral of cos(x) is just sin(x). (Note: I wish I could use integral signs but I can t find a good way to do so).",
        "video_name": "xVWCfMe97ws",
        "timestamps": [
            167
        ],
        "3min_transcript": "We still haven't fully separated the y's and the x's. Let's divide both sides of this by y, and then let's see. We get 1 over y plus 2y squared divided by y, that's just 2y, times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function the natural log of the absolute value of y. And this is just a slightly broader function, because it's domain includes positive and negative numbers, it just excludes 0. While natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice, and it's actually true that at all points other than 0, its derivative is 1 over y. It's just a slightly broader function. So that's the antiderivative of 1 over y, and we proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared, is equal to-- I'll do the plus c on this side. Whose derivative is cosine of x? Well, it's sine of x. And then we could add the plus c. We could add that plus c there. And what was our initial condition? y of 0 is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of one, e to the what power is 1? Well, 0, plus 1 is-- sine of 0 is 0 --is equal to C. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it, we figured out c is equal to 1, so we can just scratch this out, and we could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you can actually get rid of that absolute value function there. But anyway, that's just a little technicality. But this is the implicit form of the solution to this"
    },
    {
        "Q": "\nAt 0:30 it says that f of negative 6 is 7\nHow did they get 7?",
        "A": "f(x) is a function and when you put -6 in for x you get 7. He got it by finding -6 on the x-axis and going strait up to see what was the y-coordinate when the function was at -6 for x. Does that make any sense? Basically, when the function s x-coordinate is -6, the y-coordinate is 7.",
        "video_name": "uaPm3Tpuxbc",
        "timestamps": [
            30
        ],
        "3min_transcript": "We're asked to evaluate negative 2 times f of negative 6 plus g of 1. And they've defined, at least graphically, f of x and g of x here below. So let's see how we can evaluate this. Well, to do this, we first have to figure out what f of negative 6 is. So our input into our function is negative 6. And we'll assume that's along the horizontal axis. So our input is negative 6. And based on our function definition, f of negative 6 is 7. Let me write this down. f of negative 6 is equal to 7. And what is g of 1? Well, once again, here's our input axis. And then the function says that g of 1, which is right over there, is negative 5. g of 1 is equal to negative 5. So this statement simplifies to negative 2 times f of negative 6, which is 7. So times 7 plus g of 1, which is negative 5. Negative 2 times 7 is negative 14 plus negative 5, which is negative 19. And we are done."
    },
    {
        "Q": "At 7:28 , why are corrseponding altitudes at similar triangles have the same ratio as other corresponding parts such as corresponding sides? perhaps i missed the part where we proves or speaks about corresponding altitudes having the same ratio of the sides? how can we prove this?\n",
        "A": "This was proven in Sal s videos on similarity.",
        "video_name": "v5SAMuRanGM",
        "timestamps": [
            448
        ],
        "3min_transcript": "is PR, segment PR, the length of that, times the height, which is, the height is segment AQ, so the length of segment AQ, we could just write it like that, times the length of segment AQ. So how can we simplify this a little bit? Well, we could divide the 1/2 by the 1/2. Those two cancel out. But what else do we know? Well, they gave us the ratio between AC and AQ. The ratio of AC to AQ right over here is phi plus 1 to 1. Or we could just say this is equal to phi. Or we could say this is just equal to phi plus 1. So let me rewrite this. Actually, let me write it this way. This is going to be equal to-- So we have the length of segment BD over the length of segment PR, and then this part right over here we can rewrite, this is equal to phi plus 1 over 1. So I'll just write it that way. So what's the ratio of BD to PR? So the ratio of the base of the larger triangle to the base of the smaller triangle. So let's think about it a little bit. What might jump out at you is that the larger triangle and the smaller triangle, that they are similar triangles. They both obviously have angle A in common, and since PR is parallel to BD, we know that this angle corresponds to this angle. So these are going to be congruent angles. And we know that this angle corresponds to this angle right over here. So now we have three correspondingly angles are congruent. This is congruent to itself, which is in both triangles. This is congruent to this. This is congruent to that. You have three congruent angles, you're dealing with two similar triangles. And what's useful about similar triangles are the ratio between corresponding parts. of the similar triangles are going to be the same. And they gave us one of those ratios. They gave us the ratio of the altitude of the larger triangle to the altitude of the smaller triangle. AC to AQ is phi plus 1 to phi. But since this is true for one corresponding part of the similar triangles, this is true for any corresponding parts of the similar triangle, that the ratio is going to be phi plus 1 to 1. So the ratio of BD, the ratio of the base of the larger triangle to the base of the smaller one, that's also going to be phi plus 1 over 1. Let me just write it this way. This could also be rewritten as phi plus 1 over 1. So what does this simplify to? Well, we have phi plus 1 over 1 times phi plus 1 over 1. Well, we could just divide by 1."
    },
    {
        "Q": "\nat 1:19 what dose sal say?",
        "A": "Here s what Sal is saying between 1:17 and 1:42: Well, we can t subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don t we take a hundred from the six hundred, so that becomes five hundred, and give that hundred to the ten s place. If we give that hundred to the ten s place, what is a hundred plus twenty? Well, it s going to be \u00e2\u0080\u00a6 120. (He is explaining how regrouping or borrowing works during subtraction.)",
        "video_name": "QOtam19NQcQ",
        "timestamps": [
            79
        ],
        "3min_transcript": "I've written the same subtraction problem twice. Here we see we're subtracting 172 from 629. And all I did here is I expanded out the numbers. I wrote 629 as 600 plus 20 plus 9, and I rewrote 172, the one is 100. So that's there. This is 7/10. It's in the tens place, so it's 70. And then the 2 is 2 ones, so it just represents 2. And we'll see why this is useful in a second. So let's just start subtracting, and we'll start with the ones place. So we have 9 minus 2. Well, that's clearly just 7. And over here we could also say, well, 9 minus 2, we have the subtraction out front. That is going to be 7. Pretty straightforward. But then something interesting happens when we get to the tens place. We're going to try to subtract 2 minus 7, or we're going to try to subtract 7 from 2. And we haven't learned yet how to do things like negative numbers, which we'll learn in the future, so we have a problem. How do you subtract a larger number from a smaller number? regrouping, sometimes called borrowing. And that's why this is valuable. When we're trying to subtract a 7 from a 2, we're really trying to subtract this 70 from this 20. Well, we can't subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don't we take 100 from the 600, so that becomes 500, and give that 100 to the tens place? If we give that 100 to the tens place, what is 100 plus 20? Well, it's going to be 120. So all I did, I didn't change the value of 629. I took 100 from the hundreds place and I gave it to the tens place. Notice 500 plus 120 plus 9 is still 629. We haven't changed the value. So how would we do that right over here? Well, if we take 100 from the hundreds place, give that hundred to the tens place, it's going to be 10 hundreds. So this will now become a 12. This will now become a 12. But notice, this 12 in the tens place represents 12 tens, or 120. So this is just another way of representing what we've done here. There's no magic here. This is often called borrowing, where you say hey, I took a 1 from the 6, and I gave it to the 2. But wait, why did this 2 become a 12? Why was I able to add 10? Well, you've added 10 tens, or 100. You took 100 from here, so 600 became 500, and then 20 became 120. But now we're ready to subtract. 12 tens minus 7 tens is 5 tens. Or you could say 120 minus 70 is 50. And then finally, you have the hundreds place."
    },
    {
        "Q": "\nat 1:27, do you have to take away 100? or can you take away just 1?",
        "A": "Well,it depends on what strategy you are using to subtract.If you re just doing standard algorithm then, you would just write that you are borrowing 1 but if you want to say the accurate value of the number then you would say that you re borrowing 100, which is what you are actually borrowing! Hope this helps!",
        "video_name": "QOtam19NQcQ",
        "timestamps": [
            87
        ],
        "3min_transcript": "I've written the same subtraction problem twice. Here we see we're subtracting 172 from 629. And all I did here is I expanded out the numbers. I wrote 629 as 600 plus 20 plus 9, and I rewrote 172, the one is 100. So that's there. This is 7/10. It's in the tens place, so it's 70. And then the 2 is 2 ones, so it just represents 2. And we'll see why this is useful in a second. So let's just start subtracting, and we'll start with the ones place. So we have 9 minus 2. Well, that's clearly just 7. And over here we could also say, well, 9 minus 2, we have the subtraction out front. That is going to be 7. Pretty straightforward. But then something interesting happens when we get to the tens place. We're going to try to subtract 2 minus 7, or we're going to try to subtract 7 from 2. And we haven't learned yet how to do things like negative numbers, which we'll learn in the future, so we have a problem. How do you subtract a larger number from a smaller number? regrouping, sometimes called borrowing. And that's why this is valuable. When we're trying to subtract a 7 from a 2, we're really trying to subtract this 70 from this 20. Well, we can't subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don't we take 100 from the 600, so that becomes 500, and give that 100 to the tens place? If we give that 100 to the tens place, what is 100 plus 20? Well, it's going to be 120. So all I did, I didn't change the value of 629. I took 100 from the hundreds place and I gave it to the tens place. Notice 500 plus 120 plus 9 is still 629. We haven't changed the value. So how would we do that right over here? Well, if we take 100 from the hundreds place, give that hundred to the tens place, it's going to be 10 hundreds. So this will now become a 12. This will now become a 12. But notice, this 12 in the tens place represents 12 tens, or 120. So this is just another way of representing what we've done here. There's no magic here. This is often called borrowing, where you say hey, I took a 1 from the 6, and I gave it to the 2. But wait, why did this 2 become a 12? Why was I able to add 10? Well, you've added 10 tens, or 100. You took 100 from here, so 600 became 500, and then 20 became 120. But now we're ready to subtract. 12 tens minus 7 tens is 5 tens. Or you could say 120 minus 70 is 50. And then finally, you have the hundreds place."
    },
    {
        "Q": "\nat 1:33 it just dose not make sence",
        "A": "Ok. If you have a calculator on your computer (what you are using right now) I want you to multiply 10 x 32. It will equal 320. If you multiply 100 x 32, you will get 3200. Instead of multiplying by 10 you are multiplying by 100. There are two zeroes in 100, so you take 32 and add zeroes, 3200. Izzy, if this does not help comment below, and I will give you a more detailed answer.",
        "video_name": "tHQOAvbyRL0",
        "timestamps": [
            93
        ],
        "3min_transcript": "- [Voiceover] Let's multiply four times 80. So we can look at this a few ways. One way is to say four times, we have the number 80. So we have the number 80 one time, two times, three times, four times. Four times we have the number 80. And we could do this computation, add all of these, and get our solution. But let's look at it another way. Let's try to stick with multiplication. And one way we can do that is to break up this 80. We know a pattern for multiplying by 10, so let's try to break up this 80 to get a 10. So if we have four times, and instead of 80, let's say eight times 10. Because 80 and eight times 10 are equal; those are equivalent; so we can replace our 80 with eight times 10. which is super helpful 'cause there's a nice neat pattern in math that we can use to help us with the times 10 part. So let's start to solve this. Four times eight is 32. And then we still have 32 times 10. And then we can use our pattern for multiplying by 10, which is that anytime we multiply a whole number times 10 we take that whole number, in this case 32, and we add a zero to the end. So 32 times 10 is 320. And there's a reason that pattern works; we went into it in another video, but here just real quickly, 32 times 10 is 32 tens. And we can do a few examples. If we had, say, three times 10, that would be three tens, or a 10 plus another 10 plus another 10, which equals 30: If we had something like 12 times 10, well, that would be 12 tens. And if we listed out 10 12 times and counted 'em up, there would be 120, it would add up to 120, which again is our whole number with a zero on the end, or 12 with a zero on the end. So we can use that pattern here to see that 32 times 10 is 32 with a zero on the end. Let's try another one. Let's do something like, let's say 300 this time, we'll do hundreds instead of tens, times six. 300 we can break up, like we did with 80 in the last one, and we can say that 300 is 100 three times, or 100 times three. And then we still have our times six after that."
    },
    {
        "Q": "At 23:58 he mentions that parametric equations are the only way to define a line in 3D. Would parametric equations make a line in any dimension from R2 to Rn?\n",
        "A": "Yes. Parametric equations will always define a line, assuming you have a parametric equation for each of the dimensions.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            1438
        ],
        "3min_transcript": "It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically."
    },
    {
        "Q": "At 20:24 you are making a \"general definition\" in R*3 for a parametric equations for a line.\nL= {P1 + t(P1 - P2)}, t in R but could there also be L={P2 + t(P1 - P2)} . Is that the same line and does it have the same direction? If not then the \"general definition\" is not so general(?).\n",
        "A": "They re both the same line with the only difference being which point on the line you re at given a specific value for t. This definition also isn t restricted just to R3. Any line in any R^n vector space can be written in this form.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            1224
        ],
        "3min_transcript": "can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2 plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time, this was convoluted. You have to define these sets and all that. But now I'm going to show you something that you probably-- that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1."
    },
    {
        "Q": "\nAt 24:20, the line is represented in terms of x,y and z. And x,y and z are defined in terms of t. If we assume that t is time then each value of t will give a single points on R3 at different times. Are we saying that the points on a 3-dimensional line can be represented in terms of a 4th dimension? t is a scalar, what is the interpretation of t?",
        "A": "t is a free variable, getting all values in R.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            1460
        ],
        "3min_transcript": "It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically."
    },
    {
        "Q": "why is it vector b - vector a ? at 13:43 ?\n",
        "A": "It s to get the slope of the line. The vector b\u00e2\u0083\u0097 - a\u00e2\u0083\u0097 (or a\u00e2\u0083\u0097 - b\u00e2\u0083\u0097) point in the direction of the line we want to represent.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            823
        ],
        "3min_transcript": "all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which"
    },
    {
        "Q": "At 0:04, why does it say \"4,5000 equals 3 thousands plus how many\" in the words when it says \"4,500 = 3 thousands + ? hundreds\" on the board?\n",
        "A": "He is correct it at 0:04 it says 45,000 with subtitles on.",
        "video_name": "a_mzIWvHx_Y",
        "timestamps": [
            4
        ],
        "3min_transcript": "We have 4,5000 equals 3 thousands plus how many hundreds, question mark hundreds? So let's write this left-hand side, but I'm going to write it out in terms of thousands and hundreds. So I'll write the thousands in orange. So this is equal to 4 thousands, which is the same thing as just 4,000, plus 500, which you could also view as 5 hundreds. So this is the left-hand side. Now let's look at the right-hand side. We have 3 thousands. So it's 3 thousands. Now let's not even look at this right now. Let's just think about what do we have to add to this right-hand side in order to get the same thing that we have on the left-hand side? Well, if you compare the 3,000 and the 4,000, you see you have an extra 1,000 over here. So let's add an extra 1,000 on the right. So we're going to add one extra 1,000. And now we just have 3,000 plus 1,000. This makes it the 4,000. But then, of course, we also need another 500. So we're going to need plus a 500 right over here. we need to say 4,000 plus 500 is equal to 3,000 plus 1,500. Now, the way they've set this up, we need to express-- so it almost looks the same. On the left-hand side, this is 4,500. So this right over here, this is the same thing as 4,500. This is this right over here. And on the right-hand side, we have 3 thousands. So that's this right over here. That's the 3 thousands. And then we just need to express this as hundreds. So 1,500, this is the same thing as 15 hundreds. So let's rewrite everything. We can rewrite this as saying 4,500, just to get the exact same form that they wrote it over there. So we could write 4,500 is equal to 3 thousands plus-- now This is 15 hundreds. Literally, if you took 15 times 100, it's going to be equal to 1,500. So this could be viewed as 15 hundreds, so plus 15 hundreds. So in this situation, the question mark is equal to 15."
    },
    {
        "Q": "\nAt 6:51 Sal says that it doesn't have to defined at that point, what does he mean by that?",
        "A": "He meant that the denominator does not have to be 0.",
        "video_name": "igJdDN-DPgA",
        "timestamps": [
            411
        ],
        "3min_transcript": "here, this is g of x. So this is my g of x. And we know that as g of x approaches-- so the g of x could look something like that, right? And we know that the limit as x approaches a of g of x is equal to L. So that's right there. So this is g of x. That's g of x. Let me do h of x in a different color. So now h of x could look something like this. Like that. So that's h of x. And we also know that the limit as x approaches a of h of x -- let's see, this is the function of x axis. So you can call it h of x, g of x, or f of x. That's just the dependent access, and this is the x-axis. So once again, the limit as x approaches a of h of x, well Or at least the limit is equal to that. And none of these functions actually have to even be defined at a, as long as these limits, this limit exists and this limit exists. And that's also an important thing to keep in mind. So what does this tell us? f of x is always greater than this green function. It's always less than h of x, right? So any f of x I draw, it would have to be in between those two, right? So no matter how I draw it, if I were to draw a function, it's bounded by those two functions just by definition. So it has to go through that point. Or at least it has to approach that point. Maybe it's not defined at that point, but the limit as we approach a of f of x also has to be at point L. And maybe f of x doesn't have to be defined right there, but the limit as we approach it is going to be L. And hopefully that makes a little bit of sense, and hopefully my calories example made a little So let's keep that in the back of our mind, And now we will use that to prove that the limit as x approaches 0 of sine of x over x is equal to 1. And I want to do that, one, because this is a super useful limit. And then the other thing is, sometimes you learn the squeeze theorem, you're like, oh, well that's obvious but when is it useful? And we'll see. Actually I'm going to do it in the next video, since we're already pushing 8 minutes. But we'll see in the next video that the squeeze theorem is tremendously useful when we're trying to prove this. I will see you in the next video."
    },
    {
        "Q": "7:34 Sal said that x could not equal -1. But it also cannot equal 2. Did Sal forget it?\n",
        "A": "Sal did not forget the condition that x cannot be equal to 2. You know from the resulting equation not to use x = 2 but you would not know that x = -1 would produce a wrong answer. So, if you were to report to someone that she could use either equation, he would know not to use +2 in the first equation but would not know that -1 will cause an error.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            454
        ],
        "3min_transcript": "multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say"
    },
    {
        "Q": "at 1:39 in the video, why do you factor out a 3 from the numerator? javascript:%20void%200\n",
        "A": "When working with rational expressions (fractions), we always simplify both numerator and denominator so that each of their factors is exposed. This then allows us easily to see what can be divided out in order to further simplify the expression.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            99
        ],
        "3min_transcript": "When we first started learning about fractions or rational numbers, we learned about the idea of putting things in lowest terms. So if we saw something like 3, 6, we knew that 3 and 6 share a common factor. We know that the numerator, well, 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3/3, and they would cancel out. And in lowest terms, this fraction would be 1/2. Or just to kind of hit the point home, if we had 8/24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out and we get this in lowest terms as 1/3. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator be an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now, this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares."
    },
    {
        "Q": "At 10:57 why did he write (3x-6) before (x+3)? He didn't do it in previous problems and I wasn't sure if it would have made a difference if he would have written (x+3) (3x-6)\n",
        "A": "nope it would not had made a difference how you wrote it cuz in the muliplication property you can switch terms around and get the same answer for example: 2*3=6 3*2=6 or: 3x*4x=12x^2 4x*3x=12x^2 hope this helps:)",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            657
        ],
        "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1."
    },
    {
        "Q": "at 8:59 you did a harder way then what I was told, could you not just factor the 3 out from the beggining?\n",
        "A": "Certainly! That would be easier, faster, and would give you a more thoroughly factored result. (He does some odd things with factoring throughout the videos -- certainly his approach works, but it isn t always the simplest.)",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            539
        ],
        "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3."
    },
    {
        "Q": "I didn't understand any one thing he explained in this video and it's getting me frustrated!\n\nAt 4:55 you see the whole x squared - 9 = (x + 3 ) (x-3)! How does this make sence? How did u get x+3 and minus 3 in relation to x squared - 9. Help me!\n",
        "A": "(x + 3)\u00e2\u0080\u00a2(x - 3) FOIL: x\u00e2\u0080\u00a2x + 3\u00e2\u0080\u00a2x - 3\u00e2\u0080\u00a2x + 3\u00e2\u0080\u00a2(-3) x^2 - 9",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            295
        ],
        "3min_transcript": "And in the denominator we can factor a 5 out. This is 5 times x plus 3. So once again, a common factor in the numerator and in the denonminator, we can cancel them out. But we touched on this a couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5, but we have to exclude the values of x that would have made this denominator equal to 0, that would have made the entire expression undefined. So we could write this as being equal to x minus 3 over 5, but x cannot be equal to negative 3. Negative 3 would make this zero or would make this whole thing zero. So this and this whole thing are equivalent. This is not equivalent to this right here, because this is defined that x is equal to negative 3, while this isn't So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So"
    },
    {
        "Q": "At 9:00, what numbers represent a and b?\n",
        "A": "a and b represent the coefficients of what you have to split your middle term into when you factor, in order to get the quadratic to factor.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            540
        ],
        "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3."
    },
    {
        "Q": "in the example of minute 10:45 wouldn't the proper way to factorize be (3x-3)(x+6) ? sal made (3x-6)(x+3)\n",
        "A": "No, Sal was right. The expression 3x(x-3) -6(x-3) has a common factor of x-3. Factoring that out yields (x-3)(3x-6), which is the same pair of factors Sal had..",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            645
        ],
        "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1."
    },
    {
        "Q": "\nThank you for the lesson. At 15:13, only one restriction is included. Should we also say that x cannot be 1/2 since it would make (2x-1) a zero as well?",
        "A": "Yes that would be appropriate.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            913
        ],
        "3min_transcript": "And a plus b needs to be equal to 5. So in this situation, it looks like if we went with 6 and negative 1, that seems to be a better situation. 6 minus 1 is 5. 6 times negative 1 is negative 6. So that would have been a horrible mistake. So we can rewrite this up here as 2x squared, and I'll group the 6 with the 2x squared because they share a common factor. So plus 6x minus x, this is the same thing as 5x minus 3. I just had to find the numbers to split this 5x into. But 6x minus x is 5x. And if I put some parentheses here, I can factor out of 2x out of this first term. I get 2x times x plus 3. times x plus 3. And then our grouping was successful. We get 2-- let me do this in a different color-- we get 2x minus 1 times x plus 3. So our denominator here is equal to 2x minus 1 times x plus 3. And once again, we have a common factor in our numerator and our denonminator, the x plus 3. But we have to add the condition that x cannot be equal to negative 3, because that would make this whole thing equal to zero. Or not equal to zero, it would make us divide by zero, which is undefined. So we have to say that x cannot be equal to negative 3. over 2x minus 1, granted that we also imposed the condition that x does not equal negative 3. Hopefully, you found that interesting."
    },
    {
        "Q": "I stopped understanding at around 4:00 when you started talking about conditions and why x can't equal -3, why can't it equal -3 when both of the (x+3)'s have been cancelled out??\n",
        "A": "By being cancelled out, that meant that the (x+3) was divided out of the expression. But if x = - 3, then (x+3) would have equalled zero ---- and division by zero is never permitted. That s why the only way the (x+3) s are allowed to be cancelled out is if the statement that x cannot equal -3 is included with the simplified fraction.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            240
        ],
        "3min_transcript": "So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares. And in the denominator we can factor a 5 out. This is 5 times x plus 3. So once again, a common factor in the numerator and in the denonminator, we can cancel them out. But we touched on this a couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5, but we have to exclude the values of x that would have made this denominator equal to 0, that would have made the entire expression undefined. So we could write this as being equal to x minus 3 over 5, but x cannot be equal to negative 3. Negative 3 would make this zero or would make this whole thing zero. So this and this whole thing are equivalent. This is not equivalent to this right here, because this is defined that x is equal to negative 3, while this isn't So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to"
    },
    {
        "Q": "Near 8:36 couldn't you just factor out a 3 and get 3(x^2+x-6) and then factor the rest of it using the quadratic formula or other methods?\n",
        "A": "Yes, you could",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            516
        ],
        "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3."
    },
    {
        "Q": "\n@ 7:17 ish you said the restriction was x cannot equal -1 but there is another restriction too it is x cannot equal 2",
        "A": "The only restrictions that need to be stated separately are the ones that are fully divided out of the problem. Because ( x - 2 ) is still visible in the denominator, it is already clear that x cannot equal 2 (since division by zero is never allowed).",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            437
        ],
        "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it."
    },
    {
        "Q": "\nAt 7:29 don't we also have to add that x does NOT equal 2. Wouldn't 2 make the equation undefined?",
        "A": "Yes, but that is obvious from the problem since x-2 is in the denominator. In this case we are reminding ourselves that x cannot be equal to -1 as well. We need the reminder because after we factored the expression, we had in the denominator (x+1) and (x-2), but the (x+1) term cancelled out with a like term in the numerator. So the original expression could not have x=2 and x=-1, but we lost the visual clue when we cancelled the (x+1) terms (see 6:44).",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            449
        ],
        "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it."
    },
    {
        "Q": "At 8:57, couldn't Sal have made a= -6 and b=+9 instead of the other way round as both of them are divisable by 3 and 18?\n",
        "A": "Yes, he could ve had. It doesn t matter whether they are a or b as those are just place holders. As long as the numbers are correct you end up with the same answer.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            537
        ],
        "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3."
    },
    {
        "Q": "\nWhat did he do at 1:50 when it was 4/3 pi 27",
        "A": "Alegbra. 4/3 * \u00cf\u0080 * 27 or (4 * \u00cf\u0080 * 27)/3. If you multiply it out, 108\u00cf\u0080/3=36\u00cf\u0080 or, what Sal does, is simplify through division because he recognizes 27/3 in his head, so he gets to 4 * \u00cf\u0080 * 9=36\u00cf\u0080 by skipping a bit.",
        "video_name": "IXRMVcoqRRQ",
        "timestamps": [
            110
        ],
        "3min_transcript": "Frank wants to fill up a spherical water balloon with as much water as possible. The balloons he bought can stretch to a radius of 3 inches-- not too big. If the volume of a sphere is-- and this is volume as a function of radius-- is equal to 4/3 pi r cubed, what volume of water in cubic inches can Frank put into the balloon? So this function definition is going-- if you give it a radius in inches, it's going to produce a volume in cubic inches. So let's rewrite it. Volume as a function of radius is equal to 4/3 pi r cubed. Now, they say the balloons he bought can stretch to a radius of 3 inches. So let's think about, if the radius gets to 3 inches, what the volume of that balloon is going to be. So we essentially would just input 3 inches into our function definition. So everywhere where we see an r, we would replace it with a 3. let me rewrite it in the same color. V of-- that's not the same color. We do it in that brownish color right over here. So V of 3 is equal to 4/3 pi-- and instead of r cubed, I would write 3 cubed-- 4/3 pi 3 cubed. This is how the function definition works. Whatever we input here, it will replace the r in the expression. So V of 3 is going to be equal to-- so this is going to be equal to 4/3 pi times-- 3 to the third power is 27. 27 divided by 3 is 9, so this is 9. 9 times 4 is 36 pi. So this is equal to 36 pi. And since this was in inches, our volume is going to be in inches cubed or cubic inches. can put in the balloon-- 36 pi cubic inches."
    },
    {
        "Q": "\nwhile he was rearranging the operations to solve for \"h\" ,at 1:51 why did he divide \"b\" from \"b\" instead of subtracting \"b\" from \"b\" to eliminate it?",
        "A": "Yes, you would subtract it from both sides instead of dividing in order to rearrange the equation. That would result in 2A = b + h becoming 2A -b = h",
        "video_name": "eTSVTTg_QZ4",
        "timestamps": [
            111
        ],
        "3min_transcript": "The formula for the area of a triangle is A is equal to 1/2 b times h, where A is equal to area, b is equal to length of the base, and h is equal to the length of the height. So area is equal to 1/2 times the length of the base times the length of the height. Solve this formula for the height. So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what b and h are. b would be the length of the base. So this distance right over here is b. And then this distance right here is our height. That is the height of the triangle-- let me do that at a lower case h because that's how we wrote it in the formula. Now, they want us to solve this formula for the height. So the formula is area is equal to 1/2 base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right-hand side. So let's get rid of everything else on the right-hand side. We could kind of skip steps if we wanted to. But let's see if we can get rid of this 1/2. So the best way to get rid of a 1/2 that's being multiplied by h is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2/1 or by 2. So let's do that. So let's multiply-- remember anything you do to one side of the equation, you also have to do to the other side of the equation. Now, what did this do? Well, the whole point behind multiplying by 2 is 2 times 1/2 is 1. So on the right-hand side of the equation, we're just going to have a bh. And on the left-hand side of the equation, we have a 2A. And we're almost there, we have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well, the right-hand side, the b's cancel out. So we get h-- and I'm just swapping the sides here. h is equal to 2A over b. And we're done. We have solved this formula for the height. And I guess this could be useful. If someone just gave you a bunch of areas and a bunch of base lengths, and they said keep giving me the height for those values, or for those triangles."
    },
    {
        "Q": "0:22 how should you divide if your number is a fraction? Like 11 4/6 divided by 5\n",
        "A": "To divide 11 4/6 by 5, make 11 4/6 into ana improper fraction, 70/6. Then When diving a fraction like 70/6 divided by 5, Remember KEEP CHANGE FLIP. To do the strategy, first turn teh divide symbol into a multiplying symbol, then make 5 into 5/1. Then flip 5/1 into 1/5, then multiply. Then Simplify, and you SHOULD have the answer.",
        "video_name": "Z_NHrwK6ALE",
        "timestamps": [
            22
        ],
        "3min_transcript": "Let's see if we can divide 5.005 by-- let's divide it by 7, and see what we get. So we can rewrite this as 5.005 divided by 7. And the key here is to keep track of the decimal. But other than that, you're really treating it like a traditional long division problem. So you want to put the decimal, it's to the right of the ones place. So it's right over there. I just put the decimal right above the decimal. And now we would treat it just like a traditional long division problem. So how many times does 7 go into 5? Well, it goes into 5 zero times, and we could write 0 times 7 is 0. Actually, let's do that, just to show that it works to kind of do the process all the way through. 0 times 7 is 0. You subtract, you get 5. So you get this 5 right over here. So you get 5, and then we can bring down another 0. So now we ask ourselves, how many times does 7 go into 50? Well, 7 times 7 is 49, so it's going to go seven times. 7 times 7, 49. Subtract, you get 1. And now we can bring down this 0. Let's do that. We bring down that 0. 7 goes into 10 one time. 1 times 7 is 7. Subtract, get a 3. And now finally, we can bring down this 5. We can bring down the 5. And we ask ourselves, how many times does 7 go into 35? Well, 5 times 7 is 35. This goes 5 times. 5 times 7 is 35. Subtract, and we are done. We have divided it completely. And so 5.005 divided by 7 is equal to 0.715."
    },
    {
        "Q": "at 4:13pm, Solve the following system of linear equations:\n-2x + 4y - z = 8\nx + 7y + 2z = 5\n3x + 3y + 3z = -3\n\nelimination does not seem to work :(\n",
        "A": "The first equation, is equal to the second minus the third. So you really have 2 equations and 3 unknowns, and there is not a unique solution.",
        "video_name": "GWZKz4F9hWM",
        "timestamps": [
            253
        ],
        "3min_transcript": "Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left hand side. On the left hand side, we're just left with the -11x and then on the right hand side we are left with -22. Now we can divide both sides by -11. And we are left with x is equal to 22 divided by 11 is 2, and the negatives cancel out. x = 2. So we are not quite done yet. We've done, I guess you can say the hard part, we have solved for x but now we have to solve for y. We could take this x value to either one of these equations and solve for y. But this second one has already explicitly solved for y and instead of x, we now know that the x value where these two intersect, you could view it that way is going to be equal to 2, so 2 * 2 - 5 let's figure out the corresponding y value. So you get y=2(2)-5 and y = 4 - 5 so y = -1. And you can verify that it'll work in this top equation If y = -1 and x=2, this top equation becomes -3(2) which is -6-4(-1) which would be plus 4. And -6+4 is indeed -2. So it satisfies both of these equations and now we can type it in to verify that we got it right, So, let's type it in... x=2 and y=-1. Excellent, now we're much less likely to be embarassed by talking birds."
    },
    {
        "Q": "At 14:10 why can't sqrt of 39/3 be simplified to sqrt of 13?\n",
        "A": "Because the 3 is not under the radical. It is \u00e2\u0085\u0093 \u00e2\u0088\u009a39, it is not \u00e2\u0088\u009a(39\u00c3\u00b73)",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            850
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nOn the second problem (answer at 9:16) couldn't you turn the negative square root into \u00e2\u0088\u009a(84)I?",
        "A": "negative square roots are insolvable, and you cannot change intergers at free will. All we can derive is that the equation is insolvable.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            556
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 13:36, what happened to the 2 outside of the radical?",
        "A": "He simplified the fraction by dividing the top and bottom by 2.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            816
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 14:35, why does he divide -12 by 2 as well? Is it because it's a different number than all the stuff with the square root? I thought it would just stay the same because you already divided by two from the top.\n",
        "A": "Because (a+b)/c = a/c + b/c. Concrete example (2 + 4)/2 = 6/2 = 3 Right? Now distribute: (2 + 4)/2 = 2/2 + 4/2 = 1 + 2 = 3 Same! What you are trying to do is a/c + b Using our example your way: (2 + 4)/2 = 2/2 + 4 = 1 + 4 = 5.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            875
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 4:37, on the final step, the problem is (4 + or - 10)/2. sal divides both terms by two. i thought he could only divide 1 term, and then the 2 would be gone. how did he do that?\n",
        "A": "The whole formula must be divided by 2. Think of the quadratic formula as this. [-b \u00c2\u00b1 sqrt(b^2-4ac)] / 2 or in your case it could have been written as (-4\u00c2\u00b110)/2 doing the parentheses first would result in the same answer as dividing both the numbers by two first.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            277
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:16, Sal mentioned caveat. What does caveat mean?",
        "A": "It is a limitation or condition, so he says if you remember this, then you should also remember how to prove it. The condition is the then part of it.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            16
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 13:37, how did -12\u00c2\u00b12\u00e2\u0088\u009a39/-6 become -6\u00c2\u00b1\u00e2\u0088\u009a39/-3 ? I don't understand that division/simplification process.\n",
        "A": "Think of it as (-12\u00c2\u00b12\u00e2\u0088\u009a39)/-6. If I want to factor out a 2 on the top, I would end up with 2(-6 \u00c2\u00b1 \u00e2\u0088\u009a39)/6, if I divide 2/6, I am left with 1/3, so we would have (-6 \u00c2\u00b1 \u00e2\u0088\u009a39)/3. Be careful how you write it because you do not show the -12 as being part of the numerator, and that is why I added the extra parentheses.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            817
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 12:38, why did he chose to stop factoring? Why didn't he continue?\n",
        "A": "Because it cant be factored anymore",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            758
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 10:51 with the yellow matrix Sal just finished, could you add the pivot entries with free variables and finish the reduced row echelon form?\n\nSo something like:\n| 1 1 0 0 | 2 |\n| 0 1 0 0 | s |\n| 0 0 1 2 | 5 |\n| 0 0 0 1 | t |\nInto:\n| 1 0 0 0 | 2 - s |\n| 0 1 0 0 | s |\n| 0 0 1 0 | 5 - 2t |\n| 0 0 0 1 | t |\n\nOr wouldn't schools and universities be looking for something so detailed?",
        "A": "Yes, you can reduce it all the way. Don t quote me but... the first form is called Gaussian elimination the second form is called Gauss-Jordan elimination The nice thing about fully reducing the matrix is now you have you re entire solution on the right side.",
        "video_name": "JVDrlTdzxiI",
        "timestamps": [
            651
        ],
        "3min_transcript": "of parallel equations, they won't intersect. And you're going to get, when you put it in reduced row echelon form, or you just do basic elimination, or you solve the systems, you're going to get a statement that zero is equal to something, and that means that there are no solutions. So the general take-away, if you have zero equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations-- let me write this down, this is good to know. if you have zero is equal to anything, then that means no solution. If you're dealing with r3, then you probably have parallel planes, in r2 you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, this I think you get the idea. That equals a, b, c, d. Then you have a unique solution. Now if, you have any free variables-- so free variables look like this, so let's say we have 1, 0, 1, 0, and then I have the entry 1, 1, let me be careful. 0, let me do it like this. 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of zeroes over here. And then this has to equal zero-- remember, if this was a bunch of zeroes equaling some variable, then I would have no solution, or equalling some constant, let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free Because it has no pivot entries. These are the pivot entries. So this is variable x2 and that's variable x4. Then these would be free, we can set them equal to anything. So then here we have unlimited solutions, or no unique solutions. And that was actually the first example we saw. And these are really the three cases that you're going to see every time, and it's good to get familiar with them so you're never going to get stumped up when you have something like 0 equals minus 4, or 0 equals 3. Or if you have just a bunch of zeros and a bunch of rows. I want to make that very clear. Sometimes, you see a bunch of zeroes here, on the left-hand side of the augmented divide, and you might say, oh maybe I have no unique solutions, I have an infinite number of solutions. But you have to look at this entry right here. Only if this whole thing is zero and you have free variables, then you have an infinite number of solutions. If you have a statement like, 0 is equal to a, if this is equal to 7 right here, then all of the sudden you would"
    },
    {
        "Q": "\nAt 1:58 why is the square root of 74,74",
        "A": "That square root of 74 is squared, so you have sqrt(74) and you square it, by definition you get 74.",
        "video_name": "T0IOrRETWhI",
        "timestamps": [
            118
        ],
        "3min_transcript": "A carpet measures 7 feet long and has a diagonal measurement of square root of 74 feet. Find the width of the carpet. So let's draw ourselves a carpet here. So let's draw a carpet. It has a length of 7 feet, so let's say that that is 7 feet, right there. And it's going to be a rectangle of some kind. So let's say that we're looking down on the carpet like that. That's our carpet. And then it has a diagonal measurement of square root of 74 feet. So that means that this distance, right here-- draw it a little bit neater than that --this distance right here, the diagonal of the carpet, is the square root of 74 feet. And what they want to know is the width of the carpet. Find the width of the carpet. So let's say that this is the width of the carpet. That is w, right there. Now, you might already realize that what I have drawn here is a right triangle. Let me make sure you realize it. And since that is a triangle that has a 90 degree angle, it's a right triangle. The side opposite the right angle, or the 90 degrees, is a hypotenuse, or the longest side. It is the square root of 74. And the shorter sides are w and 7. And the Pythagorean Theorem tells us that the sum of the squares of the shorter side will be equal to the square of the hypotenuse, so the square of the longer side. So we get w squared, this side squared. plus 7 squared, this other side squared, is going to be equal to the hypotenuse squared, square root of 74 squared. And then we get w squared plus 49 is equal to the square root of 74 squared. Well, that's just going to be 74. It is equal to 74. We can subtract 49 from both sides of this equation. Subtract 49 from both sides. The left side-- these guys are going to cancel out, we're just going to be left with a w squared --is equal to-- What's 74 minus 49? 74 minus 49, well, we can do a little bit of regrouping or borrowing here, if we don't want to do it in our head. We can make this a 14. This becomes a 6. 14 minus 9 is 5. 6 minus 4 is 2. And we have w squared is equal to 25. So w is going to be equal to the square root of 25, the positive square root. So let's take the square root of both sides, the positive square root, and we will get w is equal to 5. Because we obviously we don't want it to be negative 5. That wouldn't be a realistic distance. So the width of the carpet is 5. And we're done."
    },
    {
        "Q": "\nAt 00:27, wouldn't we cross off 7 and 49 and simplify them like Sal did at 00:56? I'm a bit confused.",
        "A": "Yes, you can. Doing so makes the problem easier as well.",
        "video_name": "pi3WWQ0q6Lc",
        "timestamps": [
            27,
            56
        ],
        "3min_transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign."
    },
    {
        "Q": "\nin the captions of 5:23 - 5:26 it says \"maybeyou\"",
        "A": "It might have just been a mistake. It s been sorted out now (it says maybe you)",
        "video_name": "DPuK6ZgBGmE",
        "timestamps": [
            323,
            326
        ],
        "3min_transcript": "Once again, this is two, this is three. She deviates. Her absolute deviation is three. And then we wanna take the mean of the absolute deviation. That's the M in MAD, in Mean Absolute Deviation. This is Manueala's absolute deviation, Sophia's absolute deviation, Jada's absolute deviation, Tara's absolute deviation. We want the mean of those, so we divide by the number of datapoints, and we get zero plus one, plus two, plus three, is six over four. Six over four, which is the same thing as 1 1/2. Or, lemme just write it in all the different ways. We could write it as three halves, or 1 1/2, or 1.5. Which gives us a measure of how much do these datapoints vary from the mean of four. I know what some of you are thinking. \"Wait, I thought there was a formula \"associated with the mean absolute deviation. \"It seems really complex. \"It has all of these absolute-value signs That's all we did. When we write all those absolute-value signs, that's just a fancy way of looking at each datapoint, and thinking about how much does it deviate from the mean, whether it's above or below. That's what the absolute value does. It doesn't matter, if it's three below, we just say three. If it's two above, we just say two. We don't put a positive or negative on. Just so you're comfortable seeing how this is the exact same thing you would've done with the formula, let's do it that way, as well. So the mean absolute deviation is going to be equal to. Well, we'll start with Manueala. How many bubbles did she blow? She blew four. From that you subtract the mean of four, take the absolute value. That's her absolute deviation. Of course, this does evaluate to this zero, to zero here. Then you take the absolute value. Sophia blew five bubbles, and the mean is four. Then you do that for Jada. Jada blew six bubbles; the mean is four. And then you do it for Tara. Then you divide it by the number of datapoints you have. Lemme make it very clear. This right over here, this four, is the mean. This four is the mean. You're taking each of the datapoints, and you're seeing how far it is away from the mean. You're taking the absolute value 'cause you just wanna figure out the absolute distance. Now you see, or maybe you see. Four minus four, this is. Four minus four, that is a zero. That is that zero right over there. Five minus four, absolute value of that? That's going to be. Lemme do this in a new color. This is just going to be one. This thing is the same thing as that over there. We were able to see that just by inspecting this graph, or this chart. And then, six minus four, absolute value of that, that's just going to be two. That two is that two right over here, which is the same thing as this two right over there. And then, finally, our one minus four, this negative three,"
    },
    {
        "Q": "at 2:03, what does Sal mean by deviate?\n",
        "A": "Sorry I am late but deviate basically means how far the number is from the mean. Where sal uses it, Sophia (I think that was her name) blew 5 bubbles and the mean is 4, so when you deviate, you find the distance away from the mean.",
        "video_name": "DPuK6ZgBGmE",
        "timestamps": [
            123
        ],
        "3min_transcript": "- This bar graph here tells us bubbles blown by each gum-chewer. We have four different gum-chewers, and they tell us how many bubbles each of them blew. What I wanna do is, I wanna figure out first the mean of the number of bubbles blown, and then also figure out how dispersed is the data, how much do these vary from the mean. I'm gonna do that by calculating the mean absolute deviation. Pause this video now. Try to calculate the mean of the number of bubbles blown. And then, after you do that, see if you can calculate the mean absolute deviation. Step one, let's figure out the mean. The mean is just going to be the sum of the number of bubbles blown divided by the number of datapoints. Manueala blew four bubbles. She blew four bubbles. Sophia blew five bubbles. Jada blew six bubbles. Tara blew one bubble. So let's divided by four. And so, this is going to be equal to four plus five is nine, plus six is 15, plus one is 16. So it's equal to 16 over four, which is 16 divided by four is equal to four. The mean number of bubbles blown is four. Lemme actually do this with a bold line right over here. This is the mean number of bubbles blown. Now what I wanna do is I wanna figure out the mean absolute deviation. Mean. MAD: Mean Absolute Deviation. What we wanna do is we wanna take the mean of how much do each of these datapoints deviate from the mean. I know I just used the word mean twice in a sentence, so it might be a little confusing, but as we work through it, hopefully, it'll make a little bit of sense. How much does Manueala's, the number of bubbles she blew, Well, Manueala actually blew four bubbles, and four is the mean. So her deviation, her absolute deviation from the mean is zero. Is zero. Actually, lemme just write this over here. Absolute deviation, that's AD, absolute deviation from the mean. Manueala didn't deviate at all from the mean. Now let's think about Sophia. Sophia deviates by one from the mean. We see that right there, she's one above. Now, we would say one whether it's one above or below, 'cause we're saying absolute deviation. Sophia deviates by one. Her absolute deviation is one. And then, we have Jada. How much does she deviate from the mean? We see it right over here. She deviates by two. She is two more than the mean. And then, how much does Tara deviate from the mean? She is at one, so that is three below the mean."
    },
    {
        "Q": "\nFor this problem I looked at just the equation and solved like in the \"Factoring 5th degree polynomials to find real zeroes\" (pervious) video. I took y=x^3+3x^2+x+3 and split it into two, just like in 4:20, x^3+3x^2 and x+3. When I got to the (x^2+1) (x+3) I factored out the (x^2+1) again and got (x-1) (x+1). Now I have three solutions, x= -3, -1, 1.\nWhy is this not applicable?",
        "A": "Because x^2 + 1 is not equal to (x - 1)(x + 1). If you multiply that out, you will get x^2 - 1.",
        "video_name": "uFZvWYPfOmw",
        "timestamps": [
            260
        ],
        "3min_transcript": "see that the real function does not equal 0 at 4 or 7. Another giveaway that this is not going to be the function is that you are going to have a total of three roots. Let me write this down. So you're going to have a total of three roots. Now, , those three roots could be real or complex roots. And the big key is complex roots come in pairs. So you might have a situation with three real roots. And this is an example with three real roots, although we know this actually isn't the function right over here. Or if you have one complex root, you're going to have another complex root. So if you have any complex roots, the next possibility is one real and two complex roots. And this right over here has two real roots. That would somehow imply that you have only one complex root, which that is not a possibility. Now another way that you could have thought about this-- and this would have been the longer way. But let's say you didn't have the graphs here for you, and someone asked you to just find the roots-- well, you could have attempted to factor this. And this one actually is factorable. y is equal to x to the third plus 3x squared plus x plus 3. As mentioned in previous videos, factoring things of a degree higher than 2, there is something of an art to it. But oftentimes, if someone expects you to, you might be able to group things in interesting ways, especially when you see that several terms have some common factors. So for example, these first two terms right over here have the common factor x squared. So if you were to factor that out, you would get x squared times x plus 3, which is neat because that looks a lot like the second two terms. And then you can factor the x plus 3 out. We could factor the x plus 3 out, and we would get x plus 3 times x squared plus 1. And now, your 0's are going to happen, or this whole y-- remember this is equal to y-- y is going to equal 0 if either one of these factors is equal to 0. So when does x plus 3 equal 0? Well, subtract 3 from both sides. That happens when x is equal to negative 3. When does x squared plus 1 equal 0, I should say? Well, when x squared is equal to negative 1. Well, there's no real x's, no real valued x's."
    },
    {
        "Q": "\nAt around 2:20 Sal says that the polynomial function has three roots. Can someone explain how we know that ? :^)",
        "A": "Because the polynomial is a 3rd degree polynomial (x^3). 2nd degree polynomials (x^2) have 2 solutions and 1st degree polynomials (x^1) have only 1 solution. Hope this helps. Good Luck.",
        "video_name": "uFZvWYPfOmw",
        "timestamps": [
            140
        ],
        "3min_transcript": "Use the real 0's of the polynomial function y equal to x to the third plus 3x squared plus x plus 3 to determine which of the following could be its graph. So there's several ways of trying to approach it. One, we could just look at what the 0's of these graphs are or what they appear to be and then see if this function is actually 0 when x is equal to that. So for example, in graph A-- and first of all, as always, I encourage you to pause this video and try it before I show you how to solve it. So I'm assuming you've given a go at it. So let's look at this first graph here. Its 0, it clearly has a 0 right at this point. And just by trying to inspect this graph, it looks like this is at x is equal to negative 3, if I were to estimate. So that looks like the point negative 3, 0. So let's see, if we substitute x equals negative 3 here, whether we get y equaling 0. So let's see, negative 3 to the third power plus 3 What does this give us? This gives us negative 27. This gives us positive 27. This gives, of course, negative 3. This is plus 3. These two cancel out. These two cancel out. This does indeed equal 0. So this was actually pretty straightforward. Graph A does indeed work. You could try graph B right here, and you would have to verify that we have a 0 at, this looks like negative 2. Another one, this looks like at 1, another one that looks at 3. And since we already know that A is the answer, none of these-- if you were to input x equals negative 2, x equals 1, or x equals 3 into this function definition right over here, you should not get 0. And you'll see that this doesn't work. Same thing for this one. If you tried 4 or 7 for your x's, you see that the real function does not equal 0 at 4 or 7. Another giveaway that this is not going to be the function is that you are going to have a total of three roots. Let me write this down. So you're going to have a total of three roots. Now, , those three roots could be real or complex roots. And the big key is complex roots come in pairs. So you might have a situation with three real roots. And this is an example with three real roots, although we know this actually isn't the function right over here. Or if you have one complex root, you're going to have another complex root. So if you have any complex roots, the next possibility is one real and two complex roots. And this right over here has two real roots."
    },
    {
        "Q": "\nAt 5:23, you say that x^3 divided by x^2 =x , why would that be?",
        "A": "Note that: \u00f0\u009d\u0091\u00a5\u00c2\u00b3 = \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00f0\u009d\u0091\u00a5\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 Therefore: \u00f0\u009d\u0091\u00a5\u00c2\u00b3/\u00f0\u009d\u0091\u00a5\u00c2\u00b2 = (\u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5)/(\u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u00a5 We could also use exponent properties: \u00f0\u009d\u0091\u00a5\u00c2\u00b3/\u00f0\u009d\u0091\u00a5\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5\u00c2\u00b9 = \u00f0\u009d\u0091\u00a5 Comment if you have questions.",
        "video_name": "MZl6Mna0leQ",
        "timestamps": [
            323
        ],
        "3min_transcript": "But what do these simplify to? So this first term over here, this simplifies to 2x squared times-- now you get 4 divided by 2 is 2, x to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y, and then you have minus 2x squared times, 8 divided by 2 is 4. x to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared time-- this right here simplifies to 1-- times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared Right, if you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y, minus 4xy, and then you have minus 1, minus 1, and we're done. We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say OK, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2, so let me put that 2, let me factor 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. this guy divided by 2x squared? Well 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1, there's no other y degree that we factored out, so it's just going to be a y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And then you have y divided by say, 1, is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared 1, so 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head, but I really want you to understand what we did here. There is no magic. And to realize that there's no magic, you could just use the distributive property to multiply this out again, to multiply it out again, and you're going to see that you get exactly this."
    },
    {
        "Q": "At about \"8:00\" you prove that the linear combination of the two vectors can represent any vector in R^2.And then you set to prove that they are linearly independent.\n\nIsn't it self proving that if two vectors span R^2 (and n vectors span R^n) they HAVE to be linearly independent?\nJust my intuition speaking.\n",
        "A": "If n vectors spans R^n, then n vectors are linearly independant and a basis for R^n. So you are right in your intuition.",
        "video_name": "zntNi3-ybfQ",
        "timestamps": [
            480
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at the end of the video 18:47 he says that in order for a set to be a basis it has to have the minimum or the most efficient set o vectors that can span R2.\nWhat does he mean for most efficient?\n",
        "A": "It means the set with the smallest possible number of vectors. So the set containing <1, 0>, <0, 1>, and <1, 1> is not a basis, since <1, 0> and <1, 1> already span R\u00c2\u00b2. The <1, 0> vector is unnecessary.",
        "video_name": "zntNi3-ybfQ",
        "timestamps": [
            1127
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In the example that Sal provides around 22:58, is it obvious from the start that the rank of the 3x2 matrix can be at most 2 and therefore less than 3, and therefore not \"onto\" R3?\n",
        "A": "That would be a correct observation, yes! :) In his case, the function is not surjective (therefore not invertible), but injective- because rank(S) = n.",
        "video_name": "eR8vEdJTvd0",
        "timestamps": [
            1378
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 3:15 on the video Sal starts by doing the subtraction, per order of operations shouldn't the division of /c be done first? If not, why not?",
        "A": "Given: 1 1 \u00e2\u0094\u0080 - \u00e2\u0094\u0080 a b 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d The 1/a - 1/b is treated has a parenthetical term, which gives it order precedence: \u00e2\u0094\u008c \u00e2\u0094\u0090 \u00e2\u0094\u00821 1\u00e2\u0094\u0082 \u00e2\u0094\u0082\u00e2\u0094\u0080 - \u00e2\u0094\u0080\u00e2\u0094\u0082 \u00e2\u0094\u0082a b\u00e2\u0094\u0082 \u00e2\u0094\u0094 \u00e2\u0094\u0098 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d \u00e2\u0094\u008c \u00e2\u0094\u0090 \u00e2\u0094\u0082 b a \u00e2\u0094\u0082 \u00e2\u0094\u0082\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 - \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0082 \u00e2\u0094\u0082a\u00e2\u0080\u00a2b a\u00e2\u0080\u00a2b\u00e2\u0094\u0082 \u00e2\u0094\u0094 \u00e2\u0094\u0098 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d b - a \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d b - a 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c d b - a d \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00e2\u0080\u00a2 \u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c 1 d\u00e2\u0080\u00a2(b - a) \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c d\u00e2\u0080\u00a2b - d\u00e2\u0080\u00a2a \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            195
        ],
        "3min_transcript": "And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so I don't confuse you, that d was purple, times d over c, times d over c and then it reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab,"
    },
    {
        "Q": "At 5:40, why can't you simplify the expression (db-da)/(abc) by cancelling the a variable or the b variable to make it become (d-d)/(c)?\n",
        "A": "You cannot do that because the db and the da are being subtracted. If there was a multiplication sign (ex. (db*da)/(abc) ) you would be fine to do that.",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            340
        ],
        "3min_transcript": "so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c. we can distribute this d, and we're going to be left with, we deserve a minor drum roll at this point, we can write this as d times b, d times b minus d, woops, I want to do that in the same green color so you really see how it got distributed, minus d times a, all of that over ab, abc. And we are done."
    },
    {
        "Q": "\nat 4:10, why must we multiply by the reciprocal of c??",
        "A": "Multiplication by the reciprocal of c is the same as dividing by c.",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            250
        ],
        "3min_transcript": "Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c."
    },
    {
        "Q": "At 3:25 Sal quickly glosses over how he transforms 1/a - 1/b into b/ba - a/ba. I think an explanation about how these two statements are equivalent would be useful!\n",
        "A": "For the 1/a term, multiply it by b/b. Now you have: (1/a)(b/b) = b/ab For the 1/b term, multiply it by a/a. Now you have (1/b)(a/a) = a/ab So you have converted 1/a - 1/b into b/ab - a/ab.",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            205
        ],
        "3min_transcript": "And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so I don't confuse you, that d was purple, times d over c, times d over c and then it reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab,"
    },
    {
        "Q": "\nin the video at 5:00 d is multiplied by (b-a) and c multiplied by ab so whenc is multiplied why is the answer abc instead of acbc",
        "A": "The distributive property multiplies across addition or subtraction. d (b-a) = bd - ad But, with c (ab), there is no addition or subtraction. Thus, the distributive property does not apply. You just get abc Note: think about what happens with numbers. 2 (3*5) = 2*3*5 = 6*5 = 30 You just have 3 numbers to multiply, not 4. Hope this helps.",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            300
        ],
        "3min_transcript": "so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c. we can distribute this d, and we're going to be left with, we deserve a minor drum roll at this point, we can write this as d times b, d times b minus d, woops, I want to do that in the same green color so you really see how it got distributed, minus d times a, all of that over ab, abc. And we are done."
    },
    {
        "Q": "At about 8:00, he turns A^2=h^2-h^2/4\nto\nA^2=h^2(1-1/4)\n\nI don't get how he did that.\n",
        "A": "That confused me at first too, but then I realised that since h^2/4 is 1/4 of h^2, h^2 - h^2/4 is equal to three quarters of h^2, which is the same as h^2 times 1 - 1/4.",
        "video_name": "Qwet4cIpnCM",
        "timestamps": [
            480
        ],
        "3min_transcript": "Because that's h over 2, and this is also h over 2. Right over here. So if we go back to our original triangle, and we said that this is 30 degrees and that this is the hypotenuse, because it's opposite the right angle, we know that the side opposite the 30 degree side is 1/2 of the hypotenuse. And just a reminder, how did we do that? Well we doubled the triangle. Turned it into an equilateral triangle. Figured out this whole side has to be the same as the hypotenuse. And this is 1/2 of that whole side. So it's 1/2 of the hypotenuse. So let's remember that. The side opposite the 30 degree side is 1/2 of the hypotenuse. Let me redraw that on another page, because I think this is getting messy. So going back to what I had originally. This is a right angle. This is the hypotenuse-- this side right here. If this is 30 degrees, we just derived that the side opposite that this is equal to 1/2 the hypotenuse. If this is equal to 1/2 the hypotenuse then what is this side equal to? Well, here we can use the Pythagorean theorem again. We know that this side squared plus this side squared-- let's call this side A-- is equal to h squared. So we have 1/2 h squared plus A squared is equal to h squared. This is equal to h squared over 4 plus A squared, is equal to h squared. Well, we subtract h squared from both sides. We get A squared is equal to h squared minus h squared over 4. This is equal to 3/4 h squared. And once going that's equal to A squared. I'm running out of space, so I'm going to go all the way over here. So take the square root of both sides, and we get A is equal to-- the square root of 3/4 is the same thing as the square root of 3 over 2. And then the square root of h squared is just h. And this A-- remember, this is an area. This is what decides the length of the side. I probably shouldn't have used A. But this is equal to the square root of 3 over 2, times h. We've derived what all the sides relative to the hypotenuse are of a 30-60-90 triangle. So if this is a 60 degree side."
    },
    {
        "Q": "I'm pretty sure this is true, but just want to be sure. When he make the denominators the same in the last problem around 3:30, does the five still remain? For example does 5 4/9 = 5 28/63? This makes sense to me as to why it would be right, I just want to know if actually am.\n",
        "A": "Yes. The 5 stays put because you re only changing the 4/9 by multiplying it. The fraction is manipulated, but the whole number is not.",
        "video_name": "R8YKuGJ0plI",
        "timestamps": [
            210
        ],
        "3min_transcript": "3 and 9/10 is clearly a larger number. We have a 3 out here instead of a 1, so we will write less than. And the way I always remember it is, the opening always faces the larger number. And the point is small. It always points to the smaller number. Now let's do this next one. 4 and 7/8 versus 49/9. So let's convert this to a mixed number. 9 goes into 49 5 times, and 5 times 9 is 45. So the remainder is going to be 4. The remainder is 4, so this is 5 and 4/9. Once again, we can literally just look at the whole number parts. 5 is clearly larger than 4, so once again, less than. Point facing the smaller number, opening facing the larger number. Now 2 and 1/2 versus 11/10. 10 goes into 11 only 1 time. And if you care about the remainder, it's 1. Which is clearly smaller than 2 and 1/2. You just look at the whole number parts. 2 is clearly larger than 1. So we want our opening of our less than or greater than sign to face the larger number. So we would write it like this. And this is greater than, so 2 and 1/2 is greater than 11/10. The little point facing the smaller number. 5 and 4/9 versus 40/7. 7 goes into 40, so let me rewrite this, 7 goes into 40 5 times. And then you're going to have a remainder of 5, because 7 times 5 is 35. You have a remainder of 5 to get to 40. So it's 5 and 5/7. And if that looks like I'm doing some type of voodoo, just remember, I'm really just breaking it up. I'm just really saying that 40/7 is the same thing as 35 plus 5/7. The largest multiple of 7 that is less than this number. And this is the same thing as 35/7 plus 5/7. And 5/7 is just 5/7 there. This one is interesting because we have the same whole number out front on our mixed numbers. 5 versus 5. So now we actually do have to pay attention to the fractional part of our mixed number. We essentially have to compare 4/9 to 5/7. And there's a couple of ways to do this. You could get them to have the same denominator. That's probably the easiest way to do it. So you could rewrite-- so what's the least common multiple of 9 and 7? They share no factors, so really the least common multiple is going to be their product. So if we want to rewrite 4/9 we would write 63 in the denominator, that's 9 times 7. If we multiply the denominator by 7 we also have to multiply the numerator by 7. So that will be 28. Now 5/7, we're going to make the denominator 63. We're multiplying the denominator times 9. Then we have to multiply the numerator times 9 as well. 5 times 9 is 45."
    },
    {
        "Q": "\nWhen you factor it out at 1:46 what becomes of that lonely +x in the middle of the equation?",
        "A": "When (x+3)(x-2) is multiplied you distribute the first and last numbers separately, meaning that you have (x^2 - 2x +3x -6) as your answer. After combining the like terms of -2x and +3x we get +1x or simply put, + x. (x^2 + x - 6)",
        "video_name": "EAa3J_nDkoI",
        "timestamps": [
            106
        ],
        "3min_transcript": "Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x, as x approaches 2, is equal to. Now the first attempt that you might want to do right when you see something like this, is just see what happens what is f of 2. Now this won't always be the limit, even if it's defined, but it's a good place to start, just to see if it's something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator, we get 2 squared plus 2 minus 6. So it's going to be 4 plus 2, which is 6, minus 6, so you're going to get 0 in the numerator and you're going to get 0 in the denominator. So we don't have, the function is not defined, so not defined at x is equal 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, but that doesn't necessarily mean the case. But we see very clearly the function is not defined here. So let's see if we can simplify this and also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we can rewrite the top expression. And this just goes back to your algebra one, two numbers whose product is negative 6, whose sum is positive 3, well that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all X's except for x is equal to 2. So that's another way of looking at it. Another way we could rewrite our f of x, we could rewrite f of x, this is the exact same function, f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is, that is not anywhere near being a straight line, that is much better. So let's call this the y-axis call it y equals f of x. And then let's, over here, let me make a horizontal line, that is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3 and then the slope is 1. And it's defined for all X's except for x is equal to 2. So this is x is equal to 1, x is equal to 2."
    },
    {
        "Q": "at 2:55 - when it says f(x)= x +3 why did you graph on the y axis... u went up 3 and then u added the limit of 2 in there and said it would be 5 do we do that for all problems?\nso when it says f(x) its only really saying y= x+3 so we go up 2? is that the B in Y=MX +B ??\n",
        "A": "In the example f(x)=x+3 the vertical axis is the f(x) axis. Generally f(x ) is the same thing as y. So f(x)=x+3 is the same thing as y=x+3.",
        "video_name": "EAa3J_nDkoI",
        "timestamps": [
            175
        ],
        "3min_transcript": "but that doesn't necessarily mean the case. But we see very clearly the function is not defined here. So let's see if we can simplify this and also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we can rewrite the top expression. And this just goes back to your algebra one, two numbers whose product is negative 6, whose sum is positive 3, well that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all X's except for x is equal to 2. So that's another way of looking at it. Another way we could rewrite our f of x, we could rewrite f of x, this is the exact same function, f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is, that is not anywhere near being a straight line, that is much better. So let's call this the y-axis call it y equals f of x. And then let's, over here, let me make a horizontal line, that is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3 and then the slope is 1. And it's defined for all X's except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So let me make sure I can, so it's undefined right over there. So this is what f of x looks like. Now given this, let's try to answer our question. What is the limit of f of x as x approaches 2. Well, we can look at this graphically. As x approaches 2 from lower values in 2, so this right over here is x is equal to 2, if we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5 our f of x is right over there."
    },
    {
        "Q": "at 1:08 sal mentioned radians. what are they ??\n",
        "A": "They are another unit of angle measure like degrees. In higher level math, they are much easier to use. You can learn about them in the Trigonometry section of KA.",
        "video_name": "1BH2TNzAAik",
        "timestamps": [
            68
        ],
        "3min_transcript": "A circle has a circumference of 20 pi. It has an arc length of 221/18 pi. What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about, that's 221/18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference-- let me write that down-- the ratio of this arc length, which is 221/18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle. This will give us our theta in degrees. in terms of 2 pi radians around the circle, but it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So if we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over-- let's see, we have 18 times 20 times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36/2, which is the same thing as 18. So this all simplified to 221 degrees. Theta is 221 degrees."
    },
    {
        "Q": "\nAt 1:25, how can you put fraction on top of another fraction?",
        "A": "It looks messy, but you can do it. When you put a fraction on top of another fraction it helps to think of it as the numerator fraction divided by the denominator fraction.",
        "video_name": "1BH2TNzAAik",
        "timestamps": [
            85
        ],
        "3min_transcript": "A circle has a circumference of 20 pi. It has an arc length of 221/18 pi. What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about, that's 221/18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference-- let me write that down-- the ratio of this arc length, which is 221/18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle. This will give us our theta in degrees. in terms of 2 pi radians around the circle, but it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So if we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over-- let's see, we have 18 times 20 times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36/2, which is the same thing as 18. So this all simplified to 221 degrees. Theta is 221 degrees."
    },
    {
        "Q": "\nAt 10:14, couldn't have Sal simplified 10y and 4y by dividing it?\nSo if my reasoning is correct (you can correct me if I'm wrong, here to learn :)),\nthe answer should be 5y\u00e2\u0088\u009ay+25y divided by 2y - 25.",
        "A": "No, Sal can not reduce the fraction. To reduce fractions we can only cancel common factors (items being multiplied). You are trying to cancel out parts of terms (items being added/subtracted). Since they aren t factors, you can reduce. Hope this helps.",
        "video_name": "gY5TvlHg4Vk",
        "timestamps": [
            614
        ],
        "3min_transcript": "So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want. All of those are equivalent. Plus 25y. Anyway, hopefully you found this rationalizing the denominator interesting."
    },
    {
        "Q": "At 10:11 why didn't he add 10y and 25y to make 35y square root y as the numerator?\n",
        "A": "You can t add unlike terms. The expression in the numerator is: 10y\u00e2\u0088\u009a(y) + 25y. These are unlike because y\u00e2\u0088\u009a(y) is not the same as just y . Its like having 10yz + 25y. Hope this helps.",
        "video_name": "gY5TvlHg4Vk",
        "timestamps": [
            611
        ],
        "3min_transcript": "So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want. All of those are equivalent. Plus 25y. Anyway, hopefully you found this rationalizing the denominator interesting."
    },
    {
        "Q": "9:23, Why would I be multiplying as x * x, sqy * sqy , -5*5\nfor the denominator. but in the numerator its\n\n2 * 5, sqrY*y 5*5\nhow come the numerator is being applied to all terms and the denominator is being multiplied by like terms?\n",
        "A": "In the numerator since we are multiplying ( 2*sqrt(y)-5 ) by its complement we are using a shortcut. If you multiply a factor by its complement like (x+a)*(x-a) you get x^2-a^2 Proof: (x+a)*(x-a)=x*x+x*(-a)+a*x-a*a=x^2-a*x+a*x-a^2=x^2-a^2 so we can use shortcut on the numerator: ( 2*sqrt(y)-5 )*(2*sqrt(y)+5)=(2*sqrt(y))^2 - (5)^2= 4y-25 or we can do it the long way like: ( 2*sqrt(y)-5 )*(2*sqrt(y)+5)=2*sqrt(y)*2*sqrt(y)+2*sqrt(y)*5-5*2*sqrt(y)-5*5 =4y-25 Both give you the same answer. Hope this makes sense.",
        "video_name": "gY5TvlHg4Vk",
        "timestamps": [
            563
        ],
        "3min_transcript": "It actually made it look a little bit better. And you know, I don't if I mentioned in the beginning, this is good because it's not obvious. If you and I are both trying to build a rocket and you get this as your answer and I get this as my answer, this isn't obvious, at least to me just by looking at it, that they're the same number. But if we agree to always rationalize our denominators, we're like, oh great. We got the same number. Now we're ready to send our rocket to Mars. Let's do one more of this, one more of these right here. Let's do one with variables in it. So let's say we have 5y over 2 times the square root of y minus 5. So we're going to do this exact same process. We have a binomial with an irrational denominator. It might be a rational. We don't know what y is. But y can take on any value, so at points it's going to be So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want."
    },
    {
        "Q": "\nAt @2:54 isn't the minus symbol after 8 supposed to be in the green color because of how it's a -12?",
        "A": "mabey and do you want to be friends",
        "video_name": "GmD7Czmol0k",
        "timestamps": [
            174
        ],
        "3min_transcript": "Then I have plus five. And then I'm gonna subtract. I am subtracting eight times 0.25. 0.25, this is 1/4. I could rewrite this if I want. 0.25, that's the same thing as 1/4. Eight times 1/4, or another way to think about it is eight divided by four is gonna be equal to two. So this whole thing over here is going to be equal to two. So it's gonna be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. Alright. And we'll, just like before, try to work through it on your own before we do it together. Alright, now let's do it together. plus eight minus 12 N, when M is equal to 30 and N is equal to 1/4. Alright. So everywhere I see an M I want to replace with a 30. And everywhere I see an N I want to replace with a 1/4. So this is going to be equal to 0.1 times M. M is 30. Times 30 plus eight minus 12 times N, where N is 1/4. N is 1/4. So what is, what is 1/10, This right over here, 0.1, that's the same thing as 1/10 of 30? Well 1/10 of 30, that's going to be three. So this part is three. And we have three plus eight. And then we're gonna have minus. Well what is 12 times 1/4? That's gonna be 12/4, or 12 divided by four, And now when we evaluate this, so that is equal to this, we have three plus eight minus three. Well, threes are going to, you know positive three, then you're gonna subtract three, and you're just going to be left with, you're just going to be left with an eight. And you're done. This expression when M is equal to 30 and N is equal to 1/4 is equal to eight."
    },
    {
        "Q": "At 3:16, can you always just cross out all the numbers that are congruent in the same equation, or is it better, more accurate, to solve the equation all the way through?\n",
        "A": "He did not cross out numbers that are congruent, he crossed out additive inverses, so 3 - 3 = 0. If they were congruent, you would get 3 + 3 to get 6",
        "video_name": "GmD7Czmol0k",
        "timestamps": [
            196
        ],
        "3min_transcript": "Then I have plus five. And then I'm gonna subtract. I am subtracting eight times 0.25. 0.25, this is 1/4. I could rewrite this if I want. 0.25, that's the same thing as 1/4. Eight times 1/4, or another way to think about it is eight divided by four is gonna be equal to two. So this whole thing over here is going to be equal to two. So it's gonna be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. Alright. And we'll, just like before, try to work through it on your own before we do it together. Alright, now let's do it together. plus eight minus 12 N, when M is equal to 30 and N is equal to 1/4. Alright. So everywhere I see an M I want to replace with a 30. And everywhere I see an N I want to replace with a 1/4. So this is going to be equal to 0.1 times M. M is 30. Times 30 plus eight minus 12 times N, where N is 1/4. N is 1/4. So what is, what is 1/10, This right over here, 0.1, that's the same thing as 1/10 of 30? Well 1/10 of 30, that's going to be three. So this part is three. And we have three plus eight. And then we're gonna have minus. Well what is 12 times 1/4? That's gonna be 12/4, or 12 divided by four, And now when we evaluate this, so that is equal to this, we have three plus eight minus three. Well, threes are going to, you know positive three, then you're gonna subtract three, and you're just going to be left with, you're just going to be left with an eight. And you're done. This expression when M is equal to 30 and N is equal to 1/4 is equal to eight."
    },
    {
        "Q": "\nAt 2:30 when i solved for the equation 0.7^0 multiplied 0.3^6 my answer was 0.000729 not 0.001.",
        "A": "Your calculation is correct but he is rounding to the nearest 1/100. Thus, 0.000729 ~ 0.001.",
        "video_name": "eL965_Lscb8",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] Now that we've spent a couple of videos exploring a scenario where I'm taking multiple free throws and figuring out the probability of making K of the scores and six attempts or in N attempts. Let's actually define a random variable using this scenario and see if we can construct it's probability distribution and we'll actually see that it's a binomial distribution. So, let's define the random variable X. So, let's say that X is equal to the number, the number of made shots, number of made free throws when taking, when taking six free throws. So, it's how many of the six do you make? And we're going to assume what we assumed in the first video in this series of this, these making free throws. So, we're gonna assume the seventy percent free throw probability right over here. So, assuming assumptions, assuming seventy percent free throw, free throw percentage. So, let's figure out the probabilities of the different values that X could actually take on. So, let's see, what is the probability, what is the probability that X is equal to zero? That even though you have a seventy percent free throw percentage that you make none of the shots and actually you could calculate this through probably some common sense without using all of these fancy things but just to make things consistent, I'm gonna write it out. So, this is going to be, this is going to be, it's going to be equal to six choose zero times zero point seven to the zeroth power times zero point three to the sixth power, and this right over here is gonna end up being one. This over here's going to end up being one. And so, you're just gonna be left with zero point three to the sixth power and I calculated it ahead of time. if we do, if we round our percentages to the nearest tenth, this is going to give you approximately, approximately, well, if we round the decimal to the nearest, to the nearest thousandth you're gonna get something like that which is approximately equal to zero point one percent chance of you missing all of them. So, roughly I'm speaking, roughly here a one in a thousand, one in a thousand chance of that happening, of missing all six free throws. Now, let's keep going, this is fun. So, what is the probability that our random variable is equal to 1? Well, this is going to be six choose one times zero point seven to the first power times zero point three to the six minus first power. So, it's a fifth power. And I calculated this and this is approximately zero point zero one"
    },
    {
        "Q": "\nI think you made a mistake at the end (12:18):\ny = (2/5)*x + (2/5) should be y = (2/5)*x + (4/5)",
        "A": "you are correct 4/5 is b*",
        "video_name": "QkepM8Vv3kw",
        "timestamps": [
            738
        ],
        "3min_transcript": "We can say 6 times m star-- This is just straight Algebra 1. So 6 times our m star, so 6 times 2 over 5, plus 2 times our b star is equal to 4. Enough white, let me use yellow. So we get 12 over 5 plus 2b star is equal to 4, or we could say 2b star-- let me scroll down a little bit --2b star is equal to 4. Which is the same thing as 20 over 5, minus 12 over 5, which is equal to-- I'm just subtracting the 12 over 5 from both sides --which is equal to 8 over 5. And you divide both sides of the equation by 2, you get b star is equal to 4/5. And just like that, we got our m star and our b star. Our least squares solution is equal to 2/5 and 4/5. And remember, the whole point of this was to find an equation of the line. y is equal to mx plus b. Now we can't find a line that went through all of those points up there, but this is going to be our least squares solution. This is the one that minimizes the distance between a times our vector and b. No vector, when you multiply times that matrix a-- that's not a, that's transpose a --no other solution is going to give us a closer solution to b than when we put our newly-found x star into this equation. This is going to give us our best solution. It's going to minimize the distance to b. So let's write it out. y is equal to mx plus b. So y is equal to 2/5 x plus 2/5. y is equal to 2/5 x plus 2/5. So its y-intercept is 2/5, which is about there . This is at 1. 2/5 is right about there. And then its slope is 2/5. Let's think of it this way: for every 2 and 1/2 you go to the right, you're going to go up 1. So if you go 1, 2 and 1/2, we're going to go up 1. We're going to go up 1 like that. So our line-- and obviously this isn't precise --but our line is going to look something like this. I want to do my best shot at drawing it because this is the fun part. It's going to look something like that."
    },
    {
        "Q": "\nAt around 2:50, do we always have to multiply by negative one? in every system?",
        "A": "Not 100% on the context, but it looks like he s eliminating a variable. The answer is no. It depends on the coefficient of the variable you re tying to eliminate. If the first equation was -3x + ... and the bottom was x + ...., you would just multiply by 3. In general terms: You want A + B = 0, where A and B are coefficients of the first term. To do that, you choose some c such that A + cB = 0. The key thing to remember is to multiply both sides of the equation by c. ex) c*(Ax + By) = 10*c",
        "video_name": "z1hz8-Kri1E",
        "timestamps": [
            170
        ],
        "3min_transcript": "going to be equal to 62.5 pounds. That's exactly what this first statement is telling us. The second statement, the weight of 3 televisions and 2 DVD players, so if I have 3 televisions and 2 DVD players, so the weight of 3 televisions plus the weight of 2 DVD players, they're telling us that that is 52 pounds. And so now we've set up the system of equations. We've done the first part, to create a system that represents the situation. Now we need to solve it. Now, one thing that's especially tempting when you have two systems, and both of them have something where, you know, you have a 3t here and you have a 3t here, what we can do is we can multiply one of the systems by some factor, so that if we were to add this equation to that equation, we would get one of the terms to cancel out. And that's what we're going to do right here. equations to each other, because remember, when we learned this at the beginning of algebra, anything you do to one side of an equation, if I add 5 to one side of an equation, I have to add 5 to another side of the equation. So if I add this business to this side of the equation, if I add this blue stuff to the left side of the equation, I can add this 52 to the right-hand side, because this is saying that 52 is the same thing as this thing over here. This thing is also 52. So if we're adding this to the left-hand side, we're actually adding 52 to it. We're just writing it a different way. Now, before we do that, what I want to do is multiply the second, blue equation by negative 1. And I want to multiply it by negative 1. So negative 3t plus-- I could write negative 2d is equal to negative 52. So I haven't changed the information in this equation. I just multiplied everything by negative 1. The reason why I did that is because now if I add these two equations, these 3t terms are going to cancel out. So let's do that. Let's add these two equations. And remember, all we're doing is we're adding the same thing We're adding essentially negative 52 now, now that we've multiplied everything by a negative 1. This negative 3t plus negative 2d is the same thing as negative 52. So let's add this left-hand side over here. The 3t and the negative 3t will cancel out. That was the whole point. 5d plus negative 2d is 3d. So you have a 3d is equal to 62.5 plus negative 52, or 62.5 minus 52 is 10.5. And now we can divide both sides of this equation by 3. And you get d is equal to 10.5 divided by 3. So let's figure out what that is. 3 goes into 10.5-- it goes into 10 three times. 3 times 3 is 9. Subtract. Get 1. Bring down the 5. Of course, you have your decimal point right here. 3 goes into 15 five times."
    },
    {
        "Q": "\nAt 0:25 does that mean you can change it around for every operation?",
        "A": "No. It only works for multiplication and addition because they are commutative. Communitive means that no matter where your numbers are, you get the same answer. Helpful?",
        "video_name": "SfxULALs_u8",
        "timestamps": [
            25
        ],
        "3min_transcript": "Let's try to compute 6 times 37. And I'll show you one way of doing this. And then in future videos, we can look at other ways of doing this and think about why this is actually working. So what I like to do-- and this is often called the standard method-- is take the larger of the two numbers. It doesn't matter if you're doing 6 times 37 or 37 times 6. They equal each other. 6 times 37 is the same thing as 37 times 6. So what I like to do is I take the larger of the two numbers, and I write it on top. So I'll write 37. And then the smaller of the two numbers, which is 6, I'll write it on the bottom. And I'll align it by the correct place. This only has one digit. It's in the ones place, obviously. So I can write the 6 right over here. And I'll write the multiplication symbol like that. And this is just another way of expressing 37 times 6, which is the same thing as 6 times 37. Now, what we do is we go with this, the first place in this lower number. And there's only one place here. It's only the number 6 right over here. And we're going to multiply that times each of the digits So first, we will start with 6 times 7. So we're going to first multiply 6 times 7. Well, you remember from your multiplication tables, 6 times 7 is equal to 42. Now, we don't just write 42 here. At least, not in the standard method we wouldn't write 42 here. We'd write the 2 in 42 in the ones place. So we'd write that right over there. And then we'd carry the 4 in 42 up to the tens place. Now we need to think about what 6 times 3 is. Well, once again, we know 6 times 3. 6 times 3 is equal to 18. But we can't just write an 18 down here. We still have this 4 to deal with. So 6 times 3 is 18, but we've got to then add the 4. So 6 times 3 is 18, plus 4 is 22. So it's 6 times 3, and then we're And that's how we get our answer-- so 222."
    },
    {
        "Q": "\n9:20, couldn't he also have used HL Theorem to prove DBE and FBE are congruent?",
        "A": "Yes, that is what Sal is calling the RSH theorem at around 9:00. Either strategy is effective for showing that the two triangles are congruent.",
        "video_name": "yj4oS-27Q3k",
        "timestamps": [
            560
        ],
        "3min_transcript": "between the point and those 2 respective sides. We've just proven the first case, if a point lies on an angle bisector, it is equidistant from the 2 sides of the angle, now let's go the other way around, let's say that I have, so let me draw another angle here, so let me draw another angle over here and let's call this A, B, and C and let's pick some arbitrary point E, let's point some arbitrary point E right over here and let's say we start off with the assumption that E is equidistant to BC and BA and what we want to do is prove that E must be on the angle bisector, so here, if you're on the angle bisector you're equidistant, over here we're going to show if you're equidistant you're on the angle bisector, so if it's equidistant to BC and BA, then this perpendicular perpendicular right over there, and let me give these points' labels, so let call this point D, and let's call this point right over F, and let's just draw segment BE here, let's just draw segment BE right over here, so once again, we have 2 right triangles, we already know that 2 of the legs are congruent to each other, they both share the hypotenuse this hypotenuse is equal to its self, we know from the pythagorean theorm if you know 2 sides of a triangle, it determines the third side, so and we know 2 sides of both of these, so the third sides must be the same, so this side must be equal to that side so you could invoke SSS, Side Side Side, to show these 2 triangles are congruent or you actually didn't need to -have to good there you could have used a special case, the RSH case, where if you have a right triangle, so if you have the right triangle then you're also okay, you could use RSH to prove congruency as well, and so either way we know that triangle EBD, triangle EBD, is congruent to triangle EBF congruent to triangle EBF, we used Side Side Side here but you could have used RSH, let me write that RSH, which is -we know that Angle Side Side can be used for any general triangle but it can be, -RSH is essentially Angle Side Side for right triangles, if you have 2 sides of a right triangle and- in common, --if 2 sides of a right triangle are congruent then the 2 triangles are definitely congruent but once you know that two triangles are congruent, then they're corresponding angles have to be congruent, and angle EBD, angle EBD"
    },
    {
        "Q": "At 7:55, Sal begins the proof of the reverse case, where: if a point is equidistant from both sides of an angle, then that point is on the angle bisector. So that is clear, but when I went on to the videos about medians, I thought, if the median bisects the side opposite the vertex, then that point is also equidistant from both sides of the angle from that vertex, so that makes it an angle bisector, right? Well, apparently not, so I'm confused. What is the relationship of angle bisectors and medians?\n",
        "A": "As far as I know a median is set to see if the sides are equal to each other, where an angle bisector splits a angle in half making said angle into 2 equal angles. If I am wrong sorry its been awhile seance i went over medians.",
        "video_name": "yj4oS-27Q3k",
        "timestamps": [
            475
        ],
        "3min_transcript": "EBG and we can use AAS, by Angle Angle Side congruency or you could say hey if two angles, -if corresponding angles are the same then the third angle is also going to be the same so this angle right over here could also be the same and you could use ASA, but either way these two things are going to be congruent but if these two things are congruent, then the corresponding sides are going to be congruent so then, then length of EF, segment EF, is going to be congruent to segment EF is going to be congruent to segment EG which is the same thing as the length of EF is equal to the length of EG, these are really, these are really equivalent statements right over there, so the length of EF, the length of EF between the point and those 2 respective sides. We've just proven the first case, if a point lies on an angle bisector, it is equidistant from the 2 sides of the angle, now let's go the other way around, let's say that I have, so let me draw another angle here, so let me draw another angle over here and let's call this A, B, and C and let's pick some arbitrary point E, let's point some arbitrary point E right over here and let's say we start off with the assumption that E is equidistant to BC and BA and what we want to do is prove that E must be on the angle bisector, so here, if you're on the angle bisector you're equidistant, over here we're going to show if you're equidistant you're on the angle bisector, so if it's equidistant to BC and BA, then this perpendicular perpendicular right over there, and let me give these points' labels, so let call this point D, and let's call this point right over F, and let's just draw segment BE here, let's just draw segment BE right over here, so once again, we have 2 right triangles, we already know that 2 of the legs are congruent to each other, they both share the hypotenuse this hypotenuse is equal to its self, we know from the pythagorean theorm if you know 2 sides of a triangle, it determines the third side, so and we know 2 sides of both of these, so the third sides must be the same, so this side must be equal to that side so you could invoke SSS, Side Side Side, to show these 2 triangles are congruent or you actually didn't need to -have to good there you could have used a special case, the RSH case, where if you have a right triangle, so if you have the right triangle"
    },
    {
        "Q": "At 7:07, Vi says that the cardioid is like an \"Anti-parabola\", and I don't quite understand what that mean. Please explain.\n",
        "A": "If you turn positive to negative, infinity to zero while graphing a parabola you get a cardioid",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            427
        ],
        "3min_transcript": "both are like extreme ellipses, because a circle is like taking one focus of an ellipse and putting the other focus zero distance away. And a parabola is like an ellipse where one focus is infinity distance away. Which is why everyone lies to you and says throwing balls or shooting birds is all about parabolas when really it's about ellipses because the earth is a sphere and gravity doesn't actually go straight down. And the other focus of the elliptical orbit of your thrown object of choice may be very far away, but very far away is a great bit closer than infinity. So let's not fool ourselves. You can't look at everything that seems kind of parabolic and call it a parabola. Sure, if you connect two dots with a hanging string or chain, it looks parabolic, and so do structurally strong arches, but they're actually catenary arches and maybe you can't tell by sight, but if you're an architect you'd better know the difference. Though caternarys are quite related to parabolas, you get them by rolling around a parabola and tracing the focus which makes them a cousin of the ellipse and even a hyperbola is like an ellipse that got turned inside out or whose focus went through infinity and came out the other side or something. And of course parabolas and hyperbolas and ellipses they all come from a line that got spun around. And a line is just what happens when a couple dots get connected. Or maybe what happens when your circle is so big that, like the extreme ellipse that becomes a parabola, the extreme circle is broken at infinity and becomes a line before getting larger than infinitely big which brings it back to the other side. This linear circle, the infinite in-between. Or maybe a line is what happens when you roll around a circle and trace the focus. Or rather the 2 foci, which are zero distance apart, in ellipse terms. Which makes you wonder what you get when you roll around an ellipse and trace the foci. In fact, there's lots of great shapes you can get by rolling around shapes on other shapes, like if you roll a circle around a circle and trace the focus, you get just another circle. But, if you trace a point on the edge, you get our awesome friend the cardioid again. So now it's related to circles in three ways , which means it's a close cousin of the ellipse and a second cousin to the infinite ellipse or, parabola. Except, not just that, if you take a parabola and invert it around the unit circle, reversing inside and outside, one half becomes two, one hundred becomes a hundredth, you get once again, the cardioid. The cardioid is the anti-parabola which is good because parabolas make you sad but you heart cardioids. And of course, any time you want to connect two dots on a piece of paper, instead of drawing the line you could fold the line. Here's the thing about connecting dots. You can have all the steps laid out for you, taking whatever next step is easiest and closest and be sure of what you're getting the whole time. This way is safe and comfortable. Or, you can try new ways of connecting dots and not know what you're going to get. Maybe it will be something great, maybe it will fail. And when it fails it will be your fault. You can't blame anyone else, not mathematics or the system or the check-boxes. But if I am to have faults I would rather they be my own."
    },
    {
        "Q": "at 3:00 whats a prolabala?\n",
        "A": "Did you mean parabola? A parabola is basically the opposite of a hyperbola. If you took two identical and perfectly symmetrical cones and lined then up perfectly, one on top of the other, meaning one right-side-up and the one on top upside-down, so it should look some what related to the shape of an hour glass. In a hyperbola, no matter how far you draw out the lines, the ends will never meet. In a parabola it is quite the opposite, or a least that s what I think.",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            180
        ],
        "3min_transcript": "and how many dots you skip. But then there are other shapes. Circles are good friends with sine waves. And sine waves are good friends with square waves. And let's admit it, that's pretty cool looking. In fact, just two simple straight lines of dots connecting the dots from one line to the other in order somehow gives you this awesome woven curve shape. Another student is asking the teacher when he's ever going to need to know how to graph a parabola \u2013 even as he hides his multi-million dollar enterprise of a parabola graphing game under his desk. If your teacher thought about it, he would probably think shooting birds at things is a great reason to learn about parabolas because he's come to understand that education is about money and prestige and not about becoming a better human able to do great things. You yourself haven't done anything really great yet but you figure the path to your future greatness lies more in inventing awesome new connect-the-dot arrangements than in graphing parabolas or shooting birds at things. And that's when you begin to worry. What if this cool liney curvey thing you drew approximates a parabola? As if your teacher doesn't realize everyone has their phones under their desks, his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius,"
    },
    {
        "Q": "how does she do the plaid thing at 4:27\n",
        "A": "The circles overlapped over each other, and created a pseudo-grid of shapes. She just alternated colors between them.",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            267
        ],
        "3min_transcript": "his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius, and all of the others define radii. But then you just get concentric circles, which I suppose should have been obvious. But what if you did it the other way around and said one dot always stays on the circle and all the other dots are centers, like this. Looks more promising. So you try putting all the dots in a circle and using them as circle centers and choose just one dot for the circles to go through and you get this awesome shape that looks kind of like a heart. So let's call it, oh I don't know, a cardioid. Which happens to be the same curve that you get when parallel lines like rays of light reflect off a circle the same heart of sunshine in a cup. Or maybe instead of circle centers you could have points all on the curve of a circle, which means you need three points to define a circle, maybe just a point and its two closest neighbors to start with. And of course, any collection of circles is two-colorable, which means you can contrast light and dark colors for a classy color scheme. Or maybe you could throw down some random points to make all possible circles. Only that would be a lot of circles, so you choose just ones you like. And then, against your will, you begin to wonder how many points it takes to define the boring old parabola."
    },
    {
        "Q": "3:37 - 3: 40 why would you divide by 10 when the scientific notion is 10^-2??\n",
        "A": "Actually, Sal was multiplying by 10, so 10^-3 * 10 is 10^-2.",
        "video_name": "ios3QL9t9LQ",
        "timestamps": [
            217
        ],
        "3min_transcript": "So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4. times 10 to the -3 power. x 10 to the -3 power. Now, this is what the product, or this is what the difference of these two numbers is but this is no longer in scientific notation. In order to be scientific notation this number right over here has to be between 1, has to be greater than or equal to 1 and less than 10. So what we could do is we could divide this number right over here by 10. We can divide this by 10 and then we can multiply, then we can multi...so we could do this. We could, kind of the opposite of what we did up here. We could divide this by 10 and then multiply by 10. So if you divide by 10 and multiply by 10 you're not changing the value. So 38.4 divided by 10 is 3.84 and then all of this business, 10 to the -3, 10 to the -3 x 10 is 10 to the -2 power."
    },
    {
        "Q": "\nWhen you move the decimal point in 4.1 to the right, does it increase or decrease the exponent of 10? How about moving it to the left? At 1:41, Sal is doing it a way I wasn't taught so I am getting confused.",
        "A": "When you move the decimal to the right in 4.1, it decreases the exponent by I or how many spaces you move the decimal to the right. When you move it to the left, the exponent of 10 will increase. This is one way you can think about it, when you are moving the decimal point to the right, you are putting a SMALLER number into scientific notation. When you are moving the decimal point to the left , you are putting a LARGER number into Scientific notation.",
        "video_name": "ios3QL9t9LQ",
        "timestamps": [
            101
        ],
        "3min_transcript": "- [Voiceover] What I want to do in this video is get a little bit of practice subtracting in scientific notation. So let's say that I have 4.1 x 10 to the -2 power. 4.1 x 10 to the -2 power and from that I want to subtract, I want to subtract 2.6, 2.6 x 10 to the -3 power. Like always, I encourage you to pause this video and see if you can solve this on your own and then we could work through it together. All right, I'm assuming you've had a go it. So the easiest thing that I can think of doing is try to convert one of these numbers so that it has the same, it's being multiplied by the same power of ten as the other one. What I could think about doing, well can we express 4.1 times 10 to the -2? Can we express it as something times 10 to the -3? So we have 4.1 times 10 to the -2. we would divide by 10, but we can't just divide by 10. That would literally change the value of the number. In order to not change it, we want to multiply by 10 as well. So we're multiplying by 10 and dividing by 10. I could have written it like this. I could have written 10/10 times, let me write this a little bit neater. I could have written 10/10 x this and then you take 10 x 4.1, you get 41, and then 10 to the -2 divided by 10 is going to be 10 to the -3. So this right over here, this is equal to 10 x 4.1 is 41 times 10 to the -3. And that makes sense. 41 thousandths is the same thing as 4.1 hundredths and all we did is we multiplied this times 10 and we divided this times 10. So let's rewrite this. We can rewrite it now as 41 X 10 to the -3 So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4."
    },
    {
        "Q": "\nat 3:50, why does he divide by 10?",
        "A": "You Cannot have a scientific notation over 10 so he divides 38.4 by 10 to make it valid for a scientific notation, giving you the answer of 3.84 x 10^-2",
        "video_name": "ios3QL9t9LQ",
        "timestamps": [
            230
        ],
        "3min_transcript": "So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4. times 10 to the -3 power. x 10 to the -3 power. Now, this is what the product, or this is what the difference of these two numbers is but this is no longer in scientific notation. In order to be scientific notation this number right over here has to be between 1, has to be greater than or equal to 1 and less than 10. So what we could do is we could divide this number right over here by 10. We can divide this by 10 and then we can multiply, then we can multi...so we could do this. We could, kind of the opposite of what we did up here. We could divide this by 10 and then multiply by 10. So if you divide by 10 and multiply by 10 you're not changing the value. So 38.4 divided by 10 is 3.84 and then all of this business, 10 to the -3, 10 to the -3 x 10 is 10 to the -2 power."
    },
    {
        "Q": "At 3:13, did Sal just invert and multiply? Also can I cross multiply and then transfer the 4 to divide the other side by 4?\n",
        "A": "Yeah it looks to me like he just multiplied the reciprocals of the two numbers. I don t think you could just transfer the 4 to the other side, I think you d have to divide BOTH sides by four which would get X alone, and then you could divide the other side by 4. Hope this helps!",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            193
        ],
        "3min_transcript": "So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2, So the change in x from A to C is going to be 10. So what does that tell us about C's x-coordinate? So we could start with A's x-coordinate, which is negative 6, add 10 to it, negative 6 plus 10 is 4. So we figured out the x-coordinate, now we just have to do the same thing for the y. So what is the change in y going from A to B? Well here, we go from 9 to 3, we've gone down 6. Another way you could say it is, 3 minus 9 is negative 6. To find the change, you could think, I'm just taking the end point and subtracting from that the starting point. Negative 2 minus negative 6 was positive 4. 3 minus 9 is negative 6. Or you could just look at it."
    },
    {
        "Q": "\nWhat about when im trying to solve with fractions or is the form 2:5 the same as 2/5",
        "A": "No, it isn t. It should be 2/7 because you re dividing into 7 sections. If you need more explanation, just ask me.",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            125
        ],
        "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2,"
    },
    {
        "Q": "So in one of my questions it says the ratio between AC and BC is 2:3 and to find point B. Then when I click on hint it says the ratio between AC and BC is 2:3 therefore the ratio between AB and AC is 2:5, how does that work?\n",
        "A": "AC to BC is 2 : 3 and since they are the only two parts of the line segment AC, the segment has a total of 2 + 3 = 5 parts. Therefore, AB to the whole thing (AC) is 2 : 5.",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            123,
            123,
            125
        ],
        "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2,"
    },
    {
        "Q": "At 3:29,How did Sal used the chain rule and got the derivative to be (e^((ln a)x) * ln a).\nShouldn't we use the power rule? I mean, if Sal is using the chain rule, (d [f(g(x))]) / dx =f \u00e2\u0080\u00b2(g(x)) g \u00e2\u0080\u00b2(x), then what is the f(x) and g(x)?\nThank you for you reply\n",
        "A": "He let \u00f0\u009d\u0091\u0094(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u00a5ln \u00f0\u009d\u0091\u008e and \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u0092\u00cb\u00a3. Power rule only works when the power is a constant, not a variable which we are differentiating WRT.",
        "video_name": "gHzLHknEk1M",
        "timestamps": [
            209
        ],
        "3min_transcript": "you could view a as being equal to e. Let me write it this way. Well all right, a as being equal to e to the natural log of a. Now if this isn't obvious to you, I really want you to think about it. What is the natural log of a? The natural log of a is the power you need to raise e to, to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e too to get to a. Well then you're just going to get to a. So really think about this. Don't just accept this as a leap of faith. It should make sense to you. And it just comes out of really what a logarithm is. And so we can replace a with this whole expression here. If a is the same thing as e to the natural log of a, well then this is going to be, with respect to x of e to the natural log, I keep writing la (laughs), to the natural log of a and then we're going to raise that to the xth power. We're going to raise that to the x power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color-coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a times x power. Times x power. And now we can use the chain rule to evaluate this derivative. we will first take the derivative of the outside function. So e to the natural log of a times x with respect to the inside function, with respect to natural log of a times x. And so, this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well natural log of a, it might not immediately jump out to you, but that's just going to be a number. So that's just going to be, so times the derivative. If it was the derivative of three x, it would just be three. If it's the derivative of natural log a times x, it's just going to be natural log of a. And so this is going to give us the natural log of a times e to the natural log of a. And I'm going to write it like this."
    },
    {
        "Q": "At 3:51, how would you write 8^ -1/3 as a radical? Would it be the negative third root (with the three over the root symbol being negative)?\n",
        "A": "Any value raised to a negative exponent is the reciprocal of that value raised to the same positive exponent. For example: 8^(-1/3)=1/(8^(1/3))=1/2",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            231
        ],
        "3min_transcript": "Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2. do I have to raise 8 to to get to 1/2-- is negative 1/3. Equal to negative 1/3. I hope you enjoyed that as much as I did."
    },
    {
        "Q": "\nWow, that was a lot of information to soke in. I don't really get the last example. At 3:36, Sal was giving another example that was very complicate. Can someone help me with it?",
        "A": "I got lost when Sal talked about he reciprocal of 2 and then he quickly moved on to 1/3 being -3.",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            216
        ],
        "3min_transcript": "Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2. do I have to raise 8 to to get to 1/2-- is negative 1/3. Equal to negative 1/3. I hope you enjoyed that as much as I did."
    },
    {
        "Q": "\nAt 1:37 how did you get 1/3 im not really understanding how I would get that?",
        "A": "Look ! it was a good question but hear him again . Sal said you need to think in a different manner so, if you look at the question with log 2 with base 8 you have an anti-log or the exponential form as 2= 8^x so for cube Root we have 8 to the power 1/3 . For example 2 to the power 1/2 means an under root of 2 . so same way is for cube root i.e, power 1/3. hope you like the answer",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            97
        ],
        "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3."
    },
    {
        "Q": "\nat 1:29 he says since 2x2x2=8 so there fore the answer is 1/3 i dont really get this can you help me? ( im not even learning this yet I'm in 7th grade algebra 1 honors, i just want to be prepared)",
        "A": "Logx(2)=8 can also be told as 8^x=2. For 8 to be less than itself as 2, the exponent must be a fraction or negative as in powers, fractions and negative integers make the number smaller while whole numbers make the number bigger in this situation. This is not true for all situations. The reason x=1/3 is because it means 8^(1/3)=2. The cube root of 8 is 2 because 2x2x2=8. So that means log 1/3(2)=8, which is true. I m also in 7th grade algebra 1 honors.",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            89
        ],
        "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3."
    },
    {
        "Q": "At 1:27, how did he get 1/3?\n",
        "A": "The question is effectively what power do you need to put 8 to, to get 2. As 2^3=8, 8^(1/3) must equal 2.",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            87
        ],
        "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3."
    },
    {
        "Q": "At around 5:39 he put 5x-5. But why did he do that? Why didn't he distribute all the way?\n",
        "A": "Sal is multiplying the trinomial with a fraction that has the trinominal in the denominator. Thetrinomials cancel out. This is just like if you multiply: 7 * 5/7, the 7 s cancel out leaving just the 5 behind. Hope this helps.",
        "video_name": "pLCmwHsDYqo",
        "timestamps": [
            339
        ],
        "3min_transcript": "but we're saying it's not. It's now a lower degree than this down here. So we could say it's x plus 1 plus whatever this remainder is, divided by this thing over here. So our answer-- I'm going to write it one more time. It's x plus 1 plus 5x minus 5, over x squared minus x plus 1. And we can check that this works. If we take this thing over here, and we multiply it by this thing over here, we should get the x to the third plus 5x minus 4. So let's do that. Let's multiply this thing by x squared minus x plus 1. And to do that, let's just distribute this whole trinomial times each of these terms. When we do the first term, we have So that's going to be x times x squared, which is x to the third; x times negative x, which is negative x squared; x times 1, which is plus x. Then we can multiply this whole thing times 1. So it's going to be plus x squared minus x plus 1. I'm just multiplying all of these times 1. And then we can multiply this whole thing times this thing. Now, this is the same as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left with the numerator over here, so plus 5x minus 5. And now we can try to simplify it. We only have one third-degree term, the x to the third. So we have x to the third here. Second-degree terms-- we have a negative x squared. And then we also have a positive x squared. So they cancel out with each other. First-degree terms-- let's see. We have a positive x and a negative x. Those cancel out with each other. So we're just going to have this 5x. So then we have plus 5x. And then we have the 0-th degree terms, or the constant terms. We have a positive 1 and a negative 5. You get negative 4. So you get x to the third plus 5x minus 4, which is exactly what we had over here."
    },
    {
        "Q": "\nat 5:36, Sal canceled the denominator and is only left with the numerator. Why he doesn't distribute the x^2-x+1 into 5x-5 as well? Thanks!",
        "A": "The entire fraction is one term, so he is distributing x^2-x+1 into the entire fraction, not just the denominator. Let s say 5x-5 = a and x^2-x+1 = b. Distributing into the last term would leave you with (a/b) * b. Simplifying this would just leave you with a, or 5x-5. This question was asked last year, so my answer is a bit late - you may have already figured it out yourself - but maybe it ll help with someone else who has this question, since there are no other answers here.",
        "video_name": "pLCmwHsDYqo",
        "timestamps": [
            336
        ],
        "3min_transcript": "but we're saying it's not. It's now a lower degree than this down here. So we could say it's x plus 1 plus whatever this remainder is, divided by this thing over here. So our answer-- I'm going to write it one more time. It's x plus 1 plus 5x minus 5, over x squared minus x plus 1. And we can check that this works. If we take this thing over here, and we multiply it by this thing over here, we should get the x to the third plus 5x minus 4. So let's do that. Let's multiply this thing by x squared minus x plus 1. And to do that, let's just distribute this whole trinomial times each of these terms. When we do the first term, we have So that's going to be x times x squared, which is x to the third; x times negative x, which is negative x squared; x times 1, which is plus x. Then we can multiply this whole thing times 1. So it's going to be plus x squared minus x plus 1. I'm just multiplying all of these times 1. And then we can multiply this whole thing times this thing. Now, this is the same as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left with the numerator over here, so plus 5x minus 5. And now we can try to simplify it. We only have one third-degree term, the x to the third. So we have x to the third here. Second-degree terms-- we have a negative x squared. And then we also have a positive x squared. So they cancel out with each other. First-degree terms-- let's see. We have a positive x and a negative x. Those cancel out with each other. So we're just going to have this 5x. So then we have plus 5x. And then we have the 0-th degree terms, or the constant terms. We have a positive 1 and a negative 5. You get negative 4. So you get x to the third plus 5x minus 4, which is exactly what we had over here."
    },
    {
        "Q": "@ 0:51 How He Knows That Slope is -4 by just seeing the Equation of Function?\n",
        "A": "The equation of a line is y = mx + b. m is the slope of the line. In the video the equation is g(x) = -4x + 7, so -4 is the slope of the line.",
        "video_name": "nGCW5teACC0",
        "timestamps": [
            51
        ],
        "3min_transcript": "Let g of x equal negative 4x plus 7. What is the value of the limit as x approaches negative 1 of all of this? So before we think about this, let's just visualize the line. And then we can think about what they're asking here. So let me draw some axes here. So this is my vertical axis and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would say y-intercept. or vertical axis intercept. It's going to have a slope of negative 4, so it's going to look something like this. Let me draw my best. So it's going to look something like that. And we already know the slope here is going to be negative 4. We get that right from this slope intercept form of the equation, slope is equal to negative 4. And they ask us, what is the limit as x approaches negative 1 of all of this kind of stuff? So let's plot the point negative 1. this point right over here. And this point right over here would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of y is equal to g of x. So what they're doing right over here is they're finding the slope between an arbitrary point x, g of x, and this point right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be your change in the vertical axis. That would be g of x minus g of negative 1. And then that's over-- actually, let me write it this way all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is"
    },
    {
        "Q": "\nAt 3:38, Sal says that the slope will be the same no matter what points are picked in the line. Does that mean the derivative of f or f'(x) will always equal to the slope of the secant line in the function? I just thought even if f(x) and f(x+h) are extremely close (derivative), they would have the same slope as the general slope of the function, therefore derivative equals to the slope.",
        "A": "The function Sal is talking about here is linear, so its graph is a straight line. In this case we can t take a secant in the normal sense of the word. Every line that passes through two points on the graph of this function will be the same as the line that graphs the function. This is the only situation where the slope will be the same no matter what points are picked.",
        "video_name": "nGCW5teACC0",
        "timestamps": [
            218
        ],
        "3min_transcript": "all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is It's going to be the slope of the line. It's going to be equal to negative 4. This thing is going to be equal to negative 4. It's going to be equal to negative 4. Doesn't matter how close x gets, and weather x comes from the right or whether x comes from the left. So this thing, taking the limit of this, this just gets you to negative 4. It's really just the slope of the line. So even if you were to take the limit as x approaches negative 1, as x gets closer and closer and closer to negative 1, well then, these points are just going to get closer and closer and closer. But every time you calculate the slope, it's just going to be the slope of the line, which Now, you could also do this algebraically. And let's try to do it algebraically. So let's actually just take the limit as x approaches negative 1 of g of x. Well, they already told us what g of x is. It is negative 4x plus 7, minus g of negative 1. Negative 1 times 4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1, all of that over x plus 1. And that's really x minus negative 1, is you want to think of it that way. But I'll just write x plus 1 this way here. So this is going to be equal to the limit as x approaches negative 1 of, in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus 1, all of that over x plus 1. And then since we're just trying to find the limit as x approaches negative 1, so we can cancel those out. And this is going to be non-zero for any x value other than negative 1. And so this is going to be equal to negative 4."
    },
    {
        "Q": "at 0:40 I dont get how did he draw that line...he says the line has a slope of -4 but i dont get how that works...pls help meunderstand this :/\n",
        "A": "The slope of a line can be found from its coefficient before x in its equasion. For example, the line y = -4x + 7 has a slope of -4 because it is the coefficient before x. It also represents the tangent of the angle created by the line and the positive direction of the X axis. However, most curves don t behave like that.",
        "video_name": "nGCW5teACC0",
        "timestamps": [
            40
        ],
        "3min_transcript": "Let g of x equal negative 4x plus 7. What is the value of the limit as x approaches negative 1 of all of this? So before we think about this, let's just visualize the line. And then we can think about what they're asking here. So let me draw some axes here. So this is my vertical axis and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would say y-intercept. or vertical axis intercept. It's going to have a slope of negative 4, so it's going to look something like this. Let me draw my best. So it's going to look something like that. And we already know the slope here is going to be negative 4. We get that right from this slope intercept form of the equation, slope is equal to negative 4. And they ask us, what is the limit as x approaches negative 1 of all of this kind of stuff? So let's plot the point negative 1. this point right over here. And this point right over here would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of y is equal to g of x. So what they're doing right over here is they're finding the slope between an arbitrary point x, g of x, and this point right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be your change in the vertical axis. That would be g of x minus g of negative 1. And then that's over-- actually, let me write it this way all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is"
    },
    {
        "Q": "Regarding, the y constraint at 2:30 - 2:44. I don't really get it why it is >= 1...\n",
        "A": "If y is a number less than 1, it will make the value sqrt (x - 1) not able to be graphed in the Cartesian plane. (eg you cannot plot sqrt (-4) etc) There s no x-value that would then satisfy the condition sqrt (x - 1) = x + 2. So the constraint for y >= 1.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            150,
            164
        ],
        "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is"
    },
    {
        "Q": "\nAt 04:28, why don't we take the positive and negative sqrt of ( y -1 ) ?",
        "A": "As Sal explained earlier in the video, (x+2)^2 can only be positive, due to the restriction on the function x>=2. Therefore, the square-root of (y-1) can only be positive. The next steps after this are purely algebra, and have nothing to do with a positive or negative square root.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            268
        ],
        "3min_transcript": "the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is the the x's and y's. So let's just leave that there. So here we haven't explicitly solved for x and y. But we can write for y is greater than or equal to 1, this is going to be the domain for our inverse, so to speak. And so here we can keep it for y is greater This y constraint's going to matter more. Because over here, the domain is x. But for the inverse, the domain is going to be the y-value. And then, let's see. We have the square root of y minus 1 is equal to x plus 2. Now we can subtract 2 from both sides. We get the square root of y minus 1 minus 2, is equal to x for y is greater than or equal to 1. And so we've solved for x in terms of y. Or, we could say, let me just write it the other way. We could say, x is equal to, I'm just swapping this. x is equal to the square root of y minus one minus 2, for y is So you see, now, the way we've written it out. y is the input into the function, which is going to be the inverse of that function. x the output. x is now the range. So we could even rewrite this as f inverse of y. That's what x is, is equal to the square root of y minus 1 minus 2, for y is greater than or equal to 1. And this is the inverse function. We could say this is our answer. But many times, people want the answer in terms of x. And we know we could put anything in here. If we put an a here, we take f inverse of a. It'll become the square root of a minus 1 minus 2, 4. Well, assuming a is greater than or equal to 1. But we could put an x in here. So we can just rename the the y for x. So we could just do a renaming here. So we can just rename y for x. And then we would get -- let me scroll down a little bit. We would f inverse of x."
    },
    {
        "Q": "At 1:42, Sal said that if x was greater than or equal to -2 then the underlined expression would be positive, but zero isn't positive.\n",
        "A": "So the inverse of y is x it s opposite.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            102
        ],
        "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is"
    },
    {
        "Q": "\nat 7:12 what is the range of the inverse O_o ?",
        "A": "The range of the inverse is the domain of the original function.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            432
        ],
        "3min_transcript": "You could rename it with anything really, is equal to the square root of x minus 1. Of x minus 1. Minus 2 for, we have to rename this to, for x being greater than or equal to 1. And so we now have our inverse function as a function of x. And if we were to graph it, let's try our best to graph it. Maybe the easiest thing to do is to draw some points here. So the smallest value x can take on is 1. If you put a 1 here, you get a 0 here. So the point 1, negative 2, is on our inverse graph. So 1, negative 2 is right there. And then if we go to 2, let's see, 2 minus 1 is 1. The principle root of that is 1. Minus 2. So it's negative 1, so the point, 2, negative 1 is right there. And let's think about it. If we did 5, I'm trying take perfect squares. 5 minus 1 is 4, minus 2. So the point 5, 2 is, let me make sure. 5 minus 1 is 4. Square root is 2. Minus 2 is 0. So the point 5, 0 is here. And so the inverse graph, it's only defined for x greater than or equal to negative 1. So the inverse graph is going to look something like this. It's going to look something like, I started off well, and it got messy. So it's going to look something like that. Just like that. And just like we saw, in the first, the introduction to function inverses, these are mirror images around the line y equals x. Let me graph y equals x. y equals x. y equals x is that line right there. Notice, they're mirror images around that line. Over here, we map the value 0 to 5. If x is 0, it gets mapped to 5. Here we go the other way. So that's why they're mirror images. We've essentially swapped the x and y."
    },
    {
        "Q": "ok,...at 1:04 does anyone understand moving the decimal?\n",
        "A": "You multiply like you always do, & when you are done, take that answer and put in the decimal. Oh you want me to also tell you where the decimal goes? Sure, just count how many numbers where behind the decimal in the 2 numbers that you multiplied, and make sure that your answer has the same amount of numbers behind the new decimal. (Hint, when you get harder problems, your going to have to get comfortable putting extra zeros right behind the decimal, before your answer, sometimes!) -Cheers!",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            64
        ],
        "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense."
    },
    {
        "Q": "\n@1:26, what do you do if there is a digit instead of a zero? Like say 32.12 multiplied by 35. I'm stuck on that part!",
        "A": "You just add as many zero where there are no numbers for example 63.46 +02.59",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            86
        ],
        "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense."
    },
    {
        "Q": "at 0:07 the problem has a dot in between 32.12 and 0.5 what does that dot mean?\n",
        "A": "It is the same as the multiplication sign. Once you start using variables, it gets really confusing.",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            7
        ],
        "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense."
    },
    {
        "Q": "\nAt 1:00 she says \"two separate... um, two separate...\". What does actually result from cutting it like so?",
        "A": "its a mobius strip.",
        "video_name": "AmN0YyaTD60",
        "timestamps": [
            60
        ],
        "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles,"
    },
    {
        "Q": "did anyone notice the mistake in 1:01 - 1:02\n",
        "A": "I honestly couldn t because she was talking too fast for me to understand.",
        "video_name": "AmN0YyaTD60",
        "timestamps": [
            61,
            62
        ],
        "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles,"
    },
    {
        "Q": "1:29 shows that vi hart is just awesome\n",
        "A": "This should be in Tips and Thanks, not Questions.",
        "video_name": "AmN0YyaTD60",
        "timestamps": [
            89
        ],
        "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles,"
    },
    {
        "Q": "I calculated the answer to the formula at 12:33, and I got 0.004998. can someone explain the intuition here? I also tried calculating what would be p(x=9), and I got ~0.13\n",
        "A": "OK, so I used excel to graph all possible results from X=0 => X=20. I got a nice bell curve, and the sum inside is indeed 1...pretty cool. Indeed 13% (for 9, 10) seems to be the most likely outcome.",
        "video_name": "Jkr4FSrNEVY",
        "timestamps": [
            753
        ],
        "3min_transcript": "As this approaches infinity, this term's going So you have 1 to the minus k. 1 to any power is 1, so that term becomes 1. So we have another 1 there. So there you have it. We're done. The probability that our random variable, the number of cars that passes in an hour, is equal to a particular number. You know, it's equal to 7 cars pass in an hour. Is equal to the limit as n approaches infinity of n choose k times-- well, we said it was lambda over n to the k successes times 1 minus lambda over n to the n minus k failures. And we just showed that this is equal to lambda to the kth power over k factorial times e to the minus lambda. And that's pretty neat because when you just see it in kind of a vacuum-- if you have no context for it, you wouldn't binomial theorem. I mean, it's got an e in there. It's got a factorial, but a lot of things have factorials in life, so not clear that that would make it But this is just the limit as you take smaller and smaller and smaller intervals, and the probability of success in each interval becomes smaller. But as you take the limit you end up with e. And if you think about it it makes sense because one of our derivations of e actually came out of compound interest and we kind of did something similar there. We took smaller and smaller intervals of compounding and over each interval we compounded by a much smaller number. And when you took the limit you got e again. And that's actually where that whole formula up here came from to begin with. But anyway, just so that you know how to use this thing. So let's say that I were to go out, I'm the traffic engineer, and I figure out that on average, 9 cars passed per hour. And I want to know the probability that-- so this is my expected value. In a given hour, on average, 9 cars are passing. pass in a given hour, exactly 2 cars pass. That's going to be equal to 9 cars per hour to the 2'th power or squared, to the 2'th power. Divided by 2 factorial times e to the minus 9 power. So it's equal to 81 over 2 times e to the minus 9 power. And let's see, maybe I should just get the graphing Well, I'll let you do that exercise to figure out what that is, but I'll see you in the next video."
    },
    {
        "Q": "At 1:50 you said that 800 + 0 + 4 is 804. That is correct, but what is the reason to say 800 + 0 + 4 if the 0 doesn't have any value?\n",
        "A": "He s trying to reinforce the concept that the numbers have different values based on their place, and how we keep track while we multiply. If it was 231 instead, it would work out to 800 + 30 + 4.",
        "video_name": "4BtXvopHXI8",
        "timestamps": [
            110
        ],
        "3min_transcript": "Let's multiply 4 times 2,012. Actually, let's make it a little bit simpler. Let's multiply 4 times 201 just to simplify things a little bit. So 4 times 201. So as we've seen in previous videos, I like to write the larger number on top. This is just one of many ways of tackling a calculation like this. I'll write the 201. And then I'll write the 4 right below it, and I'll write it right below the ones place. And so I have 201 times 4. Now, just like we did when we were multiplying a one digit times a two digit, we do essentially the same process. We first multiply 4 times the 1. Well, 4 times 1 we know is equal to 4. So we put a 4 right over there in the ones place. Then we can multiply our 4 times the digit that we have in the tens place. In this case, we have a 0 in the tens place. So 4 times 0, well, that's just 0. 4 times 0 is 0. We put the 0 in the tens place right over here. And then last, we have 4 times this 2 right over here. And we put the 8 right over here. And we get our answer-- 804. Now, why did this work? Well, remember, when we multiplied 4 times 1, that was literally just 4. And we've got that 4 right over here. When we multiply 4 times 0, that's 0 tens. So we've got 0 tens right over here. And when we multiplied 4 times 2, this was actually a 200. It's in the hundreds place. So 4 times 200 is 800. So what we're essentially doing by writing it in the right place is we're saying, 4 times 201, that's the same thing as 4 times 200, which is 800, plus 4 times 0 tens, which is 0 tens, plus 4 times 1, which is 4. So 800 plus 0 plus 4 is 804."
    },
    {
        "Q": "\nAt 8:13, he says 1/2^2 is just 1/2. but isn't that 1/4 since it's 1/2 x 1/2?? and he did the same thing for 1/3. shouldn't the answer have been 1/27?",
        "A": "it s only squaring one in the numerator, not the quotient as a whole. so (1^2)/2 is just 1/2",
        "video_name": "vhMl755vR5Q",
        "timestamps": [
            493
        ],
        "3min_transcript": "to be equal to the value of this inner function, which is just x. And so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely-thin depth right over here. So it's going to be dx. And so the volume our, kind of our truffle with a cone carved out, is going to be this integral minus this integral right over here. And we could evaluate it just like that. Or we could even say, OK we could factor out a pi out of both of them. There are actually, there's multiple ways But let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. You take the pi outside. Square root of x squared is going to be x dx minus the integral, we can factor the pi out. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi, times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to-- and I'm going to arbitrarily switch colors just because the green's getting monotonous-- pi times 1 squared over 2 minus 0 squared over 2. I could write squared. 1 squared over 2 minus 0 squared over 2, minus pi times 1 to the third over 3 minus 0 to the third over 3. me do it in that same blue color-- so this is this simplified. This is just 0 right over here. This is 1 squared over 2, which is just 1/2. So it's just pi over 2, 1/2 times pi minus-- well this is just 0, this is 1/3, minus pi over 3. And then to simplify this, it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with, we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the volume of that wacky kind"
    },
    {
        "Q": "AT 0:56 on the video you multiply all numbers and how do you get the second number\n",
        "A": "When Sal does 6x8x7, instead of doing 48x7, he does 40x7, which is 280, and 8x7, which is 56. Adding the two up gets 336. You get the same answer doing 6x8x7, or 48x7.",
        "video_name": "I9efKVtLCf4",
        "timestamps": [
            56
        ],
        "3min_transcript": "What is the volume of this box? Drag on the box to rotate it. So this is pretty neat. We can actually sit and rotate this box. And here it looks like everything's being measured in meters. So we want to measure our volume in terms of cubic meters. That's going to be our unit cube here. So when we want to think about how many cubic meters could fit in this box, we've already seen examples. You really just have to multiply the three different dimensions of this box. So if you wanted the number of cubic meters that could fit in here, it's going to be six meters times 8 meters times 7 meters which is going to give you something in cubic meters. So let's think about what that is. 6 times 8 is 48. Let me see if I can do this in my head. 48 times 7, that's 40 times 7, which is going to be 280 plus 8 times 7, which is 56, Let's check our answer. Let's do one more of these. So what's the volume of this box? We'll once again, we have its height at six feet. Now everything is being measured in feet. We have it's width being four feet. So we could multiple the height times the width of four feet. And then we can multiply that times its depth of two feet. So 6 times 4 is 24 times 2 is 48 feet. 48, and I should say cubic feet. We're saying how many cubic feet can fit in here? When we multiply the various dimensions measured in feet, we're counting almost how many of those cubic feet can fit into this box."
    },
    {
        "Q": "At 2:08, if the x-axis is inches, and later on it is shown that the area under the curve is the probability, what is the y-axis representative of, as far as units or meaning?\n",
        "A": "Y is the rate of change of the area as you change the range of desired outcomes (x). E.g. if y is .5 then probability (area) is increasing at the rate of .5 per additional inch of rain in the interval of chosen outcomes.",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            128
        ],
        "3min_transcript": "In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5."
    },
    {
        "Q": "\nAt 2:00, is .5 a bad example for the height? In a continuous random variable, it's very unlikely any observation is this common. If the y axis is not probability of a particular rainfall level, what is it?",
        "A": "the rainfall amounts are on the x axis. The probability is for an interval (range of possible outcomes) and is the area under the curve for that interval. Y is the rate of change of probability (area) for an interval as it changes limits (as x changes).",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5."
    },
    {
        "Q": "\n8:50 This might be a stupid question, but is a probability distribution function related only to discrete random variables and a probability density function is for continuous random variables?",
        "A": "From Wikipedia: The terms probability distribution function and probability function have also sometimes been used to denote the [continuous] probability density function. However, this use is not standard among probabilists and statisticians. I don t believe that a probability distribution of a discrete random variable is referred to as a function. Maybe it could be made into a step function, but I don t know if that is common.",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            530
        ],
        "3min_transcript": "is pretty much 0. That you really have to say, OK what's the probably that we'll get close to 2? And then you can define an area. And if you said oh, what's the probability that we get someplace between 1 and 3 inches of rain, then of course the probability is much higher. The probability is much higher. It would be all of this kind of stuff. You could also say what's the probability we have less than 0.1 of rain? Then you would go here and if this was 0.1, you would calculate this area. And you could say what's the probability that we have more than 4 inches of rain tomorrow? Then you would start here and you'd calculate the area in the curve all the way to infinity, if the curve has area all And hopefully that's not an infinite number, right? Then your probability won't make any sense. But hopefully if you take this sum it comes to some number. And we'll say there's only a 10% chance that you have more than 4 inches tomorrow. And all of this should immediately lead to one light events that might occur can't be more than 100%. Right? All the events combined-- there's a probability of 1 that one of these events will occur. So essentially, the whole area under this curve has to be equal to 1. So if we took the integral of f of x from 0 to infinity, this thing, at least as I've drawn it, dx should be equal to 1. For those of you who've studied calculus. For those of you who haven't, an integral is just the area under a curve. And you can watch the calculus videos if you want to learn a little bit more about how to do them. And this also applies to the discrete probability distributions. Let me draw one. The sum of all of the probabilities have to be equal to 1. And that example with the dice-- or let's say, since it's faster to draw, the coin-- the two probabilities have So this is 1, 0, where x is equal to 1 if we're heads or 0 if we're tails. Each of these have to be 0.5. Or they don't have to be 0.5, but if one was 0.6, the other would have to be 0.4. They have to add to 1. If one of these was-- you can't have a 60% probability of getting a heads and then a 60% probability of getting a tails as well. Because then you would have essentially 120% probability of either of the outcomes happening, which makes no sense at all. So it's important to realize that a probability distribution function, in this case for a discrete random variable, they all have to add up to 1. So 0.5 plus 0.5. And in this case the area under the probability density function also has to be equal to 1. Anyway, I'm all the time for now. In the next video I'll introduce you to the idea of an expected value. See you soon."
    },
    {
        "Q": "At 4:27 it is said that the probability of something happening is \"actually zero\". Which makes sense when you are talking about area under a curve, but it doesn't make intuitive sense because doesn't that mean that the probability of ANYTHING happening in a continuous variable is zero? That doesn't make sense because things are still going to happen, even though we are saying the probability is zero. So if it rains in California, the exact amount of rain that came down had a zero probability of happening?\n",
        "A": "This is quite an interesting question! While all impossible events definitely have probability zero of occurring, the reverse is not always true! An event that has probability zero of occurring could still be possible, though extremely unlikely. The probability that a continuous random variable equals any specific, exact value is always zero, but the probability that a continuous random variable lies in a specific interval of real numbers could be greater than zero. Have a blessed, wonderful day!",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            267
        ],
        "3min_transcript": "this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5. And I would say no, it is not a 0.5 chance. And before we even think about how we would interpret it visually, let's just think about it logically. What is the probability that tomorrow we have exactly 2 inches of rain? Not 2.01 inches of rain, not 1.99 inches of rain. Not 1.99999 inches of rain, not 2.000001 inches of rain. Exactly 2 inches of rain. I mean, there's not a single extra atom, water molecule above the 2 inch mark. And not as single water molecule below the 2 inch mark. It's essentially 0, right? It might not be obvious to you, because you've probably heard, oh, we had 2 inches of rain last night. But think about it, exactly 2 inches, right? Normally if it's 2.01 people will say that's 2. But we're saying no, this does not count. We want exactly 2. 1.99 does not count. Normally our measurements, we don't even have tools that can tell us whether it is exactly 2 inches. No ruler you can even say is exactly 2 inches long. At some point, just the way we manufacture things, there's going to be an extra atom on it here or there. So the odds of actually anything being exactly a certain measurement to the exact infinite decimal point is actually 0. The way you would think about a continuous random variable, you could say what is the probability that Y is almost 2? So if we said that the absolute value of Y minus is 2 is less than some tolerance? Is less than 0.1. And if that doesn't make sense to you, this is essentially just saying what is the probability that Y is greater than 1.9 and less than 2.1?"
    },
    {
        "Q": "\nat 1:55, why does he do it like a(a+b) + b(a+b) instead of (a x a) + (ab) + (ab) + (b x b)?",
        "A": "He is just showing a simpler way of doing it..it looks a little less confusing, :) Hope that helps!",
        "video_name": "xjkbR7Gjgjs",
        "timestamps": [
            115
        ],
        "3min_transcript": "We're asked to simply, or expand (7x + 10) ^ 2 Now the first thing I will show you is exactly what you should NOT do, well there's this huge temptation. A lot of people will look at this and say oh, that's (7x)^2 + 10^2. This is WRONG. And I'll write it in caps. This is WRONG! What your brain is doing is thinking if I had 7x times 10 and I squared that, this would be (7x)^2 times 10^2. We aren't multiplying here, we're adding 7x to 10. So you can't just square each of these terms. I just wanted to highlight, this is completely wrong, and to see why it's wrong, you have to remind yourself that (7x + 10)^2 is the exact same thing as (7x + 10)(7x + 10). That's what it means to square something. So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have"
    },
    {
        "Q": "\nAt 0:23 isn't pretty redundant to point this out this late in the course wouldn't it help to emphasize this point earlier?",
        "A": "Well not everyone watches all the videos on order. If people are just watching this specific video because they re not so strong at this topic, you don t know what they already know so it might be a helpful reminder for them",
        "video_name": "xjkbR7Gjgjs",
        "timestamps": [
            23
        ],
        "3min_transcript": "We're asked to simply, or expand (7x + 10) ^ 2 Now the first thing I will show you is exactly what you should NOT do, well there's this huge temptation. A lot of people will look at this and say oh, that's (7x)^2 + 10^2. This is WRONG. And I'll write it in caps. This is WRONG! What your brain is doing is thinking if I had 7x times 10 and I squared that, this would be (7x)^2 times 10^2. We aren't multiplying here, we're adding 7x to 10. So you can't just square each of these terms. I just wanted to highlight, this is completely wrong, and to see why it's wrong, you have to remind yourself that (7x + 10)^2 is the exact same thing as (7x + 10)(7x + 10). That's what it means to square something. So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have"
    },
    {
        "Q": "At 3:15, what was wrong with leaving 7x^2 as is? When he simplified it, he still got 49x^2 so it didn't get rid of the x or the square, so i don't know why he did it. But it seems like it changes the answer, so could someone help me understand?\nThanks in advance.\n",
        "A": "He was solving not simplifying. Distribute the ^2 in (7x)^2 and you have 7 squared, which is 49, and x squared, which is x^2.",
        "video_name": "xjkbR7Gjgjs",
        "timestamps": [
            195
        ],
        "3min_transcript": "So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have And this comes out when you are multiplying all the different combinations of the terms here. if we simplify this, if we simplify seven x squared That's seven squared times x squared. So seven squared is 49 times x squared When you multiply this part out 2 times 7 times 10 which is 140 and then we have our x. No other x there. And then plus 10 squared. So plus 100. And we are done."
    },
    {
        "Q": "\nFrom 6:30 to 6:56, Sal lists non linear equations Why are they nonlinear?",
        "A": "A linear equation is only one where x is to the power of 1, generally some form of y=x Its graph is a straight line. In all the equations listed, x was being raised to a power that was either less than or greater than 1 - for example, x^2, or 1/x, which is the same as x^-1. If you look at the graphs of any of those equations, they re curves, not straight lines.",
        "video_name": "AOxMJRtoR2A",
        "timestamps": [
            390,
            416
        ],
        "3min_transcript": "And it's indeed-- that's indeed the case. Two times five is ten, minus three is seven. The point-- the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line. That is why it is called a linear equation. Now, this isn't the only way that we could write a linear equation. You could write a linear equation like-- let me do this in a new color. You could write a linear equation like this: Four x minus three y is equal to twelve. This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this we would once again get a line. X and y. If x is equal to zero, then this goes away and you have negative three y is equal to twelve. Let's see, if negative three y equals twelve then y would be equal to negative four. You can verify that. Four times zero minus three times negative four well that's gonna be equal to positive twelve. And let's see, if y were to equal zero, if y were to equal zero then this is gonna be four times x is equal to twelve, well then x is equal to three. And so you have the point zero comma negative four, zero comma negative four on this line, and you have the point three comma zero on this line. Three comma zero. Did I do that right? So zero comma negative four and then three comma zero. These are going to be on this line. Three comma zero is also on this line. So this is, this line is going to look something like-- something like, I'll just try to hand draw it. Something like that. So once again, all of the xy-- all of the xy pairs that satisfy this, Now what are some examples, maybe you're saying \"Wait, wait, wait, isn't any equation a linear equation?\" And the simple answer is \"No, not any equation is a linear equation.\" I'll give you some examples of non-linear equations. So a non-- non-linear, whoops let me write a little bit neater than that. Non-linear equations. Well, those could include something like y is equal to x-squared. you will see that this is going to be a curve. it could be something like x times y is equal to twelve. This is also not going to be a line. Or it could be something like five over x plus y is equal to ten. This also is not going to be a line. So now, and at some point you could-- I encourage you to try to graph these things, they're actually quite interesting. But given that we've now seen examples of linear equations and non-linear equations,"
    },
    {
        "Q": "\nat 0:36 he explains that negitive 3p minus p is equal to negative 4 p. how is that possible.",
        "A": "To combine like terms, you add / subtract the coefficients (the numbers in front of the variables). The coefficient of -3p is -3 The coefficient of - p is -1 -3 - 1 = -4 If you don t get this part: -- use a number line: got to -3. To take away 1, you move 1 unit to the left. You will now be on -4. -- and, you need to review the lessons on adding and subtracting negative numbers. Thus, -3p - p = -4p Hope this helps.",
        "video_name": "SgKBBUFaGb4",
        "timestamps": [
            36
        ],
        "3min_transcript": "We're asked to solve for p. And we have the inequality here negative 3p minus 7 is less than p plus 9. So what we really want to do is isolate the p on one side of this inequality. And preferably the left-- that just makes it just a little easier to read. It doesn't have to be, but we just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side. And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p-- that's negative 4p. And then we still have a minus 7 up here-- is going to be less than p minus p. Those cancel out. It is less than 9. Now the next thing I'm in the mood to do is get rid of this negative 7, or this minus 7 here, so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it will just cancel out to 0. So let's add 7 to both sides of this inequality. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 equals 16. And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember, since this is an inequality, not an equation. If you're dealing with an inequality and you multiply or divide both sides of an equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than, since we're dividing by a negative number. And so negative 4 divided by negative 4-- those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, negative 1, 0. Let me write that a little bit neater. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.9999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one,"
    },
    {
        "Q": "\nat 1:32 why did sal attach the rope of the mast all the way at the top\nhow does he know where to put the rope on the mast?\nthanks for whoever answers my question in the future!!\nthanks future answers",
        "A": "The question clearly states that the rope is attached to the top of the mast.",
        "video_name": "JVrkLIcA2qw",
        "timestamps": [
            92
        ],
        "3min_transcript": "The main mast of a fishing boat is supported by a sturdy rope that extends from the top of the mast to the deck. If the mast is 20 feet tall and the rope attaches to the deck 15 feet away from the base of the mast, how long is the rope? So let's draw ourselves a boat and make sure we understand what the deck and the mast and all of that is. So let me draw a boat. I'll start with yellow. So let's say that this is my boat. That is the deck of the boat. And the boat might look something like this. It's a sailing boat. This is the water down here. And then the mast is the thing that holds up the sail. So let me draw ourselves a mast. And they say the mast is 20 feet tall. So this distance right here is 20 feet. I can draw it as a pole so it's a little bit clearer. Even shade it in if we like. And then they say a rope attaches to the deck 15 feet away from the base of the mast. So this is the base of the mast. This is the deck right here. The rope attaches 15 feet away from the base of the mast. So if this is the base of the mast, we go 15 feet, might be about that distance right there. Let me mark that. And the rope attaches right here. From the top of the mast all the way that base. So the rope goes like that. And then they ask us, how long is the rope? So there's a few things you might realize. We're dealing with a triangle here. And it's not any triangle. We're assuming that the mast goes straight up and that the deck is straight left and right. So this is a right triangle. This is a 90 degree angle right here. we can always figure out the third side of a right triangle using the Pythagorean theorem. And all that tells us is it the sum of the squares of the shorter sides of the triangle are going to be equal to the square of the longer side. And that longer side is call the hypotenuse. And in all cases, the hypotenuse is the side opposite the 90 degree angle. It is always going to be the longest side of our right triangle. So we need to figure out the hypotenuse here. We know the lengths of the two shorter sides. So we can see that if we take 15 squared, that's one of the short sides, I'm squaring it. And then add that to the square of the other shorter side, to 20 feet squared."
    },
    {
        "Q": "i don't understand how this works does the dot such as in : 4'5 ( the ' means dot . look at 1:11 to 3:30 ) does that dot system work fo more than 1 digit numbers ??\n\nThanks\nYaz Lightning\n",
        "A": "The dot is used for any number, large or small. It is mainly used in algebra, where x could be a variable, not an operator. Hope it helped!",
        "video_name": "Yw3EoxC_GXU",
        "timestamps": [
            71,
            210
        ],
        "3min_transcript": "Rewrite 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 as a multiplication expression. And then they want us to write the expression three times using different ways to write multiplication. So let's do the first part. Let's write it as a multiplication expression. So how many times have we added 5 here? Well, we've got it at one, two, three, four, five, six, seven. So one way to think of it, if I just said what is here? How many 5's are there? You'd say, well, I added 5 to itself seven times, right? You could literally say this is 7 times 5. We could literally write, this is 7 times 5, or you could view it as 5 seven times. I'm not even writing it mathematically yet. I'm just saying, look, if I saw seven of something, you would literally say, if these were apples, you would say apples seven times, or you'd say seven times the apple, whatever it is. Now, in this case, we're actually adding the number to each other, and we could figure out what that is, and But the way we would write this mathematically, we would say this is 7 times 5. We could also write it like this. We could write it 7 dot 5. This and this mean the exact same thing. It means we're multiplying 7 times 5 or 5 times 7. You can actually switch the order, and you get the exact same value. You could actually write it 5 times 7. So you could interpret this as 7 five times or 5 seven times, however you like to do it, or 5 seven times. I don't want to confuse you. I just want to show you that these are all equivalent. This is also equivalent. 5 times 7. Same thing. You could write them in parentheses. You could write it like this. This all means the same thing. That's 7 times 5, and so is this. These all evaluate to the same thing: 5 times 7. So these are all equivalent, and since we've worked with it so much, let's just figure out the answer. So if we add up 5 to itself seven times, what do we get? Well, 5 plus 5 is 10. plus 5 is 35. So all of these evaluate to 35, just so you see that they're the same thing. These are all equivalent to 35. And just something to think about, this is also the exact same thing, depending on how you want to interpret this, as 7 five times. They didn't ask us to do it, but I thought I would point it out to you. 7 five times would look like this: 7 plus 7 plus 7 plus 7 plus 7, right? I have 7 five times. I added it to itself five different times. There's five 7's here added to each other. And when you add these up, you'll also get 35. And that's why 5 times 7 and 7 times 5 is the same thing."
    },
    {
        "Q": "At 1:14 that looks like a decimal point?\n",
        "A": "Yes. The dot, point or period can be used to mean the decimal point, or it can be used instead of the multiplication symbol to mean multiplied by , just like x . You can usually tell from the context what the dot is being used for. As a decimal point, it is normally written at the bottom, where commas and periods go in writing regular sentences. As a multiplication symbol, it is normally written higher up, like the hyphen as in break-dance .",
        "video_name": "Yw3EoxC_GXU",
        "timestamps": [
            74
        ],
        "3min_transcript": "Rewrite 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 as a multiplication expression. And then they want us to write the expression three times using different ways to write multiplication. So let's do the first part. Let's write it as a multiplication expression. So how many times have we added 5 here? Well, we've got it at one, two, three, four, five, six, seven. So one way to think of it, if I just said what is here? How many 5's are there? You'd say, well, I added 5 to itself seven times, right? You could literally say this is 7 times 5. We could literally write, this is 7 times 5, or you could view it as 5 seven times. I'm not even writing it mathematically yet. I'm just saying, look, if I saw seven of something, you would literally say, if these were apples, you would say apples seven times, or you'd say seven times the apple, whatever it is. Now, in this case, we're actually adding the number to each other, and we could figure out what that is, and But the way we would write this mathematically, we would say this is 7 times 5. We could also write it like this. We could write it 7 dot 5. This and this mean the exact same thing. It means we're multiplying 7 times 5 or 5 times 7. You can actually switch the order, and you get the exact same value. You could actually write it 5 times 7. So you could interpret this as 7 five times or 5 seven times, however you like to do it, or 5 seven times. I don't want to confuse you. I just want to show you that these are all equivalent. This is also equivalent. 5 times 7. Same thing. You could write them in parentheses. You could write it like this. This all means the same thing. That's 7 times 5, and so is this. These all evaluate to the same thing: 5 times 7. So these are all equivalent, and since we've worked with it so much, let's just figure out the answer. So if we add up 5 to itself seven times, what do we get? Well, 5 plus 5 is 10. plus 5 is 35. So all of these evaluate to 35, just so you see that they're the same thing. These are all equivalent to 35. And just something to think about, this is also the exact same thing, depending on how you want to interpret this, as 7 five times. They didn't ask us to do it, but I thought I would point it out to you. 7 five times would look like this: 7 plus 7 plus 7 plus 7 plus 7, right? I have 7 five times. I added it to itself five different times. There's five 7's here added to each other. And when you add these up, you'll also get 35. And that's why 5 times 7 and 7 times 5 is the same thing."
    },
    {
        "Q": "\nSal keeps on saying Thrill (cola) instead of Thrill (soda), right?\nI think its at 2:02 one of the times.",
        "A": "well really that s not the point and it doesn t really matter, but yeah.",
        "video_name": "gs-OPF3KEGU",
        "timestamps": [
            122
        ],
        "3min_transcript": "Thrill Soda hired a marketing company to help them promote their brand against Yummy Cola. The company gathered the following data about consumers' preference of soda. So they have, year by year, percentage of respondents who preferred Yummy Cola, percentage of respondents who preferred Thrill Cola, and then these are people who had no preference. So in 2006, 80% liked Yummy, only 12% liked Thrill, and 8% didn't like either one or didn't have any preference. And so actually just from here you see that many, many more people liked Yummy Cola than Thrill Cola, actually every year over year. So, Thrill Cola definitely has something. They have an uphill battle. But then they said the advertising company created the following two graphs to promote Thrill Soda. And so let's see what's happening over here. And let's think about whether this is misleading or not. So if we look at this graph over here, in 2006, sure enough, 80% liked Yummy Cola. Then in 2007, 76%. Then it keeps going to then 77%, then 73%, then 73% to 68%. It actually represents the data that's given right over here. I'll do it in the same. It actually represents this data very faithfully. Then right over here, if we look at this chart, Percentage of People who Prefer Thrill Soda, so over here in 2006, 12% preferred Thrill Soda. 2007, 19%. 2008, 19%. Then we go up to 20%, 21%, and 25%. So the graphs are actually accurate. They're not lying. These are actually the data points of the percentage who prefer Thrill Soda. Now what's misleading is if someone were to just look at these two graphs without actually looking at the scales over here, they'll see two things. They'll say, oh, look, you see a declining trend. And that's what line graphs are good for, for seeing trends. They say, look, I see a declining trend in the percentage of people who prefer Yummy Cola. And I see this increasing trend in the percentage And that's true. You have a declining trend here. And you have an increasing trend here. But what's misleading here is the way that they've plotted the scales. These scales are not the same. So when you look at this, you say not only is there an increasing trend of people who prefer Thrill Soda, but the way they set up the scale, it looks like the trend is above. The human brain is tempted to compare these and to say, look, not only is this an upward trend, but it's above this trend right over here. Even in 2006, this data point looks higher than these data points right over here. But the reality is that it's only because the scale is distorted. Now this is the oldest trick in the book when plotting line graphs. It all depends on the scale. So this just looks good because they used this scale that went from 0 to 30 as opposed to 0 to 100. The better thing to do, or the more genuine thing to do, or the more honest thing to do, would have actually been to plot them on the same graph. Although if they did that, that wouldn't have painted a very good picture for Thrill Soda. So if we plotted on the same graph Thrill Soda,"
    },
    {
        "Q": "\n@ 3:00 we play, what seems to me to be, a dirty trick and move the dx to before the e^x... term. How is that OK? If this is a valid algebraic tool then why not do that every time we have a product so that \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 b \u00e2\u0080\u00a2 c \u00e2\u0080\u00a2 dx can be simplified to \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 dx \u00e2\u0080\u00a2 b \u00e2\u0080\u00a2 c? This of course isn't right because we could just insert a 1 x in the front of any expression and play the same trick, e.g. \u00e2\u0088\u00ab a = \u00e2\u0088\u00ab 1 \u00e2\u0080\u00a2 a dx = \u00e2\u0088\u00ab 1 dx a = x \u00e2\u0080\u00a2 a + C which can't be right.",
        "A": "A notationally better version of your equation is \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 dx = \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 1 \u00e2\u0080\u00a2 dx = a \u00e2\u0080\u00a2 \u00e2\u0088\u00ab 1 \u00e2\u0080\u00a2 dx = a \u00e2\u0080\u00a2 x + C , so your example equation is essentially correct, apart from forgetting the dx in the first integral.",
        "video_name": "b76wePnIBdU",
        "timestamps": [
            180
        ],
        "3min_transcript": "But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared, and this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is 3x squared, derivative of x squared is 2x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now, what is going to be the derivative of u with respect to x? du dx. Well, we've done this multiple times. It's going to be 3x squared plus 2x. And now we can write this in differential form. And du dx, this isn't really a fraction of the differential It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? You could multiply both sides times a dx. So both sides times a dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the trouble of doing that? Well we see we have a 3x squared plus 2x, and then it's being multiplied by a dx right over here. I could rewrite this as the integral of-- and let me do it in that color-- of 3x squared plus 2x times dx times e-- let me do that in that other color-- times e to the x to the third plus x squared. Now what's interesting about this? Well the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit, it's going to be equal to-- and what I'm going to do is I'm going to change the order. I'm going to put the du, this entire du, I'm gonna stick it on the other side here, so it looks like more of the standard form that we're used to seeing our indefinite integrals in."
    },
    {
        "Q": "\nAt time 1:44 why do you swap the sign if you aren't dividing by a negative number to solve the right side of the inequality? How come the sign switches from less than or equal to, to greater than or equal to? what is the logic behind that?",
        "A": "He moved all of the constants from the right side of the inequality to the left. He left the variable on the right side if the variable is on the right side you swap the sign. I just isolate my variable and bring it to the left and all the constants on the right it s a lot less confusing! Hope that helped!!",
        "video_name": "A3xPhzs-KBI",
        "timestamps": [
            104
        ],
        "3min_transcript": "Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this \"and\" there to think about the equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would"
    },
    {
        "Q": "\nwhy does he add 4 to every equation at 1:30?",
        "A": "He does so to isolate x. By adding 4 to both sides, the x - 4 becomes x because (x - 4) + 4 = x.",
        "video_name": "A3xPhzs-KBI",
        "timestamps": [
            90
        ],
        "3min_transcript": "Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this \"and\" there to think about the equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would"
    },
    {
        "Q": "\nWait....at like 3:10, in the vid, it says 54/10=5 remainder 4. So...since 4/10=.4, can we write it as 5.4? Are they the same thing?",
        "A": "No, you can t. When someone says remainder 4 don t think it s .4. remainder 4 is going to be a whole number!",
        "video_name": "STyoP3rCmb0",
        "timestamps": [
            190
        ],
        "3min_transcript": "0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54. Now you might see a little pattern here. Between these two numbers, I had exactly one number to the right of the decimal. When I take its product, let's say I ignored the decimal. I just said 9 times 6, I would've gotten 54. But then I have to divide by 10 in order to take account of the decimal, take account of the fact this wasn't a 6. This was a 6/10. And so I have one number to the right of the decimal here. And I want to you to think about that whether that's a general principle. Can we just count the total numbers of digits to the right of the decimals and then our product is going to have the same number of digits to right of the decimal? I'll let you to think about that."
    },
    {
        "Q": "\nAt 7:57, why can we not simply (3 * X ^ 2) - (4 * X ^ -5) ? I know it's a stupid question, but I am really confused. Will the answer not be = -1X^7?",
        "A": "You cannot subtract exponents like that. In order to do a subtraction, the exponent needs to match. Take a look below: 2^5 - 2^2 You re saying this would equal: 2^3 = 8 But if you evaluate it: 2^5 - 2^2 32 - 4 = 28 The exponents have to match in order for a subtraction or addition to work.",
        "video_name": "mzOBlH32qdk",
        "timestamps": [
            477
        ],
        "3min_transcript": "being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f-- actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we'll put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you. But when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of let's say x to the third power tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule. So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus-- that's a different shade of blue-- and over here is negative 4. So it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have-- and I could just simplify a little bit. This is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's say that I have-- and I'll do it in white. Let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's just going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x, or 3 times the derivative of x to the first."
    },
    {
        "Q": "\nSurely, at 0:30, the rule does also works if n = 0, since the derivative will equal 0.x^-1, which equals 0, and that is the derivative of a constant.",
        "A": "It s pretty nitpicky, but 0 * x ^ (-1) is not always equal to 0. When x equals 0, it s undefined.",
        "video_name": "mzOBlH32qdk",
        "timestamps": [
            30
        ],
        "3min_transcript": "Now that we know the power rule, and we saw that in the last video, that the derivative with respect to x, of x to the n, is going to be equal to n times x to the n minus 1 for n not equal 0. I thought I would expose you to a few more rules or concepts or properties of derivatives that essentially will allow us to take the derivative of any polynomial. So this is powerful stuff going on. So the first thing I want to think about is, why this little special case for n not equaling 0? What happens if n equals 0? So let's just think of the situation. Let's try to take the derivative with respect to x of x to the 0 power. Well, what is x to the 0 power going to be? And we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis. This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there. f of x equals 1 is just a horizontal line. So that right over there is the graph, y is equal to f of x, which is equal to 1. Now, remember the derivative, one way to conceptualize is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change. It has a constant slope. And it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0. is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say that f of x is equal to 3. Let's say that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. Derivative with respect to x of any constant-- so let's say of a where this is just a constant, that's going to be equal to 0."
    },
    {
        "Q": "\nWhat is the reasoning behind the taking the constant out of the dy/dx at 4:08?",
        "A": "It is one of the rules/properties of derivatives so we are allowed to do that. Now, why would you do that? It is to make the computation easier. It is not very noticeable when the constant is a nice small integer and you are taking the derivative of a power. But things can get really messy when you take the derivatives of a quotient with chain rules (just an example). So it is easier to take a constant out. You do not need to take out a constant if you can compute with it.",
        "video_name": "mzOBlH32qdk",
        "timestamps": [
            248
        ],
        "3min_transcript": "is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say that f of x is equal to 3. Let's say that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. Derivative with respect to x of any constant-- so let's say of a where this is just a constant, that's going to be equal to 0. Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of-- let's use the same A. Let's say I have some constant times some function. Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to A. I didn't want to do that magenta color. It's going to be equal to A times the derivative of f of x. Let me do that blue color. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to A times this thing right over here is the exact same thing as f prime of x. but I think if I gave you an example it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative of x to the fifth, 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times-- let me write that. I want to keep it consistent with the colors."
    },
    {
        "Q": "At 4:40, why Sal doesn't make a shortcut, and use the average of the past 4 exams as is? if we solve for - ( 82 + X ) divided by 2 = 88 , we could spare ourselves a step or two. is the number of past exams really necessary to the equation?\n",
        "A": "now i noticed that Sal s way didn t work out. if we go the way i mentioned we get that X=92, and 92 + 84 = 176, then finally 176 / 2 = 88. so, if on past 4 exams you got an average of 84, to get an average of 88 - you ll have to get a score of 92 points on the next exam... isn t it?",
        "video_name": "9VZsMY15xeU",
        "timestamps": [
            280
        ],
        "3min_transcript": "8 plus 8 is 16. I just ran eight miles, so I'm a bit tired. And, 4/8, so that's 32. Plus 1 is 33. And now we divide this number by 4. 4 goes into 336. Goes into 33, 8 times. 8 times 4 is 32. 33 minus 32 is 1, 16. So the average is equal to 84. So depending on what school you go to that's either a B or a C. So, so far my average after the first four exams is an 84. Now let's make this a little bit more difficult. We know that the average after four exams, at four exams, is equal to 84. average an 88, to average an 88 in the class. So let's say that x is what I get on the next test. So now what we can say is, is that the first four exams, I could either list out the first four exams that I took. Or I already know what the average is. So I know the sum of the first four exams is going to 4 times 84. And now I want to add the, what I get on the 5th exam, x. And I'm going to divide that by all five exams. So in other words, this number is the average of my first five exams. We just figured out the average of the first four exams. We add what I got on the fifth exam, and then we divide it by 5, because now we're averaging five exams. And I said that I need to get in an 88 in the class. And now we solve for x. Let me make some space here. So, 5 times 88 is, let's see. 5 times 80 is 400, so it's 440. 440 equals 4 times 84, we just saw that, is 320 plus 16 is 336. 336 plus x is equal to 440. Well, it turns out if you subtract 336 from both sides, you get x is equal to 104. So unless you have a exam that has some bonus problems on it, it's probably impossible for you to get ah an 88 average in the class after just the next exam."
    },
    {
        "Q": "\n42:14=x:2 what does x equal",
        "A": "X would equal 6.",
        "video_name": "MaMk6-f3T9k",
        "timestamps": [
            2534
        ],
        "3min_transcript": ""
    },
    {
        "Q": "So according, to this video,is 2/3 is equal to 2:3\n",
        "A": "Yes you are understanding it very well",
        "video_name": "MaMk6-f3T9k",
        "timestamps": [
            123
        ],
        "3min_transcript": "We're told this table shows equivalent ratios to 24 to 40. Fill in the missing values. And they write the ratio 24 to 40 right over here. 24-- when the numerator is 24, the denominator is 40. So in that way, you could think of 24/40. But then they want us to write equivalent ratios where we have to fill in different blanks over here-- here in the denominator and here in the numerator. And there's a bunch of ways that we could actually tackle this. But maybe the easiest is to start with the ratio that they gave us, where they gave us both the numerator and the denominator, and then move from there. So for example, if we look at this one right over here, the numerator is 12. It is half of the 24. So the denominator is also going to be half of the denominator It's going to be half of 40. So we could stick a 20 right over there. And then we could go up here. If you compare the 3 to the 12, to go from 12 to 3, you have to divide by 4. So in the numerator, you're dividing by 4. So in the denominator, you also want to divide by 4. So 20 divided by 4 is 5. this numerator right over here. And we see from the denominator, we doubled the denominator. We went from 40 to 80. So we would double the numerator as well, and so you would get 48. And what we just did here is we wrote four equivalent ratios. The ratio 3 to 5 or 3/5 is the same thing as 12 to 20, is the same thing as 24 to 40, is the same thing as 48 to 80. Let's make sure we got the right answer. Let's do a couple more of these. The following table shows equivalent fractions to 27/75. So then they wrote all of the different equivalent fractions. This table shows ratios equivalent to 18/55. All right, so these are all equivalent to 27/75. These are all equivalent to 18/55, so all of these. Which fraction is greater, 27/75 or 18/55? What we want to do-- because you look at these two things. And you're like, well, I don't know. Their denominators are different. How do I compare them? And the best way that I can think of comparing them is look at a point where you're getting an equivalent fraction. And either the numerators are going to be the same, or the denominators are going to be the same. So let's see if there's any situation here. So you have this situation where we see 27/75 is 54/150. And over here, we see that 18/55 is 54-- and this 54 jumped out at me because it's the same numerator-- over 165. And that makes the comparison much easier. What is smaller? 54/150 or 54/165? Well, if you have the same numerator, having a larger denominator will make the number smaller. So 54/165 is smaller than 54/150, which tells us that 18/55 is smaller than 27/75."
    },
    {
        "Q": "\nWhat do you do to find out the rate of 14:10?",
        "A": "You first find the greatest common factor between the 2 numbers and divide each number by the factor. To find the missing rates your use the unit rate to find the proportional number to the number given.",
        "video_name": "MaMk6-f3T9k",
        "timestamps": [
            850
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:23, how is 5 over x (100) + 5 = 5.05? How does the 5 become 0.5?\n",
        "A": "I do not see where in the video 5 becoming 0.5, but dividing by powers of 10 moves decimal places, so 5/10 would be 0.5, 5/100 = 0.05, 5/1000 = 0.005, etc",
        "video_name": "UvDcEvDC4vg",
        "timestamps": [
            203
        ],
        "3min_transcript": "Say, x is zero, x is 50, x is 100. Well, when an x is zero, 100 - x is 100 - zero. So, it's 100. When x is 50, it's going be 100 - 50, so it's going to be 50. When x is 100, it's 100 - 100, so it's zero. So, it's pretty clear here that as x is increasing, as x is increasing, I'll just write incr. for increasing, we see that 100 - x is decreasing. We see that that is decreasing. Let's do this with a couple more expressions that have different forms. So, let's say that I have the expression. Let's say that I have the expression five over x plus five and x is decreasing, x is decreasing, but we also know that it is positive and even while it's decreasing, it's staying above zero. So, we're saying x is staying greater than zero. where x is decreasing from 10 to nine or a million to 100,000, but it's staying positive while it's decreasing. Let's think about that. We are going to be dividing by smaller and smaller positive values. So, as you have smaller and smaller positive values of the denominator, you're dividing by smaller and smaller positive values. So, if you're dividing by smaller positive values, you're going to know that this thing is going to get larger. This entire expression is going to get larger as you divide by smaller and smaller positive values. So, if that expression gets larger, then you're just adding five to it. The whole thing is going to increase. The whole thing is going to increase as, the whole thing is going to increase as x decreases while staying positive. And once again, we can make a little table to take a look at that. So, this is x, this is five over x plus five. Let's see, I'll go from x 100 to x is five, to x is one. So, this is clearly x is decreasing. X is decreasing. When x is 100, you're gonna have five divided by 100, which would be five hundredths plus five, so it would be 5.05. When x is five, you're gonna have five divided by five, which is one plus five, which is six. When x is one, you're gonna have five divided by one, which is five plus five. You just, it's 10. So notice, when x is staying positive, but decreasing, the whole expression, five over x plus five, this thing right over here is increasing. Let's do one more of these. So, let's say that we have the expression and we'll change up the variable here. 3y over 2y and I'm curious what happens"
    },
    {
        "Q": "\nAt 1:35, when we count do we always count starting from the right side if the rectangle?\nBecause if you see, when you count from the left side of the first blue rectangle, the total count is 9. But if i start counting from the right side of the firzt blue triangle it adds up to be 8.\nCan someone please explain this to me",
        "A": "There are 8 rectangles, counting either from the left or from the right. I guess you re counting the sides rather than the rectangles. By doing so, you ll get 9 walls . Try counting the upper side of each rectangle...",
        "video_name": "WeVWv_OEJsY",
        "timestamps": [
            95
        ],
        "3min_transcript": "- [Voiceover] The graph of F is shown below. A total of 24 right hand rectangles are shown. So, what do I mean by right hand rectangles? So, there's clearly 24 rectangles. You can count them. And right hand rectangle means that for each of these rectangles the height of the rectangle is defined by the value of the function on the right hand side of the rectangle. So you can see this is the right hand side of this first rectangle and if you take the value of the function of that point that is the height of the rectangle. A left hand rectangle would define the height of the rectangle by the value of the function on the left hand side So, a left handed rectangle's height, the first rectangle's height would look like that. That's what they mean by right handed rectangle Fair enough, eight in blue. We see that. 16 in red. All right. All 24 of the rectangles have the same width. Which of the statements below is or are true? They give us three expressions in sigma notation and they say, like this first one is the sum of the areas of the blue rectangles. the red rectangles. This is the sum of the areas of all of the rectangles. So, I encourage you now to pause the video and try to determine on your own, which of the statements is or are true. So, I assume you've had a go at it. Let's just go through each of these and see whether they make sense. So this first one, the sum of the areas of the blue rectangles. Well, we know we have one, two, three, four, five, six, seven, eight blue rectangles, and we're summing from one to eight. So it seems like we're summing eight things right over here. This is one, two, three, four, five, six, seven, eight. So, this is looking good right over here. And then we take F of something times one half. So, we're not even looking at this yet. It looks like this would be the height of each of the rectangles. Remember, we're taking the value of the function on the right hand side for the height, and this would be the width. So does it make sense that the width Well, the total distance between X equals negative five and X equals seven, that distance is 12. Five plus seven, that's 12, and we're dividing it into 24 rectangles of equal width. So, if you divide 12 divided by 24 each of these are going to have a width of one half. So this is checking out that the one half. Now let's think about this part. Let's think about the F of negative five plus I over two. So, let's see. When I is equal to one, so we're going to take one half times F of negative five plus one over two. Right? I is one. So negative five plus one half is going to get us to this point right over here. F of that is going to be this distance, this height right over here. This is consistent with these being right handed rectangles."
    },
    {
        "Q": "\nNewbie here, how do you decide which side to initially subtract? 1:13 he deducts the left-hand variable using the right-hand equation's first \"9X - 7X\". What is a rule of thumb for deciding which side you are supposed to use? Thanks!",
        "A": "It actually doesn t matter which side you subtract first - however, one way is easier then the other. When he subtracts 7x from 9x, he gets positive 2x. If he had subtracted 9x from 7x, he would ve gotten negative 2x, which is more confusing to work with. So, you can work out the equation and come to the same answer either way, but it is generally easier to work with positive numbers. Hope this helps!",
        "video_name": "2CZrkdtgeNU",
        "timestamps": [
            73
        ],
        "3min_transcript": "Let's say we have two intersecting lines. So that's one of the lines right over there. And then I have another line right over here. So those are my two intersecting lines. And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees, so it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. So if we want all the x-terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with-- let's see. 182 minus 194. So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller, so it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one. 182, so 7 times negative 6 is negative 42, plus 182 is going to be equal to 140 degrees. And you'll see the same thing over here. If we say 9 times negative 6, which is negative 54, plus 194, this also equals 140 degrees."
    },
    {
        "Q": "At 2:19 how in the world does it equal to 140? I get 248?\n",
        "A": "On 9x + 194, you made the mistake of adding, 54 to 194, correct? You should have done 9(-6) + 194 = -54 + 194 = 140 Instead you did 9(6) + 194 = 54 + 194 = 248 hope this helps (sorry I m a bit late)",
        "video_name": "2CZrkdtgeNU",
        "timestamps": [
            139
        ],
        "3min_transcript": "Let's say we have two intersecting lines. So that's one of the lines right over there. And then I have another line right over here. So those are my two intersecting lines. And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees, so it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. So if we want all the x-terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with-- let's see. 182 minus 194. So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller, so it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one. 182, so 7 times negative 6 is negative 42, plus 182 is going to be equal to 140 degrees. And you'll see the same thing over here. If we say 9 times negative 6, which is negative 54, plus 194, this also equals 140 degrees."
    },
    {
        "Q": "at 10:44 if we were just asked to write the quadratic function would you just subtract the 61/20 from the right side to get it equal to 0\n",
        "A": "If you wanted to set this up to use the quadratic equation, you would have just used the original problem given (10x\u00c2\u00b2 - 30x -8 = 0) and not bothered with all the complete the square stuff.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            644
        ],
        "3min_transcript": "got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just"
    },
    {
        "Q": "At 10:20 it seems as if Sal forgot to right 3/2x\nI may be worng\n",
        "A": "There is no 3/2 x. To complete the square, divide the coefficient of x by 2 and square it. It is just the coefficient that is divided by 2, not the whole middle term. Hope this helps.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            620
        ],
        "3min_transcript": "sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the"
    },
    {
        "Q": "At 8:54, why would you half the coefficient?\n",
        "A": "The easiest way to see why is to work backwards, expand (x+a)^2 to get (x+a)(x+a) using foil, you have x^2 + ax + ax + a^2 or x^2 + 2ax + a^2, so to complete the square, since b term is 2ax, we have to divide whatever coefficient is there by 2 so that 2ax/2 will leave us with ax. Hope this helps",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            534
        ],
        "3min_transcript": "we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it."
    },
    {
        "Q": "I'm confused about the part where he equates things at 01:59. If you have x2-4x+?=x2-2ax+a2\nshouldn't (-4x) correspond to (-2ax), meaning that a=2 (not -2)?\n\n(by \"x2\" I mean x squared)\n",
        "A": "You re right. The first time I watched this video I didn t notice the problem because I knew what Sal intended to say and assumed he actually said what he intended -- but there s a mistake here. What he meant to do was set up (x+a)^2 (not (x-a)^2) on the right. That s the normal way to explain the process of completing the square. When he makes the substitution you mention, he does it as if he had written (x+a)^2 on the right, which is what he meant to do.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            119
        ],
        "3min_transcript": "In this video, I'm going to show you a technique called completing the square. And what's neat about this is that this will work for any quadratic equation, and it's actually the basis for the And in the next video or the video after that I'll prove the quadratic formula using completing the square. But before we do that, we need to understand even what it's all about. And it really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. You see, completing the square is all about making the quadratic equation into a perfect square, engineering So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4"
    },
    {
        "Q": "I don't get it why does he do 5+4 instead of 9 ?\n\nat 3:00\n",
        "A": "It is 9, but he wants to show that he added 4 to both sides in order to complete the square.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            180
        ],
        "3min_transcript": "So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9."
    },
    {
        "Q": "Around 6:40, Sal divides the quadratic equation by 5. This process makes the coefficient of x^2 equal to 1. My question is does the coefficient of x^2 need to be 1 to complete the square.\n",
        "A": "no it doesn t have to. for example; 2x^2+18x+16 one can factor this by.. (x+8)(2x+2) but if you divide everthing by 2, you can make 2x^2+18x+16 to x^2+9x+8 then you can factor this to (x+8)(x+1) you see, this is the same as (x+8)(2x+1) but simpler. so to answer your question; it doesn t matter, but is s the matter of which one is simpler hope this helps :)",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            400
        ],
        "3min_transcript": "negative 1. And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about the completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey look, we could maybe divide both sides by 2. That does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get? But once again, now we have this crazy 5 in front of this coefficent and we would have to solve it by grouping which is a reasonably painful process. But we can now go straight to completing the square, and to do that I'm now going to divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go but I wanted to go to this the step first just to show you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4/5 is equal to 0. So, you might say, hey, why did we ever do that factoring we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way."
    },
    {
        "Q": "Where did Sal get the -2 two from at about 8:45?\n",
        "A": "I assume you are referring to where he wrote: a = -3/2 The -3 comes from the equation. It is the coefficient of X Dividing by 2 is part of the process of completing the square. Note: Sal used 2, not -2. The minus sign is from the -3. I suggest you rewatch the video. You will see every time Sal completes the square that he takes the coefficient of X and divides it by 2.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            525
        ],
        "3min_transcript": "we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it."
    },
    {
        "Q": "\nAround 0:47 he mentions a \"plain vanilla one for real numbers.\" Does that mean there is a square root for negative numbers?",
        "A": "There is a square root for negative numbers but it is not yet taught at this level of math.",
        "video_name": "4h54s7BBPpA",
        "timestamps": [
            47
        ],
        "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done."
    },
    {
        "Q": "At 0:34, why is the sign \"less than or equal to\"? Shouldn't it be just \"greater than\" because if x was 0, then it'd be the square root of -8?\n",
        "A": "Sal IS using the greater than or equal to sign ( not the less than or equal to sign). But he isn t saying that x can equal zero. He s saying that the quantity ( 2x - 8 ) can equal zero. Then he has to solve that inequality to finally find out that x must be GREATER than or equal to 4.",
        "video_name": "4h54s7BBPpA",
        "timestamps": [
            34
        ],
        "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done."
    },
    {
        "Q": "\nAt 2:00, why did you subtract 64 from the other side, where in other videos you added 64? What you do to one side you do to the other?",
        "A": "Sal doesn t have an equation, so there is no other side. He is working with an expression. To ensure he maintains the original value of the expression, he can only add a value that equal 0 (this is the identify property of addition). This is done by adding the 64 and then also subtracting the 64. 64 - 64 = 0. Thus, he hasn t the expression. He s just making it look different. Hope this helps.",
        "video_name": "sh-MP-dVhD4",
        "timestamps": [
            120
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square."
    },
    {
        "Q": "\nAt 3:09, Sal multiplied it by negative 1, but couldn't you just also subtract it with changing it in to a negative?",
        "A": "but the can also mess you up right if you interpret it wrong right?",
        "video_name": "-nlMXVrgtjw",
        "timestamps": [
            189
        ],
        "3min_transcript": "How did that work? The reason why we were able to mindlessly drop this 3 straight down is because we assumed, to do this basic type of synthetic division, that we had just an x right over here. We didn't have a 3x. We didn't have a 4x. We didn't have an x squared. We just had an x. And if you divide an x into whatever this highest degree term is, your coefficient is going to be the exact same thing. It's just going to be one degree lower. So you go from a 3x to the third to 3x squared-- so the exact same coefficient. And now, in our little synthetic division right over here, that right over there is the x squared term. So you went from 3x to the third to 3x squared. We essentially divided it by x. And we could blindly do it because we knew, we assumed, that we were dealing with just a 1x to the first right over here. and see why we're essentially doing the exact same thing. Now let's take this 3x squared and multiply that times x plus 4. So 3x squared times x-- I'll do it in that white color-- it's going to be 3x to the third. And 3x squared times 4 is going to be 12x squared. And now we'll want to subtract this. So now we subtract. We subtract this out. These guys cancel out, and you have 4x squared minus 12x squared. And so that will give you negative 8x squared. So once again, you're probably seeing some parallels. You had the 4x squared over here. You have the 4x squared over there. We just wrote the coefficient, but that's what it represented. 4x squared, we wrote the 4 there. Then we essentially subtracted 12x squared. And the way we got that 12, we multiplied 3 times 4, Here we're multiplying 3 times negative 4. We're essentially multiplying 3 times 4 and then subtracting. That's why we put that negative there, so we don't have to keep remembering to subtract this row. So we could just keep adding them. But that's essentially what we did. We multiplied 3 times this 4. And now we subtract it. We get that negative 12x squared. And then we subtracted, and you got negative 8x squared. And you might say, oh, is this the same negative 8 as this right over here? Not quite yet, because over here, this negative 8 literally represents negative 8x. This is actually part of our simplification. When we divide this into that, we got 3x squared minus 8x plus 30. So over here in the algebraic long division, we then say, how many times does x plus 4 go into negative 8x squared? Well x goes into negative 8x squared negative 8x times."
    },
    {
        "Q": "Can this work?\n\n1. First apply a rotation until the projected vector aligns with the vector we want to project to.\n2. Second scale the rotated vector accordingly.\n\nWill the matrix of product of the transformations in 1) and 2) be the same with the one Sal defined at at 13:01?\n",
        "A": "This would theoretically work. However, you also require an expression for the angle between every possible vector and the vector you want to project onto. By the time you figure that out and get a way to convert it back into a linear transformation, you may as well have done what Sal did.",
        "video_name": "JK-8XNIoAkI",
        "timestamps": [
            781
        ],
        "3min_transcript": "ready to take the projection of this guy. The definition of our projection is you dot this guy with our unit vector. So we dot it. We're taking the dot product of 0, 1. 0, 1 dot my unit vector dot u1, u2. I'm going to multiply that times my unit vector, times u1, u2. This seems very complicated, but it should simplify when we actually try to work out our transformation matrix. Let's do it. When I dot these two guys, what do I get. Let me write it here. My matrix A will become 1 times u1, plus 0 times u2. That's just u1. This whole thing just simplifies to u1, when I take the dot product of these two things. Times u1, u2. My second column, if I dot these two guys, I get 0 times u1 plus 1 times u2. So I'm going to get u2 times my unit vector, u1, u2. If I multiply that out, this will be equal to what. I can just write them as columns. u1 times u1 is u1 squared. u1 times u2 is u1, u2. u2 times u1 is u2 times u1. Then, u2 times u2 is u2 squared. You give me any unit vector and I will give you the transformation that gives you any projection of some other vector onto the line defined by that. That was a very long way of saying that. Let's go back to what I did before. Let's say we want to find any projection onto the line, onto the vector, I'll draw it here. We'll do the same example that we did in the last video. We said the vector v was equal to the vector 2, 1. That was my vector v. How can we find sum transformation for the projection onto the line defined by v? Onto this line right here. The line defined by v. What we can first do is convert v into a unit vector. We can convert v into a unit vector that goes in the same direction. Some unit vector u. We did that already up here. Where we essentially just divided [? bv ?] by it's length. Let's take v and divide by it's length. The unit vector is this, 1 over the square root of 5 times our vector v. It was 1 over the square root of 5 times our vector v, right there. You start with a unit vector there. You just create this matrix, and then we will have our"
    },
    {
        "Q": "\nAt 5:43, Sal says that dx/dy is the slope of the tangent line at any point. I didn't see where it was explained how/why this represents the slope prior to this point; it was just injected as a given into the explanation midstream.\n\nIs this explained elsewhere? How did we go from having d/dx represent the slope to having dx/dy represent it here?",
        "A": "I haven t watched the video but... d/dx is a differential operator. d/dx[4x] means take the derivative of the expression 4x with respect to x. If instead we write d/dx[f(x)] we would write df/dx as the derivative (or f (x)). The derivative represents the slope of the tangent line. dx means a very small change in x (infinitesimally small). and dy means a very small change in y. If you write dy/dx and y is your dependent variable then you have \u00e2\u0088\u0086y/\u00e2\u0088\u0086x which is the formula for slope.",
        "video_name": "mSVrqKZDRF4",
        "timestamps": [
            343
        ],
        "3min_transcript": "It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y, as a function of x. This might be, or y is a function of x squared, which is essentially another way of writing what we have here. This might be a little bit clearer in terms of the chain rule. The derivative of y is a function of x squared with respect to y of x. So the derivative of something squared with respect to that something, times the derivative of that something, with respect to x. This is just the chain rule. I want to say it over and over again. This is just the chain rule. So let's do that. What do we get on the right hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y, is just going to be 2 times y. And the derivative of y with respect to x? Well, we don't know what that is. So we're just going to leave that as times the derivative of y with respect to x. So let's just write this down over here. So we have is 2x plus the derivative of something squared, with respect to that something, is 2 times the something. In this case, the something is y, so 2 times y. And then times the derivative of y with respect to x. And this is all going to be equal to 0. Now that was interesting. Now we have an equation that has the derivative of a y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x. Solve this equation. And actually just so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract 2x from both sides. So we're left with 2y times the derivative of y, with respect to x, is equal to-- we're subtracting 2x from both sides-- so it's equal to negative 2x. And if we really want to solve for the derivative of y with respect to x, we can just divide both sides by 2y. And we're left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, well the 2s cancel out. We we're left with negative x over y. So this is interesting. We didn't have to us explicitly define y"
    },
    {
        "Q": "\nAt 2:24 why is 4 divided in to 36 and why does 4 become a 1 ?",
        "A": "it is because 36/4 equals 9",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            144
        ],
        "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5."
    },
    {
        "Q": "in Gulshan the ratio of private homes to apartment is 3:5.if all apartments are bricks made and 1/6 of the private homes are wooden, that is maximum portion of houses that may be brick...........??\n",
        "A": "3+5=8. so apartments are 5/8 of all houses.and all of them are brick made. 3/8 are private homes and 1-1/6 of them may be brick made(1/6 are wooden). that means the maximum of private houses from brick are 3/8 * 5/6=15/48=5/16. adding the apartments we get 5/8+5/16=15/16",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            185
        ],
        "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5."
    },
    {
        "Q": "At 0:07, what does he mean by \"mixed fraction?\"\n",
        "A": "no, this is a mixed fraction 5 1/2",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            7
        ],
        "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5."
    },
    {
        "Q": "\nAt 0:08 he meant improper fraction not mixed fraction right?",
        "A": "Hi Carolyn, He actually meant mixed fraction. You have to convert 2 mixed fractions to fractions, then multiply them, then convert the result (fraction) into mixed fraction again.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            8
        ],
        "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5."
    },
    {
        "Q": "At the end (+- 9:00) he says that one of the equations is impractical, because it would entail the calculation of all the pop, is that correct?\n",
        "A": "Exactly. It is pretty hard to calculate the heights of all the 150 million people, after all.",
        "video_name": "k5EbijWu-Ss",
        "timestamps": [
            540
        ],
        "3min_transcript": "They could call this x subscript 4. They could call this x subscript 5. And so if you had n of these you would just keep going. x subscript 6, x subscript seven, all the way to x subscript n. And so to take the sum of all of these, they would denote it as let me write it right over here. So they will say that the sample mean is equal to the sum of all my x sub i's-- so the way you can conceptualize it, these i's will change. In this case, the i started at 1. The i's are going to start at 1 until the size of our actual sample. So all the way until n. In this case n was equal to 5. So this is literally saying this is equal to x sub 1 plus x sub 2 plus x sub 3, all Once again, in this case, we only had five. Now, are we done? Is this what the sample mean is? Well, no, we aren't done. We don't just add up all of the data points. We then have to divide by the number of data points there are. So this might look like very fancy notation, but it's really just saying, add up your data points and divide by the number of data points you have. And this capital Greek letter sigma literally means sum. Sum all of the x i's, from x sub 1 all the way to x sub n, and then divide by the number of data points you have. Now let's think about how we would denote the same thing but, instead of for the sample mean, doing it for the population mean. So the population mean, they will denote it with mu, we already talked about that. but this time it's going to be the sum of all of the elements in your population. So your x sub i's-- and you'll still start at i equals 1. But it usually gets denoted that, hey you're taking the whole population, so they'll often put a capital N right over here to somehow denote that this is a bigger number than maybe this smaller n. But once again, we are not done. We have to divide by the number of data points that we are actually summing. And so this, once again, is the same thing as x sub 1 plus x sub 2 plus x sub 3-- all the way to x sub capital N, all of that divided by capital N. And once again, in this situation, we found this practical. We found this impractical. We can debate whether we took enough data points on our sample mean right over here. But we're hoping that it's at least somehow indicative of our population mean."
    },
    {
        "Q": "At 4:54 how it is 12. The equation is supposed to be 5-(- -7). The answer is supposed to be -2.\n",
        "A": "no. It s supposed to be 5-(-7), in which the minus symbol and the negative symbol cancel out each other, making the equation then 5+7, equalling 12.",
        "video_name": "XkRD9lv_y44",
        "timestamps": [
            294
        ],
        "3min_transcript": "So, actually first of all, I can figure out what three plus six is. Three plus six is equal to nine. And then I have this minus four here. So I could say nine minus four, actually I want to be careful not to skip any steps. So three plus six, three plus six, in a color that you can see, so three plus six is nine. So that's nine minus H minus four, minus four. Now I could change the order in which I do this addition or subtraction, so this is going to be the equivalent of nine minus four. Nine minus four minus H, minus H. And I just did that so I can simplify and figure out what nine minus four is. Nine minus four, of course, let me do this in blue, navy blue. Nine minus four is five. before I even did the substitution. And now I can substitute H with negative seven. So this is going to be equal to, this is going to be equal to, when I do the substitution, I'll write it up here, it's going to be five minus, I'll do the minus in that magenta color, minus and now where I see an H, I'm gonna replace it with negative seven. Five minus negative seven. You want to be very careful there, you might be tempted to say, oh I have a negative here, negative here, let me just replace H with a seven. Remember, H is negative seven, so you're subtracting H. You're gonna subtract negative seven. So this is five minus negative seven, which is the same thing, which is the same thing as five plus seven. Five plus seven, which we all know is equal to 12. And we're done. Let's do a few more of these. You can't really get enough practice here, this is some important foundational skills (laughing) Alright. So consider, I don't make you too stressed about it. Consider the following number line. Alright, so we've got a number line here and let's see, they didn't mark off all the numbers here. This is negative four, E is at this point, this is then we go to two. So it looks like we're counting by twos here, that this is negative two, this is zero, yep. Negative four, if you increase by two, negative four, negative two, zero, two, four, this would be a six. This would be a negative six. They intentionally left those numbers off, so we had to figure out that hey, look, between negative four and two, to go from negative four to two, you have to increase by six and we only have one, two, three hashmarks. So each of those hashmarks must be increasing by two. Well anyway, now we know, now we know what all the points in the number line are. Evaluate E minus F. Well we know, we know that E is equal to negative two. And we know, we know that F"
    },
    {
        "Q": "at around 1:17 I don't really understand how at least classic principal root is a defined for a negative number. Can someone please explain it to me?\n",
        "A": "Are you trying to find the square root of a negative number?",
        "video_name": "n17q8CBiMtQ",
        "timestamps": [
            77
        ],
        "3min_transcript": "Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So like always, pause this video and see if you can figure out what what is the domain of this function. Based on this function definition what is the domain of g? What is the set of all inputs for which this function is defined? Alright. So, to think about all of the inputs that would allow this function to be defined, it may be easier to state when is this function not get defined. Well if we divide by 0 then we're not going to be defined. Or if we have a negative under the radical. So if you think about it -- if if what we have under the radical 0, you can take the, you can take the principal root of 0. It's going to be 0. But then you are going to divide by 0. That's going to be undefined. And if what you have another radical is negative, the principal root isn't defined for native number, at least the classic principal root is a defined for a negative number. So if 6 minus the absolute value of x is zero or negative, this thing is not going to be defined. Or another way to think about it is it's going to be defined -- so g is is defined -- is defined if -- g is defined if 6 minus, maybe I can write if and only if. Sometimes people write if and only if with two f's right there, iff. g is defined if and only if -- this is kind of mathy way of saying if and only if -- 6 minus the absolute value of x is greater than 0 we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6."
    },
    {
        "Q": "\nHow did he get that a negative wouldn't work, at 0:56 & 1:15? If it was the absolute value of x, or whatever the number is, wouldn't the absolute value have canceled out the negative ( - ) and turned it to positive, so that it'd work? Also, How did he get -6 < x at about 3:00?? He had just said earlier that it COULDN'T be a negative number, but -6 < x < 6 implies that it could be anything from -5 to 5.",
        "A": "At 0:56 and 1:15 he s not saying that a negative value won t work for x, since you are correct -- a negative value would not affect anything since we are taking the absolute value of it anyway. Rather, he s saying that we can t have the expression in the radical -- 6-x -- be equal to a negative, since we would then be taking the square root of a negative of number. This translates to |x|<6 (or -6<x<6), since if it were greater than 6 the radicand would be negative and thus undefined.",
        "video_name": "n17q8CBiMtQ",
        "timestamps": [
            56,
            75,
            180
        ],
        "3min_transcript": "Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So like always, pause this video and see if you can figure out what what is the domain of this function. Based on this function definition what is the domain of g? What is the set of all inputs for which this function is defined? Alright. So, to think about all of the inputs that would allow this function to be defined, it may be easier to state when is this function not get defined. Well if we divide by 0 then we're not going to be defined. Or if we have a negative under the radical. So if you think about it -- if if what we have under the radical 0, you can take the, you can take the principal root of 0. It's going to be 0. But then you are going to divide by 0. That's going to be undefined. And if what you have another radical is negative, the principal root isn't defined for native number, at least the classic principal root is a defined for a negative number. So if 6 minus the absolute value of x is zero or negative, this thing is not going to be defined. Or another way to think about it is it's going to be defined -- so g is is defined -- is defined if -- g is defined if 6 minus, maybe I can write if and only if. Sometimes people write if and only if with two f's right there, iff. g is defined if and only if -- this is kind of mathy way of saying if and only if -- 6 minus the absolute value of x is greater than 0 we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6."
    },
    {
        "Q": "at 1:03 sal says that g(x) cannot be defined if the number under the radical is 0 or a negative. But can't it be a negative and make the equation g(x)= (6-|x|)I?\n",
        "A": "The equation is 1/sqrt(6-|x|), The reason 6-|x| cannot be 0 is because then the denominator of g(x) would be 0 and that is undefined. The reason 6-|x| cannot be negative is because then the argument of the square root would take on a negative number and this is undefined when we are only using the real numbers.",
        "video_name": "n17q8CBiMtQ",
        "timestamps": [
            63
        ],
        "3min_transcript": "Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So like always, pause this video and see if you can figure out what what is the domain of this function. Based on this function definition what is the domain of g? What is the set of all inputs for which this function is defined? Alright. So, to think about all of the inputs that would allow this function to be defined, it may be easier to state when is this function not get defined. Well if we divide by 0 then we're not going to be defined. Or if we have a negative under the radical. So if you think about it -- if if what we have under the radical 0, you can take the, you can take the principal root of 0. It's going to be 0. But then you are going to divide by 0. That's going to be undefined. And if what you have another radical is negative, the principal root isn't defined for native number, at least the classic principal root is a defined for a negative number. So if 6 minus the absolute value of x is zero or negative, this thing is not going to be defined. Or another way to think about it is it's going to be defined -- so g is is defined -- is defined if -- g is defined if 6 minus, maybe I can write if and only if. Sometimes people write if and only if with two f's right there, iff. g is defined if and only if -- this is kind of mathy way of saying if and only if -- 6 minus the absolute value of x is greater than 0 we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6."
    },
    {
        "Q": "\nWhy at 4:05 does he put in pi into the function as default for x being equal to 5?",
        "A": "Because he wants to. When defining a function, you can make it be anything. He could have put 5=6,903,576,912 if he wanted to. Hope this helps.",
        "video_name": "n17q8CBiMtQ",
        "timestamps": [
            245
        ],
        "3min_transcript": "we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6. we could write the domain of g is going to be x, all the x's that are a member of real numbers such that negative 6 is less than x, which is less than 6, and we're done. Let's do another one. And this was gonna get even a little bit a little bit hairier, just for, just for kicks. Alright. So let's say that I have -- let's say that I have -- h of x is equal to -- and I have a kind of composite definition here. So let's say it's x plus 10 over x plus 10 times x minus 9 times x minus 5 times x minus 5 and it's this if it -- h of x is this if x does not equal 5 and it's equal to -- it's equal to if x is equal to 5. So once again, pause this video. Think about what is the domain of h, or another way to think about is what made h not defined? So let's think about it. So what would make h not defined? So if some -- if x is anything other than 5, we go this clause. If it's 5, we go this clause. So in this clause up here, what would make this thing undefined? Well the most obvious thing is if we divide by 0. So what's going to cause us to divide by 0? So if -- if x is equal to if -- let me write it here -- so we're going to divide by 0 -- divide by 0. That would happen if x is equal to 9. That would happen if x is equal to negative 10. x is equal to negative 10. Now we have to be careful, would that happen"
    },
    {
        "Q": "Around 4:40 Sal writes down that the result of 2(0.25*AC*BD) is 0.5*AC*BD. Why is that? Personally I feel 2(0.25*AC*BD) should be equal to 0.5*2AC*2BD, or 2*AC*BD.\n",
        "A": "The distributive property distributed multiplication across ADDITION. a(b + c) = ab + ac OR a(b - c) = ab - ac. You re trying to apply it to distributed multiplication across multiplication. There is no such thing. For example, if you were asked to multiply: 2 (3 * 5 * 6), you multiply 2 numbers at a time until everything gets multiplied. 2 (3 * 5 * 6) = 2(15 * 6) = 2(90) = 180 2(0.25*AC*BD) is the same problem. Clean up inside the parentheses. 2 (0.25ACBD) = 0.5ACBD or 0.5ABCD. Hope this helps.",
        "video_name": "3FManXv4mZM",
        "timestamps": [
            280
        ],
        "3min_transcript": "What is the height here? Well we know that this diagonal right over here, that it's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE, that is the height. This is an altitude. It intersects this base at a 90-degree angle. Or we could say BE is the same thing as 1/2 times BD. So this is-- let me write it. This is equal to, so it's equal to 1/2 times AC, that's our base. And then our height is BE, which we're saying is the same thing as 1/2 times BD. So that's the area of just ABC, that's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we decided that the area of the whole thing So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC, where the area of ABC is this thing right over here. So 2 times the area of ABC, area of ABC is that right over there. So 1/2 times 1/2 is 1/4 times AC times BD. And then you see where this is going. 2 times 1/4 fourth is 1/2 times AC times BD. Fairly straightforward, which is a neat result. And actually, I haven't done this in a video. I'll do it in the next video. There are other ways of finding the areas of parallelograms, It's essentially base times height, but for a rhombus we could do that because it is a parallelogram, but we also have this other neat little result that we proved in this video. That if we know the lengths of the diagonals, the area of the rhombus is 1/2 times the products"
    },
    {
        "Q": "\nAt 1:43, what does SSS mean or 'side side side'?",
        "A": "SSS (side side side) is one way to prove that two triangles are congruent, if three sides of one triangle are congruent to corresponding sides of another triangle, then the triangles are congruent Sal is using this intermediate step to prove the area of a rhombus",
        "video_name": "3FManXv4mZM",
        "timestamps": [
            103
        ],
        "3min_transcript": "So quadrilateral ABCD, they're telling us it is a rhombus, and what we need to do, we need to prove that the area of this rhombus is equal to 1/2 times AC times BD. So we're essentially proving that the area of a rhombus is 1/2 times the product of the lengths of its diagonals. So let's see what we can do over here. So there's a bunch of things we know about rhombi and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know the diagonals bisect each other. So we know that this length-- let me call this point over here B, let's call this E. We know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know, because this is a rhombus, and we proved this in the last video, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward. So we can see that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side. This side is congruent to that side, and they both share a C right over here. So this is by side-side-side. And so we can say that the area-- so because of that, we know that the area of ABCD is just we could pick either one of these. We could say 2 times the area of ABC. Because area of ABCD-- actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well area of a triangle is just 1/2 base times height. So area of ABC is just equal to 1/2 times the base of that triangle times its height, which is equal to 1/2. What is the length of the base? Well the length of the base is AC. So it's 1/2-- I'll color code it. The base is AC."
    },
    {
        "Q": "\ni dont get how it simplifies to one after 4:33",
        "A": "Rewatch the firsst 10 seconds. Sal explains that as you go to higher and higher powers, i cycles through 1,i,-1,-i. Every 4th power it is back at 1, so i^7320 is going to be 1.",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            273
        ],
        "3min_transcript": "So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? So this is the same thing as i to the 96th power times i to the third power, right? If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the fourth, and then that to the 16th power. So that's just 1 to the 16th, so this is just 1. And then you're just left with i to the third power. And you could either remember that i to the third power is equal to-- you can just remember that it's equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more just for the fun of it. Let's take i to the 38th power."
    },
    {
        "Q": "At 4:56,can't we write i^99 as i^100/i\nthen the answer is i^-1\n\n\n^ denotes exponential power\n",
        "A": "MY method is based on the cycle of real numbers +1 & -1. Since i^2 = -1 and i^4 = +1 Simply choose the EVEN exponent before. \u00e2\u0080\u00a2If an EVEN exponent IS divisible by 4, then => +1 BUT If NOT => -1. \u00e2\u0080\u00a2Secondly, any ODD exponent is either -i or +i. i^99 IS i^98 * i => 98 IS NOT divisible by 4 so => -i",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            296
        ],
        "3min_transcript": "So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? So this is the same thing as i to the 96th power times i to the third power, right? If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the fourth, and then that to the 16th power. So that's just 1 to the 16th, so this is just 1. And then you're just left with i to the third power. And you could either remember that i to the third power is equal to-- you can just remember that it's equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more just for the fun of it. Let's take i to the 38th power."
    },
    {
        "Q": "\nAt 4:10 sal wrote i^i ?",
        "A": "He wrote i^i but said i^1, so he meant i^1.",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            250
        ],
        "3min_transcript": "as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99?"
    },
    {
        "Q": "At 4:09, Sal wrote i^i. Just out of curiosity, what is i^i?\n",
        "A": "Euler s formula says that e^(ix)=cos(x)+isin(x) Let x=\u00cf\u0080/2 and we get e^(i\u00cf\u0080/2)=cos(\u00cf\u0080/2)+isin(\u00cf\u0080/2) e^(i\u00cf\u0080/2)=0+i e^(i\u00cf\u0080/2)=i Now raise both sides to the power of i and we get [e^(i\u00cf\u0080/2)]^i=i^i e^(i\u00c2\u00b2\u00cf\u0080/2)=i^i e^(-\u00cf\u0080/2)=i^i",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            249
        ],
        "3min_transcript": "as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99?"
    },
    {
        "Q": "\nAt 1:35-\n5 is also a multiple of 100, so why would i^100 =1 and not i?",
        "A": "i^100 can be written as (i^4)^25 which is equal to 1^25 = 1. For what you re asking - even though 5 is a factor of 100 (not multiple) - we could take it like this: i^100 = (i^5)^20 which can written again as (i^5)^4 = i^4 = 1.",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            95
        ],
        "3min_transcript": "Now that we've seen that as we take i to higher and higher powers, it cycles between 1, i, negative 1, negative i, then back to 1, i, negative 1, and negative i. I want to see if we can tackle some, I guess you could call them, trickier problems. And you might see these surface. And they're also kind of fun to do to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on a back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, as i to the fourth power raised to the 25th power. If you have something raised to an exponent, and then that is raised to an exponent, that's the same thing as multiplying the two exponents. that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k"
    },
    {
        "Q": "\nAt 3:02, Sal mentions that K \u00e2\u0089\u00a5 0. I understand that for now K can't be a negative number, but can it be a positive decimal? For example, is it possible to find and answer to i to the 7.89th power? (It is between the 7th and 8th power, so it couldn't follow any of the principles making the value either 1, i, -1, or -i.)",
        "A": "You can surely raise any number to a decimal.. I ll tell you what that means... You can convert any decimal number into a fractional number very easily and when you do that understanding the power of decimals becomes meaningful. For eg. If u r saying 2^1.5, you can write it as 2^3/2 which means it is square root of 2^3.. Right? Similarly you can convert any decimal power into a fractional number and understand its meaning...",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            182
        ],
        "3min_transcript": "that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over."
    },
    {
        "Q": "At 9:27 , how did you come up with how many songs and game he would buy?\n",
        "A": "Since the shaded area (the darker blueish color) is the solution to both equations (the first being the one about how much money he has and the second being how many songs and games he can buy) any coordinates in that region would satisfy the problem. So 4 games and 14 songs would work, as Sal mentioned. Obviously, the two values need to be integers because you can never buy a half game or song. Hope that helps. : )",
        "video_name": "BUmLw5m6F9s",
        "timestamps": [
            567
        ],
        "3min_transcript": "Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy"
    },
    {
        "Q": "\nAt 9:33, Sal says that if you buy 4 games and 14 songs that would work. Sure, on the equation it makes sense but on the graph it looks like it's an impossible point. Or is it just the line....I don't know.",
        "A": "no it seems like it could work on the graph too. its just that there is not 14 mark on his graph. but we know 14 is halfway between 12 and 16 so he put his point there.... Hope This Helps =)",
        "video_name": "BUmLw5m6F9s",
        "timestamps": [
            573
        ],
        "3min_transcript": "Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy"
    },
    {
        "Q": "Hi,\nI am with you until 2:14 in the video,\nWhere pi \"f\" 1/u du,\nThen that is = to pi ln u (but what happens to the du?)...+C\nAs i understand it there is two term after the \"f\" (1/u) x (du),\nRegards\nPs: thank you very much for all the math help :)\n",
        "A": "When you integrate with respect to u, the du disappears, just as dx disappears when you integrate with respect to x. For example, the integral of 2x dx is x\u00c2\u00b2 (not x\u00c2\u00b2 dx). There are different ways to think about why this happens. One is to consider dx or du as part of the integral symbol. It s probably more helpful, especially when doing u-substitution, to understand dx or du as a distinct element, but one that gets incorporated into the answer when you integrate.",
        "video_name": "OLO64d4Y1qI",
        "timestamps": [
            134
        ],
        "3min_transcript": "We are faced with a fairly daunting-looking indefinite integral of pi over x natural log of x dx. Now, what can we do to address this? Is u-substitution a possibility here? Well for u-substitution, we want to look for an expression and its derivative. Well, what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? du is going to be the derivative of the natural log of x with respect to x, which is just 1/x dx. This is an equivalent statement to saying that du dx is equal to 1/x. So do we see a 1/x dx anywhere in this original expression? Well, it's kind of hiding. It's not so obvious, but this x in the denominator is essentially a 1/x. And then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi. since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here."
    },
    {
        "Q": "\nI get all that happened, but wouldn't the answer at 2:48 be \u00cf\u0080(ln |lnx| + C) = \u00cf\u0080(ln |lnx|) + \u00cf\u0080C instead of \u00cf\u0080(ln|lnx|) + C?",
        "A": "It could be, but the two C s would have different values. Imagine if the pi had been left inside the integral--that s the way Sal is handling it. Typically we choose the constant to be the thing added on after computing the integral as a whole rather than at an intermediate step.",
        "video_name": "OLO64d4Y1qI",
        "timestamps": [
            168
        ],
        "3min_transcript": "since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here. this original expression was only defined for positive values of x because you had to take the natural log here, and it wasn't an absolute value. So we can leave this as just a natural log of x, but this also works for the situations now because we're doing the absolute value of that where the natural log of x might have been a negative number. For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression."
    },
    {
        "Q": "\nAt 5:33, Sal said that the area of the parallelogram is 5x6, whereas before, following the previously spoken format it would be 2x5x6. Is this right?",
        "A": "Yes! Correct! 2x5x6 is pretty much the same thing as what sal previously said.",
        "video_name": "tFhBAeZVTMw",
        "timestamps": [
            333
        ],
        "3min_transcript": "We could call that the height right over there. So if you want the total area of parallelogram ABCD, it is equal to two times one half times base times height. Well, two times one half is just one. And so you're just left with base times height. So we can call this b. So it's just b times this height over here, base times height. So that's a neat result. And you might have guessed that this would be the case. But if you want to find the area of any parallelogram, and if you can figure out the height, it is literally you just take one of the bases, because both bases are going to be the-- opposite sides are equal, so it could have been either that side or that side, times the height. So that's one way you could have found the area. Or you could have multiplied. The other way to think about it is you could have multiplied. So if I were to turn this parallelogram over, it would look something like this. It would look something like this. So if I were to rotate it like that, let me draw the points-- this would be point A. Let me make sure I'm doing this right. This would be point A. This would be point D. This would be point C. And then this would be point B. You could also do it this way. You could say it's 1-- sorry, not 1/2. That would be for a triangle. The area of this would be base times height. So you could say it's h times DC. So you could say this is going to be equal to h times the length DC. That's one way to do it. That's this base times this height. Or you could say it's equal to AD times, I'll call this altitude right here, I'll call this height 2. Times h2. Maybe I'll call this h1. h1, h2. So you could take this base times this height, or you could take this base times This is h2. Either way. So if someone were to give you a parallelogram, just to make things clear, obviously, you'd have to be have some way to be able to figure out the height. So if someone were to give you a parallelogram like this, they would tell you this is a parallelogram. If they were to tell you that this length right over here is 5, and if they were to tell you that this distance is 6, then the area of this parallelogram would literally be 5 times 6. I drew the altitude outside of the parallelogram. I could have drawn it right over here as well. That would also be 6. So the area of this parallelogram would be 30."
    },
    {
        "Q": "\nat 1:22 doesn't that sign mean delta as well?",
        "A": "You re right, it s capital delta (at least it looks like it). Here it just means triangle of course.",
        "video_name": "tFhBAeZVTMw",
        "timestamps": [
            82
        ],
        "3min_transcript": "We know that quadrilateral ABCD over here is a parallelogram. And what I want to discuss in this video is a general way of finding the area of a parallelogram. In the last video, we talked about a particular way of finding the area of a rhombus. You could take half the product of its diagonals. And a rhombus is a parallelogram. But you can't just generally take half the product of the diagonals of any parallelogram. It has to be a rhombus. And now we're just going to talk about parallelograms. So what do we know about parallelograms? Well, we know the opposite sides are parallel. So that side is parallel to that side, and this side is parallel to this side. And we also know that opposite sides are congruent. So this length is equal to this length, and this length is equal to this length over here. Now, if we draw a diagonal-- I'll draw a diagonal AC-- we can split our parallelogram into two triangles. And we've proven this multiple times. These two triangles are congruent. But we can do it in a pretty straightforward way. Obviously, AD is equal to BC. We have DC is equal to AB. And then both of these triangles share this third side right over here. They both share AC. So we can say triangle-- let me write this in yellow-- we could say triangle ADC is congruent to triangle-- let me get this right. So it's going to be congruent to triangle-- so I said A, D, C. So I went along this double magenta slash first, then the pink, and then I went D, and then I went the last one. So I'm going to say CBA. Because I went the double magenta, then pink, then the last one. So CBA, triangle CBA. And this is by side, side, side congruency. All three sides, they have three corresponding sides that are congruent to each other. So the triangles are congruent to each other. And what that tells us is that the areas of these two So if I want to find the area of ABCD, the whole parallelogram, it's going to be equal to the area of triangle-- let me just write it here-- it's equal to the area of ADC plus the area of CBA. But the area of CBA is just the same thing as the area of ADC because they are congruent, by side, side, side. So this is just going to be two times the area of triangle ADC. Which is convenient for us, because we know how to find the areas of triangles. Area of triangles is literally just 1/2 times base times height. So it's 1/2 times base times height of this triangle. And we are given the base of ADC. It is this length right over here. It is DC. You could view it as the base of the entire parallelogram. And if we wanted to figure out the height, we could draw an altitude down like this."
    },
    {
        "Q": "At 0:54, Khan says to factor out the 2. Is that necessary?\n",
        "A": "We factor out 2 to bring expressions in the brackets to similar form. Thus, you can easily see that parts inside the brackets are similar on both sides of the equation. In this form it is quite easy to identify and make a substitution. You don t have to do this step if you can identify similarities right away, remember (4x-6)=2(2x-3).",
        "video_name": "ZIqW_sXymrM",
        "timestamps": [
            54
        ],
        "3min_transcript": "- [Voiceover] So let's try to find the solutions to this equation right over here. We have the quantity two X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well let's look at this, we have two X minus three squared on the left-hand side, on the right-hand side we have four X minus six. Well four X minus six, that's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. And so this is really interesting, we have something squared is equal to two times that something. let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So let's say, let's just replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. And now we just have to solve for P. And I'll switch to just one color now. So we can write this as, if we subtract two P from both sides, we can get P squared minus two P and we can factor out a P, so we get P times P minus two is equal to zero. And we've seen this shown multiple times, if I have the product of two things and they equal to zero, at least one of them needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going to be equal to this P value, is going to be equal to zero, or two X minus three is going to be equal to this P value, is going to be equal to two. And so this is pretty straightforward to solve,"
    },
    {
        "Q": "\nAt 3:48 of the video why does he want to multiply and take 1/3 ?",
        "A": "He s trying to get the x by itself on one side of the equation and as long as you do the same thing to both sides then everything equals out. So to get from 3x to x you divide by 3 which is the same as multiplying it by 1/3. Since want to do the same thing to both sides you multiply 12 by 1/3 (which is the same as dividing it by 3). 12 * 1/3 = 12/3 = 4. So x=4.",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            228
        ],
        "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced."
    },
    {
        "Q": "\nAt 4:11, why does he multiply by 1/3? Wouldn't it be easier to divide by 3?",
        "A": "Multiplying with 1/3 is the same with dividing by 3.",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            251
        ],
        "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced."
    },
    {
        "Q": "I just noticed at 4:26, there is 6 yellow blocks on the right when there's supposed to be 4 blocks on the right.\n",
        "A": "The video has amended this. If you watch it again, you should notice the box in the lower right corner.",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            266
        ],
        "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced."
    },
    {
        "Q": "at 4:23 There were 6 boxes on the right side. WAs that on purpose or what.\n",
        "A": "that was an accident",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            263
        ],
        "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced."
    },
    {
        "Q": "\n@4:26 he left 6 on the right did anyone else notice that?",
        "A": "Yeah. I noticed it.",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            266
        ],
        "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced."
    },
    {
        "Q": "Any one else notice Sal repeats himself a lot? 2:40 to 2:53\n",
        "A": "Repetition is the mother of learning :)",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            160,
            173
        ],
        "3min_transcript": "We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these."
    },
    {
        "Q": "\nFor 1:40 of the video, how do you know if all three are 30 degrees? How do you find those angles if they were different angles? If they weren't the 30, 60, 90, then how would you find the angles? I get the 90 degrees part, but what about the rest?",
        "A": "The reason that Sal knows those angles all equal 30\u00c2\u00b0 is because the problem states that the 90\u00c2\u00b0 angle is trisected or split into 3 equal parts. So 90\u00c2\u00b0/3=30\u00c2\u00b0.",
        "video_name": "dgHksfBFbjk",
        "timestamps": [
            100
        ],
        "3min_transcript": "So we have this rectangle right over here, and we're told that the length of AB is equal to 1. So that's labeled right over there. AB is equal to 1. And then they tell us that BE and BD trisect angle ABC. So BE and BD trisect angle ABC. So trisect means dividing it into 3 equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is, what is the perimeter of triangle BED? So it's kind of this middle triangle in the rectangle right over here. So at first this seems like a pretty hard problem, because you're like well, what is the width of this rectangle. How can I even start on this? They've only given us one side here. But they've actually given us a lot of information, given that we do know this is a rectangle. We have four sides, and that we have four angles. The sides are all parallel to each other and that the angles are all 90 degrees. Which is more than enough information And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now we know what the measure of this angle is. It was a right angle, it was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees, this angle right over here is 30 degrees, and then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90, so this other side right over here needs to be 60 degrees. This triangle right over here, you have 30, you have 90, so this one has to be 60 degrees. They have to add up to 180, 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, So for example, here we have the shortest side. We have the side opposite of the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If that was a little bit mysterious, how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So this is 1, this is a 30 degree side, this is going to be square root of 3 times that. And the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times this side right over here. So 2 times 1 is just 2. So that's pretty interesting."
    },
    {
        "Q": "I understand this topic until the last part. I thought that the opposite of the 60 degree angle was supposed to be sqrt(3) and that the side opposite the 90 degree angle is 2 times the number. I don't understand how he got 2/sqrt(3) and 1/sqrt(3). I become confused around 3:16.\n",
        "A": "Yes, you multiply the short side by sqrt(3) to find the side opposite the 60\u00c2\u00b0 angle. This problem is different because they gave the you side opposite the 60\u00c2\u00b0 angle and you need to find the short side. So essentially you are saying that the short side times sqrt(3) = 1; therefore to find the length of the short side you have to divide by sqrt(3) to find the short side. This gives you the 1/sqrt(3) for the short side and you multiply this by 2/sqrt(3).",
        "video_name": "dgHksfBFbjk",
        "timestamps": [
            196
        ],
        "3min_transcript": "And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now we know what the measure of this angle is. It was a right angle, it was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees, this angle right over here is 30 degrees, and then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90, so this other side right over here needs to be 60 degrees. This triangle right over here, you have 30, you have 90, so this one has to be 60 degrees. They have to add up to 180, 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, So for example, here we have the shortest side. We have the side opposite of the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If that was a little bit mysterious, how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So this is 1, this is a 30 degree side, this is going to be square root of 3 times that. And the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times this side right over here. So 2 times 1 is just 2. So that's pretty interesting. with this side right over here. Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over This is the 60, remember this 1, this is the 60 degree side. So this has to be 1 square root of 3 of this side. Let me write this down, 1 over the square root of 3. And the whole reason, the way I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3 I should get the shortest side, the 30 degree side. So if I start with the 60 degree side, divide by the square root of 3, I get that right over there. And then the hypotenuse is always"
    },
    {
        "Q": "How did Sal get 3sqrt(3) from just sqrt(3) at 6:05. Can someone please explain and link the appropriate khan lesson for that.\n",
        "A": "Sal actually got 3sqrt(3)/3 because he multiplied both the numerator and denominator by 3.",
        "video_name": "dgHksfBFbjk",
        "timestamps": [
            365
        ],
        "3min_transcript": "So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2 over the square root of 3. So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3. And then all we have to really figure out is, what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3. If that entire length is square root of 3, if this AE is 1 over the square root of 3, then this length right over here, ED is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there. And how to find the perimeter is pretty straight forward. We just have to add these things up and simplify it. this, perimeter of triangle BED is equal to-- This is short for perimeter. I just didn't feel like writing the whole word.-- is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal approximation for it. Let's see, if we have 2 square root of 3 minus 1 square root of 3, that will leave us with 1 over the square root of 3. 2 over the square of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us-- this is the drum roll part now-- so one square root of 3 plus 3 square roots of 3, and all of that over 3, gives us 4 square roots of 3 over 3 plus 2. Or you could put the 2 first. Some people like to write the non-irrational part before the irrational part. But we're done. We figured out the perimeter. We figured out the perimeter of this inner triangle BED, right there."
    },
    {
        "Q": "at 3:59, why do we have to subtract 6100 from 400 & 6100?\n",
        "A": "Yes, we are isolating the cosine of theta (at 3:59) . You are right.",
        "video_name": "Ei54NnQ0FKs",
        "timestamps": [
            239
        ],
        "3min_transcript": "one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100."
    },
    {
        "Q": "\nI have a question about when you are doing the equation and you are crossing out and everything when you was getting on the part when you add on one side and then you subtracted on the other why did you do that? I have always been taught that what you do to one side that you do to the other can you please explain that to me? I do believe it is on minute 5:04 Thank you for the help in advance",
        "A": "I don t really understand what you are asking, but I think I have an idea. When the equation was -5700 = -6000 cos (Theta) It also meant -5700 = -6000 times cos (Theta). He divided -6000 from the right, so he did the same to the left. It equaled -5700/-6000 on the left, which is the same as 5700/6000. I think you were just confused by the way he drew the positive signs. I hope I helped! Good luck!",
        "video_name": "Ei54NnQ0FKs",
        "timestamps": [
            304
        ],
        "3min_transcript": "And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20."
    },
    {
        "Q": "\nAt 2:14, Sal says,\"The best way to get 12x on the left is to subtract.\" Was he implying that there's more than one way to balance the equation? If so, how?",
        "A": "It was just a figure of speech. The only way to move that 12x from one side to the other is to subtract it. However, you could add 9x to both sides instead. Your x s will end up on the right hand side instead of the left but that s OK. x = -1 means the exact same thing as -1 = x",
        "video_name": "YZBStgZGyDY",
        "timestamps": [
            134
        ],
        "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best"
    },
    {
        "Q": "At 2:12, Sal says that -12x is the best way to do it. I thought that was the only way- have I been missing something?\n",
        "A": "You could also add 9x to both sides.",
        "video_name": "YZBStgZGyDY",
        "timestamps": [
            132
        ],
        "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best"
    },
    {
        "Q": "At 6:21 how come pi as a coefficient is allowed? Isn't pi an irrational number?\n",
        "A": "Coefficients can be any real number including Pi.",
        "video_name": "Vm7H0VTlIco",
        "timestamps": [
            381
        ],
        "3min_transcript": "So, this right over here is a coefficient. It can be, if we're dealing... Well, I don't wanna get too technical. Positive, negative number. Could be any real number. We have our variable. And then the exponent, here, has to be nonnegative. Nonnegative integer. So here, the reason why what I wrote in red is not a polynomial is because here I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So, if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the one half power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer; it's one half. the square root of a minus five. This also would not be a polynomial. Or, if I were to write nine a to the a power minus five, also not a polynomial because here the exponent is a variable; it's not a nonnegative integer. All of these are examples of polynomials. There's a few more pieces of terminology that are valuable to know. Polynomial is a general term for one of these expression that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form. But there's more specific terms for when you have only one term or two terms or three terms. When you have one term, it's called a monomial. This is a monomial. which we could write as six x to the zero. Another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi. Pi. Whoops. Could be pi. So we could write pi times b to the fifth power. Any of these would be monomials. So what's a binomial? Binomial's where you have two terms. Monomial, mono for one, one term. Binomial is you have two terms. This right over here is a binomial. Binomial. You have two terms. All these are polynomials but these are subclassifications. It's a binomial; you have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here."
    },
    {
        "Q": "\nAt 3:45, Sal says that he could change it to x to the power of zero. Is this incorrect ? Anything to the power of zero is one, and 9x is not always equal to nine.",
        "A": "9x^0 does not be come 9x. 9x means 9x^1 Having x to the 1st power is not the same as x^0 power. x^0 = 1, not x 9x^0 = 9(1) = 9 Hope this helps.",
        "video_name": "Vm7H0VTlIco",
        "timestamps": [
            225
        ],
        "3min_transcript": "for what makes something a polynomial. You have to have nonnegative powers of your variable in each of the terms. I just used that word, terms, so lemme explain it, 'cause it'll help me explain what a polynomial is. A polynomial is something that is made up of a sum of terms. And so, for example, in this first polynomial, the first term is 10x to the seventh; the second term is negative nine x squared; the next term is 15x to the third; and then the last term, maybe you could say the fourth term, is nine. You can see something. Let me underline these. These are all terms. This is a four-term polynomial right over here. You could say: \"Hey, wait, this thing you wrote in red, \"this also has four terms.\" We have to put a few more rules for it to officially be a polynomial, especially a polynomial in one variable. Each of those terms are going to be made up of a coefficient. This is the thing that multiplies So in this first term the coefficient is 10. Lemme write this word down, coefficient. It's another fancy word, but it's just a thing that's multiplied, in this case, times the variable, which is x to seventh power. The first coefficient is 10. The next coefficient. Actually, lemme be careful here, because the second coefficient here is negative nine. We are looking at coefficients. The third coefficient here is 15. You can view this fourth term, or this fourth number, as the coefficient because this could be rewritten as, instead of just writing as nine, you could write it as nine x to the zero power. And then it looks a little bit clearer, like a coefficient. So, in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter A, being multiplied by a variable So, this right over here is a coefficient. It can be, if we're dealing... Well, I don't wanna get too technical. Positive, negative number. Could be any real number. We have our variable. And then the exponent, here, has to be nonnegative. Nonnegative integer. So here, the reason why what I wrote in red is not a polynomial is because here I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So, if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the one half power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer; it's one half."
    },
    {
        "Q": "\nAt 8:15 Sal mentions that the highest degree will be the degree of the polynomial. What if there was an equation where there were two terms and they both had the same degree? What will the degree of the polynomial be?",
        "A": "The degree of the entire polynomial would just be that degree",
        "video_name": "Vm7H0VTlIco",
        "timestamps": [
            495
        ],
        "3min_transcript": "which we could write as six x to the zero. Another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi. Pi. Whoops. Could be pi. So we could write pi times b to the fifth power. Any of these would be monomials. So what's a binomial? Binomial's where you have two terms. Monomial, mono for one, one term. Binomial is you have two terms. This right over here is a binomial. Binomial. You have two terms. All these are polynomials but these are subclassifications. It's a binomial; you have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here. Trinomial's when you have three terms. Trinomial. This right over here is an example. This is the first term; this is the second term; and this is the third term. Now, the next word that you will hear often in the context with polynomials is the notion of the degree of a polynomial. You might hear people say: \"What is the degree of a polynomial?\", or \"What is the degree of a given term of a polynomial?\" Let's start with the degree of a given term. Let's go to this polynomial here. We have this first term, 10x to the seventh. The degree is the power that we're raising the variable to. So this is a seventh-degree term. The second term is a second-degree term. The third term is a third-degree term. And you could view this constant term, which is really just nine, you could view that as, sometimes people say the constant term. If people are talking about the degree of the entire polynomial, they're gonna say: \"What is the degree of the highest term? \"What is the term with the highest degree?\" That degree will be the degree of the entire polynomial. So, this first polynomial, this is a seventh-degree polynomial. This one right over here is a second-degree polynomial because it has a second-degree term and that's the highest-degree term. This right over here is a third-degree. You could even say third-degree binomial because its highest-degree term has degree three. If this said five y to the seventh instead of five y, then it would be a seventh-degree binomial. This right over here is a 15th-degree monomial. This is a second-degree trinomial. A few more things I will introduce you to is the idea of a leading term and a leading coefficient."
    },
    {
        "Q": "At 1:40 why did he multiply 4 * 0 ?\n",
        "A": "In the original equation it was 4*Y and since he was solving for the X intercept, Y would be equal to 0 and so 4*0",
        "video_name": "xGmef7lFc5w",
        "timestamps": [
            100
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 1:40 what is the somthing ?\n",
        "A": "The something is x , what you re solving for.",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            100
        ],
        "3min_transcript": "Let's do a few more examples of solving equations. And I think you're going to see that these equations require a few more steps than the ones we did in the last video. But the fun thing about these is that there's more than one way to do it. But as long as you do legitimate steps, as long as anything you do to the left-hand side, you also do to the right-hand side, you should move in the correct direction, or you shouldn't get the wrong answer. So let's do a couple of these. So the first one says-- I'll rewrite it-- 1.3 times x minus 0.7 times x is equal to 12. Well, here the first thing that my instinct is to do, is to merge these two terms. Because I have 1.3 of something minus 0.7 of that same something. This is the same variable. If I have 1.3 apples minus 0.7 apples, well, why don't I subtract 0.7 from 1.3? And I will get 1.3 minus 0.7 x's, or apples, or whatever So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12."
    },
    {
        "Q": "\nAt 12:34 Sal said ten minus -2.35 but he had a plus sign next to -2.35? Im confused, can someone help me?",
        "A": "adding a negative number and subtracting a positive number are the same thing :)",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            754
        ],
        "3min_transcript": "So the total cost of a cab ride is going to be equal to just the initial hire charge, which is $2.35, plus the $0.75 per mile, times the number of miles. We're letting m is equal to the miles she travels. Miles traveled. So this is the equation. We know that she only has $10 to get home. So her cost has to be $10. So we have to say, the cost has to be $10. So 10 is equal to 2.35 plus 0.75m. So how do we solve for m, or the number of miles Jade can travel? Well, we can get rid of the 2.35 on this right-hand side by subtracting that amount from both So let's do that. So let's subtract minus 2.35 from both sides. These will cancel out. That was the point. The left-hand side-- what is 10 minus 2.35? Now, these will cancel out. Now what is 10 minus 2.35? 10 minus 2 is 8. 10 minus 2.3 is 7.7. So it's going to be 7.65. If you want to believe me, let's do it. 10 minus 2.35. 7.65. And that is going to be equal to 0.75m. Let me write that in that same color. It's nice to see where different things came from. I have, like, five shades of this purple here. so this is that, that is that, and then these two guys canceled out. Now to solve for m, I can just divide both sides by 0.75. So if I divide that side by 0.75, I have to do it to the left-hand side as well. 0.75. That cancels out, so on the right-hand side, I'm left with just an m. And on the left-hand side-- I'll have to get my calculator out for this one-- I have 7.65 divided by 0.75, which is equal to 10.2. m is 10.2, so Jade can travel 10.2 miles."
    },
    {
        "Q": "\nin 8:18 he say 7 can be written 35/5 , how is that?",
        "A": "35 divided by five equals seven, he was simply just showing another way to write whole numbers. But by using fractions! Like 25/5 equals 5, so that s a way of representing five. In my mind, it looks like you know pretty cool things then you put this stuff on a test!",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            498
        ],
        "3min_transcript": "inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3. that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both"
    },
    {
        "Q": "at 3:34 how did he get the negative 2\n",
        "A": "here is the full equation: 5x -1(3x + 2) = so then you distribute the -1 and you get: 5x - 3x - 2 = hope this helps!",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            214
        ],
        "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?"
    },
    {
        "Q": "At 5:34 why did you or how did you rewrite it to 8/8?\n",
        "A": "So that he could subtract his number that was over 8. You probably know that 1=8/8, as well as 4/4 etc... So by making the denominators the same, therefore being able to subtract a number from 1. Hope this helps! :)",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            334
        ],
        "3min_transcript": "And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something. inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3."
    },
    {
        "Q": "@ 4:43 how did he end up with 3/2?\n",
        "A": "The problem was 5x-(3x+2)=1. He first, to get rid of the parentheses, reversed the sign inside (because the sign outside was negative). He then subtracted 3x from 5x, and got 2x. So far it is 2x-2=1. You can then add 2 to both sides, to get 2x=3, and divided both sides by 2, to get x=3/2. You can look at the previous videos if you re still confused. Hope it helped!",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            283
        ],
        "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something."
    },
    {
        "Q": "\nAt 3:35, where does he get the -2 from? Using the Distributive property doesn't change the sign, right? Why would re-arranging the numbers turn a +2 into a -2?",
        "A": "Because, he is having to multiply EVERYTHING in the parenthesis by a negative. There is no number written, but in between the negative and parenthesis symbol is an imaginary number 1. So it is a negative 1.",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            215
        ],
        "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?"
    },
    {
        "Q": "\nat 5:28 why does he change it to 1S to S(1-3/8)",
        "A": "He is using the distributive property to combine like terms. He isn t changing 1S to S(1-3/8) he s changing 1S-3/8S to S(1-3/8).",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            328
        ],
        "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something."
    },
    {
        "Q": "\nAt 3:35, how did you change (3x+2) into -3x-2?",
        "A": "He distributed -1 to the expression. There was a - before the 3x+2. That s actually a hidden -1. -1(3x+2)=-3x-2. Hope this helps!",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            215
        ],
        "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?"
    },
    {
        "Q": "And why 8/8 on 5:43?\n",
        "A": "Because he was subtracting 3/8 from one so 1 - 3/8 = 1/1 - 3/8, since he needs a common denominator he multiplied the denominator by 8 and whatever you do to the denominator you must do to the nummerator thus he got 8/8 - 3/8",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            343
        ],
        "3min_transcript": "And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something. inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3."
    },
    {
        "Q": "\nAt around 4:50, you say that x^l * x^m = x^(l+m)\nI thought it would be 2x^(l+m)\n\nWHat am I missing?",
        "A": "Where would you get the 2 from? You are multiplying, not adding. Try this with some actual number, maybe that might make the concept more clear: 4\u00c2\u00b2 * 4\u00c2\u00b3 = 4*4 * 4*4*4 = 4\u00e2\u0081\u00b5 = 1024",
        "video_name": "FP2arCfAfBY",
        "timestamps": [
            290
        ],
        "3min_transcript": "It's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Well, let's start with with this right here. x to the n is equal to a times b. So, how could we rewrite this? Well, a is this. And b is this, right? So let's rewrite that. So we know that x to the n is equal to a. a is this. x to the l. x to the l. And what's b? Times b. Not doing anything fancy right now. But what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors, because I think that's useful. l, l plus m. That's kind of onerous to keep switching colors, but. You get what I'm saying. So, x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus n. So what do we know? We know x to the n is equal to x to the l plus m. Right? Well, we have the same base. So we know that n is equal to l l plus m. What does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere? I think you'll see that I am. Well, what's another way of writing n? So we said, x to the n is equal to a times b -- oh, I actually skipped a step here. So that means -- so, going back here, x to the n That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything. I just forgot to write that down when I first did it. But, anyway. So, what's n? What's another way of writing n? Well, another way of writing n is right here. Log base x of a times b."
    },
    {
        "Q": "at 7:08 Sal does not multiply (a+b) squared and then on the right sides multiplies ab x 2. Why?\n",
        "A": "You can t multiply two different variables.",
        "video_name": "EINpkcphsPQ",
        "timestamps": [
            428
        ],
        "3min_transcript": "The average of the bottom and the top gives you the area of the trapezoid. Now, how could we also figure out the area with its component parts? Regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is one half a times b, but there's two of them. Let me do that b in that same blue color. But there's two of these right triangle. So let's multiply by two. So two times one half ab. That takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side."
    },
    {
        "Q": "at around 7:00 how and where did he get the 2's with?\n",
        "A": "Those are power of twos, the formula for pythagorean theorem is a^2+B^2= C^2",
        "video_name": "EINpkcphsPQ",
        "timestamps": [
            420
        ],
        "3min_transcript": "The average of the bottom and the top gives you the area of the trapezoid. Now, how could we also figure out the area with its component parts? Regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is one half a times b, but there's two of them. Let me do that b in that same blue color. But there's two of these right triangle. So let's multiply by two. So two times one half ab. That takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side."
    },
    {
        "Q": "\nAt 7:44 in the video when you were solving the equation you managed to get a \"2ab\" after solving (a+b)^2. Why is that? Please explain.",
        "A": "Try rewriting (a+b)^2 as (a+b)(a+b) instead. Then, use FOIL and see what happens. First: a times a equals a^2. Outer: a times b equals ab Inner: b times a equals ab Last: b times b equals b^2 Add them all up, and you get a^2 + 2ab + b^2. That s where the 2ab came from.",
        "video_name": "EINpkcphsPQ",
        "timestamps": [
            464
        ],
        "3min_transcript": "Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side. How can we simplify this? Is there anything that we can subtract from both sides? Well, sure there is. You have a 2ab on the left-hand side. You have a 2ab on the right-hand side. Let's subtract 2ab from both sides. If you subtract 2ab from both sides, what are you left with? You are left with the Pythagorean theorem. So you're left with a squared plus b squared is equal to c squared. Very, very exciting. And for that, we have to thank the 20th president of the United States, James Garfield. This is really exciting. The Pythagorean theorem, it was around for thousands of years before James Garfield, and he was able to contribute just kind of fiddling around while he was a member of the US House of Representatives."
    },
    {
        "Q": "at 2:16 what is that symbol he drew and what does it do?\n",
        "A": "It s a greek symbol called theta",
        "video_name": "EINpkcphsPQ",
        "timestamps": [
            136
        ],
        "3min_transcript": "What we're going to do in this video is study a proof of the Pythagorean theorem that was first discovered, or as far as we know first discovered, by James Garfield in 1876, and what's exciting about this is he was not a professional mathematician. You might know James Garfield as the 20th president He was elected president. He was elected in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is that it shows that Abraham Lincoln was not the only US politician or not the only US President who was really into geometry. And what Garfield realized is, if you construct a right triangle-- so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say this side right over here is length b. Let's say this side is length a, and let's say has length c. So I've just constructed enough a right triangle, and let me make it clear. It is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that. So we're going to have length b, and it's collinear with length It's along the same line, I should say. They don't overlap with each other. So this is side of length b, and then you have a side of length-- let me draw a it so this will be a little bit taller-- side of length b. And then, you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then, you have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves that it really is what we think it looks like. If we look at this original triangle and we call this angle \"theta,\" what's this angle over here, the angle that's between sides of length a and length c? What's the measure of this angle going to be? Well, theta plus this angle have to add up to 90. Because you add those two together, they add up to 90. And then, you have another 90. You're going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90. This angle is going to be 90 minus theta. Well, if this triangle appears congruent-- and we've constructed it so it is congruent-- the corresponding angle to this one is this angle right over here. So this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively go 180 degrees."
    },
    {
        "Q": "what do you mean at 2:19 ?\n",
        "A": "She protested with a sign. That s all",
        "video_name": "sxnX5_LbBDU",
        "timestamps": [
            139
        ],
        "3min_transcript": "[SINGING] On the first day of Christmas, my true love gave to me, the multiplicative identity. I always hated the song the \"12 days of Christmas\" when I was younger. Not that the tune or the words are any worse than any other Christmas song, but it's just so long and repetitive. Singing it sucks too, because like math class, it seems no matter how hard you try to pay attention, you lose focus out of the sheer tedium of it all and forget where you are. Unless you keep vigorous records by drawing complicated graphs. [SINGING] On the second day of Christmas my true love gave to me, the only even prime, and the absolute value of e to the i pi. See? It doesn't have to be repetitive, but the \"12 days of Christmas\" is more fun to think about, I do like these kinds of visualizations. And you can think of other ways to visualize the process of singing the \"12 Days.\" Which might come in useful next time you're at a family Christmas party, and someone insists on singing through the whole thing. [SINGING] On the third day of Christmas my true love gave to me, the number of spatial dimensions, at least macroscopically-- don't yell at me string theorists-- the number of points that define a line, and the limit of sine x over x as x goes to zero. Unlike a normal, reasonable songs, adding 12 more verses wouldn't just make it twice as long, because the stupid thing just grows and grows. Even 99 bottles of beer on the wall has the promise of getting to zero, unless you believe in anti-beer. But \"12 Days\" is just disheartening, because the closer you think you're getting to the end, the longer the verses get. Dragging it out to the bitter end. [SINGING] On the fourth day of Christmas my true love gave to me, the smallest possible number of points that define a plane, the divisor of even numbers, and any other number to the power of 0. If I had a time machine, and was not bitterly anti-time travel-- and yes I've actually protested with a sign-- one of the things on my list would be to go back to the year 0, pick your era, and be like, hey three kings of orient, hurry it up, will ya? Also what's myrrh? Because it sounds like the noise a camel makes. Really though, five days of Christmas would be more reasonable. Even eight I could live with. [SINGING] On the fifth day of Christmas, my true love gave to me, five golden ratio producing pentagons, the number of sides on a square, the number of sides on that rigid, functional, and beautiful creature called the triangle, and I guess a two, and the number"
    },
    {
        "Q": "at 3:06, what is a mobius strip?\n",
        "A": "A two dimensional object that can exist in three dimensional space.",
        "video_name": "sxnX5_LbBDU",
        "timestamps": [
            186
        ],
        "3min_transcript": "I do like these kinds of visualizations. And you can think of other ways to visualize the process of singing the \"12 Days.\" Which might come in useful next time you're at a family Christmas party, and someone insists on singing through the whole thing. [SINGING] On the third day of Christmas my true love gave to me, the number of spatial dimensions, at least macroscopically-- don't yell at me string theorists-- the number of points that define a line, and the limit of sine x over x as x goes to zero. Unlike a normal, reasonable songs, adding 12 more verses wouldn't just make it twice as long, because the stupid thing just grows and grows. Even 99 bottles of beer on the wall has the promise of getting to zero, unless you believe in anti-beer. But \"12 Days\" is just disheartening, because the closer you think you're getting to the end, the longer the verses get. Dragging it out to the bitter end. [SINGING] On the fourth day of Christmas my true love gave to me, the smallest possible number of points that define a plane, the divisor of even numbers, and any other number to the power of 0. If I had a time machine, and was not bitterly anti-time travel-- and yes I've actually protested with a sign-- one of the things on my list would be to go back to the year 0, pick your era, and be like, hey three kings of orient, hurry it up, will ya? Also what's myrrh? Because it sounds like the noise a camel makes. Really though, five days of Christmas would be more reasonable. Even eight I could live with. [SINGING] On the fifth day of Christmas, my true love gave to me, five golden ratio producing pentagons, the number of sides on a square, the number of sides on that rigid, functional, and beautiful creature called the triangle, and I guess a two, and the number Graphing the numbers like this may seem trivial, but I think it's nice to be reminded that these numbers aren't just squiggles on a page, symbols to be manipulated, but actually represent something. And I think it's nice that it results in such a lovely triangle. You know how I feel about triangles. [SINGING] On the sixth day of Christmas, my true love gave to me, my name in Roman numerals, number of feet in iambic pentameter, my name in Roman numerals backwards, the first Mersenne prime, the number of syllables in a foot of iambic pentameter, and sine x squared plus cosine x squared. Right. So if you want to know how many times you're going to have to sing a line about whatever stuff your true love got you, it's like counting up all the things in one of these triangles. That has however many layers, in this case 12."
    },
    {
        "Q": "At about 5:45, what's a frieze pattern?\n",
        "A": "They are patterns that have all the kinds of symmetry, I think. Go to Vi s video about freizepatterns called : Math Improv:Fruit By The Foot.",
        "video_name": "sxnX5_LbBDU",
        "timestamps": [
            345
        ],
        "3min_transcript": "number, for obvious reasons. You can also shift the things around to get an equilateral triangle, which is how triangular numbers are usually explained. So the first triangular number is 1, the second is 3, the third is 6, the fourth is 10. [SINGING] On the seventh day of Christmas, my true love gave to me, the most common lucky number, the first perfect number, the only prime ending in five. The number of colors sufficient for coloring in a map, the only prime triangular number, the highest number that is its own factorial, and 1/2 plus 1/4 plus 1/8 plus 1/16 and so on. Almost. Here's a famous story that I've heard told a few different ways, and the actual facts are fuzzy, but basically, here's the gist of it. Carl Gauss was bored in his math class. but math classes sucked 300 years ago, just as much as they suck today. And Gauss would get himself in trouble when he was bored. Maybe he also liked to escape via the window. So his teacher got fed up and was like, Gauss-- I mean, he probably called him Carl, but that's not the point-- Guass, if you're so freaking bored with my class, how about you go sit in the corner and add up all the numbers between 1 and 100. That ought to keep you busy for a while. So Gauss goes to the corner, but he's just sitting there. And the teacher gets all mad and he's like, hey Gauss. I guess that means you've already added up all those numbers, right? And Gauss is like, sure, it's 5,050. [SINGING] On the eighth day of Christmas, my true love gave to me, the only perfect cube that's a Fibonacci number besides one, the number of Frieze patterns, the number of sides on a cube, the number of platonic solids, the first composite number, the number of regular polytropes in all dimensions greater than four, the Euler characteristic of polyhedra homeomorphic to a sphere, it's undefined. His teacher of course, didn't believe him. I think the teacher spent the next 10 minutes adding up the numbers by hand in an effort to catch Gauss in a lie, and when he saw that Gauss was right, he probably gave him detention any way. Or more likely whacked him with a ruler a few times for having the gall to be smarter than him. Or it could be that the story's mostly made up. Nevertheless, here's how he did it. Instead of adding up the numbers individually, like his teacher did, which would have been super boring, he realized this fact. The numbers 1 through 100 come in pairs that add up to 101. 1 plus 100. 2 plus 99. 3 plus 98. 4 plus 97. And so on. There's 50 such pairs of numbers, and 50 times 101 is really easy to do in your head, because it's 50 times 100 plus 50, or 5050. [SINGING] On the ninth day of Christmas my true love gave to me, an upside down six, infinity sideways, flipped over L, an upside"
    },
    {
        "Q": "(0:35) - specifically the word placeholder gave me this question\nIf you were to \"say\" 1 is \"y\", and anything times 1 is the same, would 1 actually represent the \"laws of physics\"? Like, the laws of physics is based on \"common perception/observation\", which should always equal the same thing(s)... Like when you multiply by one... Is this an accurate/logical way to look at it?\n",
        "A": "And does 567x1 is. 567",
        "video_name": "6nZp2QGeQ9k",
        "timestamps": [
            35
        ],
        "3min_transcript": "We're asked to multiply 65 times 1. So literally, we just need to multiply 65-- we could write it is a times sign like that or we could write it as a dot like that-- but this means 65 times 1. And there's two ways to interpret this. You could view this as the number 65 one time or you could view this as the number 1 sixty-five times, all added up. But either way, if you have one 65, this is literally just going to be 65. Anything times 1 is going to be that anything, whatever this is. Whatever this is times 1 is going to be that same thing again. If I have just some kind of placeholder here times 1, and I could even write it as the times symbol times 1, that's going to be that same placeholder. If I have 5 times 1, I'm going to get 5, because literally, all this is saying is 5 one time. If I put-- I don't know-- 157 times 1, that'll be 157. I think you get the idea."
    },
    {
        "Q": "\nwhat is the in don't know what to call it but at 6:44 when u ad the numbers together what if its 5/4 how do u divide",
        "A": "Well when you divide a fraction you must flip the denominator (Bottom number of the fraction) , but when you flip it has to be the SECOND fraction. Ex 4/5 \u00c3\u00b7 6/7 to 4/5 \u00c3\u00b7 7/6 A lot of people forget to add the \u00c3\u0097 sign when you are dividing. If this sounds confusing then let me show you Ex 4/5 \u00c3\u00b7 6/7 to 4/5 \u00c3\u0097 7/6",
        "video_name": "GdIkEngwGNU",
        "timestamps": [
            404
        ],
        "3min_transcript": "So we say, two minus the mean. Two minus the mean, and we take the absolute value. So that's its absolute deviation. Then we have another two, so we find that absolute deviation from three. Remember, if we're just taking two minus three, taking the absolute value, that's just saying its absolute deviation. How far is it from three? It's fairly easy to calculate in this case. Then we have a four and another four. Let me write that. Then we have the absolute deviation of four from three, from the mean. Then plus, we have another four. We have this other four right up here. Four minus three. because once again, it's absolute deviation. And then we divide it, and then we divide it by the number of data points we have. So what is this going to be? Two minus three is negative one, but we take the absolute value. It's just going to be one. Two minus three is negative one. It's just gonna be one. And you see that here visually. This point is just one away. It's just one away from three. This point is just one away from three. Four minus three is one. Absolute value of that is one. This point is just one away from three. Four minus three, absolute value. That's another one. every data point was exactly one away from the mean. And we took the absolute value so that we don't have negative ones here. We just care how far it is in absolute terms. So you have four data points. Each of their absolute deviations is four away. So the mean of the absolute deviations are one plus one plus one plus one, which is four, over four. So it's equal to one. One way to think about it is saying, on average, the mean of the distances of these points away from the actual mean is one. And that makes sense because all of these are exactly one away from the mean. Now, let's see how, what results we get for this data set right over here. Let me actually get some space over here. At any point, if you get inspired, I encourage you to calculate the Mean Absolute Deviation on your own. So let's calculate it. The Mean Absolute Deviation here, I'll write MAD, is going to be equal to ... Well, let's figure out the absolute deviation of each of these points from the mean. It's the absolute value of one minus three, that's this first one, plus the absolute deviation, so one minus three, that's the second one, then plus the absolute value of six minus three, that's the six, then we have the four, plus the absolute value of four minus three. Then we have four points. So one minus three is negative two. Absolute value is two. And we see that here. This is two away from three. We just care about absolute deviation. We don't care if it's to the left or to the right. Then we have another one minus three is negative two. It's absolute value, so this is two. That's this. This is two away from the mean. Then we have six minus three. Absolute value of that is going to be three."
    },
    {
        "Q": "\nAt 1:19 when he wrote P-2w/2 = l, why can't sal divide 2w/2 to just get w?",
        "A": "The way he arranged the equation (which is hard to represent here because you can t represent a fraction with a horizontal bar) is L= (P-2w)/2....so both the P and the 2w need to be divided by 2. Think of it with just numbers... - let P=6 and w = 1 so the way you wrote it: P-2w/2 = L says that 6- 2(1)/2 = 5 but the way he represented it L= (P-2w)/2 : again let P=6 and w = 1: L = (6-2)/2 = 2 not the same expression without the ( )",
        "video_name": "fnuIT7EhAvs",
        "timestamps": [
            79
        ],
        "3min_transcript": "Solve P equals 2l plus 2w for l. So this right here, this is just the formula for the perimeter of a rectangle. Perimeter is equal to 2 times the length plus 2 times the width. But they just want us to solve this equation right here, solve for l. So let's do that. So we have P is equal to 2 times l plus 2 times w. So we need to solve for l. So let's isolate all of the l terms on one side. And the best way, we could just do that by leaving it here on the right and then getting rid of this 2w. And the best way to get rid of this 2w is to subtract it from the right. But if you're going to subtract it from the right, you also have to subtract it from the left if you want this equality to hold. So you have to also subtract it from the left. And so the left-hand side becomes P minus 2w. And the right-hand side, you get-- this 2w minus 2w cancels out. You just have a 2l. And then if you want to solve for l, you just have to divide both sides of this equation by 2. And we have isolated our l. We get l is equal to P minus 2w over 2. Or if we wanted to write it the other way, you could write l is equal to P minus 2w over 2. And we are done."
    },
    {
        "Q": "At1:09 when we divide both sides by 2. It cancells out the 2 on the right so we are left with (l). Why doesn't it cancel out on the left and leave us with (p-w)??\n",
        "A": "You are dividing both sides by 2. Another way to write the final answer would be: l = P/2 - w",
        "video_name": "fnuIT7EhAvs",
        "timestamps": [
            69
        ],
        "3min_transcript": "Solve P equals 2l plus 2w for l. So this right here, this is just the formula for the perimeter of a rectangle. Perimeter is equal to 2 times the length plus 2 times the width. But they just want us to solve this equation right here, solve for l. So let's do that. So we have P is equal to 2 times l plus 2 times w. So we need to solve for l. So let's isolate all of the l terms on one side. And the best way, we could just do that by leaving it here on the right and then getting rid of this 2w. And the best way to get rid of this 2w is to subtract it from the right. But if you're going to subtract it from the right, you also have to subtract it from the left if you want this equality to hold. So you have to also subtract it from the left. And so the left-hand side becomes P minus 2w. And the right-hand side, you get-- this 2w minus 2w cancels out. You just have a 2l. And then if you want to solve for l, you just have to divide both sides of this equation by 2. And we have isolated our l. We get l is equal to P minus 2w over 2. Or if we wanted to write it the other way, you could write l is equal to P minus 2w over 2. And we are done."
    },
    {
        "Q": "\nat 3:40 Sal writes yos does that mean yards?",
        "A": "Yes at 3:40 he wrote yos but it actually meant yards.",
        "video_name": "ZS1OZj_oWao",
        "timestamps": [
            220
        ],
        "3min_transcript": "So we took the remaining distance and we divided it into fourths. And they say, then she skipped for 1/4 of the remaining distance. So this is the remaining distance after running. She skipped for 1/4 of it. And then she walked the rest of the way home. So she walked to the rest of the way home, this is the distance that she walked. And this distance right over here is 120 yards. What is the distance between school and Judy's home? So there's a bunch of ways that you could think about this. When we talked about the remaining distance after running, so this was the remaining distance after she ran, after this orange part right over here. You see that this blue section is 3/4 of that remaining distance. 120 yards is 3/4 of this remaining distance. So this is equal to 3/4 of the remaining distance. Well since this is 3/4, 1/4 is going to be 1/3 of this. 1/3 of 120 is 40. So this distance right over here is 40. This is 40. This is 40. Notice 40 plus 40 plus 40 is 120. And this right over here is also going to be 40. These are all in terms of yards. So the remaining distance is going to be 40 times 4 or 160 So this is 160 yards. Now, what is the total distance? Well this 160 yards, the remaining distance after she ran, that's half of the total distance. So this is 160 yards, then this is going to be 160 yards as well. Gives us a total distance from school to home of 320 yards."
    },
    {
        "Q": "\nAt 2:07 Sal substitutes (4x^3 dx) for du. But why does he not have to write another dx at the end of the Integral?",
        "A": "Because of du. When you use u-substitution, you basically are writing the integral in terms of u instead of (in this case) x. du and dx ultimately serve the same function as well as representing an infinitesimal change in x or u and also being the dummy variable that integrates to +C .",
        "video_name": "Zp5z0wa0kgo",
        "timestamps": [
            127
        ],
        "3min_transcript": "So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx. So how can we tackle this? It seems like a hairy integral. Now the key inside here is to realize you have this expression x^4 + 7 and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3. Derivative of x^4 is 4x^3; derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. U-sub -- I'll just write u- -- I'll write the whole thing. U-Substitution could be the tool of choice. So given that, what would you want to set your u equal to? And I'll let you think about that 'cause it can figure out this part and the rest will just boil down to a fairly straightforward integral. Well, you want to set u be equal to the expression that you have its derivative laying around. So we could set u equal to x^4 plus 7. Now, what is du going to be equal to? so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du?"
    },
    {
        "Q": "\nIn 8:35 - isn't it supposed to be -(34/3)? So that A will equal -(5/17)?",
        "A": "Actually I think it should be -5/17, he dropped the minus sign when he subtracted 9 from -25. It should be -34/3 which reduces to -5/17.......right??? :/",
        "video_name": "hbJ2o9EUmJ0",
        "timestamps": [
            515
        ],
        "3min_transcript": ""
    },
    {
        "Q": "If you had a ratio of 3:5, would you plot it a (3,5), (6,10), etc...\n",
        "A": "Yes, because the ratio of x:y is 3/5.",
        "video_name": "dmcVzFbXMCU",
        "timestamps": [
            185
        ],
        "3min_transcript": "In a recent gum-chewing contest, Violet tried to blow the biggest blueberry-flavored bubble ever. Every 3 seconds, she added 1 piece of gum to her huge bubble. On the graph below, the x-axis represents time in seconds. So this right over here is our x-axis. And the y-axis represents pieces of gum in Violet's bubble. So this is the y-axis right over here. Plot points to show how many pieces of gum were in Violet's bubble after 3, 6, and 9 seconds. So every 3 seconds, she adds one piece of gum. So let me throw all of my-- so what we care is about 3 seconds, 6 seconds, and 9 seconds. So every 3 seconds, we know that she adds one piece of gum. This is 0 right over here. We're not going to worry about negative time. So after 3 seconds, she would have added one piece of gum to her bubble. Now, if 3 more seconds pass by, she would have added 1 more, And then if 3 more seconds pass by, now we're at 9 seconds, she would have added 1 more piece of gum, and now we're at 3 total pieces of gum. And notice, the ratio between the pieces of gum and the time that has passed has stayed constant, because she's doing it at a constant rate. 1 to 3 is the same ratio as 2 to 6, which is the same ratio as 3 to 9. So, let's check our answer. Very good."
    },
    {
        "Q": "\nat 1:20 what does numerical expression mean?",
        "A": "A numerical expression is a mathematical phrase that involves only numbers and one or more operational symbols. The expression represents a particular number. For instance the numerical expression 10 + 6 \u00e2\u0080\u0093 8 simplifies to the number 8.",
        "video_name": "arY-EUZDNfk",
        "timestamps": [
            80
        ],
        "3min_transcript": "Alan found 4 marbles to add to his 5 marbles currently in his pocket. He then had a competition with his friends and tripled his marbles. Write a numerical expression to model the situation without performing any operations. So let's think about what's going on. So he already had 5 marbles in his pocket. And then he found 4 more marbles to add to that. So we can add the 4 marbles to the 5 marbles. So 4 marbles plus the 5 marbles. So that's what happens after the first sentence. He found 4 marbles to add to his 5 marbles currently in his pocket. He then had a competition with his friends and tripled his marbles. So this is how many marbles he had before tripling. And now he's tripling his marbles. So we want to multiply 3 times the total number of marbles he has now-- times 4 plus 5. So this right over here is the numerical expression any operations. We, of course, could then actually calculate this. He has 9 marbles before tripling. And then you multiply it by 3 and he has 27. But this is what they're asking for. They want us to write this expression."
    },
    {
        "Q": "For the last example at 5:30 can I also define like that? Is it true?\n{x \u00e2\u0088\u0088 \u00e2\u0084\u009d | x = \u00cf\u0080 or x=3}\n",
        "A": "You could, but it s a bit more complicated than it needs to be since the set has only 2 values.",
        "video_name": "-DTMakGDZAw",
        "timestamps": [
            330
        ],
        "3min_transcript": "most of the real numbers except it cannot be 0 because we don't know -- this definition is undefined when you put the input as 0 So x is a member of the real numbers, and we write real numbers -- we write it with this kind of double stroke right over here. That's the set of all real numbers such that -- we have to put the exception. 0 is not a -- x equals to 0 is not a member of that domain -- such that x does not -- does not equal 0. Now let's make this a little bit more concrete by do some more examples So more examples we do, hopefully the clearer this will become. So let's say we have another function. Just be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus 6. So what is the domain here? What is the set of all inputs over which this function g is defined? So here we are in putting a y it to function g g of y. Well it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes a negative, our traditional principal root operator here is not defined. We need something that -- if this was a negative number, how would you take the principal root of a negative number? We just think this is kind of the the traditional principal root operator. So y minus 6, y minus 6 needs to be greater than or equal to 0, in order for, in order for g to be defined for that input y. Or you could say add six to both sides. y is to be greater than or equal 6. Or you could say g is defined for any inputs y that are greater than or equal to 6. So you could say the domain here, we could say the domain here is the set of all y's that are members of the real numbers such that y, such that they're also greater than or equal, such that they're also You're all used to a function that is defined this way. You could even see functions that are divided fairly exotic ways. You could see a function -- let me say h of x -- h of x could be defined as -- it literally could be defined as, well h of x is gonna be 1 if x is equal to pi and it's equal to 0 if, if, x is equal to 3. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two input. If you, we know h of -- we know h of pi -- if you input pi into it we know you're gonna output 1, and we know that if you input 3 into it h of 3, when x equals 3, you're going to -- you're going to -- put some commas here. You're gonna get 0. But if you input anything else, what's h of 4 going to be? Well, it hasn't defined. It's undefined. What's h of negative 1 going to be? It hasn't defined. So the domain, the domain here,"
    },
    {
        "Q": "Hi, Can somebody please explain in \"domain and range of a function\" video (at 5:02 minutes). How can y-6 >= 0 because greater than i understood but how can it equals to 0 as if y= 6 than the equation goes like 6-6 which is equals to 0 and 0 is not defined output.\n\nPlease answer.\n",
        "A": "sqrt(0) is defined. It = 0. So, there is no issue have a zero inside the square root. You don t want a negative inside the square root after you have simplified because it is not a real number.",
        "video_name": "-DTMakGDZAw",
        "timestamps": [
            302
        ],
        "3min_transcript": "most of the real numbers except it cannot be 0 because we don't know -- this definition is undefined when you put the input as 0 So x is a member of the real numbers, and we write real numbers -- we write it with this kind of double stroke right over here. That's the set of all real numbers such that -- we have to put the exception. 0 is not a -- x equals to 0 is not a member of that domain -- such that x does not -- does not equal 0. Now let's make this a little bit more concrete by do some more examples So more examples we do, hopefully the clearer this will become. So let's say we have another function. Just be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus 6. So what is the domain here? What is the set of all inputs over which this function g is defined? So here we are in putting a y it to function g g of y. Well it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes a negative, our traditional principal root operator here is not defined. We need something that -- if this was a negative number, how would you take the principal root of a negative number? We just think this is kind of the the traditional principal root operator. So y minus 6, y minus 6 needs to be greater than or equal to 0, in order for, in order for g to be defined for that input y. Or you could say add six to both sides. y is to be greater than or equal 6. Or you could say g is defined for any inputs y that are greater than or equal to 6. So you could say the domain here, we could say the domain here is the set of all y's that are members of the real numbers such that y, such that they're also greater than or equal, such that they're also You're all used to a function that is defined this way. You could even see functions that are divided fairly exotic ways. You could see a function -- let me say h of x -- h of x could be defined as -- it literally could be defined as, well h of x is gonna be 1 if x is equal to pi and it's equal to 0 if, if, x is equal to 3. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two input. If you, we know h of -- we know h of pi -- if you input pi into it we know you're gonna output 1, and we know that if you input 3 into it h of 3, when x equals 3, you're going to -- you're going to -- put some commas here. You're gonna get 0. But if you input anything else, what's h of 4 going to be? Well, it hasn't defined. It's undefined. What's h of negative 1 going to be? It hasn't defined. So the domain, the domain here,"
    },
    {
        "Q": "\nAt 4:58 can't ti also be 4.5 instead of 4 1/2?",
        "A": "Since 4.5 = 4 1/2, yes you can write it the way you prefer.",
        "video_name": "hq1bUM2tyg0",
        "timestamps": [
            298
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nCould you include a fraction in the answer in 4:24? Could you also get a rational number?",
        "A": "Fractions are rational numbers. Yes, there s no reason why you couldn t use fractions here.",
        "video_name": "hq1bUM2tyg0",
        "timestamps": [
            264
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 1:59 Sal took the triangle to the other side of the parallelogram to make a rectangle, can't you just multiply base times height to get the area?",
        "A": "Yes you can. Sal most likely did that move so that you could visually see why multiplying base times height works.",
        "video_name": "hm17lVaor0Q",
        "timestamps": [
            119
        ],
        "3min_transcript": "- If we have a rectangle with base length b and height length h, we know how to figure out its area. Its area is just going to be the base, is going to be the base times the height. The base times the height. This is just a review of the area of a rectangle. Just multiply the base times the height. Now let's look at a parallelogram. And in this parallelogram, our base still has length b. And we still have a height h. So when we talk about the height, we're not talking about the length of these sides that at least the way I've drawn them, move diagonally. We're talking about if you go from this side up here, and you were to go straight down. If you were to go at a 90 degree angle. If you were to go perpendicularly straight down, you get to this side, that's going to be, that's going to be our height. So in a situation like this when you have a parallelogram, you know its base and its height, what do we think its area is going to be? So at first it might seem well this isn't as obvious But we can do a little visualization that I think will help. So what I'm going to do is I'm going to take a chunk of area from the left-hand side, actually this triangle on the left-hand side that helps make up the parallelogram, and then move it to the right, and then we will see something somewhat amazing. So I'm going to take this, I'm going to take this little chunk right there, Actually let me do it a little bit better. So I'm going to take that chunk right there. And let me cut, and paste it. So it's still the same parallelogram, but I'm just going to move this section of area. Remember we're just thinking about how much space is inside of the parallelogram and I'm going to take this area right over here and I'm going to move it to the right-hand side. And what just happened? What just happened? Let me see if I can move it a little bit better. What just happened when I did that? Well notice it now looks just like my previous rectangle. by taking some of the area from the left and moving it to the right, I have reconstructed this rectangle so they actually have the same area. The area of this parallelogram, or well it used to be this parallelogram, before I moved that triangle from the left to the right, is also going to be the base times the height. So the area here is also the area here, is also base times height. I just took this chunk of area that was over there, and I moved it to the right. So the area of a parallelogram, let me make this looking more like a parallelogram again. The area of a parallelogram is just going to be, if you have the base and the height, it's just going to be the base times the height. So the area for both of these, the area for both of these, are just base times height."
    },
    {
        "Q": "\nWhat is the name of the notation \"S-like\" figure Sal describes at 2:36?",
        "A": "The integral symbol or the integral sign. (I know, it seems almost too obvious.)",
        "video_name": "MMv-027KEqU",
        "timestamps": [
            156
        ],
        "3min_transcript": "Now, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative. Now, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell-- or give you 2x-- if someone were to say 2x-- let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea. So if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x. Now, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of. So in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c. use this type of crazy notation. It'll become more obvious when we study the definite integral and areas under curves and taking sums of rectangles in order to approximate the area of the curve. Here, it really should just be viewed as a notation for antiderivative. And this notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x."
    },
    {
        "Q": "\nAt 2:13, how did he convert 3/10 into 30/100?",
        "A": "He multiplied them both by 10. don t worry, you l get used to it. its a common technique.",
        "video_name": "YZD5ifHZILE",
        "timestamps": [
            133
        ],
        "3min_transcript": "Let's see if we can add 3/10 to 7/100. And I encourage you to try adding these two fractions on your own first before I work through it, and I'll give you one hint. Right now, it's very hard to add these fractions. You're adding 3/10 to 7/100. You're adding two different fractions with two different denominators. So what I would encourage you to do is try to rewrite 3/10 in a way that it has 100 as a denominator, or so it's expressed in terms of hundredths, and then see if you can add them. Well, I'm assuming you've given a go at it. Let's see how we can rewrite 3/10, and I'll try to visualize it. So what I've done here, so this you could consider a whole, and this is a whole as well. And this whole is divided into tenths-- 1,2, 3, 4, 5, 6, 7, 8, 9, 10. So what would 3/10 look like? Well, it would be one, two, and three of the tenths. and you divided them into 10 more sections? So you're essentially taking each of those tenths and you're dividing them into 10 sections. Well, you have 10 sections and then each of them will have 10 subsections in it. So you're going to have hundredths. Then the sections are going to describe hundredths. So 10-- let me write it this way. 10 times 10. You are then going to have hundredths. And 3/10 would be equivalent to how many of these hundredths? Well, each of these tenths will now become 10. So you're going to have 10, 20, and-- let me color them in better. So this part right over here. This-- let me get my tool right-- this part right over here, this is going to be 3 times 10, which is going to be equal to 30. 10 times 10 is equal to 100. So this is how we're going to change the denominator. Instead of thinking in terms of tenths, we're going to think in terms of hundredths. And now our numerator, 3/10, is equivalent to 30 hundredths so we can rewrite this fraction. We essentially multiplied the numerator by 10, and we multiplied the denominator by 10, which didn't change the value of the fraction. It still represents the same-- it still represents 3/10 of the whole. So when you do that, you end up-- we can rewrite this thing as 30 over 100."
    },
    {
        "Q": "\nAt 5:24, how would you solve the C part? Do you spread it out like you did? Or is there another way?\nCan anyone help me quick? Sal... you there?",
        "A": "When you say the C part, do you mean the formula for a combination? Sal actually explains that in the following video. But perhaps you meant something else?",
        "video_name": "iKy-d5_erhI",
        "timestamps": [
            324
        ],
        "3min_transcript": "I'm going to be a little bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12 of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which chair they're sitting in. This would also be the same set of three people. And so this question. If I have six people sitting in three chairs, how many ways can I choose three people out of the six And I encourage you to pause the video, and try to think of what that number would actually be. Well a big clue was when we essentially wrote all of the permutations where we've picked a group of three people. We see that there's six ways of arranging the three people. When you pick a certain group of three people, that turned into six permutations. And so if all you want to do is care about well how many different ways are there to choose three from the six? You would take your whole permutations. You would take your number of permutations. You would take your number of permutations. And then you would divide it by the number of ways to arrange three people. Number of ways to arrange, arrange three people. And we see that you can arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal to 120 divided by six, So there are 120 permutations here. If you said how many different arrangements are there of taking six people and putting them into three chairs? That's 120. But now we're asking another thing. We're saying if we start with 120 people, and we want to choose. Sorry if we're starting with six people and we want to figure out how many ways, how many combinations, how many ways are there for us to choose three of them? Then we end up with 20 combinations. Combinations of people. This right over here, once again, this right over here is just one combination. It's the combination, A, B, C. I don't care what order they sit in. I have chosen them. I have chosen these three of the six. This is a combination of people. I don't care about the order. This right over here is another combination. It is F, C, and B. Once again I don't care about the order. I just care that I've chosen these three people."
    },
    {
        "Q": "can anyone explain what sal tried to do after 4:00 ??\n\u00e2\u0096\u00baQ2 whats the difference between permutations and combinations.. ?\n",
        "A": "He was showing how to find the number of ways to choose a subset of a set of things. In this case how many ways you can choose 3 people from a pool of 6 people (where order doesn t matter). In permutations order matters, in combinations it does not.",
        "video_name": "iKy-d5_erhI",
        "timestamps": [
            240
        ],
        "3min_transcript": "Permutations. Now lemme, permutations. Now it's worth thinking about what permutations are counting. Now remember we care, when we're talking about permutations, we care about who's sitting So for example, for example this is one permutation. And this would be counted as another permutation. And this would be counted as another permutation. This would be counted as another permutation. So notice these are all the same three people, but we're putting them in different chairs. And this counted that. That's counted in this 120. I could keep going. We could have that, or we could have that. So our thinking in the permutation world. We would count all of these. Or we would count this as six different permutations. These are going towards this 120. And of course we have other permutations where we would involve other people. Where we have, it could be F, B, C, F, C, B, F, A, C, F, F. I'm going to be a little bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12 of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which chair they're sitting in. This would also be the same set of three people. And so this question. If I have six people sitting in three chairs, how many ways can I choose three people out of the six And I encourage you to pause the video, and try to think of what that number would actually be. Well a big clue was when we essentially wrote all of the permutations where we've picked a group of three people. We see that there's six ways of arranging the three people. When you pick a certain group of three people, that turned into six permutations. And so if all you want to do is care about well how many different ways are there to choose three from the six? You would take your whole permutations. You would take your number of permutations. You would take your number of permutations. And then you would divide it by the number of ways to arrange three people. Number of ways to arrange, arrange three people. And we see that you can arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal to 120 divided by six,"
    },
    {
        "Q": "\nI have a word problem it says \"the angle of the roof on Wendy's dollhouse is 56 degrees. She built a scale model of the dollhouse with a scale ratio of 1:4. What is the angle of the roof of the model she built?\" I'm curious on how I would work this out. Do I do the same as on this video?",
        "A": "Well If You think it is correct then you do it but I think you should do the video",
        "video_name": "tOd2T72eJME",
        "timestamps": [
            64
        ],
        "3min_transcript": "The scale on a map is 7 centimeters for every 10 kilometers, or 7 centimeters for 10 kilometers. If the distance between two cities is 60 kilometers-- so that's the actual distance-- how far apart in centimeters are the two cities on the map? Well, they give us the scale. For every 10 kilometers in the real world, the map is going to show 7 centimeters. Or another way to think about it is if you see 7 centimeters on the map, that represents 10 kilometers in the real world. Now, they're saying that the distance between two cities is 60 kilometers. So it's essentially 6 times 10 kilometers, so times 6. So if you have to 6 times 10 kilometers on the map, you're going to have 6 times 7 centimeters, so times 6. And 6 times 7 centimeters gets you to 42 centimeters. Well, 42 centimeters."
    },
    {
        "Q": "At 1:00 Sal talks of \"g of negative 6.999.\" Does he mean to say \"negative?\" Isn't the 6.999 positive?\n",
        "A": "He said g of negative 6.999 but it should be g of 6.999 . I m guessing what he was trying to say that g of 6.999 is still negative but got mixed up. Report it so he can add an annotation for correction.",
        "video_name": "hWJLd6bRthI",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Voiceover] So, here we have the graph y is equal to g of x. We have a little point discontinuity right over here at x is equal to seven, and what we want to do is figure out what is the limit of g of x as x approaches seven. So, since you say, well, what is the function approaching as the inputs in the function are approaching seven? So, let's see, so if we input as the input to the function approaches seven from values less than seven, so if x is three, g of three is here, g of three is right there, g of four is right there, g of five is right there, g of six looks like it's a little bit more than, or a little bit less than negative one, g of 6.5 looks like it's around negative half, g of negative 6.9 is right over there, looks like it's a little bit less than zero, g of negative 6.999 looks like it's a little bit, closer to zero so it looks like we're getting closer as x gets closer and closer but not quite at seven, it looks like the value of our function is approaching zero. Let's see if that's also true from values for x values greater than seven, so g of nine is up here, looks like it's around six, g of eight looks like it's a little bit more than two, g of 7.5 looks like it's a little bit more than one, g of 7.1 looks like it's a little bit more than zero, the g of 7.1 looks like it's a little bit more than zero, g of 7.01 is even closer to zero, g of 7.0000001 looks like it'll be even closer to zero so once again it looks like we are approaching zero as x as approaches seven, in this case as we approach from larger values of seven. And this is interesting because the limit as x approaches seven of g of x is different than the function's actual value, g of seven, When we actually input seven into the function, we can see the graph tells us that the value of the function is equal to three. So we actually have this point discontinuity, or sometimes called a removable discontinuity, right over here, and this is, I'm not gonna do a lot of depth here, but this is starting to touch on how we, one of the ways that we can actually test for continuity is if the limit as we approach a value is not the same as the actual value of the function of that point, well, then we're probably talking about, or actually we are talking about, a discontinuity."
    },
    {
        "Q": "\nAt 3:40 ish wouldn't the probability be a dependent event since there is a set amount of fish in the pond/lake?",
        "A": "Fish after catching are being released back into the pond that is why the probability is independent of what was caught.",
        "video_name": "86nb02Bx_5w",
        "timestamps": [
            220
        ],
        "3min_transcript": "Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half. of the 20 fish. If he wasn't putting the fish back in then the second sunfish you would have a nine out of 20 chance of the second one being a sunfish. In this case they keep replacing the fish every time they catch it. There is a one eighth chance that Jeremy catches three sunfish, so this right over here is one eighth. And one minus one eighth, this is seven eighths. You have a one eighth chance of paying 100 dollars and a seven eights chance of getting twenty dollars so this gets us to ... Your expected profit here, there's a one eighths chance, one eighth probability, that you lose 100 dollars here, so times negative 100. But then there is a seven eighths chance that you get -- I think the order of operations of the calculator would have taken care of it but I'll just do it so that it looks the same. Seven eighths, there's a seven eighths chance that you get 20 dollars. Your expected payoff here is positive five dollars. Your expected payoff here is equal to five dollars. This is your expected value from bet one. Now let's think about bet two. If you catch at least two sunfish of the next three fish you catch he will pay you 50, otherwise you will pay him 25. Let's think about the probability of catching at least two sunfish of the next three fish that you catch. There's a bunch of ways to think about this but since there's only three times that you're trying to catch the fish and there's only one of two outcomes you could actually"
    },
    {
        "Q": "At 1:12, Sal defines the random variable X as \"what your profit is from bet 1.\" He then goes on to define a function E(X) that gives the expected profit from taking the first bet. Shouldn't X be defined as the outcome of Jeremy's next three fish?\nAgain at 6:55, Sal similarly defines Y as the expected profit from the second bet, which is what E(Y) is. Shouldn't Y be defined as the outcome of my next three fish?\n",
        "A": "He makes a small mistake at 6:55 when he says Y is the expected profit. Y should be defined the same as X as just the profit of the bet. E(Y) is the expected profit.",
        "video_name": "86nb02Bx_5w",
        "timestamps": [
            72,
            415
        ],
        "3min_transcript": "You and your friend Jeremy are fishing in a pond that contains ten trout and ten sunfish. Each time one of you catches a fish you release it back into the water. Jeremy offers you the choice of two different bets. Bet number one. We don't encourage betting but I guess Jeremy wants to bet. If the next three fish he catches are all sunfish you will pay him 100 dollars, otherwise he will pay you 20 dollars. Bet two, if you catch at least two sunfish of the next three fish that you catch he will pay you 50 dollars, otherwise you will pay him 25 dollars. What is the expected value from bet one? Round your answer to the nearest cent. I encourage you to pause this video and try to think about it on your own. Let's see. The expected value of bet one. The expected value of bet one where we'll say bet one is -- just to be a little bit better about this. Let's say x is equal what you pay, or I guess you could say , because you might get something, what your profit is from bet one. It's a random variable. The expected value of x is going to be equal to, let's see. What's the probability, it's going to be negative 100 dollars times the probability that he catches three fish. The probably that Jeremy catches three sunfish, the next three fish he catches are Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half."
    },
    {
        "Q": "\nAt about 0:50, give or take, in the video, Sal figures out the square root so easily. Now, I know this is an easy example, but it there a formula for finding the square root?",
        "A": "a square root of a number a is a number y such that y^2 = a, in other words, a number y whose square (the result of multiplying the number by itself, or y \u00c3\u0097 y) is a. For example, 4 and \u00e2\u0088\u00924 are square roots of 16 because 4^2 = (\u00e2\u0088\u00924)^2 = 16. Wikipedia",
        "video_name": "ROIfbUQrSY4",
        "timestamps": [
            50
        ],
        "3min_transcript": "We're asked to find the square root of 100. Let me write this down bigger. So the square root is this big check-looking thing. The square root of 100. When you see it like this, this means the positive square root. If you're familiar with negative numbers, you know that there's also a negative square root, but when you just see this symbol, that means the positive square root. So let's think about what this is saying. This is asking us find the number, the positive number, that when I multiply that number by itself, I get 100. So what number when I multiply it by itself do I get 100? Well, let's see, if I multiply 9 by itself, that's only going to be 81. If I multiply 10 by itself, that is 100. So this is equal to-- and let me write it this way. Normally, you could skip this step. But you could write this as the square root of-- and instead of 100, 100 is the same thing as 10 times 10. And then you know, the square root of something times itself, that's just going to be that something. This is just equal to 10. Or another way you could write, I guess, this same truth is that 10 squared, which is equal to 10 times 10, is equal to 100."
    },
    {
        "Q": "\nAt 0:45 Sal explains that the square root of 100 is 10 times 10. This is pretty simple to do figure out because if you know your times tables, you will have this memorized. But what if I'm trying to find the square root of a larger number? Any tried and true methods OTHER than a calculator? Thanks!",
        "A": "Use factoring. Break the number down into prime factors. For example: 2025 = 3*3*3*3*5*5 Each pair of matching factors is a perfect square. Group the pairs. 2025 = (3*3)*(3*3)*(5*5) OR, you can think of this as 2025 = 9*9*25 This number contains 3 perfect squares. You can take the square root of each perfect square, then multiply the results. sqrt(2025) = sqrt(9) * sqrt(9) * sqrt(25) = 3*3*5 = 45 Hope this helps.",
        "video_name": "ROIfbUQrSY4",
        "timestamps": [
            45
        ],
        "3min_transcript": "We're asked to find the square root of 100. Let me write this down bigger. So the square root is this big check-looking thing. The square root of 100. When you see it like this, this means the positive square root. If you're familiar with negative numbers, you know that there's also a negative square root, but when you just see this symbol, that means the positive square root. So let's think about what this is saying. This is asking us find the number, the positive number, that when I multiply that number by itself, I get 100. So what number when I multiply it by itself do I get 100? Well, let's see, if I multiply 9 by itself, that's only going to be 81. If I multiply 10 by itself, that is 100. So this is equal to-- and let me write it this way. Normally, you could skip this step. But you could write this as the square root of-- and instead of 100, 100 is the same thing as 10 times 10. And then you know, the square root of something times itself, that's just going to be that something. This is just equal to 10. Or another way you could write, I guess, this same truth is that 10 squared, which is equal to 10 times 10, is equal to 100."
    },
    {
        "Q": "At 0:46 why does Sal say the answer would be a positive number? B could be -3 and A could be -2 and the answer would we -1 which is NOT a positive value.\n",
        "A": "We always measure distances in positive numbers. We have no rulers with negative numbers. We don t say we just walked -3 miles. We don t measure the objects and come up with negative inches or negative centimeters. The absolute value of any number represents its distance from zero. We use it to help ensure we create distances in positive values. If A = -2 and B = -3, the distance between them is 1 unit (a positive number. You can find this by doing: | -3 -(-2)| = |-1| = 1 or |-2 -(-3)| = |1| = 1 Hope this helps.",
        "video_name": "t4xOkpP8FgE",
        "timestamps": [
            46
        ],
        "3min_transcript": "- [Voiceover] Let's say that I have two numbers on a number line. So let me draw a little quick number line right over here. The two numbers on my number line that I care about, the number a and the number represented by b here. The way I've drawn it, b is to the right of a on our number line, and by our convention, b is going to be greater than a. So if I were to figure out the distance between a and b, what is this distance going from a, I want to draw a straight line here, this distance going from a to b, so this distance right over there, how would I figure it out? Well I could just take the larger of these two numbers, which is going to be b, and then subtract out the smaller. So I subtract out a, and I'll be left with this distance. This will give me a positive value. When I want a distance, I just think in terms of a positive value. How far apart are these two things? But I was only able to know to do b minus a because I knew that b was greater than a. This was going to give me a positive value. What if I knew that a was greater than b? So let me draw that again. Let me draw another number line right over here. In this world, in this world, I'm going to make a greater than b. This is b, that is a, and if I wanted to calculate the distance between b and a here, well now I would take the larger of the two, a, remember I want the positive distance here, and then I would subtract out the smaller. I would do a minus b. Well so here I did b minus a, here I did a minus b, but what if I didn't know which one was greater? If I didn't know whether b or a was greater, what could I do? Well what you could do is just take either a minus b or b minus a and take the absolute value. If you do that, it doesn't matter if you take b minus a or a minus b. It turns out that regardless of whether a is greater than b, or b is greater than a, or they're equivalent, that the absolute value of b minus a, and this is equivalent, either of these expressions is the distance between these numbers. I encourage you to play around with the negatives to see if you can factor out some negatives and think about the absolute value. It will actually make a lot of sense why this is true. In another video, I might do a little bit more of a rigorous justification for it. for this video is to see that this is actually true. So let's say we're in a world, let's get a number line out, and let's look at some examples. So let's say that we want to figure out the distance between, between, let's say negative two, the distance between negative two and positive three. So we can look at the number line and figure out what that distance is. To go from negative two to positive three, or the distance between them, we see is one, two, three, four, five."
    },
    {
        "Q": "\nAt 6:05 - 6:10, I believe you meant to say that the subject terms are equivalent to the Identity Matrix time vector 'a' as opposed to the Identity Matrix times 'a1'.",
        "A": "yup, i m pretty sure he meant matrix a",
        "video_name": "PErhLkQcpZ8",
        "timestamps": [
            365,
            370
        ],
        "3min_transcript": "of these-- so let's say e1, e2, all the way to en-- this is called the standard basis for Rn. So why is it called that? Well, the word basis is there, so two things must be true. These things must span Rn and they must be linearly It's pretty obvious from inspection they're linearly If this guy has a 1 here and no one else has a 1 there, there's no way you can construct that 1 with some combination of the rest of the guys. And you can make that same argument for each of the ones So it's clearly linearly independent. And then to see that you can span, that you can construct any vector with a linear combination of these guys, you just really have to-- you know, whatever vector you want to construct, if you want to construct x1-- let me put it this way. Let me pick a different one. Let's say you want to construct the vector a1, a2, a3 all the way down to an. So this is some member of Rn, you want to construct this vector. Well, the linear combination that would get you this is literally a1 times e1 plus a2 times e2 plus all the way to an times en. This scalar times this first column vector will essentially just get you-- what will this look like? This will look like a1 and then you'd have a bunch of zeroes. You'd have n minus 1 zeroes plus 0 and you'd have an a2 and then you'd have a bunch of zeroes. And then you'd keep doing that, and then you would have a bunch of zeroes, and then you would have an an. Obviously, by our definition of vector addition, you add all these things up, you get this guy right here. And it's kind of obvious, because this right here is the I just wanted to expose you to that idea. Now, let's apply what we already know about linear transformations to what we've just learned about this identity matrix. I just told you that I can represent any vector like this. Let me rewrite it in maybe terms of x. I can write any vector x as a linear combination of the standard basis, which are really just the columns of the identity matrix. I can write that as x1 times e1 plus x2 times e2, all the way to xn times en. And remember, each of these column vectors right here, like for e1, is just 1 in the first entry and then all the rest are zeroes. e2 is a 1 in the second entry and everything else is 0. e5 is a 1 in the fifth entry and everything else is 0."
    },
    {
        "Q": "At 0:45, why does Sal draw the Vertical Line?\n",
        "A": "It s kind of working as the = signs of the equations, so it separates the variable constants from the number on the other side of the equation. I think it s just so you remember which are which.",
        "video_name": "lP1DGtZ8Wys",
        "timestamps": [
            45
        ],
        "3min_transcript": "I figure it never hurts getting as much practice as possible solving systems of linear equations, so let's What I'm going to do is I'm going to solve it using an augmented matrix, and I'm going to put it in reduced row echelon form. So what's the augmented matrix for this system of equations? Three unknowns with three equations. I just have to do the coefficents. So the coefficients of x terms are just 1, 1, 1. Coefficients of the y terms are 1, 2, and 3. Coefficients of the z terms are 1, 3, and 4. And let me show that it's augmented. And then they equal 3, 0, and minus 2. Now, I want to get this augmented matrix into reduced row echelon form. So the first thing, I have a leading 1 here that's a pivot entry. Let me make everything else in that column equal to a 0. So I'm not going to change my first row. and then I have a 3. Now, to zero this out, let me just replace the second row with the first row minus the second row. So 1 minus 1 is 0. 1 minus 2 is-- actually, a better thing to do, because I eventually want this to be 1 anyway, let me replace this row with this row, with the second row minus the first row instead of the first row minus the second row. I can do it either way. So the second row minus the first row. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. And then 0 minus 3 is minus 3. Now I want to also zero this out. So let me replace this guy with this equation minus that equation. So 1 minus 1 is 0. 4 minus 1 is 3. Minus 2 minus 3 is minus 5. Fair enough. So I got my pivot entry here. I have another pivot entry here. It's to the right of this one, which is what I want for reduced row echelon form. Now, I need to target this entry and that entry. I need to zero them out. So let's do it. So I'm going to keep my second row the same. My second row is 0, 1, 2, and then I have a minus 3, the augmented part of it. And to zero this guy out, what I can do is I can replace the first row with the first row minus the second row. So I get 1 minus 0 is 1. 1 minus 1-- there's a bird outside. Let me close my window. So where was I? I'm replacing the first row with the first row minus the"
    },
    {
        "Q": "5:05 Which ordered pair I know that is on the y or x axis?\n",
        "A": "For points on a graph, you define them by (distance on the x-axis, distance on the y-axis). So even if you didn t know which point it was, you would figure out how far it was on the different axis, put them in x, y order and in parentheses, and you ve got the point. Hopefully that helps!",
        "video_name": "WkspBxrzuZo",
        "timestamps": [
            305
        ],
        "3min_transcript": "over change in x, which is-- our change in y is 14. And our change in x is negative 7. And then if we want to simplify this, 14 divided by negative 7 is negative 2. Now, what I want to show you is, is that we could have done it the other way around. We could have made this the starting point and this the endpoint. And what we would have gotten is the negative values of each of these, but then they would've canceled out and we would still get negative 2. Let's try it out. So let's say that our start point was negative 3 comma 16. And let's say that our endpoint is the 4 comma 2. 4 comma 2. So in this situation, what is our change in x? Our change in x. If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. 4 minus negative 3. And what is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. So if you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. So notice, these are just the negatives of these values from when we swapped them. So once again, this is equal to negative 2. And let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. So let me draw our two points. So this is my x-axis. That is my y-axis. So this point over here, 4 comma 2. So let me graph it. So let me save some space here. So we have 1, 2, 3, 4. It's 4 comma-- 1, 2. So 4 comma 2 is right over here. 4 comma 2. Then we have the point negative 3 comma 16. So let me draw that over here. So we have negative 1, 2, 3. And we have to go up 16. So this is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. So it goes right over here. So this is negative 3 comma 16. Negative 3 comma 16. So the line that goes between them is going to look something like this. Try my best to draw a relatively straight line. That line will keep going. So the line will keep going. So that's my best attempt. And now notice, it's downward sloping. As you increase an x-value, the line goes down. It's going from the top left to the bottom right. As x gets bigger, y gets smaller."
    },
    {
        "Q": "\nwhat does the triangle mean at 2:10.",
        "A": "The triangle is denotes change. So, slope equals the change in the y values divided by the change in the x values.",
        "video_name": "WkspBxrzuZo",
        "timestamps": [
            130
        ],
        "3min_transcript": "Find the slope of the line that goes through the ordered pairs 4 comma 2 and negative 3 comma 16. So just as a reminder, slope is defined as rise over run. Or, you could view that rise is just change in y and run is just change in x. The triangles here, that's the delta symbol. It literally means \"change in.\" Or another way, and you might see this formula, and it tends to be really complicated. But just remember it's just these two things over here. Sometimes, slope will be specified with the variable m. And they'll say that m is the same thing-- and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. And this notation tends to be kind of complicated, but all this means is, is you take the y-value of your endpoint and subtract from it the y-value of your starting point. That will essentially give you your change in y. And it says take the x-value of your endpoint And that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at-- and actually, we could do it both ways. We could start at this point and go to that point and calculate the slope or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4 comma 2. And let's say that our endpoint is negative 3 comma 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was it's change? You have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. And notice, I implicitly use this formula over here. Our change in x was this value, our endpoint, our end x-value minus our starting x-value. Let's do the same thing for our change in y. Our change in y. If we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y-value and subtract from that your starting y-value and you get 14. So what is the slope over here? Well, the slope is just change in y over change in x."
    },
    {
        "Q": "At the 4:24 mark in the video, shouldn't you say \"this is our rise\" instead of \"this is our run\"?\n",
        "A": "Yes. If you look at it again, you ll see a box pop up in the lower right corner with the correction.",
        "video_name": "WkspBxrzuZo",
        "timestamps": [
            264
        ],
        "3min_transcript": "Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. And notice, I implicitly use this formula over here. Our change in x was this value, our endpoint, our end x-value minus our starting x-value. Let's do the same thing for our change in y. Our change in y. If we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y-value and subtract from that your starting y-value and you get 14. So what is the slope over here? Well, the slope is just change in y over change in x. over change in x, which is-- our change in y is 14. And our change in x is negative 7. And then if we want to simplify this, 14 divided by negative 7 is negative 2. Now, what I want to show you is, is that we could have done it the other way around. We could have made this the starting point and this the endpoint. And what we would have gotten is the negative values of each of these, but then they would've canceled out and we would still get negative 2. Let's try it out. So let's say that our start point was negative 3 comma 16. And let's say that our endpoint is the 4 comma 2. 4 comma 2. So in this situation, what is our change in x? Our change in x. If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. 4 minus negative 3. And what is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. So if you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. So notice, these are just the negatives of these values from when we swapped them. So once again, this is equal to negative 2. And let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. So let me draw our two points. So this is my x-axis. That is my y-axis. So this point over here, 4 comma 2. So let me graph it."
    },
    {
        "Q": "\nAt 6:22, Sal writes r(theta). Does he mean f(theta) as r = f(theta) or am I getting mixed up?",
        "A": "Yup he just used both r (theta) and f (theta) as representations of the polar function.",
        "video_name": "qVn_Lfec-Ac",
        "timestamps": [
            382
        ],
        "3min_transcript": "So each of these things that I've drawn, let's focus on just one of these wedges. I will highlight it in orange. So instead of the angle being theta let's just assume it's a really, really, really small angle. We'll use a differential although this is a bit of loosey-goosey mathematics but the important here is to give you the conceptual understanding. I could call it a delta theta and then eventually take the limit as our delta theta approaches zero. But just for conceptual purposes when we have a infinitely small or super small change in theta, so let's call that d theta, and the radius here or I guess we could say this length right over here. You could view it as the radius of at least the arc right at that point. It's going to be r as a function of the thetas that to call that our r right over there. And so what is going to be the area of this little sector? Well the area of this little sector is instead of my angle being theta I'm calling my angle d theta, this little differential. So instead of one half r squared it's going to be, let me do that in a color you can see. This area is going to be one half r squared d theta. Notice here the angle was theta, here the angle was d theta, super, super small angle. Now if I wanted to take the sum of all of these from theta is equal to alpha to theta is equal to beta and literally there is an infinite number of these. This is an infinitely small angle. Well then for the entire area right over here I could just integrate all of these. So that's going to be the integral from alpha to of course, is a function of theta. So you could even write it this way, you could write it as the integral from alpha to beta of one half r of theta squared d theta. Just to remind ourselves or assuming r is a function of theta in this case."
    },
    {
        "Q": "At 0:00, we learn?\n",
        "A": "Yes, we learn!",
        "video_name": "daCT_24RnIY",
        "timestamps": [
            0
        ],
        "3min_transcript": "I have this figure here. You could call it a rectangular prism. And I want to measure its volume. And I'm defining my unit cube as being a 1 centimeter by 1 centimeter by 1 centimeter cube. It has 1 centimeter width, 1 centimeter depth, 1 centimeter And I will call this, this is equal to 1 cubic centimeter. So I want to measure this volume in terms of cubic centimeters. We've already seen that we can do that by saying, hey, how many of these cubic centimeters can fit into this figure without them overlapping in any way? So if we had this in our hands, we could kind of try to go around it and try to count it, but it's hard to see here because there's some cubes that we can't see behind the ones that we are seeing. So I'm going to try different tactics at it. So first, let's just think about what we can observe. So we see that this one, if we measure its different dimensions, its width, it's 2 of the unit length wide. So it's 2 centimeters wide. It's 4 of our unit length-- we're defining our unit length as a centimeter-- it's 4 of our unit length high. And it is 3 of our unit length deep. So this dimension right over here is 3 centimeters. So I want to explore if we can somehow use these numbers to figure out how many of these cubic centimeters would fit into this figure. And the first way I'm going to think about it is by looking at slices. So I'm going to take this slice right over here of our original figure. And let's think about how using these numbers, we can figure out how many unit cubes were in that slice. Well this is 2 centimeters wide and it is 4 centimeters high. And you might be saying, hey Sal, I could just count these things. I could get 8 squares here. But what if there was a ton there? It would be a lot harder. And you might realize well I could just the area of this surface right over here. And it's only 1 deep so that also will give me the number of cubes. Let's find the area here. Well that's going to be 2 centimeters times 4 centimeters. That gives us the area of this. And then if we want to find out the number of cubes, well that's also going to be equivalent to the number of cubes. So we have 8 square centimeters is this area, and the number of cubes is 8. And if we want the number of cubes in the whole thing, we just have to multiply by the number of slices. And we see that we need one, two, three slices. This is 3 centimeters deep. So we're going to multiply that times 3. So we took the area of one surface. We took the area of this surface right over here. And then we multiply by the depth, that essentially gives us the number of cubes because the area of this surface gives us the number of cubes in an slice that is 1 cube deep."
    },
    {
        "Q": "I already know what wide is, but what does it mean by 'deep'? Sal was saying it at 5:05.\n",
        "A": "anything with 3 dimensions. A rectangle has length and width , a box has depth as well. Think of a swimming pool , depth would be the bottom to the top although normally a pool isn t filled to the brim. In khan you will see swimming pool questions and you have to watch out for the depth of the water as well as the depth of the container",
        "video_name": "daCT_24RnIY",
        "timestamps": [
            305
        ],
        "3min_transcript": "So we would have to have this is 1 slice. We would have to have another slice, another slice and then another slice in order to construct the original figure. So 2 centimeters times 4 centimeters times 3 centimeters would give us our volume. Let's see if that works out. 2 times 4 is 8 times 3 is 24. Let me do that in that pink color. 24 centimeters cubed, or I could say cubic centimeters. So that's one way to measure the volume. Now there's multiple surfaces here. I happened to pick this surface, but I could have picked another one. I could have picked this surface right over here and done the exact same thing. So let's pick this surface and do the exact same thing. This surface is 3 centimeters by 4 centimeters. Let me do that in that blue color. So its area is going to be 12 square centimeters is the area of this surface. And 12 is also the number of cubes that we have in that slice. And so how many slices do we need like this in order to construct the original figure? Well we need, it's 2 centimeters deep. This is only 1 centimeter deep so we need two of them to construct the original figure. So we can essentially find the area of that first surface which was 3 times 4, and then multiply that times the width, times how many of those slices you need, so times 2. And once again, this is going to be 3 times 4 is 12 times 2 is 24. I didn't write the units this first time. But that's going to give us the count of how many cubic centimeters we have, how many unit cubes we can fit. So once again, this is 24 cubic centimeters. And you could imagine, you could do the same thing, but with the top surface. The top surface is 3 centimeters deep. And 2 centimeters wide. So you could view its area or its area is going to be 3 centimeters times 2 centimeters. So that area is-- let me do it in the same colors-- 3 centimeters times 2 centimeters which is 6 square centimeters. And that also tells you that there's going to be 6 cubes in this one cube deep slice. But how many of these slices do you need? Well you have this whole thing is 4 centimeters tall, and this thing is only 1 centimeter so you're going to need four of them. So that's 2, 3, try to draw it as neatly as I can, and 4. You're going to need 4 of these."
    },
    {
        "Q": "Why does Sal include a number line at 2:30 to represent is process to tackle this problem when he could just do count to 73 from 68 to show that we added 5 to get there?\n",
        "A": "Because Sal just wanna show how we get 73, by breaking up 5 into 3+2, you can add 2 to 68 first,and then add 3 to 70 to get 73. This is an easy method of solving the problem like that, technically. If we just count from 68 to 73, how could we do if we need to add 68 to 29? continue to count?",
        "video_name": "DzJvR56Suss",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] What I wanna show you now is a way to add numbers, or a way that I sometimes add numbers, when I'm doing it in my head. So let's say I wanna add five to 68. And we've already seen other ways of doing it, so what I'm showing in this video isn't the only way to do it but it might be helpful when you're doing it in your head. Well when I look at a number like 68 I think well gee, If I added just two to that I would get to 70. So why don't I break up five into three and two, add the two to 68 and then I have to add the three. So what's that going to get me? Well if I break up the five into three plus two, and the whole reason why I broke it up into three and two is so I have this two here to add to 68. So five is three plus two, and to that I'm going to add 68, 68. And once again the reason why I took this two out is because I said hey, what do I have to add to 68 to get to 70? So now I could just rewrite this as, But now I can add the two to the 68. So two plus 68 is going to be Two plus 68 is 70. 70 and I still have this three here, so I have three plus 70 which is equal to 73. 73. Now it might seem like this really long way of doing it, but this is just a way to think about it. In your head you'd say okay five plus 68, let's see if I add two to 68 I'll get to 70 and so let's see, five is three plus two, I add the two to 68, I get to 70, and I have three left over. So it's going to be 73. Another way you could think about it is on a number line. So let's draw ourselves a number line here. Let's draw a number line and let me draw some And let's say that this right over here, this right over here is 68, and so this is going to be 70, this is going to be 70, and we're gonna add five. We're gonna add five. So if you add five you say well let me add two first. So if I add two first I get to 70. So plus two, that's what I did right over here to get to 70, and then if I wanna add five then I have to add three more. Then I have to add three more. And so it's going to be 70 plus three. Which of course is 73. Hopefully you found that interesting."
    },
    {
        "Q": "\nAt 1:28 into the video you wrote a symbol which had a = sign and something that looks like a vertical line and a greater than symbol. Would you please explane what it is? Thanks",
        "A": "I think Sal is just drawing an arrow to show that he is changing the expression into a simplified form.",
        "video_name": "Q1vMNyIP4Us",
        "timestamps": [
            88
        ],
        "3min_transcript": "What I want to do in this video is write the algebraic expressions that represent the same thing that these statements are saying. So this first statement, they say the sum of negative 7 and the quantity 8 times x. So the sum-- so we're going to have an addition here-- of negative 7 and the quantity 8 times x. So the quantity 8 times x, well, that's just 8x. So I can just write 8x over there. So it is negative 7 plus 8x. Or you could view this as the sum of negative 7 and the quantity 8 times x. Let's do the next one. Take the quantity negative 3 times x and then add 1. So the quantity negative 3 times x, we can write that as negative 3x. And then we need to add 1 to that. So that's going to be plus 1. Negative 6 plus-- so we can write negative 6 plus something-- the product of negative 1 and x. So the product of negative 1 and x, that's just going to be negative 1x, which is the same thing as negative x. So we can write this as negative 6 plus negative x. Or we can just write this-- this is the exact same thing as negative 6 minus x. And we are done."
    },
    {
        "Q": "\nWhat the heck is a reciprocal? 4:00",
        "A": "Example : 1/2 and 2/1= 2/2=1",
        "video_name": "f3ySpxX9oeM",
        "timestamps": [
            240
        ],
        "3min_transcript": "We could write times 3 like that. Or, if we want to write 3 as a fraction, we know that 3 is the same thing as 3/1. And we already know how to multiply fractions. Multiply the numerators. 8 times 3. So you have 8-- let me do that that same color. You have 8 times 3 in the numerator now, 8 times 3. And then you have 3 times 1 in the denominator. Which would give you 24/3, which is the same thing as 24 divided by 3, which once again is equal to 8. Now let's see if this still makes sense. Instead of dividing by 1/3, if we were to divide by 2/3. So let's think about what 8/3 divided by 2/3 is. if we wanted to break up this section from 0 to 8/3 into sections of 2/3, or jumps of 2/3, how many sections, or how many jumps, would I have to make? Well, think about it. 1 jump-- we'll do this in a different color. We could make 1 jump. No, that's the same color as my 8/3. We could do 1 jump. My computer is doing something strange. We could do 1 jump, 2 jumps, 3 jumps, and 4 jumps. So we see 8/3 divided by 2/3 is equal to 4. Now, does this make sense in this world right over here? Well, if we take 8/3 and we do the same thing, saying hey, look, dividing by a fraction is the same thing as multiplying by a reciprocal. Well, let's multiply by 3/2. So we swap the numerator and the denominator. So we multiply it times 3/2. And then what do we get? In the numerator, once again, we get 8 times 3, which is 24. And in the denominator, we get 3 times 2, which is 6. So now we get 24 divided by 6 is equal to 4. Now, does it make sense that we got half the answer? If you think about the difference between what we did here and what we did here, these are almost the same, except here we really just didn't divide. Or you could say you divided by 1, while here you divided by 2. Well, does that make sense? Well, sure. Because here you jumped twice as far. So you had to take half the number of steps. And so in the first example, you saw why it makes sense to multiply by 3. When you divide by a fraction, for every whole, you're making 3 jumps. So that's why when you divide by this fraction,"
    },
    {
        "Q": "\nat 0:28,why is 5+5+8,5+8+5,and 8+5+5 exactly the same?",
        "A": "how do you have 100,00 and you joined 2 m ago??",
        "video_name": "HwSszh3L358",
        "timestamps": [
            28
        ],
        "3min_transcript": "Use the commutative law of addition-- let me underline that-- the commutative law of addition to write the expression 5 plus 8 plus 5 in a different way and then find the sum. Now, this commutative law of addition sounds like a very fancy thing, but all it means is if you're just adding a bunch of numbers, it doesn't matter what order you add the numbers in. So we could add it as 5 plus 8 plus 5. We could order it as 5 plus 5 plus 8. We could order it 8 plus 5 plus 5. These are all going to add up to the same things, and it makes sense. If I have 5 of something and then I add 8 more and then I add 5 more, I'm going to get the same thing as if I had took 5 of something, then added the 5, then added the 8. You could try all of these out. You'll get the same thing. Now, they say in a different way, and then find the sum. The easiest one to find the sum of-- actually, let's do But the easiest one, just because a lot of people immediately know that 5 plus 5 is 10, is to maybe start with So if you have 5 plus 5, that's 10, plus 8 is equal to 18. Now, let's verify that these two are the same exact thing. Up here, 5 plus 8 is 13. 13 plus 5 is also 18. That is also 18. If we go down here, 8 plus 5 is 13. 13 plus 5 is also equal to 18. So no matter how you do it and no matter what order you do it in-- and that's the commutative law of addition. It sounds very fancy, but it just means that order doesn't matter if you're adding a bunch of things."
    },
    {
        "Q": "At 2:30 he says that the shape can be a rhombus. I thought rhombuses could not have any right angles!?\n",
        "A": "Definition of rhombus is all sides are equal, and opposite sides are parallel. So a square is a particular type of rhombus.",
        "video_name": "wPZIa3SjPF0",
        "timestamps": [
            150
        ],
        "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right."
    },
    {
        "Q": "\nAt 1:36, Sal says the lines are parallel. How do you know for sure that they are parallel?",
        "A": "Great point. We do not know for sure. However, we can take it as a given from the problem.",
        "video_name": "wPZIa3SjPF0",
        "timestamps": [
            96
        ],
        "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right."
    },
    {
        "Q": "At 1:38 she said that they found the highest prime number so far? How have they not found more?\n",
        "A": "Because finding primes is hard. I could explain exactly why it is hard, but I will instead ask you a question: is 1000000007 (one billion seven) prime? If you try to find it out without a computer, you will soon see the difficulties behind finding primes.",
        "video_name": "Yhlv5Aeuo_k",
        "timestamps": [
            98
        ],
        "3min_transcript": "Pretend you're me and you're in math class. Actually... nevermind, I'm sick so I'm staying home today so pretend you are Stanislaw Ulam instead. What I am about to tell you is a true story. So you are Stan Ulam and you're at a meeting but there's this really boring presentation so of course you're doodling and, because you're Ulam and not me, you really like numbers... I mean <i>super</i> like them. So much that what you're doodling is numbers, just counting starting with one and spiralling them around. I'm not too fluent in mathematical notation so so i find things like numbers to be distracting, but you're a number theorist and if you love numbers who am I to judge? Thing is, because you know numbers so intimately, you can see beyond the confusing, squiggly lines you're drawing right into the heart of numbers. And, because you're a number theorist, and everyone knows that number theorists are enamoured with prime numbers( which is probably why they named them \"prime numbers\"), the primes you've doodled suddenly jump out at you like the exotic indivisible beasts they are... So you start drawing a heart around each prime. Well... it was actually boxes but in my version of the story it's hearts because you're not afraid to express your true feelings about prime numbers. You can probably do this instantly but it's going to take me a little longer... I'm all like - \"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just"
    },
    {
        "Q": "\nat 2:31 what is a prespictipal",
        "A": "Simply put, a reciprocal of a fraction is when you flip the fraction upside down. For example, the reciprocal of 5/7 is 7/5. The reciprocal of 3/4 is 4/3.",
        "video_name": "d8vvVjfTbYY",
        "timestamps": [
            151
        ],
        "3min_transcript": "Odie the octopus swam home. He swam at a constant speed. He plotted how far he'd gone every 1/7 of a second on the graph below. Not all octopi can create graphs, but Odie was no ordinary octopus. And actually, octopi are some of the smartest creatures on this planet. What was Odie's speed-- after probably dolphins and great apes and probably most primates, but they're still fairly smart, smarter than most fish. What was Odie's speed in meters per second? Note we use m/s to show the units of meters per second. All right. So to figure out speed in meters per second, we essentially need to divide distance in meters divided by time in seconds. So the easiest thing-- let me find a point that I can clearly read how far Odie has traveled. So it looks like this point right over here. It's clearly that the distance in meters is 4/7. So the distance in meters is 4/7, so I'll do that in the numerator. the distance in meters. Gee, I don't like that con-- well, I'll You know, the contrast with the white background is a little hard on the eye. And It took him 5/7 of a second to get there. So 4/7 meters-- actually, let me make sure I put the units here. 4/7 meters in 5/7 seconds. And so literally, we just need to divide 4/7 by 5/7, and we're going to get it in meters per second. So this is going to be the same thing as 4/7 over 5/7 meters per second. So let me write that down. So 4/7 times 7/5. Dividing by a fraction is the same thing as multiplying by its reciprocal. So times 7/5, and we get this-- well, this 7 and this 7 cancel out, and you get 4 over 5. You could have also done that by multiplying the numerator and the denominator here by 7. If you multiply it by 7, multiply by 7, you would get 4/5. But either way, his speed in meters per second is 4/5 meters per second."
    },
    {
        "Q": "At 8:10, you integrate (5/2)(1/x+1) to 5/2*ln(x+1) explaining that the derivative of the denominator (x+1) is equal to the numerator 1 . Could you please explain this further as I am still very confused?\n",
        "A": "F(x) = \u00e2\u0088\u00ab(5/2\u00e2\u0080\u00a21/(x + 1))dx F(x) = 5/2\u00e2\u0080\u00a2\u00e2\u0088\u00ab(1/(x + 1))dx u = x + 1 (d/dx(x + 1))dx = du (d/dx(x) + d/dx(1))dx = du (d/dx(x) + 0)dx = du (d/dx(x))dx = du (dx/dx)dx = du (1)dx = du dx = (1)du dx = du F(u) = 5/2\u00e2\u0080\u00a2\u00e2\u0088\u00ab(1/u)du F(u) = 5/2\u00e2\u0080\u00a2ln(u) + C u = x + 1 F(x) = 5/2\u00e2\u0080\u00a2ln(x + 1) + C",
        "video_name": "7IkufOBIw5g",
        "timestamps": [
            490
        ],
        "3min_transcript": "I'll do it up here since I have a little bit of real estate. A plus B is going to be equal to one, and B minus A or I could write that as negative A plus B is equal to negative four. We could add the left hand sides and add the right hand sides and then the A's would disappear. We would get two B is equal to negative three or B is equal to negative three halves. We know that A is equal to one minus B, which would be equal to one plus three halves, since B is negative three halves, which is equal to five halves. A is equal to five halves. B is equal to negative three halves. And just like that we can rewrite this whole integral in a way that is a little bit easier to take the anti or this whole expression so it's easier to integrate. So it's going to be the integral of A is five halves and so I could just write that as, let me write it this way, five halves times one over X plus one. I wrote it that way because it's very straight forward to take the antiderivative of this. Then plus B over X minus one. Which is going to be negative three halves. So I'll just write it as minus three halves times one over X minus one. That was this right over here, DX. Notice all I did is I took this expression right over here and I did a little bit of partial fraction expansion into these two, I guess you could say, expressions or terms right over there. It's fairly straight forward to integrate this. Antiderivative of one, it's just going to be X. is going to be plus five halves, the natural log of the absolute value of X plus one. We're able to do that because the derivative of X plus one is just one, so the derivative is there so that we can take the antiderivative with respect to X plus one. You could also do u-substitution like we've done in previous examples, U is equal to X plus one. And over here, this is going to be minus three halves times the natural log of the absolute value of X minus one, by the same exact logic with how we were able to take the antiderivative there. And of course we cannot forget our constant. And there we have it. We've been able to integrate, we were able to evaluate this expression."
    },
    {
        "Q": "\nAt 1:42, Sal says \"the fewest number of candy bars we can buy is zero candy bars\". Why is this so? The definition of a purchase is receiving goods in exchange for money. Surely no money and no candy bars equals no purchase. Do all functions just have to accept zero as an input?",
        "A": "Not all functions have the same words like purchase so this makes it easier to understand so in a less mathy way of saying it is you can buy No candy bars for $0.00( my keyboard does not have the cent symbol). So manly all functions have different meaning of the words so they can all accept 0 to make it easier to understand p(0) completely means 0 candy bars so $0.00 was spent for it. Hope this helps..........Sorry if you already had it. It been a month xD.",
        "video_name": "AiW7syKXfJM",
        "timestamps": [
            102
        ],
        "3min_transcript": "Thomas has 400 candy bars in his shop and each cost 50 cents. Let p of b denote the price, p, measured in dollars of a purchase of b candy bars. Alright I input b, the number candy bars I wanna buy, and p(b) will tell me what's the purchase price is really just taking the number of candy bars multiplied by 50 cents, but we won't have to worry about that just yet. Which number type is more appropriate for the domain of the function? So just to remind ourselves, what is the domain of a function? A domain is a set of all inputs over which the function is defined. So it is the set of all b's. It is the set of all inputs over which p of b will produce a defined response So let's think about it. Is it integers or real numbers? So I could buy -- b could be 0 candy bars, 1 candy bars, 2 candy bars, all up to 400 candy bars. Could I -- Could I have a fractional can-- Could be b 0.372 of a candy bar? Well, this is a normal candy shop. chunk. You're not going to be able to buy 0.372 of a candy bar. You can either buy a one more or none more, so you buy your 1, 2, 3 all the way up to 400. So I would say integers -- that the domain of this function is going to be is going to be a subset of integers. It's not -- you not, you can't have a real, all real number, but integers are obviously a subset real numbers. But you can't say, hey, I'm gonna buy pi candy bars, or I'm gonna buy the square root of two candy bars. You're gonna buy integer number candy bars. Now they say, define the interval of the domain. So the fewest candy bars I could are 0 candy bars, and I have to decide whether I put a bracket or I put a parenthesis. I can actually buy 0 candy bars so I'm gonna put a bracket. If I put a parentheses, that means I could have values above zero but not including 0, but I want to include 0 so I'm gonna put the bracket there. So the least I could buy can buy. The most I could buy are 400 candy bars, and I can buy 400. So I would put brackets there as well. So the interval of the domain, I would want to select integers. So b is a member of integers such that b is also a member of this interval. It could be as low as 0 including 0, and as high as 400 including 400. Got it right."
    },
    {
        "Q": "At 3:26 why did Sal put the x's on the right and not the left?\n",
        "A": "It doesn t matter which side you move the x s to. Either side will work. Sal likely moved the x s to the right side because it keeps the value positive (which some people prefer).",
        "video_name": "EHR-YDwrrhM",
        "timestamps": [
            206
        ],
        "3min_transcript": "x is negative five, five times negative five, plus two y. See, negative three times negative five is positive 15, plus seven y, is equal to negative 25 plus two y. And now, to solve for y, let's see, I could subtract two y from both sides, so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I'm going to be left with 15 minus 15, that's zero, that's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y. Seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40. Or, if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratchpad here. And I'll take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven, now y is going to be eight, y is eight, seven times eight is equal to five times x, y is eight, two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now, if we wanna get all of our constants on one side, and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all the xs on this side, and put 'em all on this side. So we're gonna add three x to both sides. And, let's see, if we want to get all the constants on the left hand side, we'd wanna get rid of the 16, so we could subtract 16 from the right hand side, if we do it from the right, we're gonna have to do it from the left as well. And we're gonna be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40."
    },
    {
        "Q": "\nAt 6:43, when you did 8/5 times -5 how did you end up with -8?",
        "A": "8/5 (-5) = 8/5 * (-5/1 ) = 8 (-5) / (5 * 1) = -40/5 = -8 hope this helps.",
        "video_name": "EHR-YDwrrhM",
        "timestamps": [
            403
        ],
        "3min_transcript": "We can check our answer, if we like. We got it right. Now, I said there was two ways to tackle it, I kind of just did it, I guess you could say, the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this, and then solved for x. Another way that I could have done it, that actually probably would have been, or, it would for sure, would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done, right from the get-go, is said, \"Hey, let's put all my xs on one side, \"and all my ys on the other side.\" So this is negative three x plus seven y is equal to five x plus two y. Now let's say I wanna get all my ys on the left and all my xs on the right. So I don't want this negative three x on the left, so I'd wanna add three x. Adding three x would cancel this out, but I can't just do it on the left hand side, I have to do it on the right hand side as well. And then, if I wanna get rid of this two y on the right, I could subtract two y from the right, but, of course, I'd also wanna do it from the left. So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8/5 x. So, this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this, would satisfy this, and vice versa. And this is much easier. Because if x is now negative five, if x is negative five, y would be 8/5 times negative five, well, that's going to be negative eight. And when y is equal to eight, well, you actually could even do this up here, you could say five times eight is equal to eight x, and then you could see, well five times eight the same thing as eight times five, so x would be equal to five. So I think this would actually have been You see it all, I was able to do the entire problem instead of having to do all of this, slightly, slightly hairier, algebra."
    },
    {
        "Q": "Hello! Okay, so at 1:03 in the video, how did he know that y>= -2? I get how he got it from the graph, but how would one find that algebraically? Please respond quickly, thank you so much!\n",
        "A": "Algebraically, the equation was y = ( x - 1 )^2 - 2. Whatever value of x is substituted, the quantity ( x - 1 )^2 will be positive or zero; it will never be negative. Therefore, we ll always be adding a positive number (or zero) to -2, so the total value will be getting larger than (or staying equal to) -2.",
        "video_name": "Bq9cq9FZuNM",
        "timestamps": [
            63
        ],
        "3min_transcript": "We have the function f of x is equal to x minus 1 squared minus 2. And they've constrained the domain to x being less than or equal to 1. So we have the left half of a parabola right here. They've constrained so that it's not a full U parabola. And I'll let you think about why that would make finding the inverse difficult. But let's try to find the inverse here. And a good place to start -- let's just set y being equal to f of x. You could say y is equal to f of x or we could just write that y is equal to x minus 1 squared minus 2. We know it's 4. x is less than or equal to 1. But right now we have y solved for in terms of x. Or we've solved for y, to find the inverse we're going to want to solve for x in terms of y. And we're going to constrain y similarly. We could look at the graph and we could say, well, in this graph right here, this is defined for y being greater than or equal to negative 2. than or equal to negative 2. Because this is going to be -- right now this is our range. But when we swap the x's and y's, this is going to be our domain. So let's just keep that in parentheses right there. So let's solve for x. So that's all you have to do, to find the inverse. Solve for x and make sure you keep track of the domains and the ranges. So, let's see. We could add 2 to both sides of this equation. We get y plus 2 is equal to x minus 1 squared. Minus 2, plus 2, so those cancel out. And then, I'm just going to switch to the y constraint. Because now it's not clear what we're -- whether x is the domain or the range. But we know by the end of this problem, y is going to be the domain. So let's just swap this here. So, 4 for y is greater than or equal to negative 2. And we could also say, in parentheses, x is less than 1. This is -- we haven't solved it explicitly for either one, so we'll keep both of them around right now. Now, to solve for x, you might be tempted to just take the And you wouldn't be completely wrong. But we have to be very, very, very careful here. And this might not be something that you've ever seen before. So this is an interesting point here. We want the right side to just be x minus 1. That's our goal here, in taking the square root of both sides. We want to just have an x minus 1 over there. Now, is x minus 1 a positive or a negative number? Well, we've constrained our x's to being less than 1. So we're dealing only in a situation where x is less than or equal to 1. So if x is less than or equal to 1, this is negative. This is negative. So we want to take the negative square root. Let me just be very clear here. If I take negative 3 -- I take negative 3 and I were to square it, that is equal to 9. Now, if we take the square root of 9 -- if we take the square"
    },
    {
        "Q": "At 6:53 in, shouldn't the point be (3,-1), not (2,-1)?\n",
        "A": "The point must be (3,1), not (-1,2).",
        "video_name": "Bq9cq9FZuNM",
        "timestamps": [
            413
        ],
        "3min_transcript": "You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just Is equal to the negative square root of x plus 2 plus 1 for, I'm just renaming the y, for x is greater than or equal to negative 2. And if we were to graph this, let's see. If we started at x is equal to negative 2, this is 0. So the point negative 2, 1 is going to be on our graph. So negative 2, 1 is going to be on our graph. Let's see, if we go to negative 1, negative 1 this will become a negative 1. Negative 1 is 0 on our graph. Negative 1, 0 is on our graph. And then, let's see. If we were to do, if we were to put x is equal to 2 here. So x is equal to 2 is 4. 4 square root, principle root is 2. It becomes a negative 2 So it becomes 2, negative 1. So that's on our graph right there. So the graph is going to look something like this, of f inverse. It's going to look something like that right there. f of x along the line y is equal to x. Along the line y is equal to x. Because we've essentially just swapped the x and the y. This is about as hard of an inverse problem that I expect you to see. Especially in a precalculus class because it really is tricky to realize that you have to take the negative square root here. Because the way our domain was constrained, this value right here is going to be negative. So to solve for it, you want to have the negative square root."
    },
    {
        "Q": "At 5:05 Sal mentions interval. Can someone explain to me what this 'interval' means? I keep on hearing this term more and more often.\n",
        "A": "In this context, interval just refers to the section of points (along the x-axis) between x1 and x2. If it were an interval of time, for example, it could be the segment of time between two events. Hope this helps!",
        "video_name": "8r8Vp_1iB4k",
        "timestamps": [
            305
        ],
        "3min_transcript": "You could have done that. Then you would have just gotten the negative of each of these values in the numerator and denominator, but they would have canceled out. The important thing is that you're consistent. If you're subtracting you're starting value from your ending value in the numerator, you have to subtract your starting value from your ending value in the denominator as well. So this right here you probably remember from algebra class. The definition of slope is the rate of change of y with respect to x. Or it's the rate of change of our vertical axis, I should say, with respect to our horizontal axis. Or change in y, or change in our vertical axis over change in a horizontal axis. Now I'm going to introduce a little bit of a conundrum. So let me draw another axis right over here. Scroll over a little bit just so we have some space to work with. So that was for a line. And a line, by definition, has a constant slope. If you calculate this between any two points on the line, it's going to be constant for that line. But what happens when we start dealing with curves? So let's imagine a curve that looks something like this. So what is the rate of change of y with respect to x of this curve? Well, let's look at it at different points. And we could at least try to approximate what it might be in any moment. So let's say that this is one point on a curve. Let's call that x1, and then this is y1. And let's say that this is another point on a curve right over here, x2. And let's call this y2. So this is a point x1, y1, this is a point x2, y2. So we don't have the tools yet. And this is what's exciting about calculus, we will soon have the tools to figure out, what is the rate of change of y with respect to x at exactly this point? But we don't have that tool yet. But using just the tools from algebra, what is the average rate of change over the interval from x1 to x2? Well, what's the average rate of change? Well, that's just how much did my y change-- so that's my change in y-- for this change in x. And so we would calculate it the same way. y2 minus y1 over x2 minus x1. So our change in y over this interval is equal to y2 minus y1, and our change in x is going to be equal to x2 minus x1. So just like that we were able to figure out the rate of change between these two points. Or another way of thinking about it is, this is the average rate of change for the curve between x equals x1, and x is equal to x2. This is the average rate of change of y with respect to x over this interval."
    },
    {
        "Q": "\nWhy is he multiplying the -(3x - 4) @ 1:10? Can he not also just add that?",
        "A": "The minus in front of the (3x - 4) says to subtract the entire binomial. The minus sign must be distributed to accomplish that subtraction. If you just drop the parentheses and add, then the -4 is being added, not subtracted (you have +(-4) when you need -(-4)). Hope this helps.",
        "video_name": "DMyhUb1pZT0",
        "timestamps": [
            70
        ],
        "3min_transcript": "We're asked to simplify this huge, long expression here. x to the third plus 3x minus 6-- that's in parentheses-- plus negative 2x squared plus x minus 2. And then minus the quantity 3x minus 4. So a good place to start, we'll just rewrite this and see if we can eliminate the parentheses in this step. So let's just start at the beginning. We have the x to the third right over there. So x to the third and then plus 3x-- I'll do that in pink-- plus 3x. And then we have a minus 6. And we don't have to put the parentheses around there, those don't really change anything. And we don't have to even write these-- do anything with these parentheses. We can eliminate them. Just because there's a positive sign out here we don't have to distribute anything. Distributing a positive sign doesn't do anything to these numbers. So then plus, we have a negative 2x squared. So this term right here is negative 2x squared, or minus x squared. And then we have a plus x. We have a plus x. Then we have a negative sign times this whole expression. So we're going to have to distribute the negative sign. So it's a positive 3x, but it's being multiplied by negative 1. So it's really a negative 3x. So minus 3x, then you have a negative-- you can imagine this is a negative 1 implicitly out here-- negative 1 times negative 4. That's a positive 4. So plus 4. Now, we could combine terms of similar degree, of the same degree. Now, first we have an x to the third term and I think it's the only third degree term here, because we have x being raised to the third power. So let me just rewrite it here. We have x to the third. And now let's look at our x squared terms. Looks like we only have one. We only have this term right here. So we have minus 2x squared. And then what about our x terms? We have a 3x plus an x minus a 3x again. So that 3x minus the 3x would cancel out, and you're just So plus x. And then finally our constant terms. Negative 6 minus 2 plus 4. Negative 6 minus 2 gets us to negative 8. Plus 4 is negative 4. We have simplified the expression. Now we just have a four term polynomial."
    },
    {
        "Q": "\nAt 3:08, Vertically, I don't see 2 blocks.\nThanks\n---- Moon trainer",
        "A": "I code on Scratch too! My username is Song_of_the_Cats",
        "video_name": "gkifo46--JA",
        "timestamps": [
            188
        ],
        "3min_transcript": "And so instead, let me see if I can get away with making a triangle just like that. Now, I just did that. So let me try to raise this up to see if I can make another triangle where it would be easy to figure out its dimensions. So once again, I don't want to go all the way up here because now I'm not at a whole unit. Instead, let me take a right and go right over here. And notice, both of these are very easy to figure out its dimensions. This is 1, 2, 3, 4, 5 units long and 1 unit high. This one right over here is 1, 2, 3, 4 units long and 1, 2 units wide. So let's see if we can cover the entire quadrilateral, if we can break it up, I should say, into a bunch of figures like this. And then I could drop this down, and then we're done. All of these are pretty straightforward to figure out what their dimensions are. This is 5 by 1. This is 4 by 2. This is 1, 2, 3, 4, 5, 6 by 1, 2. And this is 1 by 1, 2, 3, 4, 5. So what is the area of this figure? And of course, we have this center rectangle right over here. Well, a triangle that is 5 units long and 1 unit high, its area is going to be 1/2 times 1 times 5. Or I could write it 1/2 times 1 times 5, depending on what multiplication symbol you are more comfortable with. Well that's just going to be 1/2 times 5, which is going to be equal to 2.5. So that's 2.5 right over there. This one is going to be 1/2 times 4 times 2. This one is going to be 1/2 times 2 times 1, 2, 3, 4, 5, 6. Well, 1/2 half times 2 is 1 times 6 is just 6. And then this one's going to be 1/2 times 1 times 1, 2, 3, 4, 5. So once again, the area of this one is going to be 2.5. And then finally, this is a 3 by 4 rectangle. And you could even count the unit squares in here. But it has 12 of those unit square, so it has an area of 12. So if we want to find the total area, we just add all of these together. So 2.5 plus 2.5 is 5, plus 4 is 9, plus 6 is 15, plus 12 is 27. So it has a total area of 27."
    },
    {
        "Q": "\n@ 4:50 - I would have thought that since we are solving for where the greater change occurs, g'(4) < g'(6) would be correct, as the change at g'(6) is far more 'steep'. Its obviously that -1 > -3, however I thought that we're really solving for here is the rate of change.",
        "A": "It is descending more at 6 but the derivative gives how much something increases as we move along the graph. So if it decreases then it will be negative.",
        "video_name": "S-dcMvJlMJs",
        "timestamps": [
            290
        ],
        "3min_transcript": "the slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these. So alright, so they're telling us to compare the derivative of G at four to the derivative of G at six and which of these is greater and like always, pause the video and see if you can figure this out. Well this is just an exercise let's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that. So now that wouldn't, that doesn't do a good job so right over here at that looks like a I think I can do a better job than that no that's too shallow to see not shallow's not the word, that's too flat. So let me try to really okay, that looks pretty good. seems to be indicative of the rate of change of Y with respect to X or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is, it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it's in the negative direction. As we increase, think of it this way as we increase X one here it looks like we are decreasing Y by about one. So it looks like G prime of four G prime of four, the derivative when X is equal to four is approximately, I'm estimating it negative one while the derivative here when we increase X if we increase X by if we increase X by one so G prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative so it's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in Y with respect to X then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts, it's still decreasing but it's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase, increase, so on and so forth"
    },
    {
        "Q": "\nAt 0:33, what does translate mean?",
        "A": "Moving the entire figure, without changing its size or orientation.",
        "video_name": "6p1lweGactg",
        "timestamps": [
            33
        ],
        "3min_transcript": "These two quadrilaterals, EFGH and ABCD, these are similar. If through some combination of translations, rotations, reflections, and dilations, I can make them sit on top of each other. And similarity allows us to use dilations, which essentially means scale up one of the figures up or down. If we're talking about congruence, then we wouldn't be scaling the figures up or down. We would only be translating, rotating, and reflecting. But let's see if we can do it. So let's first translate it. Let's first translate this figure. So let's make that point on top of that point. And now let's rotate it. I'm going to rotate it around point E right over here. So I get right over there. So I got two of the sides to kind of match up. But now let me dilate it down. Let me scale it down. And so this is-- let me put it on what I want to not change-- and then let me see if I can scale this down. And I was able to. So through just purely translations, rotations-- rotations, and dilations, I was able to make these sit on top of each other. So these two figures are similar. So yes, the rectangles are similar. Let's do one more of these. So these two triangles-- and just eyeballing it, this one looks kind of taller than this one does right over here. So they don't feel similar, but let me at least try. So let me translate, maybe make this point C might correspond to point F, although it already looks pretty clear that it won't. But now let me dilate it. So I'll keep that point where it is. And let me try to scale this up. And then we see pretty clearly we made segment CB sit on top of, after scaling it up, sit on top of segment FE. But then everything else is not matching up. Point D is in a very different place than where point A is once I try to scale things So these two triangles are clearly, clearly not similar."
    },
    {
        "Q": "At 5:36, the outer radius is defines as 2-y^2.\nCan anybody explain why ones takes 2 - the function. My first guess was the function + 2 so that i would reach the line where x =2. This is wrong, but i can\u00c2\u00b4t say i really understand why.\n\nAnd, taking 2-y^2 would that be the gap between the undefined point on the x-axes and x=2? Where is this on our shape?\nThanks.\n",
        "A": "At this stage in the process we re trying to find the radius of the outer disk. The radius is the distance from the center of the disk to the perimeter. At the beginning of the video we re told that the figure is created by rotation around the line x=2, so we know the center of the disk is at x=2. The outer edge of the larger disk is at the point x=y^2, so the distance between the two points is 2-y^2.",
        "video_name": "WAPZihVUmzE",
        "timestamps": [
            336
        ],
        "3min_transcript": "going to be pi times outer radius squared minus pi-- let me write it this way-- outer-- well, I'll just write it-- outer radius-- this is going to be as a function of y-- outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize-- actually, I'll try it both places. So it's this entire distance right over here, essentially, the distance between the vertical line, the horizontal distance between our vertical line, and our outer function. It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area. I'll leave the pi there. So it's going to be pi-- right over here-- it's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared-- and let me just-- well, I'll just write it. 2 minus y squared-- and we're going to square that-- squared, minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one, too. So this gives us the area of one of our rings as a function of y, the top of the ring, where I shaded in orange. And now, if we want the volume of one of those rings, we have to multiply it by its depth or its height the way we've drawn it right over here. And its height-- we've done this multiple times already right over here-- is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings."
    },
    {
        "Q": "\nAt 5:29, I was taught to do it in the reverse order like (y^2-2), is that wrong?",
        "A": "In that case the radius would be negative, but it is just a distance, it shouldn t be negative, but it has the same magnitude so when you square it, I guess it works out just the same. Anyway how Sal does it is the most intuitive and rigorously correct manner.",
        "video_name": "WAPZihVUmzE",
        "timestamps": [
            329
        ],
        "3min_transcript": "x is equal to y squared. That's our top function the way we've drawn a kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that and it all works out because we're operating in the first quadrant That's the part of it that we care about. So you're going to get x is equal to the square root of y. That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area-- let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius, and then I subtract out the area of the circle constructed by the inner radius, just kind of subtract it out. going to be pi times outer radius squared minus pi-- let me write it this way-- outer-- well, I'll just write it-- outer radius-- this is going to be as a function of y-- outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize-- actually, I'll try it both places. So it's this entire distance right over here, essentially, the distance between the vertical line, the horizontal distance between our vertical line, and our outer function. It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area."
    },
    {
        "Q": "At 5:15,what is associative property?\n",
        "A": "The associative property of multiplication just means that when you multiply the numbers, it doesn t matter what order you do it in. (4 x 5) x 6 is (20) x 6, which is 120. 4 x (5 x 6) is 4 x 30, which is also 120.",
        "video_name": "VXrn5HOQmHQ",
        "timestamps": [
            315
        ],
        "3min_transcript": "So you could do that as 2 times 3. And we have one 3 and another 3. So in this purple zone, this is another 2 times 3. We have another 2 times 3. I wrote 2 times 2. 2 times 3. We have another 2 times 3. And then finally, we have a fourth 2 times 3. So how many 2 times 3's do we have here? Well, we have one, two, three, four 2 times 3's. So this whole thing could be written as 4 times 2 times 3. Now, what's this going to be equal to? Well, it needs to be equal to 24. And we can verify 2 times 3 is 6 times 4 is, indeed, 24. So the whole idea of what I'm trying to show here is that the order in which you multiply does not matter. Let me pick a different example, a completely new example. So let's say that I have 4 times 5 times 6. You can do this multiplication in multiple ways. You could do 4 times 5 first. Or you could do 4 times 5 times 6 first. And you can verify that. I encourage you to pause the video and verify that these two things are equivalent. And this is actually called the associative property. It doesn't matter how you associate these things, which of these that you do first. Also, order does not matter. And we've seen that multiple times. Whether you do this or you do 5 times 4 times 6-- notice I swapped the 5 and 4-- this doesn't matter. Or whether you do this or 6 times 5 times 4, Here I swapped the 6 and the 5 times 4. And I encourage you to pause the video. So when we're talking about which one we do first, whether we do the 4 times 5 first or the 5 times 6, that's called the associative property. It's kind of fancy word for a reasonably simple thing. And when we're saying that order doesn't matter, when it doesn't matter whether we do 4 times 5 or 5 times 4, that's called the commutative property. And once again, fancy word for a very simple thing. It's just saying it doesn't matter what order I do it in."
    },
    {
        "Q": "At 2:00, what are they trying to say?\n",
        "A": "in the whole column there are 8 balls but it is divided into 2 groups of four. So 2 times four = 8",
        "video_name": "VXrn5HOQmHQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "So if you look at each of these 4 by 6 grids, it's pretty clear that there's 24 of these green circle things in each of them. But what I want to show you is that you can get 24 as the product of three numbers in multiple different ways. And it actually doesn't matter which products you take first or what order you actually do them in. So let's think about this first. So the way that I've colored it in, I have these three groups of 4. If you look at the blue highlighting, this is one group of 4, two groups of 4, three groups of 4. Actually, let me make it a little bit clearer. One group of 4, two groups of 4, and three groups of 4. So these three columns you could view as 3 times 4. Now, we have another 3 times 4 right over here. This is also 3 times 4. We have one group of 4, two groups of 4, and three groups of 4. So you could view these combined as 2 times 3 times 4. We have one 3 times 4. And then we have another 3 times 4. some more space-- as 2 times-- let me do that in blue-- 2 times 3 times 4. That's the total number of balls here. And you could see it based on how it was colored. And of course, if you did 3 times 4 first, you get 12. And then you multiply that times 2, you get 24, which is the total number of these green circle things. And I encourage you now to look at these other two. Pause the video and think about what these would be the product of, first looking at the blue grouping, then looking at the purple grouping in the same way that we did right over here, and verify that the product still equals 24. Well, I assume that you've paused the video. So you see here in this first, I guess you could call it a zone, we have two groups of 4. So this is 2 times 4 right over here. We have one group of 4, another group of 4. We have one group of 4, another group of 4. So this is also 2 times 4 if we look in this purple zone. One group of 4, another group of 4. So this is also 2 times 4. So we have three 2 times 4's. So if we look at each of these, or all together, this is 3 times 2 times 4, so 3 times 2 times 4. Notice I did a different order. And here I did 3 times 4 first. Here I'm doing 2 times 4 first. But just like before, 2 times 4 is 8. 8 times 3 is still equal to 24, as it needs to, because we have exactly 24 of these green circle things. Once again, pause the video and try to do the same here. Look at the groupings in blue, then look at the groupings in purple, and try to express these 24 as some kind of product of 2, 3, and 4. Well, you see first we have these groupings of 3. So we have one grouping of 3 in this purple zone,"
    },
    {
        "Q": "At 1:38, I got kinda lost.\n",
        "A": "Andrew i can t find where you got stuck on show me pls.",
        "video_name": "VXrn5HOQmHQ",
        "timestamps": [
            98
        ],
        "3min_transcript": "So if you look at each of these 4 by 6 grids, it's pretty clear that there's 24 of these green circle things in each of them. But what I want to show you is that you can get 24 as the product of three numbers in multiple different ways. And it actually doesn't matter which products you take first or what order you actually do them in. So let's think about this first. So the way that I've colored it in, I have these three groups of 4. If you look at the blue highlighting, this is one group of 4, two groups of 4, three groups of 4. Actually, let me make it a little bit clearer. One group of 4, two groups of 4, and three groups of 4. So these three columns you could view as 3 times 4. Now, we have another 3 times 4 right over here. This is also 3 times 4. We have one group of 4, two groups of 4, and three groups of 4. So you could view these combined as 2 times 3 times 4. We have one 3 times 4. And then we have another 3 times 4. some more space-- as 2 times-- let me do that in blue-- 2 times 3 times 4. That's the total number of balls here. And you could see it based on how it was colored. And of course, if you did 3 times 4 first, you get 12. And then you multiply that times 2, you get 24, which is the total number of these green circle things. And I encourage you now to look at these other two. Pause the video and think about what these would be the product of, first looking at the blue grouping, then looking at the purple grouping in the same way that we did right over here, and verify that the product still equals 24. Well, I assume that you've paused the video. So you see here in this first, I guess you could call it a zone, we have two groups of 4. So this is 2 times 4 right over here. We have one group of 4, another group of 4. We have one group of 4, another group of 4. So this is also 2 times 4 if we look in this purple zone. One group of 4, another group of 4. So this is also 2 times 4. So we have three 2 times 4's. So if we look at each of these, or all together, this is 3 times 2 times 4, so 3 times 2 times 4. Notice I did a different order. And here I did 3 times 4 first. Here I'm doing 2 times 4 first. But just like before, 2 times 4 is 8. 8 times 3 is still equal to 24, as it needs to, because we have exactly 24 of these green circle things. Once again, pause the video and try to do the same here. Look at the groupings in blue, then look at the groupings in purple, and try to express these 24 as some kind of product of 2, 3, and 4. Well, you see first we have these groupings of 3. So we have one grouping of 3 in this purple zone,"
    },
    {
        "Q": "\n10:51pm solve for w in the formula -2.4=w/8+10.4",
        "A": "-2.4 = w/8 + 10.4 -2.4 - 10.4 = w/8 + 10.4 - 10.4 -12.8 = w/8 8*(-12.8) = w/8 * 8 -102.4 = w",
        "video_name": "Aig1hkq3OsU",
        "timestamps": [
            651
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n@ 3:54, Sal says there are other ways to write w = (P-2l)/2.\nWhy can't you cancel the 2's and have w = P - l?",
        "A": "(P-2l)/2 = P/2 - (2l)/2 = P/2 - l",
        "video_name": "Aig1hkq3OsU",
        "timestamps": [
            234
        ],
        "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is"
    },
    {
        "Q": "at 3:36, sal has his formula for W written as p-2l / 2 = W\ncouldn't you just do the division and make the formula into 1/2P - L = W?\n",
        "A": "Yes, you can simplify it that way. Sal just decided to leave it as (P-2l)/2.",
        "video_name": "Aig1hkq3OsU",
        "timestamps": [
            216
        ],
        "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is"
    },
    {
        "Q": "At 4:25, what does 'factoring out a two' mean?\n",
        "A": "He explains it as reversing the distributive property. 2(L + W) = 2*L + 2*W = 2L + 2W He is reversing this and since 2 is a factor of both terms, it is called factoring out a 2.",
        "video_name": "Aig1hkq3OsU",
        "timestamps": [
            265
        ],
        "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is"
    },
    {
        "Q": "How come you do not divide by two on the other side? Like I 3:15\n",
        "A": "It s a matter of preference. Personally, I prefer Sal s way: (p-2l)/2 However, you seem to prefer this way: (p/2)-l The question is whether you want there to be one big fraction or one little fraction.",
        "video_name": "Aig1hkq3OsU",
        "timestamps": [
            195
        ],
        "3min_transcript": "that's going to give you, if you add them together, that's going to give you 2 l's. So the perimeter is going to be 2 l's plus 2 w's. They just wrote it in a different order than the way I wrote it. But the same thing, so hopefully that makes sense. Now, their question is, rewrite the formula so that it solves for width. So the formula, the way it's written now, it says P is equal to something. They want us to write it so it's, this w, right here, they want it to be w is equal to a bunch of stuff with l's and P's in it, and maybe some numbers there. So let's think about how we can do this. So they tell us that P is equal to 2 times l, plus 2 times w. We want to solve for w. Well, a good starting point might be to get rid of the l on this side of the equation. And to get rid of it on that side of the equation, we could subtract the 2l from both sides of the equation. So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer."
    },
    {
        "Q": "At 4:28, instead of 9 minus -49, Sal writes 9 minus 49. Why is this? Doesn't the -7 multiply the 7, making it -49?\n",
        "A": "I think you are using the minus sign twice and trying to get -9 -(-49)? It s not clear from your words. The problem has -9 -7 (7). There is one minus sign with the 7 s. You either use it as part of the multiplication and create -9 -49. Or, you multiply 7(7) = +49, then apply the minus and get -9 - (+49) = -9 -49. Hope this helps.",
        "video_name": "TIwGXn4NalM",
        "timestamps": [
            268
        ],
        "3min_transcript": "What is this equal to? And I'll give you a few seconds to give it a little bit of thought. Well, you could imagine. They've given three variables with one equation up here, another two variables with another equation. We have five unknowns with two equations. There's no way we're going to be able to individually figure out what a, or b, or c, or x, or y is. But maybe we can use what we saw in the last example to solve this. And so what we might want do is rearrange it so that we have the x's and the y's kind of grouped together, and the a's, b'c, and c's kind of grouped together. So let's try that out. So let's focus first on the a's, b's, and c's. So we have negative 9a, negative 9b-- I'll go in order, alphabetical order-- negative 9b, and we have negative 9c, and I think you see what's emerging. And then let's work on the x's and the y's. And then we have negative 7x, minus 7x, and then we have a minus 7y. So all I have done is rearranged this expression here. of what's going on here. These first three terms, I can factor out a negative 9, and I get negative 9 times a plus b plus c. And these second two terms, I can factor out a 7, minus 7 times x plus y. And just to verify, if you wanted to go the other way, multiply this negative 7 times x plus y, you'll get this. Multiply negative 9 times a plus b plus c, distribute it, you'll get this right over here. And so this makes it a little bit clearer. What is a plus b plus c equal to? Well, they tell us right over here. a plus b plus c is equal to negative 1. This whole expression is negative 1, at least in the parentheses. And what is x plus y equal to? Let me do this in a new color. What is x plus y equal to? Well, they tell us right over here. x plus y is equal to 7. 9 times negative 1 minus 7 times 7. And so this is equal to-- we're in the home stretch-- negative 9 times negative 1. Well, that's positive 9. And then negative 7 times 7 is negative 49. So it's 9 minus 49, which is equal to negative 40. And we are done."
    },
    {
        "Q": "at 4:38 you say you can factor out a -9 but why is it a negative? If the original equation is a+b+c = -1 then you would need to factor in a positive 9, right? I'm so confused\n",
        "A": "can i just say probably try to rewatch the video and pay attentiion close to what he is saying and watch his steps and if that still doesnt answer your question try to research it",
        "video_name": "TIwGXn4NalM",
        "timestamps": [
            278
        ],
        "3min_transcript": "What is this equal to? And I'll give you a few seconds to give it a little bit of thought. Well, you could imagine. They've given three variables with one equation up here, another two variables with another equation. We have five unknowns with two equations. There's no way we're going to be able to individually figure out what a, or b, or c, or x, or y is. But maybe we can use what we saw in the last example to solve this. And so what we might want do is rearrange it so that we have the x's and the y's kind of grouped together, and the a's, b'c, and c's kind of grouped together. So let's try that out. So let's focus first on the a's, b's, and c's. So we have negative 9a, negative 9b-- I'll go in order, alphabetical order-- negative 9b, and we have negative 9c, and I think you see what's emerging. And then let's work on the x's and the y's. And then we have negative 7x, minus 7x, and then we have a minus 7y. So all I have done is rearranged this expression here. of what's going on here. These first three terms, I can factor out a negative 9, and I get negative 9 times a plus b plus c. And these second two terms, I can factor out a 7, minus 7 times x plus y. And just to verify, if you wanted to go the other way, multiply this negative 7 times x plus y, you'll get this. Multiply negative 9 times a plus b plus c, distribute it, you'll get this right over here. And so this makes it a little bit clearer. What is a plus b plus c equal to? Well, they tell us right over here. a plus b plus c is equal to negative 1. This whole expression is negative 1, at least in the parentheses. And what is x plus y equal to? Let me do this in a new color. What is x plus y equal to? Well, they tell us right over here. x plus y is equal to 7. 9 times negative 1 minus 7 times 7. And so this is equal to-- we're in the home stretch-- negative 9 times negative 1. Well, that's positive 9. And then negative 7 times 7 is negative 49. So it's 9 minus 49, which is equal to negative 40. And we are done."
    },
    {
        "Q": "\nI am still confused about the speed and velocity thing at 7:12. What is that about?",
        "A": "Speed indicates how fast something is moving. Velocity indicates how fast something is moving in a certain direction. Hence, speed is a positive number while velocity is positive or negative depending on the direction.",
        "video_name": "ppBJWf_Wdmc",
        "timestamps": [
            432
        ],
        "3min_transcript": "then that means the speed is increasing. The magnitude is increasing in the leftward direction. So what we really need to do, beyond just evaluating the acceleration at time 5.5, we also have to evaluate the velocity to see if it's going in the left or the rightward direction. So let's evaluate the velocity. The velocity at 5.5-- and we'll just get our calculator out again. Velocity at 5.5-- this is our velocity function-- is going to be equal to 2 times the sine of-- let me write it this way. Just because I want to make sure I get my parentheses right. 2 times the sine-- let me write it this way-- 2 sine of e to the 5.5, that's our time, divided by 4. So I did that part right over here. And I'm going to close the 2 as well. And then plus 1. So our velocity at time 5.5 is negative. So negative 0.45. Negative 0.45 roughly. So this is negative. The velocity is negative. So we have this scenario where the velocity is negative, which means we're going in the left direction. And the fact that the velocity is also decreasing means that over time-- at least at this point in time-- as we go forward in time, it'll become even more negative. And it'll become even more negative if we wait a little bit longer. So that means that the magnitude of the velocity is increasing. It's just going in the leftward direction. So if the magnitude of the velocity is increasing although it's going in the leftward direction, that means that the speed is increasing. So the velocity-- so this is one of those interesting scenarios-- the velocity is decreasing. But the speed, which is what they're asking us in the question, speed is increasing. of this explanation I gave you, you would say, hey, look. What's acceleration? Is it positive or negative? You would evaluate it. You'd say, hey, it's negative. So you know velocity is decreasing. And then you would say, hey, what is velocity? Is it positive or negative? You evaluate it. You say it's negative. So you have a negative value that is decreasing. So it's becoming more negative. So that means its magnitude is increasing, or speed is increasing."
    },
    {
        "Q": "At around 1:20: What is the magnitude?\n",
        "A": "Remember, velocity is a vector quantity, so it has both direction and magnitude. In the example at 1:20, the particle said to be moving in the negative direction and with a magnitude of 5.",
        "video_name": "ppBJWf_Wdmc",
        "timestamps": [
            80
        ],
        "3min_transcript": "Problem 1. For 0 is less than or equal to t is less than or equal to 6, a particle is moving along the x-axis. The particle's position x of t is not explicitly given. The velocity of the particle is given by v of t is equal to all of this business right over here. The acceleration of the particle is given by a of t is equal to all of this business over here. They actually didn't have to give us that because the acceleration is just the derivative of the velocity. And they also gave us-- they don't tell us the position function, but they tell us where we start off, x of 0 is equal to 2. Fair enough. Now let's do part A. Is the speed of the particle increasing or decreasing at time t equals 5.5? Give a reason for your answer. It looks like they did something a little sneaky here because they gave us a velocity function and then they ask about a speed. And you might say, wait, aren't those the same thing? And I would say, no, they aren't quite the same thing. Velocity is a magnitude and a direction. It is a vector quantity. Speed is just a magnitude. It is a scalar quantity. And to see the difference, you could have a velocity-- because they don't give us the units-- but you could have a velocity of negative 5 meters per second. And maybe if we're talking about on the x-axis, this would mean we're moving leftward at 5 meters per second on the x-axis. So the magnitude is 5 meters per second. This is the magnitude. And the direction is specified by the negative number. And that is the direction. Your velocity could be negative 5 meters per second, but your speed would just be 5 meters per second. So your speed is 5 meters per second whether you're going to the left or to the right. Your velocity, you actually care whether you're going to the left or the right. So let's just keep that in mind while we try to solve this problem. So the best way to figure out whether our rate of change is increasing or decreasing is to look at the acceleration. Because acceleration is really just the rate of change of velocity. And then, we can think a little bit about this velocity versus speed question. Get the calculator out. We can use calculators for this part of the AP exam. And I assume they intend us to because this isn't something that's easy to calculate by hand. So the acceleration at time 5.5, we just have to say t is 5.5 and evaluate this function. So 1/2-- I'll just write 0.5-- times e to the t over 4. Well t is 5.5. 5.5 divided by 4. And then, times cosine of 5.5 divided by 4 gives us 0.38. Did I do that right? We have 0.5 times e to the 5.5 divided by 4,"
    },
    {
        "Q": "At 1:37 sal talks about a magenta curve, what does he mean by that\n",
        "A": "There is a box that pops up and tells you Sal is talking about the magenta line, not curve.",
        "video_name": "xR9r38mZjK4",
        "timestamps": [
            97
        ],
        "3min_transcript": "what i want to do with this video is think about the relationship between variables and then think about what the graph of that relationship should look like. So let's say these two axis, the horizontal axis over here I plot the price of a product and lets say this vertical axis over here i plot the demand for the product, and i'm only plotting the first quadrant here because i'm assuming that the price can only be positive and i'm assuming that the demand can only be positive, that people aren't going to pay someone to take the product away from them. So let's think about what would happen for the price and demand for most normal products. So, if the price is low, you would expect that a lot of poeple are willing to buy that thing they're like \"Oh it's a good price, i would like to buy it.\" So, if the price is low, then the demand would be high so maybe it would be somplace over here, all the way that you would have really high demand if the price was zero. so if the price was low the demand would be high. Now what happens is the price -- so right here the price is low, demand is high, if the price were to price went up a little bit demand went down a little bit. if the price went up a little bit more then maybe demand goes down a little bit more. as the price went up a bunch then demand would go down a bunch and so the line that represents how the demand relates to price might look something like this, and i'm just going to assume it's it a line. It might not be a line, it might be a curve. It might look something like that. Or it might look something like that. But in general is someone were to ask you, if you saw this magenta curve that as price increases what happens to demand. You just say \"Well look price increases, as price increases what happens to demand?\" Well demand is decreasing. Now let's think about a different scenario. Let's talk about the demand for real estate. For actual property, and lets say that on this axis demand for land. So when the population is very low, you can imagine, if the population is zero there is no one there that would want to buy land. So if the population is very low the demand is going to be very low. And as population increases, demand should increase. If the population increases, more people are going to want to buy land. And if the population goes up a bunch then a lot of people are going to want to buy land. So you'll see a line that looks something like this. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. It could be a curve that looks something like that, or a curve that looks something like that. We don't know but the general idea is that if someone showed you a graph that looked like this. And as population increases what happens to demand. We'll you'd say \"Look, this is population increasing, what happens to demand?\" Demand is going up. Where as price increased the demand went down. Here as population is increasing"
    },
    {
        "Q": "At 5:19, why does Sal divide 350 by 2?\n",
        "A": "(1994/2) * 350 was changed to 1994 * (350/2) for easier multiplication",
        "video_name": "0-wa7voc0uM",
        "timestamps": [
            319
        ],
        "3min_transcript": "we have 54, 6 goes into 54 nine times, 9 times 6 is 54 and we are done. So to go from negative 50 to 2,044, I have to add 6 to 349 times, so I add it once, I add it twice, and then this right over here, this is the 349th time that I'm adding 6, so how many terms do I have? Now, you might be tempted to say 349 terms, but really, you have 349 plus 1 terms. You have the 349 for every time you added 6, so this is the first time you added 6, second time you added 6, all the way to the 349th time you added 6, so let me make it clear, this, this is, oh, actually, this is the 349th time I added 6 to get to this, but we haven't counted the first term just yet, so we're going to have, so we have 300, we have the first term and then we add 6 349 times, so we have 350 terms in this sum, N is equal to 350. And so we can say the sum of the first 300... I'll do this in green, the sum of the first 350 terms is going to be equal to the average of the first and last term, so negative 50 plus 2,044 over 2, over 2, times 350, so let's see, negative 50 plus 2,044, that's going to be what? 2,094, 2,094 divided by 2 times, times 350, so let's see, if I just take, so this is going to be, 'cause right, this is 294 times, 350 divided, oh, sorry, not 294, what am I? My brain is not working. 2,000, actually, this is going to be 1994. My brain really wasn't working a little while ago, so this is going to be 1994 divided by 2 times 350 and so let's see, 350 divided by 2 is 175, so this is going to be 1,900-- 1,994 times 175. Which is equal to, and I'll use a calculator for this one, so, let me get the calculator out, so, I have 1,994 times 175 gives us 348,950. 348,950. And we could express this in sigma notation now, now that we know what the n is. We found our answer, this is what we were looking for, but just in case you're curious,"
    },
    {
        "Q": "\nAt 4:00, why doesn't he multiplicate mu with i squared?",
        "A": "i isn t squared in this instance, it remains an imaginary number. Additionally, when the general solution is written, mu plays an important role, so keeping it separate allows you to plug it in without doing extra work.",
        "video_name": "6xEO4BeawzA",
        "timestamps": [
            240
        ],
        "3min_transcript": "Well, if this expression right here-- if this B squared minus 4AC-- if that's a negative number, then I'm going to have to take the square root of a negative number. So it will actually be an imaginary number, and so this whole term will actually become complex. We'll have a real part and an imaginary part. And actually, the two roots are going to be conjugates of each other, right? We could rewrite this in the real and imaginary parts. We could rewrite this as the roots are going to be equal to minus B over 2A, plus or minus the square root of B squared minus 4AC over 2A. And if B squared minus 4AC is less than 0, this is going to be an imaginary number. So in that case, let's just think about what the roots look like generally and then we'll actually do some problems. So let me go back up here. numbers like that. The roots, we can write them as two complex numbers that are conjugates of each other. And I think light blue is a suitable color for that. So in that situation, let me write this, the complex roots-- this is a complex roots scenario-- then the roots of the characteristic equation are going to be, I don't know, some number-- Let's call it lambda. Let's call it mu, I think that's the convention that people use-- actually let me see what they tend to use, it really doesn't matter-- let's say it's lambda. So this number, some constant called lambda, and then plus or minus some imaginary number. And so it's going to be some constant mu. That's just some constant, I'm not trying to be fancy, but this is I think the convention used in most differential So it's mu times i. So these are the two roots, and these are true roots, right? Because we have lambda plus mu i, and lambda minus mu i. So these would be the two roots, if B squared minus 4AC is less than 0. So let's see what happens when we take these two roots and we put them into our general solution. So just like we've learned before, the general solution is going to be-- I'll stay in the light blue-- the general solution is going to be y is equal to c1 times e to the first root-- let's make that the plus version-- so lambda plus mu i. All of that times x, plus c2 times e to the second root. So that's going to be lambda minus mu i times x."
    },
    {
        "Q": "I got confused at 1:03, can u help?\n",
        "A": "So first, you multiply x and y which is 3 * 2 = 6. Then you subtract y which is 2. So, 6 - 2 = 4. Then you add 3 * x which is 3 * 3 = 9. So, the answer is 13 as stated at 1:59.",
        "video_name": "S_OX3ByvBSc",
        "timestamps": [
            63
        ],
        "3min_transcript": "Now, let's think about expressions with more than one variable. So say I had the expression a plus-- I'll do a really simple one, a plus b. And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x. Or another way of saying it plus 3 times x. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13."
    },
    {
        "Q": "\nAt 3:25 the equation x+180-x+z=180... Wouldn't you put 180-x in parentheses so it would look like x+(180-x)+z=180? Wouldn't that make more sense?",
        "A": "Yes, using the parenthesis makes it clearer. You will still get the same answer either way.",
        "video_name": "9_3OxtdqmqE",
        "timestamps": [
            205
        ],
        "3min_transcript": "So I'm going to assume that x is equal to y and l is not parallel to m. So let's think about what type of a reality that would create. So if l and m are not parallel, and they're different lines, then they're going to intersect at some point. So let me draw l like this. This is line l. Let me draw m like this. They're going to intersect. By definition, if two lines are not parallel, they're going to intersect each other. And that is going to be m. And then this thing that was a transversal, I'll just draw it over here. So I'll just draw it over here. And then this is x. This is y. And we're assuming that y is equal to x. So we could also call the measure of this angle x. So given all of this reality, and we're assuming in either case that this is some distance, that this line is not of 0 length. Or this line segment between points A and B. I guess we could say that AB, the length of that line segment is greater than 0. I think that's a fair assumption in either case. AB is going to be greater than 0. So when we assume that these two things are not parallel, we form ourselves a nice little triangle here, where AB is one of the sides, and the other two sides are-- I guess we could label this point of intersection C. The other two sides are line segment BC and line segment AC. And we know a lot about finding the angles of triangles. So let's just see what happens when we just apply what we already know. Well first of all, if this angle up here is x, we know that it is supplementary to this angle right over here. So this angle over here is going to have measure 180 minus x. And then we know that this angle, this angle z-- we know that the sum of those interior angles of a triangle are going to be equal to 180 degrees. So we know that x plus 180 minus x plus 180 minus x plus z is going to be equal to 180 degrees. Now these x's cancel out. We can subtract 180 degrees from both sides. And we are left with z is equal to 0. So if we assume that x is equal to y but that l is not parallel to m, we get this weird situation where we formed this triangle, and the angle at the intersection of those two lines that are definitely not parallel all of a sudden becomes 0 degrees. But that's completely nonsensical. If this was 0 degrees, that means that this triangle wouldn't open up at all, which means that the length of AB"
    },
    {
        "Q": "At around 2:55: Why do you subtract 3 sin(5x-3y) from both sides?\n",
        "A": "You want to get all values of dy/dx isolated on one side.",
        "video_name": "-EG10aI0rt0",
        "timestamps": [
            175
        ],
        "3min_transcript": "Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun-- is going to be equal to, well, we subtracted this from both sides. So on the right-hand side, this is going to go away. So we're just going to be left with a negative 5 sine of 5x minus 3y. And we're in the home stretch now. To solve for dy/dx, we just have to divide both sides of the equation by this. And we are left with dy/dx is equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus 3 sine of 5x minus 3y."
    },
    {
        "Q": "\nat 5:49 why does he choose -2 to multiply both a and b by? Is it only to be rid of the 2b?",
        "A": "simultaneous equations. You multiply one of the equations by a number, so that when you add the two equations, one of the variables is cancelled out. You can then solve for one variable, and use it s answer for solving for the other variable. You have to do the same thing on both sides of the equals sign, otherwise it wouldn t be an equation!",
        "video_name": "EdQ7Q9VoF44",
        "timestamps": [
            349
        ],
        "3min_transcript": "Nothing fancier than there. That was Algebra two. Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could say that the numerators have to equal each other, because the denominators are equal. So we have A plus Bs plus 3A plus 2B is equal to 2s plus B. So the coefficient on s, on the right-hand side, is 2. The coefficient on the left-hand side is A plus B, so we know that A plus B is equal to 2. be equal to-- oh, this is a 13. This is a 13. That's a 13. It looks just like a B, right? That was 2s plus 13. Anyway, so on the right-hand side I get, it was 3A plus 2B is equal to 13. Now we have two equations with two unknowns, and what do we get? I know this is very tiresome, but it'll be satisfying in the end. Because you'll actually solve something with the Laplace Transform. So let's multiply the top equation by 2, or let's just say minus 2. So we get minus 2A minus 2B equals minus 4. And then we get-- add the two equations-- you get A is equal to-- these cancel out-- A is equal to 9. Great. If A is equal to 9, what is B equal to? B is equal to 9 plus what is equal to 2? And we have done some serious simplification. Because now we can rewrite this whole expression as the Laplace Transform of y is equal to A over s plus 2, is equal to 9 over s plus 2, minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2, minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well hopefully, you'll recognize this was actually the second Laplace Transform we figured out. What was that? I'll write it down here just so you remember it."
    },
    {
        "Q": "\nat 4:00 how do i make it a fraction",
        "A": "have an equal shape to make an equal fration",
        "video_name": "jgWqSjgMAtw",
        "timestamps": [
            240
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:30 that golden thingy looks a lot like a hexaflexagon. Does anyone else see that?\n",
        "A": "Yep. I bet you could make an interesting snowflake pattern with a hexaflexagon.",
        "video_name": "toKu2-qzJeM",
        "timestamps": [
            150
        ],
        "3min_transcript": "So in my last video I joked about folding and cutting spheres instead of paper. But then I thought, why not? I mean, finite symmetry groups on the Euclidean plane are fun and all, but there's really only two types. Some amount of mirror lines around a point, and some amount of rotations around a point. Spherical patterns are much more fun. And I happen to be a huge fan of some of these symmetry groups, maybe just a little bit. Although snowflakes are actually three dimensional, this snowflake doesn't just have lines of mirror symmetry, but planes of mirror symmetry. And there's one more mirror plane. The one going flat through the snowflake, because one side of the paper mirrors the other. And you can imagine that snowflake suspended in a sphere, so that we can draw the mirror lines more easily. Now this sphere has the same symmetry as this 3D paper snowflake. If you're studying group theory, you could label this with group theory stuff, but whatever. I'm going to fold this sphere on these lines, and then cut it, and it will give me something with the same symmetry as a paper snowflake. Except on a sphere, and it's a mess, so let's glue it to another sphere. And now it's perfect and beautiful in every way. But the point is it's equivalent to the snowflake OK, so that's a regular, old 6-fold snowflake, but I've seen pictures of 12-fold snowflakes. How do they work? Sometimes stuff goes a little oddly at the very beginning of snowflake formation and two snowflakes sprout. Basically on top of each other, but turned 30 degrees. If you think of them as one flat thing, it has 12-fold symmetry, but in 3D it's not really true. The layers make it so there's not a plane of symmetry here. See the branch on the left is on top, while in the mirror image, the branch on the right is on top. So is it just the same symmetry as a normal 6-fold snowflake? What about that seventh plane of symmetry? But no, through this plane one side doesn't mirror the other. There's no extra plane of symmetry. But there's something cooler. Rotational symmetry. If you rotate this around this line, you get the same thing. The branch on the left is still on top. If you imagine it floating in a sphere you can draw the mirror lines, and then 12 points of rotational symmetry. So I can fold, then slit it so it can swirl around the rotation point. And cut out a sphereflake with the same symmetry as this. And you can fold spheres other ways to get other patterns. OK what about fancier stuff like this? Well, all I need to do is figure out the symmetry to fold it. So, say we have a cube. What are the planes of symmetry? It's symmetric around this way, and this way, and this way. Anything else? How about diagonally across this way? But in the end, we have all the fold lines. And now we just need to fold a sphere along those lines to get just one little triangle thing. And once we do, we can unfold it to get something with the same symmetry as a cube. And of course, you have to do something with tetrahedral symmetry as long as you're there. And of course, you really want to do icosahedral, but the plastic is thick and imperfect, and a complete mess, so who knows what's going on. But at least you could try some other ones with rotational symmetry. And other stuff and make a mess. And soon you're going to want to fold and cut the very fabric of space itself to get awesome, infinite 3D symmetry groups, such as the one water molecules follow when they pack in together into solid ice crystals. And before you know it, you'll be playing with multidimensional, quasi crystallography, early algebra's, or something. So you should probably just stop now."
    },
    {
        "Q": "At 1:38, how ca Vi say she is lazy!?\n",
        "A": "She said she was lazy, but it was a joke, because right after that she said they are the binary expansion of pi. That shows she is not lazy.",
        "video_name": "Gx5D09s5X6U",
        "timestamps": [
            98
        ],
        "3min_transcript": "Snakes. Lots and lots of snakes. These snakes are just writhing with potential, similar parts linked together. They move in a specific and limited way. Part of the potential of things is how they break. These snakes break fantastically into these snake modules. You can put them back together too, allowing the existence of the super snake. Super snakes are obviously desirable for many reasons, besides being inherently awesome. You can wear them, and put them on things, and drop them, which I find amusing for some reason. You can arrange them into a space-filling, fractal curve, if that's what you like, which I do. You can even jump snake. But let's not forget. You can make mini-snakes too, which enables the snake stash, and starting from mini-snake gives you room to grow. Snake. Snake. Snake. Snake. They like to bend into this angle. Who cares what it actually is, besides about 90 degrees? But it begs the question, how many ways can I fold this snake if it has 10 segments? I can notate the way it slithers back and forth from tail to head, left, right, right, left, left, right, left. This is a valid slither. This is an invalid one, since anyone who's played snake knows that a snake isn't allowed to run into itself. Given a slither, how can I tell whether it's-- of course, snake lets you go straight too. So you can do another version of this that allows going straight and notate it like this, and wonder whether this snake is a loser snake or not. Snake. Snake. Snake. Snake. Snake. Don't forget to try putting the snake modules together in ways they were never meant to go. You can mix colors. So in theory, I could be hiding a secret message in the color pattern of this snake. But I'm not, because I'm lazy. So it's just the digits of the binary expansion of pi. Even better, you can attach more than one segment at a point. I can have a two-headed serpent. I can play the game where I cut off the heads of a Hydra, adding two more in the old head's place and see how far that gets me. I can put the snake modules on my fingertips and have snaky fingers. That's cool too. I can even do super snaky fingers. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake."
    },
    {
        "Q": "How do those snake modules at 0:45 add to the middle of the chain? It doesn't seems like she attaching them on. It is so cool!\n",
        "A": "The part where she adds them is cut from the video.",
        "video_name": "Gx5D09s5X6U",
        "timestamps": [
            45
        ],
        "3min_transcript": "Snakes. Lots and lots of snakes. These snakes are just writhing with potential, similar parts linked together. They move in a specific and limited way. Part of the potential of things is how they break. These snakes break fantastically into these snake modules. You can put them back together too, allowing the existence of the super snake. Super snakes are obviously desirable for many reasons, besides being inherently awesome. You can wear them, and put them on things, and drop them, which I find amusing for some reason. You can arrange them into a space-filling, fractal curve, if that's what you like, which I do. You can even jump snake. But let's not forget. You can make mini-snakes too, which enables the snake stash, and starting from mini-snake gives you room to grow. Snake. Snake. Snake. Snake. They like to bend into this angle. Who cares what it actually is, besides about 90 degrees? But it begs the question, how many ways can I fold this snake if it has 10 segments? I can notate the way it slithers back and forth from tail to head, left, right, right, left, left, right, left. This is a valid slither. This is an invalid one, since anyone who's played snake knows that a snake isn't allowed to run into itself. Given a slither, how can I tell whether it's-- of course, snake lets you go straight too. So you can do another version of this that allows going straight and notate it like this, and wonder whether this snake is a loser snake or not. Snake. Snake. Snake. Snake. Snake. Don't forget to try putting the snake modules together in ways they were never meant to go. You can mix colors. So in theory, I could be hiding a secret message in the color pattern of this snake. But I'm not, because I'm lazy. So it's just the digits of the binary expansion of pi. Even better, you can attach more than one segment at a point. I can have a two-headed serpent. I can play the game where I cut off the heads of a Hydra, adding two more in the old head's place and see how far that gets me. I can put the snake modules on my fingertips and have snaky fingers. That's cool too. I can even do super snaky fingers. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake."
    },
    {
        "Q": "At 1:18 you multiply the fraction by the denominator, I understand that this in the inverse of x/4 but I don't understand how it isolates the x. It feels like the two operations would cancel eachother out eliminating the x instead of isolating it. ?\n",
        "A": "The inverse of x/4 is actually 4/x. Multiplying these together does give 1, eliminating the x, as you suggested. However, Sal is multiplying x/4 by 4/1, which gives 4x/4, allowing us to cancel the 4 leaving x by itself. On the other side -18 x 4 gives -72.",
        "video_name": "p5e5mf_G3FI",
        "timestamps": [
            78
        ],
        "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16."
    },
    {
        "Q": "\n0:28 i don't know how to solve the X over 4 fraction.",
        "A": "Using the inverse property, you can multiply any number by its inverse to get 1. Therefore, x/4 *4 = x",
        "video_name": "p5e5mf_G3FI",
        "timestamps": [
            28
        ],
        "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16."
    },
    {
        "Q": "At 1:04 how did he get -18 when subtracted -16-2? Sorry I know this is a stupid question.\n",
        "A": "When you subtract, you count down. Because it s in the negatives, you count down twice. -16 to -17 and then to -18.",
        "video_name": "p5e5mf_G3FI",
        "timestamps": [
            64
        ],
        "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16."
    },
    {
        "Q": "\nIn the video; Patterns in raising 1 and -1 to different powers, I believe there is a mistake. Starting at 5:41 we are shown that (-1) to the 1,000,000 power = 1. At 5:43 we are told that \" \"1,000,000 is an even number so the answer is positive 1\". But, to the left of the problem we see that (-1) to the 1 power = -1, my point being; 1,000,000 is an odd number, is it not?. (-1) x 1= -1 then add the six zeroes. So, (-1) to the 1,000,000 power = -1. correct?",
        "A": "Can 1,000,000 be evenly divided by 2? If so, then 1,000,000 is even. 1,000,000 /2 = 500,000 therefore 1,000,000 is even.",
        "video_name": "jYOfMszfzAQ",
        "timestamps": [
            341,
            343
        ],
        "3min_transcript": "times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times, Or you could just think of it as you're taking three negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1, times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0 power is 1. Negative 1 to the first power is negative 1. Then you multiply it by negative 1, you're going to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is if you take negative 1 to an odd power you're going to get negative 1. And if you take it to an even power, you're going to get 1 because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so that you're always going to have negative times negatives. So this right over here, this is even. Even is going to be positive 1. And then you could see that if you went to negative 1 to the fourth power. Well, you could start with a 1 and then multiply it by negative 1 four times, so a negative 1 times negative 1, times negative 1, times negative 1, which is just going to be equal to positive 1. So if someone were to ask you-- we already established that if someone were to take 1 to the, I don't know, 1 millionth power, this is just going to be equal to 1. If someone told you let's take negative 1 and raise it to the 1 millionth power, well, 1 million is an even number, so this is still going to be equal to positive 1. But if you took negative 1 to the 999,999th power, this is an odd number. So this is going to be equal to negative 1."
    },
    {
        "Q": "At 3:46, Sal says that -1 squared equals one. So every answer above squaring is positive? I don't get how it changes when its squared.\nInstead, it changes to a positive number. How can that be possible when we are squaring a negative number?\n",
        "A": "Multiplying two negative numbers always results in a positive product- think of the two negatives as canceling each other out. So, it s the same if you multiply a negative by itself.",
        "video_name": "jYOfMszfzAQ",
        "timestamps": [
            226
        ],
        "3min_transcript": "And that's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's start try some negative numbers. So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going, based on this definition, this is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1. Let's try negative 1 to the first power. Well, anything to the first power, you could view this-- and I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say hey, let's start with a 1, and then multiply it by 1 eight times. And you're still going to get a 1 right over here. But let's do this with negative 1. So we're going to start with a 1, times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times, Or you could just think of it as you're taking three negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1, times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0 power is 1. Negative 1 to the first power is negative 1. Then you multiply it by negative 1, you're going to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is if you take negative 1 to an odd power you're going to get negative 1. And if you take it to an even power, you're going to get 1 because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so that you're always going to have negative times negatives. So this right over here, this is even. Even is going to be positive 1. And then you could see that if you went to negative 1 to the fourth power."
    },
    {
        "Q": "\nAt 2:58, wouldn't (-1)^0 equal -1?. Its 1*-1 which equals -1 right?",
        "A": "if (x < 0) {Yes your right! }",
        "video_name": "jYOfMszfzAQ",
        "timestamps": [
            178
        ],
        "3min_transcript": "except for 0-- that's where we're going to-- it's actually up for debate. But anything to the 0 power is going to be equal to 1. And just as a little bit of intuition here, you could literally view this as our other definition of exponentiation, which is you start with a 1, and this number says how many times you're going to multiply that 1 times this number. So 1 times 1 zero times is just going to be 1. And that was a little bit clearer when we did it like this, where we said 2 to the, let's say, fourth power is equal to-- this was the other definition of exponentiation we had, which is you start with a 1, and then you multiply it by 2 four times, so times 2, times 2, times 2, times 2, which is equal to-- let's see, this is equal to 16. So here if you start with a 1 and then you multiply it by 1 zero times, you're And that's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's start try some negative numbers. So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going, based on this definition, this is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1. Let's try negative 1 to the first power. Well, anything to the first power, you could view this-- and I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say hey, let's start with a 1, and then multiply it by 1 eight times. And you're still going to get a 1 right over here. But let's do this with negative 1. So we're going to start with a 1, times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times,"
    },
    {
        "Q": "\nAt 0:26, how do you know which dot is x and which one is y? Would they be labeled, usually?",
        "A": "A dot (or point) isn t x or y. A point is just a location. Using the Cartesian coordinate system (the x,y-grid), we assign the point a x-coordinate and a y-coordinate, which describes the location of the point. Hope this helped.",
        "video_name": "6_9xNMtwnfs",
        "timestamps": [
            26
        ],
        "3min_transcript": "Graph a line that has a slope that is negative and greater than the slope of the blue line. So let's think for a second about what slope means. So if you use the word slope in your everyday life, you're really talking about how inclined something is, like a ski slope. So for example, this orange line isn't inclined at all. It's flat, so this one actually has a slope of 0. Another way of thinking about it is as x increases, what is happening to y? And you see here that y isn't changing at all, so this orange line right now has a slope of 0. If the orange line looked like this, it now has a positive slope. Notice, when x is negative, your y value-- or say when x is negative 5, your y value is here, and then when x is positive 5, your y value has increased. As x increases, y is increasing, so this has a positive slope. This has an even more positive slope, an even more positive This has an even more positive slope. As x is increasing, the y value is increasing really fast. as we move towards the right, so it's a very positive slope. This is less positive, less positive. This is a 0 slope, and then this is a negative slope. Notice, as x is increasing, the line is going down. Your y value is decreasing. When x is negative 5, your y is 7. Well, when x is 5, your y is 5. So x is increased, but y has gone down, so this is a negative slope. So they say graph a line that has a slope that is negative and greater than the slope of the blue line. So the blue line also has a negative slope. As x is increasing, your blue line is going down. Over here, when x has negative values, your y value is quite high. And here, when x has positive values, your y value has gone all the way down. So this has a negative slope, but we want to have a slope greater than this one. So we still want to have a-- they so my orange line currently has a negative slope-- and greater than the slope of the blue line. So it has a negative slope, but it is less negative than this blue line right over here. If I wanted to be more negative than the blue line, I'd have to do something like that. But I want to be negative but less negative than the blue line, so that would be like that. If we wanted a 0 slope, once again, something like this. If we wanted a positive slope, something like this. So once again, negative slope, less negative than the blue line. Check our answer."
    },
    {
        "Q": "At 6:45 , can't it be like we are in a building , & are trying to get to the exit of the building.\n",
        "A": "Yes, its basically like your in a building, but not quite. It does however look like you are trying to exit the building.",
        "video_name": "wRxzDOloS3o",
        "timestamps": [
            405
        ],
        "3min_transcript": "That's the only way to get there. Or I could go two to the right there. And that's the only way to get there. And now if you watched the two dimensional path counting brain teaser, you know that there's two ways to get here. And the logic is, well you could draw it out. You could go like that. One, two. And that's the same thing as saying, one, two. Though it's easier to visualize here. But the general logic was, well, to know how many ways to get to any square, think about the squares that lead to it, and how many ways can I get to those two squares? And then sum them up. And by the same logic-- so there's two ways to get here. That's that cell. Three ways to get here, right? Two plus one is three. One plus two is three. And three plus three is six. So there were six ways to get to this cube right there, from that one. So this isn't too different from the two dimensional problem so far. But now it gets interesting. done it in the color of that layer. How many ways are there to get to this cell right here? This cell is that one right there. Well, I start here. And I can just go straight down. There's only one way to be there. But I go straight down. So there's only one way to get there. Actually let's extend. There's only one way to get here, if I'm going straight down. And so there's only one way to get to this cell too. I'd have to go straight down again. So there's only one way to get there. Hopefully you understand the way we're visualizing it. This is the bottom row. And there's only one way. You go from here, straight down to there, And that's the only way to get there. Fair enough. Now this is where it gets interesting. How many ways are there to get to this cell? Well in our old example, there was only one way in two dimensions to go from this cell. But now we can go from this cell, and we could come from above. And where's above? Above is right there. So now we add this cell to that cell. How many ways to get to this? This is kind of in the back middle of this cube. How many ways to get there? Well, there's two ways to come from this direction. And I can also come from above right there. So two plus one is three. How many ways to get here? Well, one from behind. And then one from above. So that's two. And you see a little bit of symmetry. And how many ways to come here. Well there's two here. From going straight forward. Two ways to go that way. And then one way to come from above. This is two, and we're on this cell. So if we wanted to know how many ways to get to this cell? There's two ways to go from there. And then one way from above. So that's three. And now, right here, how many ways to get to this cell? There's three ways. I could come from here, from here, or from above."
    },
    {
        "Q": "in 1:49 it said plus or negative square root of 8, I see the point because you can get it either negative or positive square root of 8 but why don't you do the same thing in the left side? like + or - 4x+1\n",
        "A": "You don t need to because you would still end up with the same 2 equations. Sal shows 4x+1 = Sqrt(8) or 4x+1 = -Sqrt(8) If we did the same for the left side we would also have -(4x+1) = Sqrt(8) and -(4x+1) = -Sqrt(8), but you can see if you just multiply these though by -1 you end up with the same 2 equations we already had.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            109
        ],
        "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously."
    },
    {
        "Q": "where does the +1 come in at5:00\n",
        "A": "Colby, The original problem was (4x+1)\u00c2\u00b2 - 8 = 0 Then at about 4:50 he said he was going to put his answer ((-1+2\u00e2\u0088\u009a2)/4) in for the x in the original equation to prove the answer actually works. And the +1 he wrote at 5:00 is the +1 in the original equation (4[x}+1)\u00c2\u00b2 - 8 = 0 When he put the answer in for x he had (4[(-1+2\u00e2\u0088\u009a2)/4}+1)\u00c2\u00b2 - 8 = 0 I hope that helps.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            300
        ],
        "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else."
    },
    {
        "Q": "At 8:45, what is the step-by-step procedure to solving (x-8)(x-2)=0 when solving for x? How did Sal immediately know that the answer would be x=8 or x=2? The only way I see him working it out is if he divided the expression up into two parts like so; (x-8)=0 and (x-2)=0 and then solving for x. If so, why is this allowed? It doesn't make intuitive sense.\n",
        "A": "Basically the expression (x+8)(x+2)=0 means that one of those two factor [you have to look at (x+8) as one factor and at (x+2) as another factor] must be equal to zero to satisfy the equation. Either (x+8) must be zero or (x+2) must be zero, or both. Therefore if x+8=0 then x = -8 and this is one valid solution, similarly if x+2=0 then x= -2 and this is the second valid solution. When you tackle problems like this remember that, for a product to be equal to zero there must be a factor which is 0. Hope it helped",
        "video_name": "55G8037gsKY",
        "timestamps": [
            525
        ],
        "3min_transcript": "on factoring so far. We can only do this when this is a perfect square. If you got, like, x minus 3, times x plus 4, and that would be equal to 9, that would be a dead end. You wouldn't be able to really do anything constructive with that. Only because this is a perfect square, can we now say x minus 5 squared is equal to 9, and now we can take the square root of both sides. So we could say that x minus 5 is equal to plus or minus 3. Add 5 to both sides of this equation, you get x is equal to 5 plus or minus 3, or x is equal to-- what's 5 plus 3? Well, x could be 8 or x could be equal to 5 minus 3, or x is equal to 2. Now, we could have done this equation, this quadratic equation, the traditional way, the way you were tempted to do it. What happens if you subtract 9 from both You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9"
    },
    {
        "Q": "at about 10:53 why is (x+8)(x+10) equivalent to x= -8 and x= -10? if you add -8 and -10 together u get -18 not +18, shouldn't it be x= +8 and x= +10?\n",
        "A": "Because if you are trying to make them both equal 0 you have to do this: (x+8) if you replace x with 8 you get 16, so you replace x with -8 and you get zero. That is why you get -8 and -10",
        "video_name": "55G8037gsKY",
        "timestamps": [
            653
        ],
        "3min_transcript": "You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9 Take the square root of both sides, you get x plus 9 is equal to plus or minus the square root of 1, which is just 1. So x is equal to-- subtract 9 from both sides-- negative 9 plus or minus 1. And that means that x could be equal to-- negative 9 plus 1 is negative 8, or x could be equal to-- negative 9 minus 1, which is negative 10. And once again, you could have done this the traditional way. You could have subtracted 1 from both sides and you would have gotten x squared plus 18x, plus 80 is equal to 0. And you'd say, hey, gee, 8 times 10 is 80, 8 plus 10 is 18, so you get x plus 8, times x plus 10 is equal to 0. And then you'd get x could be equal to negative 8, or x could be equal to negative 10. That was good warm up. Now, I think we're ready to tackle completing the square."
    },
    {
        "Q": "I do not understand at 1:43 why Sal said it could be both the positive and negative square roots of 8?\n",
        "A": "When you square any real number, either negative or positive, you always get a positive number. So the sqrt of 4 can either be -2 or 2. Therefore the square root of any number can be either negative or positive. You do not know for a fact that it is the principle square root (i.e that the answer is just 2 squared), it could easily be -2!",
        "video_name": "55G8037gsKY",
        "timestamps": [
            103
        ],
        "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously."
    },
    {
        "Q": "At 1:25, he square roots both sides. If you had another equation like (4x + 1)^2 = 0,\ncan you just square root both sides into 4x + 1 = 0?\n",
        "A": "Yes you can. Another way to solve it which produces same result as square rooting both sides: (4x+1)^2 =0 is the same as (4x+1)(4x+1) = 0 One of the factors has to be 0 for the expression to equal 0 as anything multiplied by 0 is 0. So we end up with 4x+1=0, you can then use simple algebra to solve for x.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            85
        ],
        "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously."
    },
    {
        "Q": "\nAt 1:50, isn't the square root positive?",
        "A": "Square roots can be positive or negative, so he marks it with the correct symbol",
        "video_name": "55G8037gsKY",
        "timestamps": [
            110
        ],
        "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously."
    },
    {
        "Q": "At 3:19 when you subtracted 1 from both sides, how did you get -1 on the right side?\n",
        "A": "Since he can t really subtract it, he just keeps it there for when he does find out what x is equal to.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            199
        ],
        "3min_transcript": "squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously. equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to"
    },
    {
        "Q": "\nat around 3:37, sal says 56cm squared but puts the exponent 3 [A.K.A. cubed] when he should have put 2 [A.K.A. squared.]",
        "A": "I think it was an accident. I think they meant to put a 2.",
        "video_name": "b8q6i_XPyhk",
        "timestamps": [
            217
        ],
        "3min_transcript": "So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not So it's 8 centimeters times 1/2 or 4 centimeters high. And we know how to find the area of this. This is 4 centimeters times 14 centimeters. So the area is equal to 4 centimeters times 14 centimeters which is equal to-- let's see, that's 40 plus 16-- 56 square centimeters. So if you're taking the area of a kite, you're really just taking 1/2 the width times the height, or 1/2 the width times the height, any way you want to think about it."
    },
    {
        "Q": "At 0:28, Sal said that kites are symmetrical. Does this mean that all rhombuses are also kites?\n",
        "A": "Yes. A kite can become a rhombus In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus.",
        "video_name": "b8q6i_XPyhk",
        "timestamps": [
            28
        ],
        "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not"
    },
    {
        "Q": "At roughly 4:40, you cubed the centimeters instead of squaring them\n",
        "A": "The maximum length of the video tape is 3:48. How come you said at 4:40?",
        "video_name": "b8q6i_XPyhk",
        "timestamps": [
            280
        ],
        "3min_transcript": "And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not So it's 8 centimeters times 1/2 or 4 centimeters high. And we know how to find the area of this. This is 4 centimeters times 14 centimeters. So the area is equal to 4 centimeters times 14 centimeters which is equal to-- let's see, that's 40 plus 16-- 56 square centimeters. So if you're taking the area of a kite, you're really just taking 1/2 the width times the height, or 1/2 the width times the height, any way you want to think about it."
    },
    {
        "Q": "\nAt 0:21, is calling the length a diagonal proper terminology?",
        "A": "If Sal says that, then it certainly seems right.",
        "video_name": "b8q6i_XPyhk",
        "timestamps": [
            21
        ],
        "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not"
    },
    {
        "Q": "at 1:00 i don't understand how u would find the height if you copy only half of the kite\n",
        "A": "Because if you find halve of the height then you can multiply it by 2 and get the height of the full kite :3",
        "video_name": "b8q6i_XPyhk",
        "timestamps": [
            60
        ],
        "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not"
    },
    {
        "Q": "\nAt 1:00 does he mean the 0 vector in R2?",
        "A": "Yes, that was Sal s mistake.",
        "video_name": "qBfc57x_RSg",
        "timestamps": [
            60
        ],
        "3min_transcript": "I've got this matrix, A, here, it's a 2 by 3 matrix. And just as a bit of review, let's figure out its nullspace and its columnspace. So the nullspace of A is the set of all vectors x that are member of-- let's see we have 3 columns here-- so a member of R3, such that A times the vector are going to be equal to the 0 vector. So we can just set this up. Let me just-- we just need to figure out all of the x's that satisfy this in R3. So we take our matrix A 2, minus 1, minus 3, minus 4, 2, 6. Multiply them times some arbitrary vector in R3 here. So you get x1, x2, x3. And you set them equal to the 0 vector. It's going to be the 0 vector in R2. Because we have 2 rows here. You multiply a 2 by 3 matrix times a vector in R3, you're going to get a 2 by 1 vector or 2 by 1 matrix. So you're going to get the 0 vector in R3. equations-- you get 2 x1 minus x2 minus 3 x3 is equal to 0 and so on and so forth. We can just set up an augmented matrix. So we can just set up this augmented matrix right here. 2 minus 1 minus 3 minus 4, 2, 6. And then augment it with what we're trying to set it equal to to solve the system. And you know we're going to perform a bunch of row operations here to put this in reduced row echelon form. And they're not going to change the right-hand side of this augmented matrix. And that's essentially the argument as to why the nullspace of the reduced row echelon form of A is the same thing as the nullspace of A. But anyway, that's just a bit of review. So let's perform some row operations to solve this a little bit better. So, the first thing I might want to do is divide the first row by 2. So if I divide the first row by 2 I get a 1 minus 1/2 minus And let's just divide this row right here by-- I don't know just to simplify things-- let's divide it by 4. So I'm doing two row operations in one step. And you can do that. I could have done it in two separate steps. So if we divide it by 4, this becomes minus 1, 1/2 and then you get 3/2 and then you get 0. And now, let's keep my first row the same. I'm going to keep my first row the same. It's 1 minus 1/2 minus 3/2 and of course the 0 is the right-hand side. Now let's replace my second row with my second row plus my first row. So these are just linear operations on these guys. So negative 1 plus 1 is 0. 1/2 plus minus 1/2 is 0. 3/2 plus minus 3/2 is 0. And of course, 0 plus 0 is 0."
    },
    {
        "Q": "At 0:55, he says the ratio for a 30-60-90 triangle is x/2: x*square root of 3/2: x. however, my math book says the ratio is x*square root of 3/2: x: 2x....Am i misunderstanding something?\n",
        "A": "The video: x/2 : (x\u00e2\u0088\u009a3)/2 : x The book: x : x\u00e2\u0088\u009a3 : 2x Both are equivalent statements. Multiply the terms of the video s ratio by 2 and you will get the book s ratio. Multiplying by all terms by a common factor will not change the value of a ratio.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            55
        ],
        "3min_transcript": "What I want to do in this video is discuss a special class of triangles called 30-60-90 triangles. And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in trigonometry class, is the ratios between the sides of a 30-60-90 triangle. Remember, the hypotenuse is opposite the 90-degree side. If the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x/2, and that the 60 degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x/2, that's going to be its length. So that's where we're going to prove in this video. And then in other videos, we're just going to apply this. We're going to show that this is actually a pretty useful result. Now, let's start with a triangle that we're very familiar with. So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other,"
    },
    {
        "Q": "At 6:20, Sal mentions taking the \"principal root\" of something. What is a principal root, and how is it different than square root? Thanks!\n",
        "A": "The square root of a number is both positive and negative: \u00e2\u0088\u009a9 = \u00c2\u00b13 Because both 3 and -3 squared give you 9. The principal root is just the positive answer. So the principal square root of 9 is 3.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            380
        ],
        "3min_transcript": "and they add up to x-- remember, this was an equilateral triangle of length x-- we know that this side right over here, is going to be x/2. We know this is going to be x/2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle, that if the hypotenuse-- notice, and I guess I didn't point this out. By dropping this altitude, I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90-degree side is x, that the side opposite the 30-degree side is going to be x/2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60-degree side. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x/2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x/2 squared-- this is just straight out of the Pythagorean theorem.-- plus x/2 squared is going to equal this hypotenuse squared. It's going to equal x squared. And just to be clear, I'm looking at this triangle right I'm looking at this triangle right over here on the right, and I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4. x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. or x squared over 4 from both sides, you get BD squared is equal to-- 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So it's just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. The principal root of 3 is square root of 3. the principal root of x squared is just x, over the principal root of 4 which is 2. And BD is the side opposite the 60-degree side. So we're done. If this hypotenuse is x, the side opposite the 30-degree side is going to be x/2, and the side opposite the 60-degree side is going to be square root of 3 over 2 times x, or the square root of 3x over 2, depending on how you want to view it."
    },
    {
        "Q": "\nAt 5:59, how did you get (4x^2)/(4)?",
        "A": "He changed x^2 to (4x^2)/(4) for calculation purposes, he will subtract (x^2)/(4) later, and changing x^2 will make the subtraction easier to visualize. (4x^2)/(4) is the same thing as x^2 because the 4 in the numerator and 4 in the denominator cancel out (4 / 4 = 1)",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            359
        ],
        "3min_transcript": "and they add up to x-- remember, this was an equilateral triangle of length x-- we know that this side right over here, is going to be x/2. We know this is going to be x/2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle, that if the hypotenuse-- notice, and I guess I didn't point this out. By dropping this altitude, I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90-degree side is x, that the side opposite the 30-degree side is going to be x/2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60-degree side. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x/2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x/2 squared-- this is just straight out of the Pythagorean theorem.-- plus x/2 squared is going to equal this hypotenuse squared. It's going to equal x squared. And just to be clear, I'm looking at this triangle right I'm looking at this triangle right over here on the right, and I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4. x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. or x squared over 4 from both sides, you get BD squared is equal to-- 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So it's just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. The principal root of 3 is square root of 3. the principal root of x squared is just x, over the principal root of 4 which is 2. And BD is the side opposite the 60-degree side. So we're done. If this hypotenuse is x, the side opposite the 30-degree side is going to be x/2, and the side opposite the 60-degree side is going to be square root of 3 over 2 times x, or the square root of 3x over 2, depending on how you want to view it."
    },
    {
        "Q": "At 3:21, what does the side-angle side, etc. congruence mean?\n",
        "A": "That between the two triangles, 2 sides are congruent, and the angle between them is also congruent. It is referred to as SAS There is AAS, SSS, ASA, and HL These will come up in geometry typically in your first or second year of high school.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            201
        ],
        "3min_transcript": "So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other."
    },
    {
        "Q": "\nAt 3:14, what does \"postulates\" mean when Sal says \"...of our congruence postulates.\"",
        "A": "A postulate (or axiom) is just a simple statement that is accepted as true. It is similar to theorems, but theorems must be proven, while postulates do not.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            194
        ],
        "3min_transcript": "So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other."
    },
    {
        "Q": "At 1:30 does he mean equal in length\n",
        "A": "Yes. equilateral triangles have exactly the same measures of sides and the same angles.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            90
        ],
        "3min_transcript": "What I want to do in this video is discuss a special class of triangles called 30-60-90 triangles. And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in trigonometry class, is the ratios between the sides of a 30-60-90 triangle. Remember, the hypotenuse is opposite the 90-degree side. If the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x/2, and that the 60 degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x/2, that's going to be its length. So that's where we're going to prove in this video. And then in other videos, we're just going to apply this. We're going to show that this is actually a pretty useful result. Now, let's start with a triangle that we're very familiar with. So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other,"
    },
    {
        "Q": "\nI don't understand how the one comes in at about 4:51 . How does Sal get 1.08?",
        "A": "The chain is increasing at a rate of eight percent so, you have .08. You also have the original 100 percent of 200, that is where he gets the 1.",
        "video_name": "m5Tf6vgoJtQ",
        "timestamps": [
            291
        ],
        "3min_transcript": "So we have 100 times 0.965 to the sixth power, which is equal to 80.75. This is all in percentages. So it's 80.75% of our original substance. Let's do another one of these. So we have, Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is-- oh actually, there's a typo here, it should be 8%-- the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which"
    },
    {
        "Q": "\nAt 5:00, how did Sal get 1.08?",
        "A": "If in 1999 Nadia had 200 stores and it increases by 8% per year, all you have to do is turn 8% into a decimal (by moving the decimal place 2 places to the left) you get 0.08. So for one year past 1999, you have 200 + 200(0.08), or a way to simplify this because they have the same base you simply multiply 200 by 1.08.",
        "video_name": "m5Tf6vgoJtQ",
        "timestamps": [
            300
        ],
        "3min_transcript": "So we have 100 times 0.965 to the sixth power, which is equal to 80.75. This is all in percentages. So it's 80.75% of our original substance. Let's do another one of these. So we have, Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is-- oh actually, there's a typo here, it should be 8%-- the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which"
    },
    {
        "Q": "\nat 5:37, what does gist mean?",
        "A": "Here s the etymology way to understand it: From French gist en = lie in consist in ; from Latin jacet = to lie, rest (as when boards lie on a wooden beam) Now means the real point , important part . (See the Online Etymology Dictionary and Oxford English Dictionary .)",
        "video_name": "m5Tf6vgoJtQ",
        "timestamps": [
            337
        ],
        "3min_transcript": "So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which So they're asking us, how many stores does the restaurant operate in 2007? Well, 2007 is 8 years after 1999. So here n is equal to 8. So let us substitute n is equal to 8. The answer to our question will be 200 times 1.08 to the eighth power. Let's get our calculator out and calculate it. So we want to figure out 200 times 1.08 to the eighth power. She's going to be operating 370 restaurants, and she'll be in the process of opening a few more. So if we round it down, she's going to be operating 370 So 8% growth might not look like something that's so fast"
    },
    {
        "Q": "\n0:10 isn't it 2 units wide?",
        "A": "Yes. This mistake has been noticed and now a pop up on the screen informs you that it is really 2 units 0.10",
        "video_name": "1UQ5IbihJNI",
        "timestamps": [
            10
        ],
        "3min_transcript": "- [Voiceover] So we have an interesting shape right over here. The base is a rectangular prism. And the dimensions of that rectangular prism, it's three units I guess we could say tall, four units wide, and then four units long. And then on top of that, on top of that, we have what you could call a right pyramid, where the height of this right pyramid, so if you start at the center of its base right over here, and you go to the top, this height right over here is one unit. And this hasn't been drawn completely to scale, and kind of the perspective skews a little bit. But our goal here, our goal here, is to figure out what is the length? What is the length of one of these edges right over here? So either that one, or this one right over here. What is that length? And we will call that, we will call that x. And so I encourage you to pause this video and try to think about it on your own. Remember, this is a right, this is a right pyramid. that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular prism. So with that in mind, I encourage you to pause the video and see if you can figure it out. And I will give you a hint. You will have to use the Pythagorean theorum, maybe more than once. Alright, so I am assuming you've at least given it a shot. So let's work through it together. So the key here is to realize, well okay this point, this base right over here, this point right over here it's half way in this direction and half way in this direction. So we can figure out, well this entire length right over here is length four. So half way, this is going to be, That's going to be two, and that's going to be two. Just like that. And then the other thing we can figure out, we can figure out what this length is going to be. 'Cause once again it's half way in that direction. So if this whole thing is two, and we see it right over here. This is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us? Well using that information, we should be able to figure out this length. Actually, I'll keep it in this color 'cause this color's easy to see. We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle. And then the x would be the hypotenuse. So we could just apply the Pythagorean theorem."
    },
    {
        "Q": "\n0:01 ... that is ... very easy sir.",
        "A": "for some,yes but others, not as much.",
        "video_name": "4IWfJ7-CYfE",
        "timestamps": [
            1
        ],
        "3min_transcript": "Let's see if we can calculate 2.91 times 3.2. And I encourage you to pause this video and try it out on your own. So the way I'm going to think about it is 2.91 is the same thing as 291 divided by 10. Or not divided by 10, divided by 100. And we know that if you divide something by 100, you are going to move the decimal place two places to the left-- one, two. And you would end up at 2.91. It also make sense, if I take 2, and I multiply it by 100, I'd get 200. Or if I take 200 and divided by 100, I would get 2. So it makes sense that 2.91 is the same thing as 291 divided by 100. Similarlarly-- I can never say that word-- 3.2 can be rewritten. It's the same thing as 32 divided by 10. Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing."
    },
    {
        "Q": "\nAt 02:24 why not 291/100*32/10 be equal to 291*32/(100/10)=291*32/(10)",
        "A": "No, actually when you have 2 fractions being multiplied together (in this example 291/100 and 32/10) you multiply the numerators, then multiply the denominators. So in this case it would end up looking like this: (291/100)*(32/10)= (291*32)/(100*10= (291*32)*(1000)",
        "video_name": "4IWfJ7-CYfE",
        "timestamps": [
            144
        ],
        "3min_transcript": "Let's see if we can calculate 2.91 times 3.2. And I encourage you to pause this video and try it out on your own. So the way I'm going to think about it is 2.91 is the same thing as 291 divided by 10. Or not divided by 10, divided by 100. And we know that if you divide something by 100, you are going to move the decimal place two places to the left-- one, two. And you would end up at 2.91. It also make sense, if I take 2, and I multiply it by 100, I'd get 200. Or if I take 200 and divided by 100, I would get 2. So it makes sense that 2.91 is the same thing as 291 divided by 100. Similarlarly-- I can never say that word-- 3.2 can be rewritten. It's the same thing as 32 divided by 10. Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing."
    },
    {
        "Q": "\nAt 4:37, Sal said that (-48x\u00e2\u0081\u00b4-42x\u00c2\u00b3-15x\u00c2\u00b2-5x)/(8x+7)(3x+1) is the final answer. But isn't possible to factor it by grouping to simplify the answer to (-6x\u00c2\u00b3-5x) with this as the solution:\n(-48x\u00e2\u0081\u00b4-42x\u00c2\u00b3-15x\u00c2\u00b2-5x)/(8x+7)(3x+1)\n-6x\u00c2\u00b3(8x+7)-5x(3x+1)/(8x+7)(3x+1)\n=(-6x\u00c2\u00b3-5x)",
        "A": "Factors are items that are being multiplied. When you get to: -6x\u00c2\u00b3(8x+7)-5x(3x+1) you have terms (2 items being added or subtracted). Each term is made up of factors. But, you haven t fully factored the polynomial. Since you don t have factors, you can t reduce the fraction. Hope this helps.",
        "video_name": "evmDZkDvlNw",
        "timestamps": [
            277
        ],
        "3min_transcript": "in a new color, do this in green. Our common denominator, we already established, is the product of our two denominators so it is going to be 8x plus seven times 3x plus one. Now if we multiply the denominator here was 3x plus one, we're multiplying it by 8x plus seven. So that mean's we have to multiply the numerator by 8x plus seven as well. 8x plus seven times negative 6x to the third power. Notice, 8x plus seven divided by 8x plus seven is one. If you were to do that, you would get back to your original expression right over here, the negative 6x to the third over 3x plus one. And now, we're ready to add. This is all going to be equal to, I'll write the denominator in white. So we have our common denominator, 8x plus seven times 3x plus one. to distribute the negative 5x, so negative 5x times positive 3x is negative 15x squared. And then, negative 5x times one is minus 5x. And then, in the green, I would have, let's see, I'll distribute the negative 6x to the third power, so negative 6x to the third times positive 8x is going to be, negative 48x to the fourth power. And then negative 6x to the third times positive seven is going to be negative 42, negative 42x to the third. And I think I'm done because there's no more, there's, you know, I only have one fourth-degree term, one third-degree term, one second-degree term, one first-degree term, and that's it. There's no more simplification here. Some of you might want to just write it as negative 48x to the fourth minus 42x to the third minus 15x squared minus 5x. All of that over 8x plus seven times 3x plus one. But either way, we are all done. I mean, it looks like up here, yeah, there's no, nothing to factor out. These two are divisible by five. These are divisible by six but even if I were to factor that out, nothing over here, down here, no five or six to factor out. Yeah, so it looks like we are all done."
    },
    {
        "Q": "\n@6:08 Probability of A and B are independent of each other.\n\nA --> Select a Blue garment\nB --> Select a Shirt\n\nThere is a Blue Shirt in the Sample Space. So if A happens to select the Blue shirt. Then for sure the chance of B selecting a Shirt is impacted since there is one less shirt to select from.\n\nIn this case, how is it that A and B are independent events ?",
        "A": "Tomas chooses a garment at random. So if he happens to choose the blue shirt, then A and B are true. A and B are groups of possibilities, not selections. To be clearer, Tomas doesn t select a blue garment, and then select a shirt, he simply picks one at random. So A and B can be independent events. Tell me if you still have questions, or if this wasn t clear enough, please. :)",
        "video_name": "R-NeYKSEqns",
        "timestamps": [
            368
        ],
        "3min_transcript": "let's see if we can answer these questions. The probability of A given B equals the probability of A. And that does work out. Probability of A given B is 1/2. And that's the same thing as the probability of A. The probability that Tomas likes a blue garment given that he has chosen a shirt is equal to the probability that Tomas likes a blue garment. Yep, that's exactly. So I guess the words are just rephrasing what they wrote here in a more mathy notation. So this is absolutely true. The probability of B given A is equal to the probability of B. Yep, the probability of B given A is 1/3. The probability of B is 1/3. The probability that Tomas selects a shirt given that he has chosen a blue garment is equal to the probability that Tomas selects a shirt. Yep, that's right. Events A and B are independent events. Independent events. So two events are independent These are independent if the probability of A given B is equal to the probability of A. Then we can say A and B are independent. Because the probability of A, then if this is true then this means the probability of A given B isn't dependent on whether B happened or not. It's the same thing as the probability of A. This would lead to these events being indpendent. So if you had the probability of B given A is equal to the probability of B. Same argument. That would mean they are independent. Or, if we said that the probability of A and B is equal to the probability of A times the probability of B then this also means they are independent. We know that this one is true. The probability of A and B is 1/6. The probability of A times the probability of B is 1/2 times 1/3 which is 1/6. So all of these are clearly true. So we can say that A and B are independent. of whether B has happened or not. The probability of B happening is independent of whether A has happened or not. The outcome of events A and B are dependent on each other. No. That's the opposite of saying they are independent. So we can cross that out. Probably of A and B is equal to the probability of A times the probability of B We already said that to be true. 1/6 is 1/2 times 1/3. The probability that Tom selects a blue garment that is a shirt is equal to the probability that tom selects a blue garment multiplied by the probability that he selects a shirt. Yep. That is absolutely right. So actually this is, a lot of these statements are true. The only one that is not is that the outcome of events A and B are dependent on each other."
    },
    {
        "Q": "At 0:21. Why are radian angles much smaller than degree angles?\n",
        "A": "Actually, 1 radian is much larger than 1 degree. You can equate the two as: \u00cf\u0080 rad = 180\u00cb\u009a",
        "video_name": "C3HFAyigqoY",
        "timestamps": [
            21
        ],
        "3min_transcript": "Voiecover:One angle whose tangent is half is 0.46 radians. So we're saying that the tangent right over here is... So the tangent... So we're gonna write this down. So we're saying that the tangent of 0.46 radians is equal to half. And another way of thinking about the tangent of an angle is that's the slope of that angle's terminal ray. So it's the slope of this ray right over here. Yeah that makes sense that that slope is about half. Now what other angles have a tangent of 1 half? So let's look at these choices. So this is our original angle, 0.46 radians, plus pi over 2. If you think in degrees, pi is 180. pi over 2 is 90 degrees. So this one... Actually let me do in a color you're more likely to see. This one is gonna look like this. Where this is an angle of pi over 2. And just eyeballing it, you immediately see that the slope of this ray right over here. In fact they look like they are. They are perpendicular because they have an angle of pi over 2 between them. But they're definitely not going to have the same tangent. They don't have the same slope. Let's think about pi minus 0.46. So that's essentially pi is going along the positive x axis. You go all the way around. Or half way around to your pi radians. But then we're gonna subtract 0.46. So it's gonna look something like this. It's gonna look something like that where this is 0.46 that we have subtracted. Another way to think about it, if we take our original terminal ray and we flip it over the y axis, we get to this terminal ray right over here. And you could immediately see that the slope of the terminal ray is not the same as the slope of this one, of our first one, of our original, in fact they look like the negatives of each other. So we can rule that one out as well. So that's going to take us... If you add pi to this you're essentially going half way around the unit circle and you're getting to a point that is... Or you're forming a ray that is collinear with the original ray. So that's that angle right over here. So pi plus 0.46 is this entire angle right over there. And when you just look at this ray, you see its collinear is going to have the exact same slope as the terminal ray for the 0.46 radion. So just that tells you that the tangent is going to be the same. So I could check that there. And in previous videos when we explore the symmetries of the tangent function, we in fact saw that. That if you took an angle and you add pi to it, you're going to have the same tangent. And if you wanna dig a little bit deeper, I encourage you to look at that video on the symmetries of unit circle symmetries for the tangent function. So let's look at these other choices."
    },
    {
        "Q": "\nWhy does Sal move the graph 2 units to the right when it says -2 at 1:05?\n\nShouldn't he move it to the left by 2 units?",
        "A": "Think about it on a coordinate plane: if I shift 0 two to the right, I get 2. Because you have to think about it in terms of 0, not 2, you subtract 2 instead of adding 2.",
        "video_name": "5DLkB-g8Rr8",
        "timestamps": [
            65
        ],
        "3min_transcript": "- [Voiceover] We\u2019re told, \"the graph of the function f of x \"is equal to x-squared.\" We see it right over here in grey. It\u2019s shown in the grid below. \"Graph the function g of x is equal to \"x minus two-squared, minus four \"in the interactive graph.\" This is from the shifting functions exercise on Khan Academy, and we can see we can change the graph of g of x. But let\u2019s see, we want to graph it properly, so let\u2019s see how they relate. Well, let\u2019s think about a few things. Let\u2019s first just make g of x completely overlap. Now they're completely overlapping. And let\u2019s see how they\u2019re different. Well, g of x, if you look at what's going on here, instead of having an x-squared, we have an x minus two-squared. So, one way to think about it is, when x is zero, you have zero-squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two-squared is zero-squared, if we don\u2019t look at the negative four just yet. And so, we would want to shift this graph This is essentially how much do we shift to the right. It\u2019s sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well okay, for the original graph, when it was just x-squared, to get the zero-squared, I just have to put x equals zero. Now to get a zero-squared, I have to put in a two. So this is actually shifting the graph to the right. And so, what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two-squared, it's gonna shift it down by four. So what we wanna do is just shift both of these points down by four. So this is gonna go from nine, and this is gonna go from the coordinate five comma nine, to five comma, if we go down four, five comma five. to two comma negative four. Two comma negative four. Did I do that right? I think that\u2019s right. What, essentially, what we have going on is, g of x is f of x shifted two to the right and four down. Two to the right and four down. And notice, if you look at the vertex here, we shifted two to the right and four down. And I shifted this one also, this one also I shifted two to the right, and four down. And, there you have it. We have graphed g of x, which is a shifted version of f of x."
    },
    {
        "Q": "\n5:31 what is 80 divided by 2/5",
        "A": "80/2/5=80*5/2=40*5=200",
        "video_name": "xoXYirs2Mzw",
        "timestamps": [
            331
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nIn the video at 3:09 , the subtitles say c (comedy) instead of (c, d). Someone should fix that.",
        "A": "The subtitle is auto-generated by youtube I believe. Sal said, c comma d which I guess it picked up as comedy",
        "video_name": "S-agS4YaxxU",
        "timestamps": [
            189
        ],
        "3min_transcript": "with a hypotenuse of the isosceles right triangle sits along the base. Sits along the base. So, it's isosceles. So that's equal to that. It's a right triangle, and then this distance this distance between that point and this point is the same as the distance between F of X and G of X. F of X and G of X for this X value right over there. Now obviously that changes as we change our X value. To help us visualize this shape here, I've kind of drawn a picture of our coordinate plane. If we've viewed as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base. I've drawn the base of it. I've drawn the base of it right over there. Maybe I should to make it clear. Shade it in kind of parallel to these cross sections. So, I've drawn the base right over there. There's two other sides. There the side that's on ... I've drawn it here. I guess you could view it on its top side or the left side right over there. Over on this picture that would be this. When we're looking at it from above. Then you have this other side. I guess on this view this one you could call kind of the right side. Over here this is kind of the ... when you viewed over here this is the bottom side. The whole reason why I set this up, and we tempting to visualize this figure. I want to see if you can come up with a definite integral that describes the volume of this figure. That kind of almost looks like a football if you cut it in half or a rugby ball. It's skewed a little bit as well. What's an expression a definite integral that expresses the volume of this. that it intersects at the points. These functions intersect at the point zero zero and (c,d). Can you come up with some expression a definite in row of terms of zeros, and Cs, and Ds, and Fs, and Gs that describe the volume of this figure? Assuming you've paused the video, and have had a go at it, let's think about it. If we want to find the volume, one way to think about it is we could take the volume of, we could approximate the volume as the volume of these individual triangles. That would be the area of each of these triangles times some very small depth. Some very small depth. I'll just shade it in to show the depth. Some very small depth which we could call DX. Once again we could find the volume of each of these by finding the area, the cross sectional area there, and then multiplying that times a little DX."
    },
    {
        "Q": "At about the 5:40 sec mark, Khan talks about the product of dividing:\n\n9.2 * 10^5\n11.5 * 10^2\n\nI worked the equation before Khan. Instead of dividing 9.2 by 11.5 * 10^2, I said divide by\n1.15 * 10^3 (.) We came up with the same answer; 8 * 10^2.\n\nMy doing scientific math problems that way (changing any multiplication involved in the numerator or denominator to correct scientific notation, will that cause me any problems down the road (?))\n",
        "A": "I think it is good to know how to do a math question two different ways, so if you learn to do it both ways and remember how to do it, then I think you are set!",
        "video_name": "EbmgLiSVACU",
        "timestamps": [
            340
        ],
        "3min_transcript": "And now I can divide these two things. So this is going to be equal to-- we'll have to think about what 9.2 over 11.5 is. But actually let me just do that right now, get a little practice dividing decimals. Let me get some real estate here. Let me do that in the same color. 9.2 divided by 11.5-- well if we multiply both of these times 10, that's the exact same thing as 92 divided by 115. We're essentially moved the decimal to the right for both of them. And let me add some zeros here because I suspect that I'm going to get a decimal here. So let's think what this is going to be. Let's think about this. Well 115 doesn't go into 9. It doesn't go into 92. It does go into 920. Let's see if that works out. So I have my decimal here. That's a 0. 8 times-- 8 times 5 is 40. 8 times 11 is 88. And then 88 plus 4 is 92. Oh, it went in exactly, very good. So 920, we have no remainder. So 9.2 divided by 11.5 simplified to 0.80. And then 10 to the fifth divided by 10 to the second, we have the same base, and we're dividing. So we can subtract the exponents. That's going to be 10 to the 5 minus 2. So this right over here is going to be 10 to the third power-- so times 10 to the third power. Now, are we done? Well in order to be done, this number right over here needs to be greater than or equal to 1 and less than 10. It is clearly not greater than or equal to 1. that is greater than or equal to 1 and less than 10 and some power of 10? Well this 8 right over here, this is in the tenths place. It's 8/10, 8 times 1/10. So this is going to be the same thing as 8 times 10 to the negative 1 power. And then we have this 10 to the third here-- so times 10 to the third power. We'll do that in that other color. And now we have the same base. Just add the exponents. So this is going to be equal to 8 times 10 to the 3 minus 1-- so 8 times 10 squared. And we're done. We've simplified our original expression."
    },
    {
        "Q": "At 3:45, Sal talks about the \"orientation\" and \"magnitude\" of a vector. So, is \"orientation\" the direction of a vector? And is \"magnitude\" the length of a vector?\n",
        "A": "Yes, that is correct. Sometimes you will hear angle instead of orientation or direction, but it is all the same.",
        "video_name": "gsNgdVdAT1o",
        "timestamps": [
            225
        ],
        "3min_transcript": "vector b, or 10 times victor a minus 6 times vector b-- some combination of vector a and b, where I can get vector c. And vector c is the vector 7, 6. So let me see if I can visually draw this problem. So let me draw the coordinate axes. Let's see this one. 3, negative 6. That'll be in quadrant-- these are both in the first quadrant. So I just want to figure out how much of the axes I need to draw. So let's see-- Let me do a different color. That's my y-axis. I'm not drawing the second or third quadrants, because I don't think our vectors show up there. And then this is the x-axis. So first I'll do vector a. That's 3, negative 6. 1, 2, 3, and then negative 6. 1, 2, 3, 4, 5, 6. So it's there. So if I wanted to draw it as a vector, usually start at the origin. And it doesn't have to start at the origin like that. I'm just choosing to. You can move around a vector. It just has to have the same orientation and the same magnitude. So that is vector a for the green. Now let me do in magenta, I'll do vector b. That is 2, 6. 1, 2, 3, 4, 5, 6. And that's vector b. So it'll look like this. That's vector b. And then let me write down vector a down there. That's vector a. And I want to take some combination of vectors a and b. And add them up and get vector c. So what does vector c look like? It's 7, 6. Let me do that in purple. So 1, 2, 3, 4, 5, 6, 7. Comma 6. So 7, 6 is right over there. That's vector c. Vector c looks like that."
    },
    {
        "Q": "\nWhere does the 9/3 come from at 2:32 ?",
        "A": "Since you need common denominators in order to subtract two fractions, what he did is that after simplifying 12/4 to 3, he changed it into a fraction that equals 3, 9/3=3, in order to subtract the two fractions.",
        "video_name": "BOIA9wsM4ok",
        "timestamps": [
            152
        ],
        "3min_transcript": "the negative 10/3 and the positive 10/3, those cancel out to get a zero, and I'm just left with j/4. It's equal to j/4. Now you might recognize 9/3, that's the same thing as nine divided by three. So this is just going to be three. So that simplifies a little bit. Three, let me just rewrite it so you don't get confused. Three is equal to j/4. Now, to solve for j, I could just multiply both sides by four. 'Cause if I divide something by four and then multiply by four, I'm just going to be left with that something. If I start with j and I divide by four, and then I multiply, and then I multiply by four, so I'm just going to multiply by four, then I'm just going to be left with j on the right-hand side. But I can't just multiply the right-hand side by four. I have to do it with the left-hand side So I multiply the left-hand side by four as well. And what I will be left with, four times three is twelve. And then j divided by four times four, So we get j is equal to 12. And the neat thing about equations is you can verify that you indeed got the right answer. You can substitute 12 for j here, and verify that negative 1/3 is equal to 12/4 - 10/3. Does this actually work out? Well 12/4 is the same thing as three, and if I wanted to write that as thirds, this is the same thing as 9/3. And 9/3 - 10/3 is indeed equal to negative 1/3. So we feel very good about that. Let's do another example. So I have n/5 + 0.6 = 2. So let's isolate this term that involves n on the left-hand side. So let's get rid of this 0.6. So let's subtract 0.6 from the left-hand side. But I can't just do it from the left. I have to do it from both sides if I want the equality to hold true. So, subtract 0.6. I'm just going to be left with n/5, and on the right-hand side, 2 - 0.6, that;'s going to be 1.4. And if you don't want to do this in your head, you could work this out It's going to be 2.0 - 0.6. You could say, \"Oh, this is 20/10-6/10\" which is going to be 14/10, which is that there. Or if you want to do it a little bit kind of the traditional method, six from zero, let me re-group.\" That's going to be a 10. I'm going to take from the ones place. If I take a one from the one's place, and that's going to be equal to 10/10. 10/10 - 6/10 is 4/10. And then, bring down one one minus zero ones is just one. So it's 1.4. And now, to solve for n. Well on the left have n being divided by five. If I just want n here, I can just multiply by five. So, if I multiply by five,"
    },
    {
        "Q": "Why cant we directly substitute and find at 2:28 where 3y+3=y+7\n3y-y=7-3and y=2\n",
        "A": "Add 18r to -2r and add 12s to 6s",
        "video_name": "vkhYFml0w6c",
        "timestamps": [
            148
        ],
        "3min_transcript": "Alright, now we have a very interesting situation. On both sides of the scale, we have our mystery mass and now I'm calling the mystery mass having a mass of Y. Just to show you that it doesn't always have to be X. It can be any symbol as long as you can keep track of that symbol. But all these have the same mass. That's why I wrote Y on all of them. And we also have a little 1 kilogram boxes on both sides of the scale. So the first thing I wanna do, we're gonna do step by step and try to figure out what this mystery mass is. But the first thing I wanna do is, is, is, have you think about, whether you can represent this algebraically? Whether with, with a little bit of mathematic symbolry, you can represent what's going on this scale. Over here, I have three Ys and three of these boxes and their total mass is equal to this one Y. And I think I have about let's see, I have 7 boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have 3 boxes and a mass Y. So they're going to have a mass of 3Y, and then you have 3 boxes with a mass of 1 kilogram. So that's going to be my 1Y right over there. I could've written 1Y but that's, I don't need to. A Y is the same thing as 1Y. So I have the Y kilograms right there. And I have seven of these, right? 1,2,3,4,5,6,7. Yup, seven of these. So I have Y plus 7 kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This mass, total mass is equal to this total mass. So we can write an equal sign, right over there. So that's a good starting point. We were able to represent this situation to this real-life simple situation. You know, back in the day, when people actually had to figure out the mass of things if you were to go the jewelry store, whatever. They actually did had problems like this. We were able to represent it mathematically. Now the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well why won't we remove 3 of these what, of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why won't we remove 1 of these Ys from both sides? That also would be legitimate. And we could do it in either order. So let's just pick one of them. Let's say that we've first want to remove, let's say that we first want to remove the, a Y from either sides. Just so that we feel a little bit more comfortable with all of our Ys sitting on the 1 side. And so the best way, if we don't want all of our Ys to sit on the 1 side, we can remove, we can remove a Y from each side. Remember, if you removed a Y from only 1 side, that would unbalance the scale. The scale was already balanced whatever you have to do to the one side after you do the other. So I'm gonna remove a Y, I'm gonna remove Y mass from both sides. Now what will that look like algebraically?"
    },
    {
        "Q": "at 0:43 where did you get 0 from\n",
        "A": "You get the zero when you do not move either way on the x axis.",
        "video_name": "5a6zpfl50go",
        "timestamps": [
            43
        ],
        "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy"
    },
    {
        "Q": "At 0:58, Sal draws a line segment instead of a line. Is this supposed to be the case?\n",
        "A": "Should be a line, he just didn t draw the arrows.",
        "video_name": "5a6zpfl50go",
        "timestamps": [
            58
        ],
        "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy"
    },
    {
        "Q": "\nDo you get the 1 in the rise over run from 3/1? 5:37 in the video",
        "A": "3x is the same as saying go over 1 and up 3 which is like saying 3/1, so yes, you are corect",
        "video_name": "5a6zpfl50go",
        "timestamps": [
            337
        ],
        "3min_transcript": "graph both lines, both equations, and then look at their intersection. And that will be the solution to both of these equations. In the next few videos, we're going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to understand the graphical nature of solving systems of equations. Let's do another one. Let's say we have y is equal to 3x minus 6. That's one of our equations. And let's say the other equation is y is equal to negative x plus 6. And just like the last video, let's graph both of these. I'll try to do it as precisely as I can. There you go. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I should have just copied and pasted some graph paper here, but I think this'll do the job. So let's graph this purple equation here. Y-intercept is negative 6, so we have-- let me do another 1, 2, 3, 4, 5, 6. So that's y is equal to negative 6. And then the slope is 3. So every time you move 1, you go up 3. You moved to the right 1, your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line will look like this. And it looks like I intersect at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the right, we go down 1. And so this will intersect at-- well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will look like that. The graph, I want to get it as exact as possible. And so we're going to ask ourselves the same question. What is an x, y pair that satisfies both of these equations? Well, you look at it here, it's going to be this point. This point lies on both lines. And let's see if we can figure out what that point is."
    },
    {
        "Q": "Are rates and ratios the same as fractions?\nFor example, 35:1 can also be written as 35/1, which is a fraction.\n",
        "A": "Fractions explain how much a number is out of something, whereas ratios compare numbers",
        "video_name": "qGTYSAeLTOE",
        "timestamps": [
            2101
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:09, why is decameter spelled wrong? Or am I wrong?\n",
        "A": "Decimeter and Decameter are actually 2 different things. Decameters are bigger than Meters but smaller than Kilometers and are used less often in math.and science and such.",
        "video_name": "I3kQJvR7ZIg",
        "timestamps": [
            249
        ],
        "3min_transcript": "This would be even the distance from one city to another city. Even the radius of the planet is often measured in kilometers. And to once again get a sense of that, this right over here is a map of New York City. And right over here they give us the scale. This distance right over here is 5 kilometers. So 5 kilometers just to give us a sense of things-- and they wrote km for short so 5 km. This lowercase k, lowercase m is just shorthand for kilometers, this would be equal to-- well if 1 kilometer is 1,000 meters, 5 kilometers is going to be 5 times as many. It's going to be 5,000 meters is this distance right over here. If you were to try to imagine 1 kilometer, it would be 1/5 of that. So 1 kilometer on this map might look something like that. allow us to describe distances less than a kilometer but more But they're not as typically used. But I'll list them here, just so that you see that they exist. You have the \"hect-O-meter,\" or \"heck-TOM-eter\"-- it's so infrequently used that I really haven't heard a lot of people say it-- which is equal to 100 meters. And you have the \"DECK-a-meter\" or \"deck-AM-eter,\" and I actually think it's \"DECK-a-meter.\" Dekameter, which is 10 meters. And I'm going to write these in orange because they're not that frequently used. For example, it's not typical to hear someone say that the Empire State Building is 4.43 hectometers or for them to say that this is 44.3 dekameters. They would typically say that this is 443 meters. Now you're probably saying, OK, well, this is fine. Using these prefixes on the meter, are larger than a meter, that are multiples of tens of the meter or a multiple of 100 or a multiple of 1,000 of the meter. But what if I want to go smaller? Well, the metric system has units for that. And if we go just 1/10 of a meter, this is not used as typically, but I'll write it here-- the decimeter. The decimeter is 1/10 of a meter. Or another way of saying it is 1 meter is equal to 10 decimeters, lowercase dm for short. Once again, this is not so typically used. But if we go one scale even below that, we get to the centimeter, which is a heavily used unit. And here, the prefix centi means 1/100. So this is 1/100 of a meter. Another way of thinking about it-- 1 meter is equal to 100 centimeters."
    },
    {
        "Q": "At 4:40, how did someone found out all this?\n",
        "A": "As Sal says, exponential functions and logarithmic functions are inverses so they appear as reflections on the graph. Basically, the x values and y values are swapped.",
        "video_name": "K_PiPfYxtao",
        "timestamps": [
            280
        ],
        "3min_transcript": "swap these two columns. x and y, so let me just do 1/4, 1/2, one, two, four, and eight. Here now we're saying if x is 1/4, what power do we have to raise two to, to get to 1/4. We have to raise it to the negative two power. Two to the negative one power is equal to 1/2. Two to the zero power is equal to one. Two to the first power is equal to two. Two to the second power is equal to four. Two to the third power is equal to eight. Notice all we did, as we essentially swapped these two columns, so let's graph this. When x is equal to 1/4, y is equal to negative two. When x is one, y is zero. When x is two, y is one. When x is four, y is two. When x is eight, y is three. It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two? Well you'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think about, if you swap the axis you would get the other graph. It's essentially what we're doing. Notice it's symmetric about that line and that's because these are essentially the inverse functions of each other. One way to think about it is we swapped Just as this, as x becomes more and more and more and more negative you see y approaching zero. Here you see is y is becoming more and more negative as x is approaching zero, or you could say as x approaches zero y becomes more and more and more negative. The whole point of this is just to give you an appreciation for the relationship between an exponential function and a logarithmic function. They're essentially inverses of each other. You see that in the graphs, they're reflections of each other about the line y is equal to x."
    },
    {
        "Q": "at 8:10 and 9:21 why is he multiplying by 1/2 ?\n",
        "A": "He is finding the area of the triangles. The formula for finding the area of a triangle is 1/2 times base times height.",
        "video_name": "EqNzr56h1Ic",
        "timestamps": [
            490,
            561
        ],
        "3min_transcript": "because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. ABH has a right angle there. ACG has a right angle right over there. So you have two angles. Two corresponding angles are equal to each other. You're now dealing with similar triangles. So we know that triangle ABH-- I'll just write it as AHB, since I already wrote it this way. AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle. G is the right angle, and then C is the unlabeled angle. This is similar to triangle AGC. And what that does for us is now we can use the ratios to figure out what HG is equal to. So what could we say over here? BH over its corresponding side of the larger triangle-- so we say 8 over 24 is equal to 6 over not HG, but over a AG. 6 over AG, and I think you can see where this is going. You have 1/3. 1/3 is equal to 6 over AG, or we can cross-multiply here, and we can get AG is equal to 18. So this entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. But now we have proven to ourselves that this base has length of-- well, we have 18 here, and then we have another 18 here. So it has a length of 36. So the entire base here is 36. So that is 36. of the entire isosceles triangle. So the area of ACE is going to be equal to 1/2 times the base, which is 36, times 24. And so this is going to be the same thing as 1/2 times 36 is 18. 18 times 24. I'll just do that over here on the top. So 18 times 24. 8 times 4 is 32. 1 times 4 is 4, plus 3 is 7. Then we put as 0 here, because we're now dealing not with 2, but 20. You have 2 times 8 is 16. 2 times 1 is 2, plus 1. So it's 360, and then you have a 2, 7 plus 6 is 13. 1 plus 3 is 4. So the area of ACE is equal to 432. But we're not done yet."
    },
    {
        "Q": "\nWhy does Log2(2t) become zero at 3:01?So confused",
        "A": "Have a look in Top Questions for Theresa Johnson s answer (currently 14 votes) to dugee23 s question. She explains how Sal simplified log_2(2^t) and got t. (Note that there is no zero at 3:01.)",
        "video_name": "7Ig6kVZaWoU",
        "timestamps": [
            181
        ],
        "3min_transcript": "the variable that we're trying to solve for, we're trying to find what t value will make this equal that right over there. A good first step would maybe try to get this five out of the left hand side, so let's divide the left by five. If we want to keep this being in equality we have to do the same thing to both sides. We get two to the t power is equal to 1,111 over five. How do we solve for t here? What function is essentially the inverse of the exponential function? Well it would be the logarithm. If we say that a to the b power is equal to c then that means that log base a of c is equal to b. a to the b power is equal to c. Log base a of c says what power do I need Well I need to raise a to the b power, to get to c. a to the b power is equal to c. These two are actually equivalent statements. Let's take log base two of both sides of this equation. On the left hand side you have log base two of two to the t power. On the right hand side, you have log base two of 1,111 over five. Why is this useful right over here? This is what power do we have to raise two to, to get two to t power? Well to get two to the t power, we have to raise two to the t power. This thing right over here just simplifies to t. That just simplifies to t. On the right hand side, we have log base two, we have all of these business right over here. t is equal to log base two of 1,111 over five. This is an expression that gives us our t value but then the next question is well how do we figure out what this is? If you take out your calculator, you'll quickly notice that there is no log base two button, so how do we actually compute it? Here we just have to apply a very useful property of exponents. If we have log base two of well really anything. Let me write it this way, if we have log base a of c, we can compute this as log base anything of c over log base that same anything of a. This anything has to be the same thing. Our calculator is useful because it has a log, you just press log, it's log base 10."
    },
    {
        "Q": "\nAt around 4:04, why does Sal say \"pi over 2\"? It's supposed to be pi over 4, right?",
        "A": "yes",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            244
        ],
        "3min_transcript": "arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it. of what is equal to the square root of 2 over 2. And this, I think, is a much easier question for you to answer. Sine of what is square root of 2 over 2? Well I just figured out that the sine of pi over 4 is square root of 2 over 2. So, in this case, I know that the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is equal to pi over 4. Or, I could have rewritten this as, the arcsine-- sorry --arcsine of the square root of 2 over 2 is equal to pi over 4. Now you might say so, just as review, I'm giving you a value and I'm saying give me an angle that gives me, when I take the sine of that angle that gives me that value. Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to"
    },
    {
        "Q": "Can someone explain to me why at 1:33 Sal says that the unit circle is sqrt2/2!!\n\nI don't understand, looking at the circle, knowing that 180 degrees equals pi, how what looks ro ME as pi/4, is actually the sqrt2/2!\n\nHELP!\n",
        "A": "@1:33, you re right that the ANGLE is pi / 4 (or 45 degrees), but Sal was solving the equation for the length of the non-hypotenuse SIDES of the right triangle, each of which turned out to be sqrt2 / 2. @7:32, Sal indicates - sqrt3 / 2 on the Y-axis because the problem given is arcsin(-sqrt3/2). In other words we re looking for the angle whose sin (its measure on the Y-axis) is - sqrt3 / 2.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            93
        ],
        "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it."
    },
    {
        "Q": "At 6:10 the domain arcsin is restricted to the 1st and 4th quadrant. So far I follow. It starts at Pi/2 - ok I follow. the other range is - Pi/2 ? I do not understand. On the unit circle diagram I pulled up the point (0,-1) is 3pi/2. and I do not see - 2/pi any where on the unit circle. What gives?\n",
        "A": "3pi/2 is the same place as -pi/2.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            370
        ],
        "3min_transcript": "Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I?"
    },
    {
        "Q": "At 1:13, how did Sal know that radius of the circle is 1? Like where did he get the 1 for the hyponthensus\n",
        "A": "He refer to the unit circle radius 1",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            73
        ],
        "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it."
    },
    {
        "Q": "\n@ 3:22 Couldn't he factor out (s-a) (s-b) and (s-c) to just s(-a*-b*-c)?\nI'm not sure. I'm rusty with my factoring...",
        "A": "It would seem that way, but actually, if we factor it out that way, we get -s*a*b*c which is much different from (s-a) (s-b) and (s-c)",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            202
        ],
        "3min_transcript": "And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square"
    },
    {
        "Q": "at 4:22 Sal gets to the square root of \"7\" and stops and says, \"you don't deal with the negative square roots . . . so this is just the square root of 7.\" The square root of 7 is not negative, it's 2.645... So, why did he stop and not figure out the square root of 7?\n",
        "A": "He said negative not about the square root of 7 but about all of those numbers because square root of a positive number can be either positive or negative eg. square root of 4 can be either 2 or -2 ( -2 *-2=4 ). He did not want it to be negative because you are dealing with area, which is not negative. Also by the way there is no need to calculate root of 7 unless you need to find cost of anything.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            262
        ],
        "3min_transcript": "This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square The square root of 36 is just 6. This is just 3. And we don't deal with the negative square roots, because you can't have negative side lengths. And so this is going to be equal to 18 times the square root of 7. So just like that, you saw it, it only took a couple of minutes to apply Heron's Formula, or even less than that, to figure out that the area of this triangle right here is equal to 18 square root of seven. Anyway, hopefully you found that pretty neat."
    },
    {
        "Q": "For the triangle at 3:17, with lengths 9, 11 and 16, can you divide that in half and use pythagoras, so one side becomes 8,11 and x or 8,9 and x? and find the value of x, and that is your height?\n",
        "A": "When we look at the triangle at 3;17, you cannot divide it in half and use the Pythagorean Theorem, because we don t know if it is a right triangle. So with a question were they don t give you the 90\u00c2\u00b0 right angle symbol, you can never assume anything in math. Hope this helps.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            197
        ],
        "3min_transcript": "And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square"
    },
    {
        "Q": "hey isn't that any number to the 0th power is always 1 including negative number at 1:42\n",
        "A": "Yes, but that is not what was in the problem \u00e2\u0088\u0092 b\u00e2\u0081\u00b0 means the same as \u00e2\u0088\u0092 (b\u00e2\u0081\u00b0) = \u00e2\u0088\u0092 (1) = \u00e2\u0088\u00921 However (\u00e2\u0088\u0092b)\u00e2\u0081\u00b0 would be equal to 1",
        "video_name": "gR8-vRg6Yp0",
        "timestamps": [
            102
        ],
        "3min_transcript": "Let's say that the position of some particle as a function of time is given by this expression right over here. Negative d to the negative t power plus c to the fourth over c squared plus 1, where c and d are constants and both of them are greater than one. So what I want to do over the course of this video is see what can we infer based on this expression, this function definition, that we have here. And the first thing that I want you to think about is, what is the initial position? If I were to express the initial position, in terms of c's and d's, and try to simplify it. So I encourage you to pause the video and try to find an expression for the initial position. Well, the initial position is the position we're at when our time is equal to 0. So we essentially just want to find p of 0. And p of 0 is going to be equal to negative d to the negative 0. Negative 0 plus c to the fourth over c squared plus 1. Well, d to the negative 0, that's the same thing as d to the 0, and since we know that d is non-zero, we know this is defined. Anything non-zero to the 0-th power is going to be 1. And the zero is actually under debate, what 0 to the 0-th power is. But we can safely say that this right over here is going to be equal to 1. And so the numerator here simplifies 2. This is equal to-- and I'll switch the order. c to the fourth minus 1 over c squared plus 1. And now this might jump out at you as a difference of squares. We could write this as c squared, squared, minus 1 squared over c squared plus 1. And that's the same thing as c squared plus 1 times c squared minus 1. And we have a c squared plus 1 in the numerator and that denominator so we can simplify. And so our initial position is going to be c squared minus 1. So that actually simplified out quite nicely. Now the next question I'm going to ask you is-- OK, we know that the initial position that time equals zero, the particle is going to be at c squared minus 1. But what happens after that? Does the position keep increasing? Does the position keep decreasing? Or does the position maybe increase and then decrease, or decrease and then increase as in keep swapping around? So I encourage you to pause the video now and think about what happens to the position. Does it keep increasing? Does it keep decreasing? Or does it do something else? Well, let's answer that question of what's happening to the position after our initial position. We really just have to focus on this term right over here. This d to the negative t."
    },
    {
        "Q": "\nAt 3:42 when it says the radical sign means the principle square root; what's the notation for either square root, or for the negative square root, in that case?\n\nAlso when talking about the third root, you wouldn't have to distinguish between principle and negative, right? Do you have to distinguish for all even roots, but not for odd roots?",
        "A": "The notation for both square roots is \u00c2\u00b1\u00e2\u0088\u009ax. If you want only the negative square root, then you would say -\u00e2\u0088\u009ax. As for odd roots, you still have to distinguish between principal and non principal roots. The only difference is that a non principal odd root is not simply the negative of the principal root. For instance, the cube root of -8 is -2, but it can also be 1\u00c2\u00b1\u00e2\u0088\u009a3 i.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            222
        ],
        "3min_transcript": "And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0."
    },
    {
        "Q": "\nwait.. at 1:11, Sal says (principal square root = psr) psr of (a*b) = psr of a *psr of b, when in the last video he states that it does not work when the numbers are both negative, and in this case they are. so therefore, this is wrong, right?",
        "A": "Square roots generally have two roots: the principal root and the negative root. For example: \u00e2\u0088\u009a9: psr = 3; negative root = -3. The default is to use the psr. If the negative root is wanted, then a minus sign is placed in front of the square root symbol. -\u00e2\u0088\u009a9 = -3",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            71
        ],
        "3min_transcript": "In your mathematical careers you might encounter people who say it is wrong to say that i is equal to the principal square root of negative 1. And if you ask them why is this wrong, they'll show up with this kind of line of logic that actually seems pretty reasonable. They will tell you that, OK, well let's just start with negative 1. We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look just from straight up properties of the principal square root function, they'll tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so if you have the principal square root of a times the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative."
    },
    {
        "Q": "at 3:50 sal says that each number has a positive and negative square root. And i was thinking, if that was the case, couldn't the square root of -1= -i, which is i to the 3rd power. If this was true then i to the 1st power would eqaul i to the 3rd power. in essence -i=i. I'm a little cunfused so could someone please explain this to me.\n",
        "A": "Sal says that 4 has 2 square roots. Positive numbers have two square roots. Negative numbers do not.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            230
        ],
        "3min_transcript": "And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0."
    },
    {
        "Q": "\nAt 3:23, does that mean that every number actually has two square roots? Wouldn't that make algebra nearly impossible?",
        "A": "Not really, it just means you have two possible solutions if you have a function with a square root. example: x^2=9 -> x=sqrt(9) -> x=3 or x=-3. After all 3*3 = 9 and -3*-3 = 9.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            203
        ],
        "3min_transcript": "the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers--"
    },
    {
        "Q": "How did you get the porabla at 3:02\nI'm still confused..\n",
        "A": "I think I understand your question. The line describing the parabola is curved, not straight, but with only a few points drawn in, it s hard to see why. If you went through the equation and filled in x with all the points in between the points that he graphed, you would see a curve shape appear. When you only have two points it s easy to see why it looks like it should be straight. Sal just skipped over doing 2.5, and then 2.25, and so on. The more points you graph, the more it looks l",
        "video_name": "hjigR_rHKDI",
        "timestamps": [
            182
        ],
        "3min_transcript": "So that right there is the highest point of our parabola. Then, if we want, we can a graph a couple of other points, just to see what happens. So let's see what happens when x is equal to-- let me just draw a little table here-- 2, what is y? It's negative x squared plus 6. So when x is 2, what is y? You have 2 squared, which is 4, but you have negative 2 squared, so it's negative 4 plus 6-- it is equal to 2. It's the same thing when x is negative 2. You put negative 2 there, you square it, then you have positive 4, but you have a negative there, so it's negative 4 plus 6 is 2. You have both of those points there, so 2 comma 2, and then you have a negative 2 comma 2. If I were to graph it, Let's try it with 3, as well-- if we It then becomes a negative 9 plus 3, it becomes negative 3, and negative 3 will also become a negative 3. Negative 3 squared is positive 9, you have a negative out front, it becomes negative 9 plus 6, which is negative 3. You have negative 3, negative 3, and then you have 3, negative 3. So those are all good points. Now we can graph our parabola. Our parabola will look something-- I was doing well until that second part --like that, and let me just do the second part. That second part is hard to draw-- let me do it from here. It looks something like that. We connect to this dot right here, and then let me connect this. So that it looks something like that. That's what our parabola looks like, and obviously it keeps going down in that direction. Let's graph this second one over here: y is equal to negative 2x minus 2. This is just going to be a line. It's a linear equation, and the highest degree here is 1. Our y-intercept is negative 2, so 0, 1, 2. Our y-intercept is negative 2. Our slope is negative 2. If we move 1 in the x direction, we're going to go 2 in the y-direction, and if we move 2 in the x direction, we're going to move down 4 in the y direction. If we move back 2, we're going to move up 2 in the y direction, and it looks like we found one of our points of intersection. Let's just draw that line, so that line will look something like-- It's hard for my hand to draw that, but let me try as best as I can. This is the hardest part. It will look something like that right there."
    },
    {
        "Q": "\nWhy is the R^n space not straight? I thought it was made up of vectors. At 1:47?",
        "A": "It is straight . The wiggly circle that Sal drew should be viewed very abstractly. Within that wiggly loop is all the vectors in R^n.",
        "video_name": "pMFv6liWK4M",
        "timestamps": [
            107
        ],
        "3min_transcript": "We now have the tools, I think, to understand the idea of a linear subspace of Rn. Let me write that down. I'll just always call it a subspace of Rn. Everything we're doing is linear. Subspace of Rn. I'm going to make a definition here. I'm going to say that a set of vectors V. So V is some subset of vectors, some subset of Rn. So we already said Rn, when we think about it, it's really just really an infinitely large set of vectors, where each of those vectors have n components. I'm going to not formally define it, but this is just a set of vectors. I mean sometimes we visualize it as multi-dimensional space and all that, but if we wanted to be just as abstract about it as possible, it's just all the set. It's the set of all of the -- you know we could call x1, x2, xi's are a member of the real numbers for all of the i's. That was our definition of Rn. It's just a huge set of vectors. An infinitely large set of vectors. V, I'm calling that, I'm going to call that a subset of Rn, and which means it's just some -- you know, it could be all of these vectors, and I'll talk about that in a second. Or it could be some subset of these vectors. Maybe it's all of them but one particular vector. In order for this V to be a subspace-- so I'm already saying it's a subset of Rn. Maybe this'll help you. If I draw all of Rn here as this big blob. So these are all of the vectors that are in Rn. V is some subset of it. It could be all of Rn. I'll show that a second. But let's just say that this is V. V is a subset of vectors. definition, if V is a subspace, or linear subspace of Rn, this means, this is my definition, this means three things. This means that V contains the 0 vector. I'll do it really, that's the 0 vector. This is equal to 0 all the way and you have n 0's. So V contains the 0 vector, and this is a big V right there. If we have some vector x in V. So let me write this, if my vector x is in V, if x is one of these vectors that's included in my V, then when I multiply x times any member of the reals."
    },
    {
        "Q": "\nsall, in 13:40 - 13:56, I think I don't get it. -a is a subset of R^2? since R^2 is all the cartesian plain . What we are saying here is -a is not a subset of our restriction of R^2 that we made and not R^2. Is this right? So we are saying that -a is not a subset of R^2 when x_1>0 right?",
        "A": "I think I need to rethink what a subspace is, Thank you.",
        "video_name": "pMFv6liWK4M",
        "timestamps": [
            820,
            836
        ],
        "3min_transcript": "to each other, this thing is also going to be greater than 0. And we don't care what these, these can be anything, I didn't put any constraints on the second component of my vector. So it does seem like it is closed under addition. Now what about scalar multiplication? Let's take a particular case here. Let's take my a, b again. I have my vector a, b. Now I can pick any real scalar. So any real scalar. What if I just multiply it by minus 1? So minus 1. So if I multiply it by minus 1, I get minus a, minus b. If I were to draw it visually, if this is-- let's say a, b was the vector 2, 4. So it's like this. When I multiply it by minus 1, what do I get? I get this vector. Which you can be visually clearly see falls out of, if we view these as kind of position vectors, it falls out of our subspace. Or if you just view it not even visually, if you just do it mathematically, clearly if this is positive then this is going to-- and let's say if we assume this is positive, and definitely not 0. So it's definitely a positive number. So this is definitely going to be a negative number. So when we multiply it by negative 1, for really any element of this that doesn't have a 0 there, you're going to end up with something that falls out of it, right? This is not a member of this set, because to be a member of the set, your first component had to be greater than 0. This first component is less than 0. So this subset that I drew out here, the subset of R2, is not It's not closed under multiplication or scalar multiplication. Now I'll ask you one interesting question. What if I ask you just the span of some set of vectors? Let's say I want to know the span of, I don't know, let's sat I have vector v1, v2, and v3. I'm not even going to tell you how many elements each of these vectors have. Is this a valid subspace of Rn? Where n is the number of elements that each of these have. Let's pick one of the elements."
    },
    {
        "Q": "At 4:00.. why do you use 2pi for the period?\n",
        "A": "2pi is once around a circle. Going around a circle makes a sine wave or a cosine wave.",
        "video_name": "SBqnRja4CW4",
        "timestamps": [
            240
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:46, how does Sal determine what the slope is? How does he know it's -(1/2)? Isn't -(1/2) the y-intercept?\n",
        "A": "Both the slope and the y-intercept are - \u00c2\u00bd When a function of a line is in the form of mx +b, the slope is m. Since -x/2 = -\u00c2\u00bd(x) the slope is -\u00c2\u00bd",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            346
        ],
        "3min_transcript": "like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that. look something like that. It'll just keep going, so it'll look something like that, and it'll keep going in both directions. And now let's see if this really is a reflection over y equals x. y equals x looks like that, and you can see they are a reflection. If you reflect this guy, if you reflect this blue line, it becomes this orange line. But the general idea, you literally just-- a function is originally expressed, is solved for y in terms of x. You just do some algebra. Solve for x in terms of y, and that's essentially your inverse function as a function of y, but then you can rename it as a function of x."
    },
    {
        "Q": "\nhow do we come to know wheter a curve is 1:1 function and does its has an inverse?",
        "A": "The horizontal line test is the easiest way. If you can draw a perfectly horizontal line such that it touches two points on the function, then it is not 1 to 1.",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            61
        ],
        "3min_transcript": "So we have f of x is equal to negative x plus 4, and f of x is graphed right here on our coordinate plane. Let's try to figure out what the inverse of f is. And to figure out the inverse, what I like to do is I set y, I set the variable y, equal to f of x, or we could write that y is equal to negative x plus 4. Right now, we've solved for y in terms of x. To solve for the inverse, we do the opposite. We solve for x in terms of y. So let's subtract 4 from both sides. You get y minus 4 is equal to negative x. And then to solve for x, we can multiply both sides of this equation times negative 1. And so you get negative y plus 4 is equal to x. Or just because we're always used to writing the dependent variable on the left-hand side, we could rewrite this as x is equal to negative y plus 4. Or another way to write it is we could say that f inverse of y is equal to negative y plus 4. it as a function of y, but we can just rename the y as x so it's a function of x. So let's do that. So if we just rename this y as x, we get f inverse of x is equal to the negative x plus 4. These two functions are identical. Here, we just used y as the independent variable, or as the input variable. Here we just use x, but they are identical functions. Now, just out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. It's a negative x plus 4. It's the exact same function. So let's see, if we have-- the y-intercept is 4, it's going to be the exact same thing. The function is its own inverse. So if we were to graph it, we would put it right on top of this. In the first inverse function video, I talked about how a function and their inverse-- they are the reflection over the line y equals x. So where's the line y equals x here? Well, line y equals x looks like this. And negative x plus 4 is actually perpendicular to y is equal to x, so when you reflect it, you're just kind of flipping it over, but it's going to be the same line. It is its own reflection. Now, let's make sure that that actually makes sense. When we're dealing with the standard function right there, if you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2 either way, so that makes sense. For the regular function, 4 gets mapped to 0. For the inverse function, 0 gets mapped to 4."
    },
    {
        "Q": "In 5:00 onwards, why doesn't the -2 turn the whole function into -y/2 +1/2? Shouldn't it invert for both numbers?\n",
        "A": "It doesn t invert any of the numbers actually. He had y+1 = -2x, so he divided both sides by -2. Think of it as having to divide both terms on the left side (y and 1) individually by -2. So y divided by -2 is -y/2 and 1 divided by -2 is -1/2. SO you end up with -y/2 - 1/2 = x",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            300
        ],
        "3min_transcript": "Let's think about it another way. For the regular function-- let me write it explicitly down. This might be obvious to you, but just in case it's not, it might be helpful. Let's pick f of 5. f of 5 is equal to negative 1. Or we could say, the function f maps us from 5 to negative 1. Now, what does f inverse do? What's f inverse of negative 1? f inverse of negative 1 is 5. Or we could say that f maps us from negative 1 to 5. So once again, if you think about kind of the sets, they're our domains and our ranges. So let's say that this is the domain of f, this is the range of f. f will take us from to negative 1. That's what the function f does. And we see that f inverse takes us back from negative 1 to 5. like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that."
    },
    {
        "Q": "\nat 5:50 he says the slope is -1/2. he moves over in the positive direction on the x axis which i don't quite get and also slope i thought was change in Y over change in X so when he moves down by 1/2 i am at a loss. over by 1 down 1/2 doesn't make sense to me. anybody got another way of explaining by chance?",
        "A": "nevermind, i think i got it. i get the same inverse line as Sal if i go down -1 on Y then over 2 on X.",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            350
        ],
        "3min_transcript": "like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that. look something like that. It'll just keep going, so it'll look something like that, and it'll keep going in both directions. And now let's see if this really is a reflection over y equals x. y equals x looks like that, and you can see they are a reflection. If you reflect this guy, if you reflect this blue line, it becomes this orange line. But the general idea, you literally just-- a function is originally expressed, is solved for y in terms of x. You just do some algebra. Solve for x in terms of y, and that's essentially your inverse function as a function of y, but then you can rename it as a function of x."
    },
    {
        "Q": "At 8:50, how can (x+2) be considered positive when -1 still falls in the range of x>-2?\n",
        "A": "SG, When x=-1 x+2 = -1+2 = +1 so even though x is not positive, (x+2) is positive. I hope that is of some help.",
        "video_name": "ZjeMdXV0QMg",
        "timestamps": [
            530
        ],
        "3min_transcript": "I'll just keep it the same way and maybe in the next video I'll do the case where it's greater than or equal just because I really don't want to-- maybe I want to incrementally step up the level of difficulty. We're just saying x minus 1 over x plus 2 is just straight up greater than 0. Now one thing you might say is well, if I have a rational expression like this, maybe I multiply both sides of this equation by x plus 2. So I get rid of it in the denominator and I can multiply it times 0 and get it out of the way. But the problem is when you multiply both sides of an inequality by a number-- if you're multiplying by a positive you can keep the inequality the same. But if you're multiplying by a negative you have to switch the inequality, and we don't know whether x plus 2 is positive or negative. So let's do both situations. Let's do one situation where x plus 2-- let me write it this way. x plus 2 is greater than 0. And then another situation where-- let me do that in a different color. Where x plus 2 is less than 0. Actually, in those situations can x plus 2 equal 0? If x plus 2 were to be equal to 0 than this whole expression would be undefined. And so that definitely won't be a situation that we want to deal with it. It would an undefined situation. So these are our two situations when we're multiplying both sides. So if x plus 2 is greater than 0 that means that x is greater than minus 2. We can just subtract 2 from both sides of this equation. So if x is greater than minus 2, then x plus 2 is greater than 0. And then we can multiply both sides of this equation times x plus 2. So you have x minus 1 over x plus 2 greater than 0. I'm going to multiply both sides by x plus 2, which I'm assuming is positive because x is greater than minus 2. Multiply both sides by x plus 2. These cancel out. 0 times x plus 2 is it just 0. just simplified to 0. Solve for x, add 1 to both sides, you get x is greater than 1. So we saw that if x plus 2 is greater than 0, or we could say, if x is greater than minus 2, then x also has to be greater than 1. Or you could say if x is great-- well, you could go both ways in that. But we say, look, both of these things have to be true. If for x to satisfy both of these, it just has to be greater than 1. Because if it's greater than 1 it's definitely going to satisfy this constraint over here. So for this branch we come up with the solution x is greater than 1. So this is one situation where x plus 2 is greater than 0. The other situation is x plus 2 being less than 0. If x plus 2 is less than 0 that's equivalent to saying that x is less than minus 2. You just subtract 2 from both sides. Now, if x plus 2 is less than 0, what we'll have to do when we multiply both sides-- let's do that."
    },
    {
        "Q": "At around 00:58: Why is a>0 and b>0 the consequence of a/b>0?\n",
        "A": "A/B can only be greater than 0 (in other words, positive) only if A and B are both positive (A>0 and A>0) or A and B are both negative (A<0 and B<0). Any other combination will result in a negative fraction (A/B<0 instead of A/B>0).",
        "video_name": "ZjeMdXV0QMg",
        "timestamps": [
            58
        ],
        "3min_transcript": "In this video I want to do a couple of inequality problems that are deceptively tricky. And you might be saying, hey, aren't all inequality problems deceptively tricky? And on some level you're probably right. But let's start with the first problem. We have x minus 1 over x plus 2 is greater than 0. And I'm actually going to show you two ways to do this. The first way is, I think, on some level, the simpler way. But I'll show you both methods and whatever works for you, well, it works for you. So the first way you can think about this, if I have just any number divided by any other number and I say that they're going to be greater than 0. Well, we just have to remember our properties of multiplying and dividing negative numbers. In what situation is this fraction going to be greater than 0? Well, this is going to be greater than 0 only if both a and-- so we could write both a is greater than 0 and b is greater than 0. So this is one circumstance where this'll definitely be true. definitely be a positive. It'll definitely be greater than 0. Or we could have the situation where we have a negative divided by a negative. If we have the same sign divided by the same sign we're also going to be positive. So or-- a is less than 0 and b is less than 0. So whenever you have any type of rational expression like this being greater than 0, there's two situations in which it will be true. The numerator and the denominator are both greater than 0, or they're both less than 0. So let's remember that and actually do this problem. So there's two situations to solve this problem. The first is where both of them are greater than 0. If that and that are both greater than 0, we're cool. So we could say our first solution-- maybe I'll draw a little tree like that-- is x minus 1 greater than 0 and x plus 2 greater than 0. That's equivalent to this. The top and the bottom-- if they're both greater than 0 something greater than 0. The other option-- we just saw that-- is if both of them were less than 0. So the other option is x minus 1 less than 0 and x plus 2 less than 0. If both of these are less than 0 then you have a negative divided by a negative, which will be positive. Which will be greater than 0. So let's actually solve in both of these circumstances. So x minus 1 greater than 0. If we add 1 to both sides of that we get x is greater than 1. And if we do x plus 2 greater than 0, if we subtract 2 from both sides of that equation-- remember I'm doing this right now-- we get x is greater than minus 2. So for both of these to hold true-- so in this little brown or red color, whatever you want to think of it-- in order for both of these to go hold true, x has to be greater than 1 and"
    },
    {
        "Q": "\nAt around 3:40 confused me a lot. He says you can re-write the equation as \"|x - -2|=6\", and then find that the answers are 6 away from -2. While that does appear true, why does it work for that equation, and not the original equation \"|x+2|=6\"? How did Sal know to re-write the \"+2\" as \"- -2\"?",
        "A": "The minuses cancel out. Thus, subtracting negative 2 is eqall to adding 2.",
        "video_name": "u6zDpUL5RkU",
        "timestamps": [
            220
        ],
        "3min_transcript": "Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside Or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind, that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5? these are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4, or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one, and we'll do it in purple. Let's say we have the absolute value of 4x-- I'm going to change this problem up a little bit. 4x minus 1. The absolute value of 4x minus 1, is equal to-- actually, I'll just keep it-- is equal to 19. So, just like the last few problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19. Because then when you take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation-- we could do them"
    },
    {
        "Q": "\nat 1:40 how did he remove the absolute value portion? Did he just make it into an equation? Can someone please clarify.",
        "A": "Sal is solving the absolute value equation of |x-5|=10. To solve this, you have two possible choices (since we have the absolute value there). Either x-5 could be 10 or x-5 could be -10. (because | -10 | = 10. So he is solving the absolute value equation by taking what s inside of it and setting it equal to 10 and -10. x-5=10 and therefore x = 15. x-5=-10 and therefore x= -5. So our two answers are 15 and -5.",
        "video_name": "u6zDpUL5RkU",
        "timestamps": [
            100
        ],
        "3min_transcript": "Let's do some equations that deal with absolute values. And just as a bit of a review, when you take the absolute value of a number. Let's say I take the absolute value of negative 1. What you're really doing is you're saying, how far is that number from 0? And in the case of negative 1, if we draw a number line right there-- that's a very badly drawn number line. If we draw a number line right there, that's 0. You have a negative 1 right there. Well, it's 1 away from 0. So the absolute value of negative 1 is 1. And the absolute value of 1 is also 1 away from 0. It's also equal to 1. So on some level, absolute value is the distance from 0. But another, I guess simpler way to think of it, it always results in the positive version of the number. The absolute value of negative 7,346 is equal to 7,346. So with that in mind, let's try to solve some equations So let's say I have the equation the absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation, but I'll show you how to solve it systematically. Now this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside"
    },
    {
        "Q": "6:27 Why isn't it x+3=y or x+3=-y?\n",
        "A": "think of what a basic absolute value y = | x | looks like, it is a V shape with the vertex at the origin. Thus for any value of y except 0 (y values are horizontal lines), the value of x would be +y or -y. If y = 1, we would have two points (1,1) and (-1,1), We expand this property into more complicated square roots, so y = x+3 or y = - (x+3)",
        "video_name": "u6zDpUL5RkU",
        "timestamps": [
            387
        ],
        "3min_transcript": "these are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4, or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one, and we'll do it in purple. Let's say we have the absolute value of 4x-- I'm going to change this problem up a little bit. 4x minus 1. The absolute value of 4x minus 1, is equal to-- actually, I'll just keep it-- is equal to 19. So, just like the last few problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19. Because then when you take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation-- we could do them Add 1 to both sides of this equation, you get 4x is equal to negative 18. Divide both sides of this by 4, you get x is equal to 5. Divide both sides of this by 4, you get x is equal to negative 18/4, which is equal to negative 9/2. So both of these x values satisfy the equation. Try it out. Negative 9/2 times 4. This will become a negative 18. Negative 18 minus 1 is negative 19. Take the absolute value, you get 19. You put a 5 here, 4 times 5 is 20. Minus 1 is positive 19. So you take the absolute value. Once again, you'll get a 19. Let's try to graph one of these, just for fun. So let's say I have y is equal to the absolute So this is a function, or a graph, with an absolute value in it. So let's think about two scenarios. There's one scenario where the thing inside of the absolute value is positive. So you have the scenario where x plus 3-- I'll write it over here-- x plus 3 is greater than 0. And then you have the scenario where x plus 3 is less than 0. When x plus 3 is greater than 0, this graph, or this line-- or I guess we can't call it a line-- this function, is the same thing as y is equal to x plus 3. If this thing over here is greater than 0, then the absolute value sign is irrelevant. So then this thing is the same thing as y is equal to x plus 3. But when is x plus 3 greater than 0? Well, if you subtract 3 from both sides, you get x is greater than negative 3. So when x is greater than negative 3, this graph is"
    },
    {
        "Q": "\nyou lost me at around 1:35 to 2:00. I don't get how 1to the 0 power could equal to 1. isn't the power of powers multiplication? shouldn't 1x0=0?",
        "A": "Think of it this way: x\u00c2\u00b3/x\u00c2\u00b2 = x\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b2 = x\u00c2\u00b9 = x Similarly: x\u00c2\u00b3/x\u00c2\u00b3 = x\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b3 = x\u00e2\u0081\u00b0 but, of course, x\u00c2\u00b3/x\u00c2\u00b3 = 1, Thus, x\u00e2\u0081\u00b0=1 Of course, this doesn t work if x=0 , 0\u00e2\u0081\u00b0 is undefined.",
        "video_name": "NEaLgGi4Vh4",
        "timestamps": [
            95,
            120
        ],
        "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done."
    },
    {
        "Q": "As well as my previous question, at 1:19 sal said that any number to the zeroth power is going to be 1. Does that mean that 8 to the zeroth power will be 1, or that 8 to the zeroth power will be 8?\n",
        "A": "It means that 8^0 = 1, even 100^0 = 1, any number to the zero-th power is equal to 1.",
        "video_name": "NEaLgGi4Vh4",
        "timestamps": [
            79
        ],
        "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done."
    },
    {
        "Q": "At 0:00 who is the ancient philosopher?\n",
        "A": "The video starts out at 0:00 with So you, as the ancient philosopher . So is he you?",
        "video_name": "pzQY-9Nmtws",
        "timestamps": [
            0
        ],
        "3min_transcript": "So you, as the ancient philosopher in mathematics have concluded in order for the multiplication of positive and negative numbers to be consistent with everything you've been constructing so far with all the other properties of multiplication that you know so far that you need a negative number times a positive number or a positive times a negative to give you a negative number and a negative times a negative to give you a positive number and so you accept it's all consistent so far.. this deal does not make complete concrete sense to you, you want to have a slightly deeper institution than just having to accept its consistent with the distributive property and whatever else and so you try another thought experiment, you say \"well what is just a basic multiplication way of doing it?\" So if I say, two times three, one way to to conceptualize is basic multiplication is really repeating addition, so you could view this as two threes so let me write three plus three or you could view this as three twos, and so this is the same thing as two plus two plus two and there are three of them, and either way you can conceptualize as you get the same exact answer. This is going to be equal to six, fair enough! Now, you knew this before you even tried to tackle negative numbers. Now let's try to make one of these negatives and see what happens. Let's do two times negative three, I want to make the negative into a different color. Two times negative three. Well, one way you could view this is the same analogy here, it's negative three twice so it would be negative.. I'll try to color code it negative three and then another negative or, and this is the interesting thing, instead of over here there's a two times positive three you added two, three times. But since here is two times negative three you could also imagine you are going to subtract two, three times So instead of up here, I could written two plus two plus two because this is a positive two right over here, but since we're doing this over negative three we could imagine subtracting two, three times, so this would be subtracting two (repeated) subtract another two right over here, subtract another two and then you subtract another two notice you did it, once again, you did it three times, so this is a negative three, so essentially you are subtracting two, three times. And either way, you can conceptualize"
    },
    {
        "Q": "\nAt 5:07 Why does he simplify (1/x)(x-1) to (x-1)/x. Could he also have multiplied (1/x) by both halves? so (1/x)(x-1) = (x/x)-(1/x)= -(1/x). I am wondering if these are two different simplifications or if my answer is wrong.",
        "A": "You ve made an algebra error where you say (x/x)-(1/x) = -(1/x). The expression (x/x) is equal to 1, so after the equals sign you should have 1 - (1/x), which evaluates to 0 when x = 1.",
        "video_name": "MeVFZjT-ABM",
        "timestamps": [
            307
        ],
        "3min_transcript": "over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x So if we take x approaches one of natural log of x, that will give us a, well, natural log of 1 is 0. And over here, we get natural log of 1, which is 0. And then plus 1 minus 1 over plus 1 minus 1 over 1, well, that's just going to be another 0. 1 minus 1 is zero. So you're going to have 0 plus 0. So you're going to get a 0 over 0 again. 0 over 0. So once again, let's apply l'Hopital's rule again. Let's take the derivative of that, put it over the derivative of that. So this, if we're ever going to get to a limit, is going to be equal to the limit as x approaches 1 of the derivative of the numerator, 1 over x, right, the derivative of ln of x is 1/x, over the derivative of the denominator. And what's that?"
    },
    {
        "Q": "4:30 Why doesn't the lnx in the numerator cancel with the lnx in the denominator?\n",
        "A": "In the denominator, you are adding (+) two things and thus you can not cancel it out with the numerator",
        "video_name": "MeVFZjT-ABM",
        "timestamps": [
            270
        ],
        "3min_transcript": "over x minus 1, because the natural log of x's cancel out. Let me get rid of that. And then this right here is the same thing as 1 over natural log of x, because the x minus 1's cancel out. So hopefully you realize, all I did is I added these two expressions. So given that, let's see what happens if I take the limit as x approaches 1 of this thing. Because these are the same thing. Do we get anything more interesting? So what do we have here? We have one times the natural log of 1. The natural log of 1 is 0, so we have 0 here, so that is a 0. Minus 1 minus 0, so that's going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a 1 minus 1, which is 0, times the natural log of 1, which is 0, so 0 times 0, that is 0. We have indeterminate form that we need for l'Hopital's rule, over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x"
    },
    {
        "Q": "\nSo, wait at 5:05 he got rid of a 1? How can he just remove a 1 that he was adding?",
        "A": "It was +1 - 1 so he just canceled them out",
        "video_name": "MeVFZjT-ABM",
        "timestamps": [
            305
        ],
        "3min_transcript": "over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x So if we take x approaches one of natural log of x, that will give us a, well, natural log of 1 is 0. And over here, we get natural log of 1, which is 0. And then plus 1 minus 1 over plus 1 minus 1 over 1, well, that's just going to be another 0. 1 minus 1 is zero. So you're going to have 0 plus 0. So you're going to get a 0 over 0 again. 0 over 0. So once again, let's apply l'Hopital's rule again. Let's take the derivative of that, put it over the derivative of that. So this, if we're ever going to get to a limit, is going to be equal to the limit as x approaches 1 of the derivative of the numerator, 1 over x, right, the derivative of ln of x is 1/x, over the derivative of the denominator. And what's that?"
    },
    {
        "Q": "\nAt 3:49, how is the derivative of x-1 = 1. I thought the derivative of x is 1 so, that 1 minus the 1 should equal 0.",
        "A": "You re really taking the derivative of x (which is 1) and the derivative of -1 (which is 0). (1) - (0) = 1. The derivative of -1 is 0 because -1 is a constant. The derivative of any constant is 0. If you imagine a graph with y=-1 (or any constant) the line is horizontal. The tangent line of any point on the graph will also be horizontal, and the slope of a horizontal line is zero.",
        "video_name": "MeVFZjT-ABM",
        "timestamps": [
            229
        ],
        "3min_transcript": "over x minus 1, because the natural log of x's cancel out. Let me get rid of that. And then this right here is the same thing as 1 over natural log of x, because the x minus 1's cancel out. So hopefully you realize, all I did is I added these two expressions. So given that, let's see what happens if I take the limit as x approaches 1 of this thing. Because these are the same thing. Do we get anything more interesting? So what do we have here? We have one times the natural log of 1. The natural log of 1 is 0, so we have 0 here, so that is a 0. Minus 1 minus 0, so that's going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a 1 minus 1, which is 0, times the natural log of 1, which is 0, so 0 times 0, that is 0. We have indeterminate form that we need for l'Hopital's rule, over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x"
    },
    {
        "Q": "\nMy memory of math is a bit fuzzy so this might be a stupid question. At 9:00, why did Sal randomly divide the equation 2C1 + 3C1 = 0 by one half?",
        "A": "He didn t divide by 1/2, he multiplied by 1/2. He did that to get the same coefficient on the C1 terms in both equations. After that, he can subtract one equation from the other to get rid of C1 altogether. Another possibility would have been to multiply the second equation by 2.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            540
        ],
        "3min_transcript": "In order for them to be linearly dependent, that means that if some constant times 2,1 plus some other constant times this second vector, 3,2 where this should be equal to 0. Where these both aren't necessarily 0. Before I go up for this problem, let's remember what we're going to find out. If either of these are non-zero, if c1 or c2 are non-zero, then this implies that we are dealing with a dependent, linearly dependent set. If c1 and c2 are both 0, if the only way to satisfy this equation -- I mean you can always satisfy it by sitting guys 0, then we're dealing with a linearly independent set. Let's try to do some math. And this'll just take us back to our Algebra 1 days. In order for this to be true, that means 2 times c1 plus 3 times c2 is equal to -- when I say this is equal to 0, it's really the 0 vector. I can rewrite this as 0,0. So 2 times c1 plus 3 times c2 would be equal to that 0 there. And then we'd have 1 times c1 plus 2 times c2 is equal to that 0. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1/2. equal to 0. And then if we subtract the green equation from the red equation this becomes 0. 2 minus 1 and 1/2-- 3/2 is 1 and 1/2 --of this is just 1/2 c2 is equal to 0. And this is easy to solve. c2 is equal to 0. So what's c1? Well, just substitute this back in. c2 is equal to 0. So this is equal to 0. So c1 plus 0 is equal to 0. So c1 is also equal to 0. We could have substituted it back into that top equation as well. So the only solution to this equation involves both c1 and c2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors."
    },
    {
        "Q": "At 3:00 why did you not start writing from a1v1 but rather a2v2?\n",
        "A": "It is his way of showing that v1 can be made from a linear combination of the other vectors. If he included v1 in there, then that would be saying that v1 is a linear combination of itself and the other vectors, which is true for all vectors, all the time.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            180
        ],
        "3min_transcript": "this into the 0 vector. Sometimes it's just written as a bold 0, and sometimes you could just write it -- I mean we don't know the dimensionality of this vector. It would be a bunch of 0's. We don't know how many actual elements are in each of these vectors, but you get the idea. My set of vectors is linearly dependent-- remember I'm saying dependent, not independent --is linearly dependant, if and only if I can satisfy this equation for some ci's where not all of them are equal to 0. This is key, not all are 0. Or you could say it the other way. You could say at least one is non-zero. previous video where I said look, a set is linearly dependent if one of the of vectors can be represented by the combination of the other vectors? Let me write that down. In the last few I said, look, one vector can be -- Let me write it this way. One vector being represented by the some of the other vectors, I can just write it like this. I can write it a little bit more math-y. In the last video, I said that linear dependence means that-- let me just pick an arbitrary vector, v1. Let's say that v1, you know this is arbitrary, v1 one could be represented by some combination of the other vectors. Let me call them a1 times v -- let me be careful -- a2 times This is what we said in the previous video. If this is linear dependence, any one of these guys can be represented as some combination of the other ones. So how does this imply that? In order show this if and only if, I have to show that this implies that and I have to show that that implies this. So this is almost a trivially easy proof. Because if I subtract v1 from both sides of this equation I get 0 is equal to minus 1 v1 plus a2 v2 plus a3 v3 all the way to an vn. And clearly I've just said, well, this is linearly dependent. That means that I can represent this vector as a sum of the other vectors, which means that minus 1 times v1"
    },
    {
        "Q": "\nI have a question regarding the giveaway problem (around 12:00). Sal says that if there are three two-dimensional vectors in a set, the set will definitely be linearly dependent.\nWhat if:\nv1 = [1,0]\nv2=[2,0]\nv3=[7,8]\nin this case, v1 and v2 are linearly dependent since 2*v1 = v2\nBut there is no way v3 can be represented by the linear combination of v1 and v2, how could this set of these three vectors be linearly dependent?",
        "A": "That is an example of a set of three linearly dependent vectors, for the reason you describe. If you understood Sal to mean that a linearly dependent set means that ANY of the vectors is a linear combination of the others, that is incorrect. It is just necessary that at least one of the vectors is a linear combination of the others.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            720
        ],
        "3min_transcript": "of the other one. You can't represent one as a combination of the other. And since we have two vectors here, and they're linearly independent, we can actually know that this will span r2. The span of my r vectors is equal to r2. If one of these vectors was just some multiple of the other, than the span would have been some line within r2, not all of. But now I can represent any vector in r2 as some combination of those. Let's do another example. Let me scroll to the right, because sometimes this thing, when I go too far down, I haven't figured out why, when I go too far down it starts messing up. So my next example is the set of vectors. So I have the vector 2,1. I have the vector 3,2. And I want to know are these linearly dependent or linearly independent. So I go to through the same drill. I use that little theorem that I proved at the beginning of this video. In order for them to be linearly dependent there must be some set of weights that I can multiply these guys. So c1 times this vector plus c2 times this vector plus c3 times that vector, that will equal the 0 vector. And if one of these is non-zero then we're dealing with a linearly dependent set of vectors. And if all of them are 0, then it's independent. Let's just do our linear algebra. So this means that 2 times c1 plus 3 times c2 plus c3 is And then if we do the bottom rows-- Remember when you multiply a scalar times a vector you multiply it by each of these terms. So c1 times 1. 1c1 plus 2c2 plus 2c3 is equal to 0. There's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because, in the very best case, even if you assume that that vector and that vector are linearly independent, then these would span r2. Which means that any point, any vector, in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them because this is just a vector in two-dimensional space. So it would be linearly dependent. And then, if you say, well, these aren't linearly"
    },
    {
        "Q": "At 0:39, how is this about subtracting formula for a cosine? I am interested in knowing how to solve for cos(u+v)=cos u cos v + sin u sin v\n",
        "A": "cos(u + v) is actually cos(u)cos(v) - sin(u)sin(v).",
        "video_name": "D_smr0GBPvA",
        "timestamps": [
            39
        ],
        "3min_transcript": "We have triangle ABC here, which looks like a right triangle. And we know it's a right triangle because 3 squared plus 4 squared is equal to 5 squared. And they want us to figure out what cosine of 2 times angle So that's this angle-- ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC-- well, cosine is just adjacent over hypotenuse. It's going to be equal to 3/5. And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4/5. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared And we've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine-- Let me do this in a different color. Now, we know that the cosine of angle ABC is going to be equal to-- oh, sorry. It's the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3/5 squared. Cosine of angle a ABC is 3/5. So we're going to square it. And this right over here is just 4/5 squared. And so this simplifies to 9/25 minus 16/25, which is equal to 7/25. Sorry. It's negative. Got to be careful there. 16 is larger than 9. Negative 7/25. Now, one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle-- which we already know the unit circle definition of trig functions is an extension of the Sohcahtoa definition. Y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle. So this angle right over here looks like something like this."
    },
    {
        "Q": "At 9:30, was Sal supposed to write a \"t\" after the \"cos^2\"?\n",
        "A": "yes he was trying to write cos^2t",
        "video_name": "AFF8FXxt5os",
        "timestamps": [
            570
        ],
        "3min_transcript": "but let me rewrite our vector field in terms of in terms of t, so to speak. So what's our field going to be doing at any point t? We don't have to worry about every point. We don't have to worry, for example, that over here the vector field is going to be doing something like that because that's not on our path. That force never had an impact on the particle. We only care about what happens along our path. So we can find a function that we can essentially substitute y and x for, their relative functions with respect to t, and then we'll have the force from the field at any point or any time t. So let's do that. So this guy right here, if I were to write it as a function of t, this is going to be equal to y of t, right? y is a function of t, so it's sine of t, right? that's that. Sine of t times i plus-- or actually minus x, or x of So minus cosine of t times j. And now all of it seems a little bit more straightforward. If we want to find this line integral, this line integral is going to be the same thing as the integral-- let me pick a nice, soothing color. Maybe this is a nice one. The integral from t is equal to 0 to t is equal to 2pi of f dot dr. Now, when you take the dot product, you just multiply the corresponding components, and add it up. So we take the product of the minus sign and the sine of t-- or the sine of t with the minus sine of t dt, I get-- you're going to get minus sine squared t dt, and then you're going to add that to-- so you're going to have that plus. Let me write that dt a little bit. dt, and then you're going to have that plus these two guys multiplied by each other. So that's-- well, there's a minus to sign here so plus. Let me just change this to a minus. Minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going to be equal to? This is going to be equal to the integral from 0 to 2pi of, we could say, sine squared plus-- I want to put the t -- sine squared of t plus cosine squared of t. And actually, let me take the minus sign out to the front. So if we just factor the minus sign, and put a minus there, make this a plus. So the minus sign out there, and then we factor dt out. I did a couple of steps in there, but I think you got it. Now this is just algebra at this point. Factoring out a minus sign, so this becomes positive. And then you have a dt and a dt. Factor that out, and you get this."
    },
    {
        "Q": "\nat 08:25, why the equation of normal line is: y-x0^2=-(1/2x0)*(x-x0)?",
        "A": "i don t get something here if the y =2x and we want to get a perpendicular line to that one shouldn t it be (-1/2)*x ? as i know the to perpendicular lines have some equations like: m1*m2=-1 so 2*m2=-1 m2 would be -1/2 and then the new line perpendicular to the first one would be just (-1/2)*x please tell me where i am wrong",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            505
        ],
        "3min_transcript": "This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the Minus x, minus x0. This is the equation of the normal line. So let's see. And what we care about is when x0 is greater than 0, right? We care about the normal line when we're in the first quadrant, we're in all of these values right there. So that's my equation of the normal line. And let's solve it explicitly in terms of x. So y is a function of x. Well, if I add x0 squared to both sides, I get y is equal to, actually, let me multiply this guy out. I get minus 1/2 x0 times x, and then I have plus, plus, because I have a minus times a minus, plus 1/2. The x0 and the over the x0, they cancel out. And then I have to add this x0 to both sides."
    },
    {
        "Q": "At 2:10 and 4:30, when the questions says the parabola gets smaller and after that starts to increase, isn't it referring to the y values?\n",
        "A": "The problem refers to the x coordinates but have the poor use of words and Sal explains what it means by getting smaller and increase . Although, there is a relation as x is getting smaller so is y.",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            130,
            270
        ],
        "3min_transcript": "I just got sent this problem, and it's a pretty meaty problem. A lot harder than what you'd normally find in most textbooks. So I thought it would help us all to work it out. And it's one of those problems that when you first read it, your eyes kind of glaze over, but when you understand what they're talking about, it's reasonably interesting. So they say, the curve in the figure above is the parabola y is equal to x squared. So this curve right there is y is equal to x squared. Let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola. So this is the first quadrant, right here. And they're saying that a normal line is something, when the first quadrant intersection with the parabola is normal to the parabola. So if I were to draw a tangent line right there, this line is normal to that tangent line. That's all that's saying. So this is a normal line, right there. Normal line. Fair enough. 5 normal lines are shown in the figure. 1, 2, 3, 4, 5. Good enough. And these all look perpendicular, or normal to so that makes sense. For a while, the x-coordinate of the second quadrant intersection of the normal line of the parabola gets smaller, as the x-coordinate of the first quadrant intersection gets smaller. So let's see what happens as the x-quadrant of the first intersection gets smaller. So this is where I left off in that dense text. So if I start at this point right here, my x-coordinate right there would look something like this. Let me go down. my x-coordinate is right around there. And then as I move to a smaller x-coordinate to, say, this one right here, what happened to the normal line? Or even more important, what happened to the intersection of the normal line in the second quadrant? This is the second quadrant, right here. So when I had a larger x-value here, my normal line intersected here, in the second quadrant. Then when I brought my x-value in, when I lowered my x-value, my x-value here, because this is the next point, right here, their wording is bad. They're saying that the second quadrant intersection gets smaller. But actually, it's not really getting smaller. It's getting less negative. I guess smaller could be just absolute value or magnitude, but it's just getting less negative. It's moving there, but it's actually becoming a larger number, right? It's becoming less negative, but a larger number. But if we think in absolute value, I guess it's getting smaller, right? As we went from that point to that point, as we moved the x in for the intersection of the first quadrant, the second quadrant intersection also moved in a bit, from that line to that line. Fair enough. But eventually, a normal line second quadrant intersection gets as small as it can get. So if we keep lowering our x-value in the first quadrant, so we keep on pulling in the first quadrant, as we get to this point. And then this point intersects the second quadrant, right there. And then, if you go even smaller x-values in the first"
    },
    {
        "Q": "\nAt around 11:23: Why do you use the quadratic formula?",
        "A": "Sal wants to find the x coordinates of the intersections, and these are the solution of the equation he built by considering the y values of the normaline and the parabola are the same (which happens only at the intersections). But I think he could have factorized the polynomial, because we already know one solution is x0. x\u00c2\u00b2 + (1/(2.x0)).x - ((1/2)+x0\u00c2\u00b2) = (x-x0)(x+(1/(2.x0) + x0) Thus the other solution is -(1/(2.x0) + x0)",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            683
        ],
        "3min_transcript": "That's this right there. And then I have to add this to both sides of the equation, so then I have plus x0 squared. So this is the equation of the normal line, in mx plus b form. This is its slope, this is the m, and then this is its y-intercept right here. That's kind of the b. Now, what do we care about? We care about where this thing intersects. We care about where it intersects the parabola. And the parabola, that's pretty straightforward, that's just y is equal to x squared. So to figure out where they intersect, we just have to set the 2 y's to be equal to each other. So they intersect, the x-values where they intersect, x squared, this y would have to be equal to that y. Or we could just substitute this in for that y. So you get x squared is equal to minus 1 over 2 x0 times x, Fair enough. And let's put this in a quadratic equation, or try to solve this, so we can apply the quadratic equation. So let's put all of this stuff on the lefthand side. So you get x squared plus 1 over 2 x0 times x minus all of this, 1/2 plus x0 squared is equal to 0. All I did is, I took all of this stuff and I put it on the lefthand side of the equation. Now, this is just a standard quadratic equation, so we can figure out now where the x-values that satisfy this quadratic equation will tell us where our normal line and our parabola intersect. So let's just apply the quadratic equation here. So the potential x-values, where they intersect, x is equal to minus b, I'm just applying the So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this"
    },
    {
        "Q": "At 12:36, Sal says \"let's divide everything by 1/2\" shouldn't it be 2, not 1/2?\n",
        "A": "You are right, he has a pronunciation error, he should have said either divide everything by 2 or multiply everything by 1/2 . Luckily he did wrote everything correctly.",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            756
        ],
        "3min_transcript": "So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal?"
    },
    {
        "Q": "\nAt 7:02 I thought the line perpendicular to y=2x is y=(-1/2)x not y=-1/(2x). I might not be understanding something, but -1/(2x) just doesn't make sense to me.",
        "A": "Indeed. I agree entirely. I was under the impression that only the slope was affected, and thus 2x would become -x/2...",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            422
        ],
        "3min_transcript": "is below it. Right, fair enough. For example, this guy right here, this is when we had a large x-value. He intersects with the second quadrant there. Then if you lower and lower the x-value, if you lower it enough, you pass the extreme normal line, and then you get to this point, and then this point, he intersects, or actually, you go to this point. So if you pull in your x-value enough, you once again intersect at that same point in the second quadrant. So hopefully I'm making some sense to you, as I try to make some sense of this problem. Now what do they want to know? And I think I only have time for the first part of this. Maybe I'll do the second part in the another video. Find the equation of the extreme normal line. Well, that seems very daunting at first, but I think our toolkit of derivatives, and what we know about equations of a line, should be able to get us there. So what's the slope of the tangent line at any point on this curve? Well, we just take the derivative of y equals This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the"
    },
    {
        "Q": "I find it easier to think of conversion problems in terms of proportions.\nfor example: At 4:36 It make more sense to think of it as 45 degree/ x radii is in proportion to \u00cf\u0080 radii / 180 degrees. So x(180)= 45(\u00cf\u0080), then x=45/180\u00cf\u0080, but that gives 1/4\u00cf\u0080 which is not the answer, how can you solve these with proportions and how would you solve that example from the video using a proportion?\n",
        "A": "x/\u00cf\u0080 = 45/180 Is the form I normally use. x is to \u00cf\u0080 as 45 is to 180. 180x = 45\u00cf\u0080 This is the same equation you used, and it works. x = 45\u00cf\u0080/180 x = \u00cf\u0080/4 This is the right answer. 45\u00c2\u00b0 = \u00cf\u0080/4 radians",
        "video_name": "z8vj8tUCkxY",
        "timestamps": [
            276
        ],
        "3min_transcript": "pi radians for every 180 degrees or pi/180 radians/degree. This is going to get us to...we're going to get 30 times pi/180 30 times pi/180 which will simplify to 30/180 is 1/6 so this is equal to pi/6 let me write the units out this is 30 radians which is equal to pi/6 radians. Now lets go the other way lets think about if we have pi/3 radians and I wanna convert that to degrees. So what am I going to get if I convert that to degrees? Well here we're gonna want to figure out how many degrees are there per radian? think about the pi and the 180 for every 180 degrees you have pi radians. 180 degrees/pi radians these are essentially the equivalent thing essentially you're just multiplying this quantity by 1 but you're changing the units the radians cancel out and then the pi's cancel out and you're left with 180/3 degrees 180/3 is 60 and we can either write out the word degrees or you can write degrees just like that. Now lets think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out with that notation as well. How many radians will this be equal to? Well once again we're gonna want to think about how many radians do we have per degree? So we're going to multiply this times for every 180 degrees or we can even write it this way pi radians for every 180 degrees. And here this might be a little less intuitive the degrees cancel out and that's why I usually like to write out the word and you're left with 45 pi/180 radians. Actually let me write this with the words written out for me that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with when you multiply 45 times pi over 180 the degrees have canceled out and you're just left with"
    },
    {
        "Q": "\nto be included in N, doesn't \u00c2\u00b5(x) also have to be multiplied by y' ?\nAlthough this isn't in this case- at 5:25 Sal is still treating \u00c2\u00b5(x) as part of N. how can he do that?",
        "A": "It is actually part of x. \u00c2\u00b5(x) is the factor that needs to be multiplied to M and N to make it integratable not y . If you reason like that, there is no y in M then but you can get dx if you multiply through by dx becayse y = dy/dx. Please vote if that helped.",
        "video_name": "j511hg7Hlbg",
        "timestamps": [
            325
        ],
        "3min_transcript": "So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then, what's 0 times any function? Well, it's just going to be 0, right? 0 times mu of x is just going to be 0. But I did multiply the right hand side times mu of x. And remember what we're doing. This mu of x is-- when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now, if we consider this whole thing our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? We take a partial with respect to y. x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x, you could just say, times 3x plus 2y. That's the partial of this with respect to y. And what's the partial of this with respect to x? Well, here, we'll use the product rule. So we'll take the derivative of the first expression with respect to x. mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x. Well, that's just mu prime of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but it looks a lot like a U. plus xy, plus just the first expression. This is just the product rule, mu of x. Times the derivative of the second expression with respect to x. So times-- ran out of space on that line-- 2x plus y. And now for this new equation, where I multiplied both sides by mu. In order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're kind of saying, this is going to be exact. And now, we're going to try to solve for mu. So let's see if we can do that. So let's see, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y."
    },
    {
        "Q": "\nAt 5:26 ,shouldn't you take the derivative of (2x+y)? Since he is using the chain rule?",
        "A": "At 5:13 Sal mentions that he is using the product rule which states that d/dx[f(x)*g(x)] = d/dx(f(x))*g(x) + f(x)*d/dx(g(x)) in this case f(x)=mu(x) and g(x)=x^2+xy All of that to say that (2x+y) is the derivative of g(x). Notice: To find this derivative with respect to x simply split the function at the + sign and take the derivative of x^2 and xy seperatly to get (2x+y)",
        "video_name": "j511hg7Hlbg",
        "timestamps": [
            326
        ],
        "3min_transcript": "So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then, what's 0 times any function? Well, it's just going to be 0, right? 0 times mu of x is just going to be 0. But I did multiply the right hand side times mu of x. And remember what we're doing. This mu of x is-- when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now, if we consider this whole thing our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? We take a partial with respect to y. x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x, you could just say, times 3x plus 2y. That's the partial of this with respect to y. And what's the partial of this with respect to x? Well, here, we'll use the product rule. So we'll take the derivative of the first expression with respect to x. mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x. Well, that's just mu prime of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but it looks a lot like a U. plus xy, plus just the first expression. This is just the product rule, mu of x. Times the derivative of the second expression with respect to x. So times-- ran out of space on that line-- 2x plus y. And now for this new equation, where I multiplied both sides by mu. In order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're kind of saying, this is going to be exact. And now, we're going to try to solve for mu. So let's see if we can do that. So let's see, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y."
    },
    {
        "Q": "At 1:31, how does- 3x -(+2x) equal -3x+2x and result in a sum of x?\n",
        "A": "The above equation was not solved right. It should be: - 3x - (+2x) = - 3x - 2x = - 5x",
        "video_name": "8Wxw9bpKEGQ",
        "timestamps": [
            91
        ],
        "3min_transcript": "Divide x squared minus 3x plus 2 divided by x minus 2. So we're going to divide this into that. And we can do this really the same way that you first learned long division. So we have x minus 2 being divided into x squared minus 3x plus 2. Another way we could have written the same exact expression is x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are all equivalent expressions. Now, to do this type of long division-- we can call it algebraic long division-- you want to look at the highest degree term on the x minus 2 and the highest degree term on the x squared minus 3x plus 2. And here's the x, and here's the x squared. x goes into x squared how many times? Or x squared divided by x is what? Well, that's just equal to x. So x goes into x squared x times. And I'm going to write it in this column right here above all of the x terms. And then we want to multiply x times x minus 2. That gives us-- x times x is x squared. And just like you first learned in long division, you want to subtract this from that. But that's completely the same as adding the opposite, or multiplying each of these terms by negative 1 and then adding. So let's multiply that times negative 1. And negative 2x times negative 1 is positive 2x. And now let's add. x squared minus x squared-- those cancel out. Negative 3x plus 2x-- that is negative x. And then we can bring down this 2 over here. So it's negative x plus 2 left over, when we only go x times. So then we say, can x minus 2 go into negative x plus 2? Well, x goes into negative x negative one times. You can look at it right here. Negative x divided by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus 2-- you have negative 1 times x, which is negative x. Negative 1 times negative 2 is positive 2. just like you do in long division. But that's the same thing as adding the opposite, or multiplying each of these terms by negative 1 So negative x times negative 1 is positive x. Positive 2 times negative 1 is negative 2. These guys cancel out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being equal to x minus 1. And we can verify it. If we multiply x minus 1 times x minus 2, we should get this. So let's actually do that. So let's multiply x minus 1 times x minus 2. So let's multiply negative 2 times negative 1. That gives us positive 2. Negative 2 times x-- that's negative 2x. Let's multiply x times negative 1. That is negative x."
    },
    {
        "Q": "\nAt 0:52, why did you write x over the 3x? I was taught to write it over the x squared. Sorry, just a little confused.",
        "A": "you want to align all like terms wherever possible to keep it organized. if you write x_ over the _x squared , it s more difficult to keep track of your different terms",
        "video_name": "8Wxw9bpKEGQ",
        "timestamps": [
            52
        ],
        "3min_transcript": "Divide x squared minus 3x plus 2 divided by x minus 2. So we're going to divide this into that. And we can do this really the same way that you first learned long division. So we have x minus 2 being divided into x squared minus 3x plus 2. Another way we could have written the same exact expression is x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are all equivalent expressions. Now, to do this type of long division-- we can call it algebraic long division-- you want to look at the highest degree term on the x minus 2 and the highest degree term on the x squared minus 3x plus 2. And here's the x, and here's the x squared. x goes into x squared how many times? Or x squared divided by x is what? Well, that's just equal to x. So x goes into x squared x times. And I'm going to write it in this column right here above all of the x terms. And then we want to multiply x times x minus 2. That gives us-- x times x is x squared. And just like you first learned in long division, you want to subtract this from that. But that's completely the same as adding the opposite, or multiplying each of these terms by negative 1 and then adding. So let's multiply that times negative 1. And negative 2x times negative 1 is positive 2x. And now let's add. x squared minus x squared-- those cancel out. Negative 3x plus 2x-- that is negative x. And then we can bring down this 2 over here. So it's negative x plus 2 left over, when we only go x times. So then we say, can x minus 2 go into negative x plus 2? Well, x goes into negative x negative one times. You can look at it right here. Negative x divided by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus 2-- you have negative 1 times x, which is negative x. Negative 1 times negative 2 is positive 2. just like you do in long division. But that's the same thing as adding the opposite, or multiplying each of these terms by negative 1 So negative x times negative 1 is positive x. Positive 2 times negative 1 is negative 2. These guys cancel out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being equal to x minus 1. And we can verify it. If we multiply x minus 1 times x minus 2, we should get this. So let's actually do that. So let's multiply x minus 1 times x minus 2. So let's multiply negative 2 times negative 1. That gives us positive 2. Negative 2 times x-- that's negative 2x. Let's multiply x times negative 1. That is negative x."
    },
    {
        "Q": "Can a snail live underwater?@12:30pm\n",
        "A": "Snails CAN stay underwater for a while but CAN T live underwater.",
        "video_name": "gBxeju8dMho",
        "timestamps": [
            750
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 1:40, hilarious! his pineapple is BLASPHEMY! rotfl\n",
        "A": "The pineapple is blasphemy",
        "video_name": "gBxeju8dMho",
        "timestamps": [
            100
        ],
        "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart."
    },
    {
        "Q": "why is there a snail wandering around between 0:00 and 1:00\n",
        "A": "because she likes it and it is cool, and looks like spongbobs pet garey",
        "video_name": "gBxeju8dMho",
        "timestamps": [
            0,
            60
        ],
        "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart."
    },
    {
        "Q": "at 0:42, why is the first dot on x axis on zero? It said 1 question = 5 points so shouldn't it be 1 on the x axis and 5 on the y axis?\n",
        "A": "no because if you have those points (1,0) and (0,5) the slope would be negative (5-0)/(0-1), and you would quickly go into the negatives for y as x increases. If you get 0 questions right, why would get a 1 point credit? If you get 0 questions correct, you should get 0 points (0,0), if you get 1 question right, 5 points (1,5), 2 questions is 10 points (2,10) etc.",
        "video_name": "0eWm-LY23W0",
        "timestamps": [
            42
        ],
        "3min_transcript": "On your math quiz, you earn 5 points for each question that you answer correctly. In the table below, x represents the number of questions that you answer correctly, and y represents the total number of points that you score on your quiz. Fair enough? The relationship between these two variables can be expressed by the following equation-- y is equal to 5x. Graph the equation below. So you could look at a couple of your points. You could say, well, look, if I got 0 questions right I'm going to have 0 points. So you could literally graph that point-- if I got 0 questions right, I get 0 points. And then if I get one question right, and the table tells us that, or we could logically think about it, every question I would get right I'm going to get 5 more points. So if I get one question right, I'm going to get 5 more points, and we saw that in our table as well. We saw that right over here-- one question, 5 points. We could also plot it two questions, 10 points, But you only have to do two points to define a line. So it looks like we are actually done here."
    },
    {
        "Q": "Why is 1 over 25/64 equal to 64/25? (at 6:23)? Thank you.\n",
        "A": "Ah, that s 1/(25/64) or 1 divided by 25/64 which is the same as 1 * 64/25 or 64/25.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            383
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I still don't get the point of the 1 at 0:14. Can someone please help me?\n",
        "A": "its just showing that the one doesn t make a difference",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            14
        ],
        "3min_transcript": ""
    },
    {
        "Q": "i like how at 0:51 it gives that little point out XD\n",
        "A": "Me too! it s pretty cool.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            51
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhy is at 1:30 the negitave turns into the fraction?",
        "A": "The method Sal is using may be confusing, so you may want to solve a negative exponent problem this way: 4 to the negative 3rd power=1*4*4*4(find the reciprocal)> 1/64 Why? Well, we know that 4 to the positive 3rd power is 64, or 64/1, and a negative is the opposite of a positive. Since a negative is the opposite of a positive, we will have to find the opposite, or reciprocal of 64 or 64/1, which is 1/64.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            90
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:08 Why does he put the 1. (2^4 = 1 x 2 x 2 x 2 x 2) It does'nt really do anything and it confuses me because I did that on a test and i lost a point so I don't know if it supposed to be like that or not\n",
        "A": "he puts the 1 to show you 2^0 still leaves a 1 2^0 = 1 2^1 = 1 * 2 2^2 = 1 * 2 * 2",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            8
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nSal said \"negative one\" but meant \"one.\"\n0:54",
        "A": "Yes.. The black box pops up and it tells you..",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            54
        ],
        "3min_transcript": ""
    },
    {
        "Q": "So, at 6:24 Sal says that 5/8^ -2 is 64/25. Isn't that equal to 2 14/25?\n",
        "A": "yes, but no use in here by simplifying it",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            384
        ],
        "3min_transcript": ""
    },
    {
        "Q": "5:25 So, all I have to do to use a negative exponent on a fraction is swap the numerator and denominator on the fraction, and then apply a positive exponent to the faction?\n",
        "A": "Yep, that s right!",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            325
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nSo the numerator of the fraction will always be 1? At 4:05 I noticed that.",
        "A": "Correct, the numerator of the fraction will be one, when a number is raised to a negative exponent.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            245
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Isn't scalar only in matrices?? ( 0:40 )\n",
        "A": "A scalar is just a number. When you multiply a scalar by a vector or matrix, you will just get a scaled up version of the vector or matrix.",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            40
        ],
        "3min_transcript": "A vector is something that has both magnitude and direction. Magnitude and direction. So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at 5 miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving 5 miles per hour in. So this right over here, which is often referred to as a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving 5 miles per hour east. So let's say it's moving 5 miles per hour due east. So now this combined 5 miles per are due east, this is a vector quantity. And now we wouldn't call it speed anymore. So velocity is a vector. We're specifying the magnitude, 5 miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions. And what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three. And then even four, five, six, as made dimensions as we want. Our brains have trouble visualizing beyond three. But what's neat is we can mathematically deal with beyond three using linear algebra. And we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is 5 units long. We'll assume that each of our units here is miles per hour. And that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here. And I could make its length 5. The length of the arrow specifies the magnitude. And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it."
    },
    {
        "Q": "\nAt 5:21, he says that you can use the pythagorean theorem to calculate the length of the vector, but wouldn't the length be sqr21, since 3^2+4^2=21?",
        "A": "3^2 + 4^2 = 9 + 16 = 25. Where are you getting 21 from?",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            321
        ],
        "3min_transcript": "want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it. But when you're doing it in your notebook, you would typically put a little arrow on top of it. And there are several ways that you could do it. You could literally say, hey 5 miles per hour east. But that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions? So for example, this one only moves in the horizontal dimension. And so we'll put our horizontal dimension first. So you might call this vector 5, 0. It's moving 5, positive 5 in the horizontal direction. And it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in the linear algebra context, it's more typical to write it as a column vector like this-- 5, 0. represents how much we're moving in the horizontal direction. And the second coordinate represents how much are we moving in the vertical direction. Now, this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. Let's say it's moving 3 in the horizontal direction. And positive 4. So 1, 2, 3, 4 in the vertical direction. So it might look something like this. So this could be another vector right over here. Maybe we call this vector, vector a. And once again, I want to specify that is a vector. And you see here that if you were to break it down, in the horizontal direction, it's shifting three in the horizontal direction, and it's shifting positive four in the vertical direction. about how much we're moving up and how much we're moving to the right when we start at the end of the arrow and go to the front of it. So this vector might be specified as 3, 4. 3, 4. And you could use the Pythagorean theorem to figure out the actual length of this vector. And you'll see because this is a 3, 4, 5 triangle, that this actually has a magnitude of 5. And as we study more and more linear algebra, we're going to start extending these to multiple dimensions. Obviously we can visualize up to three dimensions. In four dimensions it becomes more abstract. And that's why this type of a notation is useful. Because it's very hard to draw a 4, 5, or 20 dimensional arrow like this."
    },
    {
        "Q": "\n0:36 What does \"Scalar\" mean?",
        "A": "A scalar is just a real number. This is as opposed to a vector (which is an ordered set of real numbers).",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            36
        ],
        "3min_transcript": "A vector is something that has both magnitude and direction. Magnitude and direction. So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at 5 miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving 5 miles per hour in. So this right over here, which is often referred to as a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving 5 miles per hour east. So let's say it's moving 5 miles per hour due east. So now this combined 5 miles per are due east, this is a vector quantity. And now we wouldn't call it speed anymore. So velocity is a vector. We're specifying the magnitude, 5 miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions. And what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three. And then even four, five, six, as made dimensions as we want. Our brains have trouble visualizing beyond three. But what's neat is we can mathematically deal with beyond three using linear algebra. And we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is 5 units long. We'll assume that each of our units here is miles per hour. And that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here. And I could make its length 5. The length of the arrow specifies the magnitude. And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it."
    },
    {
        "Q": "\n3:53 Why do you have to substitute the variable into an original equation?",
        "A": "So that you can get y.",
        "video_name": "vA-55wZtLeE",
        "timestamps": [
            233
        ],
        "3min_transcript": "You would get Ax plus By, plus D is equal to C plus D. And we've seen that multiple, multiple times. Anything you do to one side of the equation, you have to do to the other side. But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation. On the right-hand side, you're adding 25.5 to the equation. Aren't you adding two different things to both sides of the equation? And my answer would be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They're both 25.5. This second equation is telling me that explicitly. So I can add this to the left-hand side. I'm essentially adding 25.5 to it. And I could add 25.5 to the right-hand side. So let's do that. If we were to add the left-hand side, 3x plus 5x is 8x. And this was the whole point. When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. If you just add these two together, they are going to cancel out. They're going to be plus 0y. Or that whole term is just going to go away. And that's going to be equal to 2.5 plus 25.5 is 28. So you divide both sides. So you get 8x is equal to 28. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. That's equal to 7 over 2. That's our x value. Now we want to solve for our y value. And we could substitute this back into either of these two equations. Let's use the top one. You could do it with the bottom one as well. So we know that 3 times x, 3 times 7 over 2-- I'm just equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Let me just write that as 5/2. We're going to stay in the fraction world. So this is going to be 21 over 2 plus 4y is equal to 5/2. Subtract 21 over 2 from both sides. So minus 21 over 2, minus 21 over 2. The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. That's negative 16 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Or we could write that-- let's continue up here-- 4y-- I'm"
    },
    {
        "Q": "\nAt the time 0:19 9 1/2 ends up being reduced to 3 like you had said. But, what happens to 1/2. Since you are using 3 to divide/reduce you have to use 3 to reduce 1/2 which you can't do. So how does that work. Since 3 is bigger then 1/2 you cant do that because the difference is ginormous like ginormous like absolutely HUGE! Well not that big just 2 1/2 but i mean still please explain.......",
        "A": "9 to the exponent 1/2 means principal square root of 9, which is 3 because 3\u00c2\u00b2=9. He didn t divide 9 by 3. Perhaps you need to watch review on exponents from previous videos then come back to this one.",
        "video_name": "tn53EdOr6Rw",
        "timestamps": [
            19
        ],
        "3min_transcript": "Let's do some slightly more complicated fractional exponent examples. So we already know that if I were to take 9 to the 1/2 power, this is going to be equal to 3, and we know that because 3 times 3 is equal to 9. This is equivalent to saying, what is the principal root of 9? Well, that is equal to 3. But what would happen if I took 9 to the negative 1/2 power? Now we have a negative fractional exponent, and the key to this is to just not get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction, and just look at this negative first. Just breathe slowly, and realize, OK, I got a negative exponent. That means that this is just going to be 1 over 9 to the 1/2. That's what that negative is a cue for. This is 1 over 9 to the 1/2, and we know that 9 to the 1/2 is equal to 3. So this is just going to be equal to 1/3. What would this evaluate to? And I encourage you to pause the video after trying it, or pause the video to try it. Negative 27 to the negative 1/3 power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1/3 power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying what number, if I were to multiply it three times-- so if I have that number, so whatever the number this is, if I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I Well, we already know that 3 to the third, which is equal to 3 times 3 times 3, is equal to positive 27. So that's a pretty good clue. What would negative 3 to the third power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9. Times negative 3 is negative 27. So we've just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1/3-- this part right over here-- is equal to negative 3. So this is going to be equal to 1 over negative 3, which is the same thing as negative 1/3."
    },
    {
        "Q": "\nAt the minute 1:50 you came out with 1/(-27)^1/3 and you started looking for a number which could be multiplied by itself in order to get -27. So my question is why did not you put the -27 into the radical sign? ( I know that it will lead to imaginary number) But is there any rule to apply when dealing with negative bases?",
        "A": "If you have a negative under a 2nd radical, then the answer to the problem is no solution. However, this problem does not need the 2nd radical (square root), but instead the cube root. The cube root can be positive or negative. If your calculator has this button, you can do that, but many simple calculators don t so that s why he was explaining how to find it without a calculator.",
        "video_name": "tn53EdOr6Rw",
        "timestamps": [
            110
        ],
        "3min_transcript": "Let's do some slightly more complicated fractional exponent examples. So we already know that if I were to take 9 to the 1/2 power, this is going to be equal to 3, and we know that because 3 times 3 is equal to 9. This is equivalent to saying, what is the principal root of 9? Well, that is equal to 3. But what would happen if I took 9 to the negative 1/2 power? Now we have a negative fractional exponent, and the key to this is to just not get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction, and just look at this negative first. Just breathe slowly, and realize, OK, I got a negative exponent. That means that this is just going to be 1 over 9 to the 1/2. That's what that negative is a cue for. This is 1 over 9 to the 1/2, and we know that 9 to the 1/2 is equal to 3. So this is just going to be equal to 1/3. What would this evaluate to? And I encourage you to pause the video after trying it, or pause the video to try it. Negative 27 to the negative 1/3 power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1/3 power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying what number, if I were to multiply it three times-- so if I have that number, so whatever the number this is, if I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I Well, we already know that 3 to the third, which is equal to 3 times 3 times 3, is equal to positive 27. So that's a pretty good clue. What would negative 3 to the third power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9. Times negative 3 is negative 27. So we've just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1/3-- this part right over here-- is equal to negative 3. So this is going to be equal to 1 over negative 3, which is the same thing as negative 1/3."
    },
    {
        "Q": "Around 5:10-\nIsn't the probability to score 4 out of 10 shots the number of ways you can arrange those 4 scores (namely 10 choose 4) and then multipy by the score probability - 40% to the power equal to the total number of shots. I don't understand why we have to multiply by the probability of a miss too? (He multiplies by (1-P)^n-k I don't understand why this is necessary in order to get the probability of k scores)\n",
        "A": "It is necessary to take in to consideration the effect of the missed shots or else you are just finding the probability of making four shots in a row if you hand a probability of making a single shot is 40%. If you don t consider the missed shots its as if you only took four shots all together and not ten. Hope this helps",
        "video_name": "SqcxYnNlI3Y",
        "timestamps": [
            310
        ],
        "3min_transcript": "So I multiply probability times the number of baskets or the number of shots I'm taking, which should be equal to 4. So I know I said-- and you really shouldn't necessarily strictly view expected value as the number of shots you should expect to make because sometimes probability distributions can be kind of weird. But in the binomial distribution you can kind of view it that way. That this is the number of shots you would expect to make. Or you can kind of view it as the most likely outcome. That if you have a 40% shot percentage, and you takes 10 shots, the most likely outcome is that you'll make 4 shots. You still might make 6 shots or 3 shots, but this is going to be the most likely outcome. and in my head, the way I think about, the way it makes intuitive sense is that every time you shoot you have a 40% chance of making the shot. So you could say that you always make 40% of a shot. And if you take 10 shots you're going to make 4 whole shots. So that's one way to think about it and why this might make a little intuitive sense. true for any a random variable that's described by a binomial distribution. So in a binomial distribution what is the probability-- so if I say, what is the probability that X is equal to k? And I know it just gets a little complicated sometimes. But I'm just saying, what's the probability say in this basketball analogy. Would be you know, what's the probability that I make-- k could be 3 shots or something like that. So that's what we're talking about. And that we learned was, if we're taking n shots we're going to choose k of them. And we did that several times in the last couple of videos. And then we multiply that times the probability of any one of those particular occurrences. So if I'm making k shots, it'll be the probability of me making any one shot, Which is p to the kth power. p times itself k times. That's the probability of me making k shots. And then the rest of the shots I have to miss. And then how many shots? If I've made k shots, the rest of the shots I have to miss. So I'm going to miss n minus k shots. So in any binomial distribution this is a probability that you get k successes. Now we know that the expected value, the way you calculate an expected value of a random variable is you just take the probability weighted sum. I don't want to confuse you too much and if you main take away from this video is just this, that's good enough. You should feel good. Now it'll get a little technical, but it'll hopefully make you a little bit more comfortable with sigma and sum notation as well. It'll make you a little bit more comfortable with binomial coefficients and things like that. But just going back, the expected value is a a probability weighted sum of each of these. So what you want to do is you want to take the probability that X is equal to k, times k, and then add that up for each of the possible k's. So how would I write that? So the expected value of X, the expected value of our random"
    },
    {
        "Q": "At 0:20 where does he get -1?\n",
        "A": "the minus before the parenthesis is the same as times -1",
        "video_name": "5ZdxnFspyP8",
        "timestamps": [
            20
        ],
        "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23."
    },
    {
        "Q": "At 1:03, if you multiply 3x^2+x-9 by -1, don't you also have to multiply the 16x+14 by -1?\n",
        "A": "Don t you have to do the same thing to both sides of the equation?",
        "video_name": "5ZdxnFspyP8",
        "timestamps": [
            63
        ],
        "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23."
    },
    {
        "Q": "\nAt 2:06 I did not receive my full points but i watched the whole video and only got 100 points also at 0:25 seconds sal said their was the negative one how did he get that",
        "A": "So when you have the problem (4x^2+6x-5)-(2x^2-4x-7) to remove the parenthesis you have to multiply everything after the - sign in the middle by -1 because that is really what the - sign in the middle represents.",
        "video_name": "5ZdxnFspyP8",
        "timestamps": [
            126,
            25
        ],
        "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23."
    },
    {
        "Q": "1:12 why couldn't a2 be a3:a1?\n",
        "A": "Because a^3*a^1 = a^4 the exponents add up.",
        "video_name": "c-wtvEdEoVs",
        "timestamps": [
            72
        ],
        "3min_transcript": "We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube"
    },
    {
        "Q": "\nAt 0:30, what if the number isn't a perfect cube?",
        "A": "Both methods to do this are likely not in your skill set yet, as they are usually taught in Trig or Calc classes. However, there is the possibility of simplifying the radical down if the number under the radical is a multiple of a perfect cube, such as 16. 16 is 2*2*2*2. As such, you could pull the group of three 2 s out, and leave the last 2 behind under the radical. This would then be 2*(cubic-root 2)",
        "video_name": "c-wtvEdEoVs",
        "timestamps": [
            30
        ],
        "3min_transcript": "We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube"
    },
    {
        "Q": "At 04:08 what did you mean about, \"3 is too big\" how does it work?\n",
        "A": "3 is too big or too large of a number cuz 3x7=21 but if your trying to find out what equels or what is close to equeling 17...........2 is the right answer cuz 2x7=14 so which is closest.....21 or 14?",
        "video_name": "NcADzGz3bSI",
        "timestamps": [
            248
        ],
        "3min_transcript": "And then you bring down this 9. And you saw in the last video exactly what this means. When you wrote this 5 up here-- notice we wrote So this is really a 500. But in this video I'm just going to focus more on the process, and you can think more about what it actually means in terms of where I'm writing the numbers. But I think the process is going to be crystal clear hopefully, by the end of this video. So we brought down the 9. 4 goes into 29 how many times? It goes into at least six times. What's 4 times 7? 4 times 7 is 28. So it goes into it at least seven times. What's 4 times 8? 4 times 8 is 32, so it can't go into it eight times so it's going to go into it seven. 4 goes into 29 nine seven times. 7 times 4 is 28. 29 minus 28 to get our remainder for this step in the problem is 1. And now we're going to bring down this 2. We're going to bring it down and you get a 12. That's easy. 4 times 3 is 12. 4 goes into 12 three times. 3 times 4 is 12. 12 minus 12 is 0. We have no remainder. So 4 goes into 2,292 exactly 573 times. So this 2,292 divided by 4 we can say is equal to 573. Or we could say that this thing right here is equal to 573. Let's do a couple of more. Let's do a few more problems. So I'll do that red color. Let's say we had 7 going into 6,475. Maybe it's called long division because you write it nice and long up here and you have this line. I don't know. There's multiple reasons why it could be called long division. So you say 7 goes into 6 zero times. So we need to keep moving forward. 7 goes into 64 how many times? Let's see. 7 times 7 is? Well, that's way too small. Let me think about it a little bit. Well 7 times 9 is 63. That's pretty close. And then 6 times 10 is going to be too big. 7 times 10 is 70. So that's too big. So 7 goes into 64 nine times. 9 times 7 is 63. 64 minus 63 to get our remainder of this stage 1. Bring down the 7. 7 goes into 17 how many times? Well, 7 times 2 is 14. And then 7 times 3 is 21. So 3 is too big. So 7 goes into 17 two times. 2 times 7 is 14. 17 minus 14 is 3. And now we bring down the 5."
    },
    {
        "Q": "\nIn this video he explains how to find if a number is divisible by 3 at 6:06 but than because it is changes it at 6:36, the first number when you used his trick added up to 27, the second number it added up to 26, at the end 9:52 the remainder was 2, if you divided 26 by 3 it would be 8 r 2 (same amount remaining as with Sal's problem) so is the method to find if a number is divisible by 3 also a way to find what the remainder would be if it isn't?",
        "A": "Pretty much. Good pick up.",
        "video_name": "NcADzGz3bSI",
        "timestamps": [
            366,
            396,
            592
        ],
        "3min_transcript": "7 goes into 64 how many times? Let's see. 7 times 7 is? Well, that's way too small. Let me think about it a little bit. Well 7 times 9 is 63. That's pretty close. And then 6 times 10 is going to be too big. 7 times 10 is 70. So that's too big. So 7 goes into 64 nine times. 9 times 7 is 63. 64 minus 63 to get our remainder of this stage 1. Bring down the 7. 7 goes into 17 how many times? Well, 7 times 2 is 14. And then 7 times 3 is 21. So 3 is too big. So 7 goes into 17 two times. 2 times 7 is 14. 17 minus 14 is 3. And now we bring down the 5. That's in our 7 multiplication tables, five times. 5 times 7 is 35. And there you go. So the remainder is zero. So all the examples I did so far had no remainders. Let's do one that maybe might have a remainder. And to ensure it has a remainder I'll just make up the problem. It's much easier to make problems that have remainders than the ones that don't have remainders. So let's say I want to divide 3 into-- I'm going to divide it into, let's say 1,735,092. This will be a nice, beastly problem. So if we can do this we can handle everything. So it's 1,735,092. That's what we're dividing 3 into. And actually, I'm not sure if this will have a remainder. In the future video I'll show you how to figure out whether Actually, we can do it right now. We can just add up all these digits. 1 plus 7 is 8. 8 plus 3 is 11. 11 5 five is 16. 16 plus 9 is 25. 25 plus 2 is 27. So actually, this number is divisible by 3. So if you add up all of the digits, you get 27. And then you can add up those digits-- 2 plus 7 is 9. So that is divisible by 9. That's a trick that only works for 3. So this number actually is divisible by 3. So let me change it a little bit, so it's not divisible by 3. Let me make this into a 1. Now this number will not be divisible by 3. I definitely want a number where I'll end up with a remainder. Just so you see what it looks like. So let's do this one. 3 goes into 1 zero times. You could write a 0 here and multiply that out, but that"
    },
    {
        "Q": "\nAt 0:32 couldn't he just multiply 21*9=189 and then multiply the same for the other side\n28*6=176 and 189 > 176",
        "A": "Not more complicated then what he did",
        "video_name": "2dbasvm3iG0",
        "timestamps": [
            32
        ],
        "3min_transcript": "Use less than, greater than, or equal to compare the two fractions 21/28, or 21 over 28, and 6/9, or 6 over 9. So there's a bunch of ways to do this. The easiest way is if they had the same denominator, you could just compare the numerators. Unlucky for us, we do not have the same denominator. So what we could do is we can find a common denominator for both of them and convert both of these fractions to have the same denominator and then compare the numerators. Or even more simply, we could simplify them first and then So let me do that last one, because I have a feeling that'll be the fastest way to do it. So 21/28-- you can see that they are both divisible by 7. So let's divide both the numerator and the denominator by 7. So we could divide 21 by 7. And we can divide-- so let me make the numerator-- and we can divide the denominator by 7. We're doing the same thing to the numerator and the denominator, so we're not going to change the value of the fraction. So 21 divided by 7 is 3, and 28 divided by 7 is 4. 3/4 is the simplified version of it. Let's do the same thing for 6/9. 6 and 9 are both divisible by 3. So let's divide them both by 3 so we can simplify this fraction. So let's divide both of them by 3. 6 divided by 3 is 2, and 9 divided by 3 is 3. So 21/28 is 3/4. They're the exact same fraction, just written a different way. This is the more simplified version. And 6/9 is the exact same fraction as 2/3. So we really can compare 3/4 and 2/3. So this is really comparing 3/4 and 2/3. And the real benefit of doing this is now this is much easier to find a common denominator for than 28 and 9. Then we would have to multiply big numbers. Here we could do fairly small numbers. The common denominator of 3/4 and 2/3 And 4 and 3 don't share any prime factors with each other. So their least common multiple is really just going to be the product of the two. So we can write 3/4 as something over 12. And we can write 2/3 as something over 12. And I got the 12 by multiplying 3 times 4. They have no common factors. Another way you could think about it is 4, if you do a prime factorization, is 2 times 2. And 3-- it's already a prime number, so you can't prime factorize it any more. So what you want to do is think of a number that has all of the prime factors of 4 and 3. So it needs one 2, another 2, and a 3. Well, 2 times 2 times 3 is 12. And either way you think about it, that's how you would get the least common multiple or the common denominator for 4 and 3. Well, to get from 4 to 12, you've got to multiply by 3."
    },
    {
        "Q": "In the video you provided \"converting decimals to fractions (ex 1) \" 1:13 I have a question, if there was a number greater than zero in front of the decimal (example 4.0727) would that change the place values to something higher than ten-thousand.\n",
        "A": "The number 4.0727 is a mixed number. Any digits to the left of the decimal point are the whole number (the 4). Any digits to the right of the decimal point are the fraction. 4.0727 = 4 727/10,000 The whole number does not change the place value used to create the fraction.",
        "video_name": "EGr3KC55sfU",
        "timestamps": [
            73
        ],
        "3min_transcript": "Let's see if we can write 0.0727 as a fraction. Now, let's just think about what places these are in. This is in the tenths place. This is in the hundredths place. This 2 is in the thousandths place. And this 7 right here, this last 7, is in the ten-thousandths place. So there's a couple of ways we can do this. The way I like to think of this, this term over here is in the ten-thousandths place. We can view this whole thing right over here as 727 ten-thousandths because this is the smallest place right over here. So let's just rewrite it. This is equal to 727 over 10,000. And we've already written it as a fraction. And I think that's about as simplified as we can get. This number up here is not divisible by 2. It's not divisible by 5. which means it wouldn't be divisible by 6 or 9. It doesn't even seem to be divisible by 7. It might be a prime number. But I think we are done."
    },
    {
        "Q": "\nAt 0:00 why can't you just do this:\n31+50+64+x=180\nx=35",
        "A": "Because connecting these two triangles would mean making a bigger triangle and the angle measurements would have to add up to 180\u00c2\u00b0, this would be a way to solve this problem as well. Doing this would allow you to find the missing measurement of other part of the third angle.",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            0
        ],
        "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114."
    },
    {
        "Q": "\nAt 4:20 why is a+b=WHAT ARE THoSE",
        "A": "a and b are variables; they re merely substituting in for a number that we don t know. You technically can use any letter. That s what Sal was using them for -- a demonstration where we didn t know the numbers, just that y = a + b (in the problem he solves at 4:20.)",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            260
        ],
        "3min_transcript": "So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides. Running out of space on the right hand side. And then you're left with-- these cancel out. On the left hand side, you're left with y. On the right hand side is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees, and then you have a supplementary angles Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you could see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step or if we just knew this property from the get go, if we know that y is equal to 114 degrees-- and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle."
    },
    {
        "Q": "how can we findout the angles if they r in ratio\nAngles of a triangle are in the ratio 3:5:4 ,the samallest angle of the triangle is ?\n",
        "A": "The ratio 3 : 5 : 4 means that if the first angle is 3\u00f0\u009d\u0091\u00a5, then the second angle is 5\u00f0\u009d\u0091\u00a5, and the third angle is 4\u00f0\u009d\u0091\u00a5. The sum of these three angles is 3\u00f0\u009d\u0091\u00a5 + 5\u00f0\u009d\u0091\u00a5 + 4\u00f0\u009d\u0091\u00a5 = 180\u00c2\u00b0 \u00e2\u0087\u0094 \u00f0\u009d\u0091\u00a5 = 180\u00c2\u00b0 \u00e2\u0088\u0095 12 = 15\u00c2\u00b0 So, the first angle is 3 \u00e2\u0088\u0099 15\u00c2\u00b0 = 45\u00c2\u00b0, the second angle is 5 \u00e2\u0088\u0099 15\u00c2\u00b0 = 75\u00c2\u00b0, and the third angle is 4 \u00e2\u0088\u0099 15\u00c2\u00b0 = 60\u00c2\u00b0",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            185
        ],
        "3min_transcript": "and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides."
    },
    {
        "Q": "\nAt 3:47, do angles a + b equal to y?",
        "A": "Yes, actually, that s exactly right :)",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            227
        ],
        "3min_transcript": "So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides. Running out of space on the right hand side. And then you're left with-- these cancel out. On the left hand side, you're left with y. On the right hand side is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees, and then you have a supplementary angles Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you could see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step or if we just knew this property from the get go, if we know that y is equal to 114 degrees-- and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle."
    },
    {
        "Q": "\nat 1:51, why did he minus 114 from 114? it would equal to zero anyways!",
        "A": "He does this to show that he is subtracting the same amount from both sides, and therefore preserving the equality of the equation. Although it isn t necessary to show, it helps clarify how he is solving the equation.",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            111
        ],
        "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114."
    },
    {
        "Q": "\n<---- push the up button if i am right, press down if ur a hatr.... My answer is 35.......\nPLZ RESPOND! I paused @ 1:50",
        "A": "correct",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            110
        ],
        "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114."
    },
    {
        "Q": "at 1:50 or so in the video, he says \"one hundred thirty-one thousand, six hundred and seventy-two\"... but isn't 2 to the 17th power 131,072?\n",
        "A": "Yes, that was a mistake in the video.",
        "video_name": "UCCNoXqCGZQ",
        "timestamps": [
            110
        ],
        "3min_transcript": "Voiceover:Hey Britt. Voiceover:How are you? Voiceover:Good, looks like we have a game going on here. Yeah, kind of a challenge question for you. What I did is, I put 1 grain of rice in the first square. Voiceover:That's right. Voiceover:There's 64 squares on the board. Voiceover:Yup. Voiceover:And in each consecutive square I doubled the amount of rice. Voiceover:How much rice do you think would be on this square? Voiceover:On that square? Let me think about it a little bit. Actually, I'm going to take some ... Here you have 1 and we multiply that times 2, so this is going to be 2 times 2. No, no 2 times 1, what am I doing? Now this is 2 times that one so this is 2 times 2. Now this is 2 times that. So this is ... Okay, we're starting to take a lot of 2's here and multiplying them together. So this is 2 times 2 ... I'm trying to write sideways. Times 2. This one is going to be 5, 2's multiplied together. This is going to be 6, 2's multiplied together. 8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them"
    },
    {
        "Q": "my bad sorry at 0:13 he says it has 64 squares. But then at 2:07 he says 63 but there is 64 squares.\n",
        "A": "He says there are 63 steps. This is measured from the first square. Therefore, the first square had 0 steps, the second square had 1 step, and so on up to the sixty-fourth square, which had 63 steps to get to it.",
        "video_name": "UCCNoXqCGZQ",
        "timestamps": [
            13,
            127
        ],
        "3min_transcript": "Voiceover:Hey Britt. Voiceover:How are you? Voiceover:Good, looks like we have a game going on here. Yeah, kind of a challenge question for you. What I did is, I put 1 grain of rice in the first square. Voiceover:That's right. Voiceover:There's 64 squares on the board. Voiceover:Yup. Voiceover:And in each consecutive square I doubled the amount of rice. Voiceover:How much rice do you think would be on this square? Voiceover:On that square? Let me think about it a little bit. Actually, I'm going to take some ... Here you have 1 and we multiply that times 2, so this is going to be 2 times 2. No, no 2 times 1, what am I doing? Now this is 2 times that one so this is 2 times 2. Now this is 2 times that. So this is ... Okay, we're starting to take a lot of 2's here and multiplying them together. So this is 2 times 2 ... I'm trying to write sideways. Times 2. This one is going to be 5, 2's multiplied together. This is going to be 6, 2's multiplied together. 8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them"
    },
    {
        "Q": "\nAt 0:19, Sal says two lines are parallel if they are on the same plane and will never intersect. If two lines are on different planes and never intersect, what is that called?",
        "A": "They are called skew!",
        "video_name": "aq_XL6FrmGs",
        "timestamps": [
            19
        ],
        "3min_transcript": "Identify all sets of parallel and perpendicular lines in the image below. So let's start with the parallel lines. And just as a reminder, two lines are parallel if they're in the same plane, and all of these lines are clearly in the same plane. They're in the plane of the screen you're viewing right now. But they are two lines that are in the same plane that never intersect. And one way to verify, because you can sometimes-- it looks like two lines won't intersect, but you can't just always assume based on how it looks. You really have to have some information given in the diagram or the problem that tells you that they are definitely parallel, that they're definitely never going to intersect. And one of those pieces of information which they give right over here is that they show that line ST and line UV, they both intersect line CD at the exact same angle, at this angle right here. And in particular, it's at a right angle. And if you have two lines that intersect a third line at the same angle-- so these are actually called corresponding angles and they're the same-- then these two lines are parallel. So line ST is parallel to line UV. And we can write it like this. Line ST, we put the arrows on each end of that top bar to say that this is a line, not just a line segment. Line ST is parallel to line UV. And I think that's the only set of parallel lines in this diagram. Yep. Now let's think about perpendicular lines. Perpendicular lines are lines that intersect at a 90-degree angle. So, for example, line ST is perpendicular to line CD. So line ST is perpendicular to line CD. And we know that they intersect at a right angle or at a 90-degree angle because they gave us this little box here which literally means that the measure of this angle is 90 degrees. Let me make sure I specified these as lines. Line UV is perpendicular to CD. So I did UV, ST, they're perpendicular to CD. And then after that, the only other information where they definitely tell us that two lines are intersecting at right angles are line AB and WX. So AB is definitely perpendicular to WX, line WX. And I think we are done. And one thing to think about, AB and CD, well, they don't even intersect in this diagram. So you can't make any comment about perpendicular, but they're definitely not parallel. You could even imagine that it looks like they're about to intersect. And they give us no information that they intersect the same lines at the same angle. So if somehow they told us that this is a right angle, even"
    },
    {
        "Q": "So whenever (3:01) two lines intersect at a right angle, it's perpendicular? Is there a case where a line is perpendicular to another line without intersecting at a right angle?\n",
        "A": "No, you can t have a line that is perpendicular to another line without them intersecting at a right angle, because that is the very definition of perpendicular. The fact that two lines intersect at right angles is what makes them perpendicular. (1:37) And don t forget, right angle just means a 90 degree angle. Also a box indicated at the intersection is what denotes a right angle.",
        "video_name": "aq_XL6FrmGs",
        "timestamps": [
            181
        ],
        "3min_transcript": "then these two lines are parallel. So line ST is parallel to line UV. And we can write it like this. Line ST, we put the arrows on each end of that top bar to say that this is a line, not just a line segment. Line ST is parallel to line UV. And I think that's the only set of parallel lines in this diagram. Yep. Now let's think about perpendicular lines. Perpendicular lines are lines that intersect at a 90-degree angle. So, for example, line ST is perpendicular to line CD. So line ST is perpendicular to line CD. And we know that they intersect at a right angle or at a 90-degree angle because they gave us this little box here which literally means that the measure of this angle is 90 degrees. Let me make sure I specified these as lines. Line UV is perpendicular to CD. So I did UV, ST, they're perpendicular to CD. And then after that, the only other information where they definitely tell us that two lines are intersecting at right angles are line AB and WX. So AB is definitely perpendicular to WX, line WX. And I think we are done. And one thing to think about, AB and CD, well, they don't even intersect in this diagram. So you can't make any comment about perpendicular, but they're definitely not parallel. You could even imagine that it looks like they're about to intersect. And they give us no information that they intersect the same lines at the same angle. So if somehow they told us that this is a right angle, even then we would have to suspend our judgment based on how it actually looks and say, oh, I guess maybe those things are perpendicular, or maybe these two things are parallel. But they didn't tell us that. And that would actually be bizarre because it looks so not parallel. And actually then this would end up being parallel to other things as well if that was done. It's a good thing that wasn't because it would look very strange. But based on the information they gave us, these are the parallel and the perpendicular lines."
    },
    {
        "Q": "\nAt 3:34, Is the slope of the tangent line constant at any point along the curve?",
        "A": "The slope of the tangent line is constant on the tangent line, but only for that tangent line. Each point on the curve has a different tangent line, meaning each point on the curve will have a different slope.",
        "video_name": "fqQ6sslzyhY",
        "timestamps": [
            214
        ],
        "3min_transcript": "and so we don't know anything other than f but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent so that line has to be tangent to our function right at that point. So our function f could look something like that. So when they say, find f prime of two, they're really saying, what is the slope of the tangent line when x is equal to two? So when x is equal to two, well the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope because that is going to be the rate of change of that function right over there, It's going to be the slope of the tangent line because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, our change in x, change in x, we go from x equals two to x equals 7 so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six. So our change in y is equal to three. So our change in y over change in x is going to be three over five which is the slope of this line, which is the derivative of the function at two because this is the tangent line at x equals two. Let's do another one of these. For a function g, we are given that g of negative one equals three and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g Alright, so once again I think it will be helpful to graph this. So we have our y axis, we have our x axis and let's see. We say for function g we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one and then we have, one, two, and three. So that's that right over there. That is the point. That is the point negative one comma three, it's going to be on our function. And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line, when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line."
    },
    {
        "Q": "Did Sal mean to say -4/2=-2 instead of 4/2=2 at 4:07?\n",
        "A": "No, he s actually right. I ll explain why. At 4:07 of the video, all he does is from y-4=2x, he brings the two from that side over to this side, and divides y-4 by 2. So that way it becomes: y/2 - 4/2 = x. and -4/2 =-2. Hope that helped.",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            247
        ],
        "3min_transcript": "This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we minus 2-- 4 divided by 2 is 2-- is equal to x. Or if we just want to write it that way, we can just swap the sides, we get x is equal to 1/2y-- same thing as y over 2-- minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner. We could say f inverse as a function of y-- so we can have 10 or 8-- so now the range is now the domain for f inverse. f inverse as a function of y is equal to 1/2y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've"
    },
    {
        "Q": "\nAt 0:58, Sal says you could input any real number into the function. Couldn't you input an imaginary number as well?",
        "A": "Yes, it could be a nonreal complex number. However, it is not typical to discuss complex domains and ranges at this level of study. Some teachers might cover that topic, but it is not standard. So, for now, assume that ranges and domains are specifically referring to real numbers unless instructed otherwise.",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            58
        ],
        "3min_transcript": "Let's think about what functions really do, and then we'll think about the idea of an inverse of a function. So let's start with a pretty straightforward function. Let's say f of x is equal to 2x plus 4. And so if I take f of 2, f of 2 is going to be equal to 2 times 2 plus 4, which is 4 plus 4, which is 8. I could take f of 3, which is 2 times 3 plus 4, which is equal to 10. 6 plus 4. So let's think about it in a little bit more of an abstract sense. So there's a set of things that I can input into this function. You might already be familiar with that notion. It's the domain. The set of all of the things that I can input into that function, that is the domain. And in that domain, 2 is sitting there, you have 3 over there, pretty much you could input any real number into this function. So this is going to be all real, but we're making it a Now, when you apply the function, let's think about it means to take f of 2. We're inputting a number, 2, and then the function is outputting the number 8. It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. And we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist, but this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let's let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping. This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2?"
    },
    {
        "Q": "@2:45 sal say that inverse of F so does he mean that the inverse is a different function\n",
        "A": "Edgar Peixoto thnx a lot :)",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            165
        ],
        "3min_transcript": "Now, when you apply the function, let's think about it means to take f of 2. We're inputting a number, 2, and then the function is outputting the number 8. It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. And we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist, but this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let's let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping. This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we"
    },
    {
        "Q": "@ 0:05 Sal says that 12 is arbitrary. What does that mean?\n",
        "A": "Arbitrary means random, and how he uses it, he means he s picking a random number from his mind, there s really no specific reason he used the number 12. What he was basically saying was: let me just pick a random number, the number 12. instead of, let me just pick an arbitrary number, the number 12. Why does he say arbitrary instead of random ? Well, he s more sophisticated than you and me, and he has a bigger vocabulary than us, so he s probably used to saying arbitrary instead of random . Got it?",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            5
        ],
        "3min_transcript": "- In earlier mathematics that you may have done, you probably got familiar with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the number 12 is the product of say two and six; two times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of 12, we could also say that six is a factor of 12. You take the product of these things and you get 12! You could even say that this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things that we could use to multiply. And you probably remember from earlier mathematics the notion of prime factorization, where you break it up into all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, \"Well, this would be 12 \"in prime factored form or the prime factorization of 12,\" so these are the prime factors. is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the number and you're breaking it up into the things that when you multiply them together, you get your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed equal two plus four X. We're just going to distribute the two. is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three,"
    },
    {
        "Q": "\"You'll appreciate this even more when you get into calculus.\"-Sal, 2:53\nWas he referring to how the second derivative of x^2 is 2?\n",
        "A": "I believe what he was getting at is that the second derivative of a quadratic equation is always a constant, and because his second step in looking at the change between adjacent numbers did not produce a constant change, the formula can t be quadratic, which is why we go on to determine that it must be a cubic function.",
        "video_name": "i7iKLZQ-vCk",
        "timestamps": [
            173
        ],
        "3min_transcript": "And now, for each of these, let's see what the output of our function should be. The output of the function should be this thing. It should be the sum from i equals 0 to n of i squared. So when n is 0-- well, that's just going to be 0 squared. We'd just stop right over there. So that's just 0. When n is 1, it's 0 squared plus 1 squared. So that is 1. When n is 2, it's 0 squared plus 1 squared plus 2 squared. So that's 1 plus 4, which is 5. When n is 3, now we go all the way to 3. So it's going to be 1 plus 4, which is 5, plus 9. So 5 plus 9 is 14. And then when n is 4, we're going to add the 16, 4 squared, So this gets us to 30. And of course, we could keep going on and on and on. So let's study this a little bit to think about what type of a function that, for each of these inputs, might give us this type of an output. So the difference here is 1. The difference here is 4. And this is obvious. We added 2 squared here. We added 3 squared, or 9, here. We added 4 squared, or 16, here. And the reason why I'm doing this is if this was a linear function, then the difference between successive terms would be the same. Now, if this is a quadratic function, then the differences between the differences would be the same. Let's see if that's the case. So the difference here is 1. The difference here is 4. So the difference between those is 3. The difference here is 5. The difference here is 7. So even the difference of the differences is increasing. But if this is a cubic function, then the differences of the difference of the difference should be constant. So let's see if that's the case. And you'll appreciate this even more when you start learning calculus. The difference between 3 and 5 is 2. The difference between 5 and 7 is 2. So the fact that the difference of the difference of the difference is fixed tells us that we should be able to express this as some type of a cubic function. So this we could write as this should be equal to some function in terms of n. And we could write it as An to the third plus Bn squared plus C times n plus D. And now we can just use what the inputs are and the outputs are of these to solve for A, B, C, and D. And I encourage you to do that. Well, let's first think about when n is equal to 0. When n is equal to 0, this function evaluates to D. So this function evaluates to D. But that function needs to evaluate to 0, so D needs to be 0. So I'm just trying to fix these letters here to get the right outputs."
    },
    {
        "Q": "\nAt 3:19 sal says the sum of 1^2 + 2^2 + ... is equal to An^3 + Bn^2 + Cn + D.\nI understand why he used An^3 + Bn^2 + Cn + D but any chance there's a video which explain that in further details?",
        "A": "A^3 Bn^2 +Cn^ + D is just the general form of a cubic equation. if he had determined that the series was quadratic he would have used An^2 + Bn + C .. does that help?",
        "video_name": "i7iKLZQ-vCk",
        "timestamps": [
            199
        ],
        "3min_transcript": "And now, for each of these, let's see what the output of our function should be. The output of the function should be this thing. It should be the sum from i equals 0 to n of i squared. So when n is 0-- well, that's just going to be 0 squared. We'd just stop right over there. So that's just 0. When n is 1, it's 0 squared plus 1 squared. So that is 1. When n is 2, it's 0 squared plus 1 squared plus 2 squared. So that's 1 plus 4, which is 5. When n is 3, now we go all the way to 3. So it's going to be 1 plus 4, which is 5, plus 9. So 5 plus 9 is 14. And then when n is 4, we're going to add the 16, 4 squared, So this gets us to 30. And of course, we could keep going on and on and on. So let's study this a little bit to think about what type of a function that, for each of these inputs, might give us this type of an output. So the difference here is 1. The difference here is 4. And this is obvious. We added 2 squared here. We added 3 squared, or 9, here. We added 4 squared, or 16, here. And the reason why I'm doing this is if this was a linear function, then the difference between successive terms would be the same. Now, if this is a quadratic function, then the differences between the differences would be the same. Let's see if that's the case. So the difference here is 1. The difference here is 4. So the difference between those is 3. The difference here is 5. The difference here is 7. So even the difference of the differences is increasing. But if this is a cubic function, then the differences of the difference of the difference should be constant. So let's see if that's the case. And you'll appreciate this even more when you start learning calculus. The difference between 3 and 5 is 2. The difference between 5 and 7 is 2. So the fact that the difference of the difference of the difference is fixed tells us that we should be able to express this as some type of a cubic function. So this we could write as this should be equal to some function in terms of n. And we could write it as An to the third plus Bn squared plus C times n plus D. And now we can just use what the inputs are and the outputs are of these to solve for A, B, C, and D. And I encourage you to do that. Well, let's first think about when n is equal to 0. When n is equal to 0, this function evaluates to D. So this function evaluates to D. But that function needs to evaluate to 0, so D needs to be 0. So I'm just trying to fix these letters here to get the right outputs."
    },
    {
        "Q": "at 2:18, if your going to do a graph is it easier to do a bar graph?\n",
        "A": "It can be, but there are other ways to do it too",
        "video_name": "0ZKtsUkrgFQ",
        "timestamps": [
            138
        ],
        "3min_transcript": "- [Voiceover] What I want to do in this video is think about all of all the different ways that we can represent data. So right over here, we have a list of, and I'm just using this as one form of data, a list of students' scores on, say, the last test, so Amy got 90 percent right, Bill got 95 percent right, Cam got 100 percent right, Efra also got 100 percent right, and Farah got 80 percent right. This is one way to show data. Remember, data is just recorded information, and it could be numeric like this, it could be quantitative, so you're recording actual numbers, or it could even be things you could record data on how do they like the test, and they could have scored it based on, I really liked it, I kind of liked it, I didn't like it, or they might have rated it on a scale of zero to five, which would have been numbers, but it's numbers that are measuring peoples' opinions, as opposed to, here, we have numbers that are measuring their actual scores. So there's all different types of data, and I don't want to get into all of that, but let's just start thinking about different ways to represent this data. So this is one way, you could view this as a table where you have the name, So you have your name column, and then you have your score column. And I can construct it as a table, so clearly, it looks like a table. Like that, that's one way, one very common way of representing data, just like that. That's actually how most traditional databases record data, in tables like this. But you could also do it in other ways. So you could record it as a... Often times called a bar graph, or sometimes, a histogram, so you could put score on the vertical axis here, and then you could have your names over here. And let's see the scores, let's see, maybe we'll make this a 50. Actually, let me just mark them off. So this is 10, 20, 30, that's too big. 60, 70, 80, 90, so that's... And then 100, so that's 100. One, two, three, four, five, that would be 50 right over there, and then you can go person by person. So Amy got a 90 on the exam, so the bar will go up to 90. So that is Amy, and then you have Bill, got a 95, so it's going to be between 90 and 100, so it's going to be right over there. Bill got a 95, and so it'll look like this. Bill, so that is Bill. And then you have Cam, who got 100, on the exam. So, make sure, you see I'm hand drawing it, so it's not as precise as if I were to do it on a computer. So this right over there, that is Cam's score."
    },
    {
        "Q": "Hmm... So at 5:08...why doesn't he take the product of 3 *3 ? Both numbers have 3 in common.. Because at 6:15, Sal says that you take the product of both numbers. In that case at 6:15, he multiplies the 3 and the 5... So why doesn't that apply to the situation at 5:08 with the numbers sharing the 3 in common?\n",
        "A": "The numbers 21 and 30 have one factor in common: 3 If you tried to use 3*3, you would be saying that 9 is a common factor of 21 and 30. Or, that you can divide both numbers evenly by 9. You can t. The numbers 105 and 30 share 2 common factors: 3 and 5. You can evenly divide both numbers by 3 and 5. Thus, you can also evenly divide both of them by 3*5 = 15. So, the greatest common factor = 15. Do you see the difference? Comment back if you have more questions.",
        "video_name": "bLTfBvkrfsM",
        "timestamps": [
            308,
            375,
            375,
            308
        ],
        "3min_transcript": "Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so So let's say these two numbers were not 21 and 30, but let's say we care about the greatest common divisor not of 21, but let's say of 105 and 30. So if we did the prime factorization method, it might become a little clearer now. Actually figuring out, hey, what are all the factors of 105 might be a little bit of a pain, but if you do a prime factorization, you'd say, well, let's see, 105-- it's divisible by 5, definitely. So it's 5 times 21, and 21 is 3 times 7. So the prime factorization of 105 is equal to-- if I write them in increasing order-- 3 times 5 times 7. The prime factorization of 30, we already figured out is 30 is equal to 2 times 3 times 5. So what's the most number of factors"
    },
    {
        "Q": "at 4:59, when you say the greatest common divisor is one, that's wrong, it is actually three.\n",
        "A": "no, if you go back a couple of seconds it actually says If you see that there s nothing in common here, the GCD or GCF would be 1",
        "video_name": "bLTfBvkrfsM",
        "timestamps": [
            299
        ],
        "3min_transcript": "Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so So let's say these two numbers were not 21 and 30, but let's say we care about the greatest common divisor not of 21, but let's say of 105 and 30. So if we did the prime factorization method, it might become a little clearer now. Actually figuring out, hey, what are all the factors of 105 might be a little bit of a pain, but if you do a prime factorization, you'd say, well, let's see, 105-- it's divisible by 5, definitely. So it's 5 times 21, and 21 is 3 times 7. So the prime factorization of 105 is equal to-- if I write them in increasing order-- 3 times 5 times 7. The prime factorization of 30, we already figured out is 30 is equal to 2 times 3 times 5. So what's the most number of factors"
    },
    {
        "Q": "\nAt 4:25 Sal says the prime factorisation of 30 is 3*10\n\nBut shouldn't it be 2*15? Aren't you supposed to start with the lowest prime that goes into the number in question, which would make it 2 and not 3?",
        "A": "It doesn t matter which prime you start dividing with. The resulting prime factors will always be the same. Method 1: 30 = 3 * 10 -> 2 * 5 prime factors = 3, 2, 5 Method 2: 30 = 2 * 15 -> 3 * 5 prime factors = 2, 3, 5",
        "video_name": "bLTfBvkrfsM",
        "timestamps": [
            265
        ],
        "3min_transcript": "So what is the greatest common factor? Well, there's only one common factor here, 1. 1 is the only common factor. So the greatest common factor of 10 and 7, or the greatest common divisor, is going to be equal to 1. So let's write that down. 1. Let's do one more. What is the greatest common divisor of 21 and 30? And this is just another way of saying that. So 21 and 30 are the two numbers that we care about. So we want to figure out the greatest common divisor, and I could have written greatest common factor, of 21 and 30. So once again, there's two ways of doing this. And so there's the way I did the last time where I literally list all the factors. Let me do it that way really fast. Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so"
    },
    {
        "Q": "In 4:28, I don't understand how in the equation b=2(m - 1/2) - 1/2, the 1/2 in brackets becomes -1..... that is the equation becomes b=2m -1 - 1/2\n\nAnd also in 4:44, how does 2m -1 - 1/2 become 2m - 3/2?\n",
        "A": "First, everything within the brackets is multiplied by 2, hence b = 2m, -2/2 (2/2 = 1). Second, the same principle applies, 2/2 is the same thing as 1, so -1 (or - 2/2) plus - 1/2 equals -3/2",
        "video_name": "cNlwi6lUCEM",
        "timestamps": [
            268,
            284
        ],
        "3min_transcript": "So that's what that sentence in orange is telling us. The top shelf needs to be 1/2 a foot shorter than the length of the middle shelf. Now, what does the next statement tell us? And the bottom shelf to be-- so the bottom shelf needs to be equal to 1/2 a foot shorter than-- so it's 1/2 a foot shorter than twice the length of the top shelf. So it's 1/2 a foot shorter than twice the length of the top shelf. These are the two statements interpreted in equal equation form. The top shelf's length has to be equal to the middle shelf's length minus 1/2. It's 1/2 foot shorter than the middle shelf. And the bottom shelf needs to be 1/2 a foot shorter than twice the length of the top shelf. And so how do we solve this? Well, you can't just solve it just with these two constraints, but they gave us more information. entire 12 feet of wood? So the length of all of the shelves have to add up to 12 feet. She's using all of it. So t plus m, plus b needs to be equal to 12 feet. That's the length of each of them. She's using all 12 feet of the wood. So the lengths have to add to 12. So what can we do here? Well, we can get everything here in terms of one variable, maybe we'll do it in terms of m, and then substitute. So we already have t in terms of m. We could, everywhere we see a t, we could substitute with m minus 1/2. But here we have b in terms of t. So how can we put this in terms of m? Well, we know that t is equal to m minus 1/2. So let's take, everywhere we see a t, let's substitute it with this thing right here. That is what t is equal to. So we can rewrite this blue equation as, the length of the but we know that t is equal to m minus 1/2. And if we wanted to simplify that a little bit, this would be that the bottom shelf is equal to-- let's distribute the 2-- 2 times m is 2m. 2 times negative 1/2 is negative 1. And then minus another 1/2. Or, we could rewrite this as b is equal to 2 times the middle shelf minus 3/2. 1/2 is 2/2 minus another 1/2 is negative 3/2, just like that. So now we have everything in terms of m, and we can substitute back here. So the top shelf-- instead of putting a t there, we could put m minus 1/2. So we put m minus 1/2, plus the length of the middle"
    },
    {
        "Q": "Why is it that 6- (minus) 1/2 (one half) = 5 1/2 (five and a half) at the time of the video 7:37 ???\n",
        "A": "The reason why 6- (minus) 1/2 (one half)= 5 1/2 is because of this: You have 6 units for example this means you have six whole units. Now you want to take away 1/2 of a unit away. What is 1/2? 1/2 is half of a whole unit. So if you have six whole units and then you take away one half of a unit what will you have? 5 1/2. A whole unit is 1, half of that unit is 1/2. I hope this helped.",
        "video_name": "cNlwi6lUCEM",
        "timestamps": [
            457
        ],
        "3min_transcript": "Well, we already put that in terms of m. That's what we just did. This is the length of the bottom shelf in terms of m. So instead of writing b there, we could write 2m minus 3/2. Plus 2m minus 3/2, and that is equal to 12. All we did is substitute for t. We wrote t in terms of m, and we wrote b in terms of m. Now let's combine the m terms and the constant terms. So if we have, we have one m here, we have another m there, and then we have a 2m there. They're all positive. So 1 plus 1, plus 2 is 4m. So we have 4m. And then what do our constant terms tell us? We have a negative 1/2, and then we have a negative 3/2. So negative 1/2 minus 3/2, that is negative 4/2 or negative 2. So we have 4m minus 2. Now, we want to isolate just the m variable on one side of So let's add 2 to both sides to get rid of this 2 on the left-hand side. So if we add 2 to both sides of this equation, the left-hand side, we're just left with 4m-- these guys cancel out-- is equal to 14. Now, divide both sides by 4, we get m is equal to 14 over 4, or we could call that 7/2 feet, because we're doing everything in feet. So we solved for m, but now we still have to solve for t and b. Let's solve for t. t is equal to m minus 1/2. So it's equal to-- our m is 7/2 minus 1/2, which is equal to 6/2, or 3 feet. know it's feet there. So that's the top shelf is 3 feet. The middle shelf is 7/2 feet, which is the same thing as 3 and 1/2 feet. And then the bottom shelf is 2 times the top shelf, minus 1/2. So what's that going to be equal to? That's going to be equal to 2 times 3 feet-- that's what the length of the top shelf is-- minus 1/2, which is equal to 6 minus 1/2, or 5 and 1/2 feet. And you can verify that these definitely do add up to 12. 5 and 1/2 plus 3 and 1/2 is 9, plus 3 is 12 feet, and it meets all of the other constraints. The top shelf is 1/2 a foot shorter than the middle shelf, and the bottom shelf is 1/2 a foot shorter than 2 times the top shelf. And we are done. We know the lengths of the shelves that Devon needs to make."
    },
    {
        "Q": "Where did he get 11 from at 6:42 - when he said 11*300?\n",
        "A": "Sal was doing a rough estimate of 3520 divided by 300. 10 *300 = 3000. 11 * 300 = 3300, and 12*300 = 3600.",
        "video_name": "AGFO-ROxH_I",
        "timestamps": [
            402
        ],
        "3min_transcript": "Now we want to figure out how many laps there are. We want this in terms of laps, not in terms of yards. So we want the yards to cancel out. And we want laps in the numerator, right? Because when you multiply, the yards will cancel out, and we'll just be left with laps. Now, how many laps are there per yard or yards per lap? Well, they say the distance around the field is 300 yards. So we have 300 yards for every 1 lap. So now, multiply this right here. The yards will cancel out, and we will get 3,520. Let me do that in a different color. We will get 3,520, that right there, times 1/300. When you multiply it times 1, it just becomes 3,520 divided by 300. And in terms of the units, the yards canceled out. We're just left with the laps. So 3,520 divided by 300. Well, we can eyeball this right here. What is 11 times 300? Let's just approximate this right here. So if we did 11 times 300, what is that going to be equal to? Well, 11 times 3 is 33, and then we have two zeroes here. So this will be 3,300. So it's a little bit smaller than that. If we have 12 times 300, what is that going to be? 12 times 3 is 36, and then we have these two zeroes, so it's equal to 3,600. So this is going to be 11 point something. It's larger than 11, right? 3,520 is larger than 3,300. So when you divide by 300 you're going to get something larger than 11. But this number right here is smaller than 3,600 so when you divide it by 300, you're going to get something a little bit smaller than 12. than 12 laps. So 2 miles is a little bit lower than 12 laps. But let's make sure we're answering their question. How many complete laps would he need to do to run at least 2 miles? So they're telling us that, look, this might be, 11 point something, something, something laps. That would be the exact number of laps to run 2 miles. But they say how many complete laps does he have to run? 11 complete laps would not be enough. He would have to run 12. So our answer here is 12 complete laps. That complete tells us that they want a whole number of laps. We can't just divide this. If we divide this, we're going to get some 11 point something, something. You can do with the calculator or do it by hand if you're interested. But we have to do at least 12 because that's the smallest whole number of laps that will get us to at least this distance right here, or this number of laps, or the"
    },
    {
        "Q": "\nSo how would I write an equation for the example at 2:21 ?",
        "A": "Since the sequence is not being summed, then it can t be equated to anything so you can t write an equation (in the strictest sense of the term) for the example. If you are merely trying to notate the sequence then: 5\u00e2\u0080\u00a2(1/7)^n for n = 0 to \u00e2\u0088\u009e",
        "video_name": "W2NnNKtquaE",
        "timestamps": [
            141
        ],
        "3min_transcript": "- [Instructor] I'm going to construct a sequence. We're going to start with some number. Let's say I start with the number, a. And then each successive term of the sequence, I'm going to multiply the, to get each successive term of the sequence, I'm going to muliply the previous term by some fixed non-zero number, and I'm going to call that r. So the next term is going to be a, is going to be a times r. And then the term after that, I'm going to multipy this thing times r. So it's going to be a, if you multiply ar times r, that's going to be ar-squared. And then if you were going to multiply this term times r, you would get a times r to the third power. And you could keep going on and on and on and on. And this type of sequence or this type of progression is called a geometric. Geometric Sequence or Progression. Geometric Sequence. You start with some first value. Let me circle that in a different color, So you start with some first value, and then to get each successive term, you multiply by this fixed number. In this case, this fixed number is r. And so we call r, our common ratio. Our common ratio. Why is it called a common ratio? Well take any two successive terms. Take this term and this term, and divide this term by this term right over here, ar to the third, divided by ar-squared. So if you find the ratio between these two things. Let me rewrite this in the same colors. So if you took ar to the third power, and were to divide it by the term before it. So if you were to divide it by ar-squared, what are you going to be left with? Well a divided by a is one, r to the third divided by r-squared is just going to be r. And this is true if you divide any term by the term before it. Or if you find the ratio between any term And so that's why it's called a common ratio. And so let's look at some examples of geometric sequences. So if I start with the number, if I start with the number five, so my a is five and then each time I'm going to multiply, I'm going to multiply by, I don't know. Let's say I multiply by 1/7. So then the next term is going to be five over seven. What's the next term going to be? Well, I'm going to multiply this thing times 1/7. So that's going to be 5/7 times 1/7 is 5/49. So it's going to be five over seven-squared or 49. If I were to multiply this times 1/7, what am I going to get? I'll just change the notation. I'll get five times, I don't actually know in my head what seven to the third power is, I guess I could calculate it, 280 plus 63,"
    },
    {
        "Q": "In the video at 3:57, Sal made a red number 7 and put a line thru it. But in the video at 4:30, Sal made a green number 7 and didn't put a line thru it. Is the red number 7 a cursive 7?\n",
        "A": "Numbers cannot be written in cursive. Some people put a line through their 7s. Neither is right or wrong, and they both mean the same thing",
        "video_name": "wx2gI8iwMCA",
        "timestamps": [
            237,
            270
        ],
        "3min_transcript": "And maybe when you're first trying you might have even called it something called 37. You would just call it this, this number. This number of days since my birthday. What if there was an easier way to group the numbers? You know I have 10 fingers on my hands. What if I were to group them into 10s? And then I would say just how many groups of 10 I have and then how many ones do I have left over. Maybe that would be an easier way to represent, to represent this quantity here. And so let's do that. So one, two, three, four, five, six, seven, eight, nine, 10. So that's a group of 10 right over there. And then you have one, two, three, four, five, six, seven, eight, nine, 10. So this is another group of 10 right over here. And then let's see. We have one, two, three, four, five, six, seven, eight, nine, 10. And then finally you have one, two, three, four, five, six, seven. So you don't get a whole group of 10 so you don't circle em. So just by doing this very simple thing now all of a sudden it's much easier to realize how many days have passed. You don't have to count everything. You just have to say okay. One group of 10. Two groups of 10. Three groups of 10. Or you can say one, two, three 10s. And so that's essentially 30. And then I have another one, two, three, four, five, six, seven. And so you say oh I have 30 and then seven if you knew to use those words which we now use. Now this is essentially what our number system does using the 10 digits we know of. The 10 digits we know of are zero, one, two, three, four, five, six, seven, eight, nine. Now what our number system allows us to do we can essentially represent any number we want in a very quick way, a very easy way for our brains to understand it. So here if we want to represent three 10s, we would have put a three in what we would call the 10s place. We would put a three in the 10s place. And then we would put the ones, one, two, three, four, five, six, seven. We'd put the seven in the ones place. And so how do you know which place is which? Well the first place starting from the right, the first place is the ones place. And then you go one space to the left of it, you get to the 10s place. And as we'll see you go one more space you go to the 100s space. But we'll cover that in a future video. So this essentially tells us the exact same thing. This tells us the exact same thing as this does right over here. This tells us three 10s. One, two, three. Three 10s Three groups of 10."
    },
    {
        "Q": "\nHow do you get badges at 1:10 for asking a question.",
        "A": "The more questions you ask, the more you ll end up learning, so Khan Academy wants to encourage you to ask questions when you feel stuck.",
        "video_name": "wx2gI8iwMCA",
        "timestamps": [
            70
        ],
        "3min_transcript": "Voiceover:Let's say that you wanna count the days since your last birthday because you just wanna know how long its been. And so one day after your birthday you put a mark on a wall. Then the next day you put another mark on the wall. The day after that you put another mark on the wall. So out that day you say well how many days has it been? Well you can say look there's been one, two, three days. So one way to think about it is this set of symbols right over here represents the number three. But then you keep going. The fourth day, you put another mark. Fifth day, you put another mark. And then you keep going like that day after day each day you add another mark. And this is actually the earliest way, the most basic way of representing numbers. The number is represented by the number of marks. So after bunch of days you get here and you're like on well how many days has it been? Well you just recount everything. You say one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17 days. it took me little bit of time to realize that this is 17 but it seems to be working so you just keep going. Day after day after day after day you just keep marking off the days on your wall just to sense you're counting the days since your last birthday. But at some point you realize every time you wanna know how may days its been to count it is a little bit painful. And not only that, is this is taking up a lot of space on your wall. You wish that there was an easier way to represent whatever number this is. So first of all let's just think about what number this actually is. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37. So you wish that there was a better way And maybe when you're first trying you might have even called it something called 37. You would just call it this, this number. This number of days since my birthday. What if there was an easier way to group the numbers? You know I have 10 fingers on my hands. What if I were to group them into 10s? And then I would say just how many groups of 10 I have and then how many ones do I have left over. Maybe that would be an easier way to represent, to represent this quantity here. And so let's do that. So one, two, three, four, five, six, seven, eight, nine, 10. So that's a group of 10 right over there. And then you have one, two, three, four, five, six, seven, eight, nine, 10. So this is another group of 10 right over here. And then let's see. We have one, two, three, four, five, six, seven, eight, nine, 10."
    },
    {
        "Q": "\nat 0:56 couldnt sal just multiply 2 and 3 and get 6 instead of breaking it down",
        "A": "he could have, but I think his objective is not for you to memorize the procedure, but to gain intuition on how he came to the solution.",
        "video_name": "Of8ezQj1hRk",
        "timestamps": [
            56
        ],
        "3min_transcript": "Simplify x to the third, and then that raised to the fourth power times x squared, and then that raised to the fifth power. Now, here we're going to use the power property of exponents, sometimes called the \"power rule.\" And that just tells us if I have x to the a, and then I raise that to b-th power, this is the same thing as x to the a times b power. And to see why that works, let's try that with a-- well, I won't do it with these right here. I'll do it with a simpler example. Let's say we are taking x squared and then raising that to the third power. Well, in that situation, that literally means multiplying x squared by itself three times. So that literally means, x squared times x squared times x squared. I'm taking x squared and I'm raising it to the third power. Now, what does this mean over here? Well, there's a couple of ways to do it. You could just say, look, I have the same base, I'm taking the product, I can just add the exponents. So this is going to be equal to-- let me do it plus 2 plus 2 power, or essentially x to the 3 times 2. This right here is just 3 times 2. So we get x to the sixth power. And if you want, you say, hey, Sal, I don't understand why you can add those exponents. You just have to remember that x squared-- this thing right over here, we could rewrite as x times x times x times x-- in parentheses, I'm putting each of these x squareds-- times x times x. And this is just x times itself 6 times x times x times x times x times x times x. This is just x to the sixth power. So that's why we can add the exponents like that. So let's just use that power property on this expression right over here. We start off with we have x to the third raised to the fourth power. So that's just going to be x to the 3 times 4 power, or x Then, we're multiplying that by x squared raised to the fifth power. Well, that's just going to be x to the 2 times 5 power, or x to the 10th power. And now we have the same base, and we're taking the product, we can just add the exponents. This is going to be equal to-- this whole expression is going to be equal to x to the 12 plus 10th power, or x to the 22nd power. And we are done."
    },
    {
        "Q": "1:55, why add the two amounts when bees prefer a new solution\n",
        "A": "When he added the 2 solutions, that was the new solution.",
        "video_name": "JVlfQEhzLMM",
        "timestamps": [
            115
        ],
        "3min_transcript": "Make a table and solve. A biologist is researching the impact of three different water-based sugar drinks on bees ability to make honey. He takes 2 liters of Drink A, which contains 40% sugar. So let me write this down. Let me make our table and then we can solve it. So let's take amount of drink. And then we'll say percent sugar. And then we can say sugar quantity, so the actual physical quantity of sugar. Maybe I should say sugar amount, or amount of sugar. Now this first drink, Drink A, it says he takes 2 liters of Drink A, which contains 40% sugar. The first column will be which drink we're talking about, so It's 40% sugar. So if we want the actual amount of sugar in liters, we just multiply 2 liters times 40%, or times 0.4. Let me write times with a dot so you don't think it's an x. 2 times 0.4, which is equal to 0.8 liters of sugar. So you have 0.8 liters of sugar. 1.2 liters of I guess the other stuff in there is water. But it's 0.8 of the 2 liters is sugar, which is 40%. Now,, he adds 1.2 liters of Drink B. He finds that bees prefer this new solution, Drink C. So when you add these two together, you end And we end up with how much of Drink C? 2 plus 1.2 is 3.2 liters of Drink C, which has 25% sugar content. So this is 25% sugar, which also says we know the amount of sugar in it. Because if we have 3.2 liters of it and it's 25% sugar, or it's 1/4 sugar, that means that we have 0.8 liters of sugar here. So this is 0.8 liters of sugar. Well, that I already wrote in the column name. That's the amount of sugar. It's 25% sugar. We have 3.2 liters of it. Now, they want to know what is the percentage of sugar in Drink B? So let's just call that x. So that's right over here. Now, if it's x percent sugar here, or this is the decimal"
    },
    {
        "Q": "why at 1:40 did you say you were not going to put x because we might think it was a variable, when we could still mistake the dot for a decimal point? you could have put an asterisk...\n",
        "A": "because you put the dot in the middle like this 9 * 34",
        "video_name": "JVlfQEhzLMM",
        "timestamps": [
            100
        ],
        "3min_transcript": "Make a table and solve. A biologist is researching the impact of three different water-based sugar drinks on bees ability to make honey. He takes 2 liters of Drink A, which contains 40% sugar. So let me write this down. Let me make our table and then we can solve it. So let's take amount of drink. And then we'll say percent sugar. And then we can say sugar quantity, so the actual physical quantity of sugar. Maybe I should say sugar amount, or amount of sugar. Now this first drink, Drink A, it says he takes 2 liters of Drink A, which contains 40% sugar. The first column will be which drink we're talking about, so It's 40% sugar. So if we want the actual amount of sugar in liters, we just multiply 2 liters times 40%, or times 0.4. Let me write times with a dot so you don't think it's an x. 2 times 0.4, which is equal to 0.8 liters of sugar. So you have 0.8 liters of sugar. 1.2 liters of I guess the other stuff in there is water. But it's 0.8 of the 2 liters is sugar, which is 40%. Now,, he adds 1.2 liters of Drink B. He finds that bees prefer this new solution, Drink C. So when you add these two together, you end And we end up with how much of Drink C? 2 plus 1.2 is 3.2 liters of Drink C, which has 25% sugar content. So this is 25% sugar, which also says we know the amount of sugar in it. Because if we have 3.2 liters of it and it's 25% sugar, or it's 1/4 sugar, that means that we have 0.8 liters of sugar here. So this is 0.8 liters of sugar. Well, that I already wrote in the column name. That's the amount of sugar. It's 25% sugar. We have 3.2 liters of it. Now, they want to know what is the percentage of sugar in Drink B? So let's just call that x. So that's right over here. Now, if it's x percent sugar here, or this is the decimal"
    },
    {
        "Q": "at 4:35 why did you multiply by 100 but then not put the decimal back in its original place (divide by 100)? Isn't the answer then incorrect?\n",
        "A": "If you think about it 4 / 2 = 400 / 200 so there is no reason to adjust the final answer. Since he multiplied both parts of the expression (the divisor and the dividend) they end up cancelling out. This is the same as multiplying the top and bottom of a fraction by the same number when trying to get matching denominators.",
        "video_name": "DAikW24_O0A",
        "timestamps": [
            275
        ],
        "3min_transcript": "can do all that, or you can just say, OK, that's not going to give us anything. So then how many times does 15 go into 78? So let's think about it. 15 goes into 60 four times. 15 times 5 is 75. That looks about right, so we say five times. 5 times 15. 5 times 5 is 25. Put the 2 up there. 5 times 1 is 5, plus 2 is 7. 75, you subtract. 78 minus 75 five is 3. Bring down a zero. 15 goes into 30 exactly two times. 2 times 15 is 30. Subtract. No remainder. Bring down the next zero. We're still to the left of the decimal point. The decimal point is right over here. If we write it up here, which we should, it's right over there, so we have one more place to go. So we bring down this next zero. 15 goes into 0 zero times. Subtract. No remainder. So 78 divided by 0.15 is exactly 520. So x is equal to 520. So 78 is 15% of 520. And if we want to use some of the terminology that you might see in a math class, the 15% is obviously the percent. 520, or what number before we figured out it was 520, that's what we're taking the percentage of. This is sometimes referred to as the base. And then when you take some percentage of the base, you get what's sometimes referred to as the amount. So in this circumstance, 78 would be the amount. You could view it as the amount is the percentage of the base, but we were able to figure that out. It's nice to know those, if that's the terminology you use just answer this question. And it makes sense, because 15% is a very small percentage. If 78 is a small percentage of some number, that means that number has to be pretty big, and our answer gels with that. This looks about right. 78 is exactly 15% of 520."
    },
    {
        "Q": "At around 5:21, is it ok to write \"obvious from diagram\" in formal proofs?\n",
        "A": "At 5:21, I would say the reason should be the reflexive property since I was taught that was used in this situation.",
        "video_name": "fSu1LKnhM5Q",
        "timestamps": [
            321
        ],
        "3min_transcript": "Their corresponding sides have the same length, and so we know that they're congruent. So we know that triangle CDA is congruent to triangle CBA. And we know that by the side side side postulate and the statements given up here. Actually, let me number our statements just so we can refer back to this 1, 2, 3, and 4. And so side side side postulate and 1, 2, and 3-- statements 1, 2, and 3. So statements 1, 2, and 3 and the side side postulate let us know that these two triangles are congruent. And then if these are congruent, then we know, for example, we know that all of their corresponding angles are equivalent. So for example, this angle is going to be equal to that angle. So let's make that statement right over there. going to be statement 5-- we know that angle DCE, that's this angle right over here, is going to have the same measure, we could even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. And this comes straight out of statement 4, congruency, I could put it in parentheses. Congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they're going to have the exact same measure. Now it seems like we could do something pretty interesting with these two smaller triangles at the top left and the top right of this, looks like, a kite like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. So let's first just establish that they So I'll just write statement 6. We have CE, the measure or the length of that line, is equal to itself. Once again, this is just obvious. It's the same. Obvious from diagram it's the same line. Obvious from diagram. But now we can use that information. So we don't have three sides, we haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB. But we do have a side, an angle between the sides, and then another side. And so this looks pretty interesting for our side angle side postulate. And so we can say, by the side angle side postulate, we can say that triangle DCE is congruent to triangle BCE."
    },
    {
        "Q": "at 2:59 how do you get absolute value\n",
        "A": "by figuring out how far the number is from 0.",
        "video_name": "Oo2vGhVkvDo",
        "timestamps": [
            179
        ],
        "3min_transcript": "temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8"
    },
    {
        "Q": "\nAt 3:25, So, in general, in a equation you must have 2 positive and 1 negative, right?",
        "A": "No. An equation has two mathematical expressions that are equal. Those expressions can have any numbers or variables as long as they are equal.",
        "video_name": "Oo2vGhVkvDo",
        "timestamps": [
            205
        ],
        "3min_transcript": "temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8"
    },
    {
        "Q": "\n@ 2:00 when he starts making the problem, would it matter if you put the smaller number first rather than the big number?",
        "A": "For this particular problem no it wouldn t. You would get the same answer either way.",
        "video_name": "Oo2vGhVkvDo",
        "timestamps": [
            120
        ],
        "3min_transcript": "One of the coldest temperatures ever recorded outside was negative 128 degrees Fahrenheit in Antarctica. One of the warmest temperatures ever recorded outside was 134 degrees Fahrenheit in Death Valley, California. How many degrees difference are there between the coldest and warmest recorded outside temperatures? So let's think about this a little bit. Now, what I'll do is I'll plot them on a number line. But I'm going to plot it on a vertical number line that has a resemblance to a thermometer, since we're talking about temperature. So I'm going to make my number line vertical right over here. So there's my little vertical number line. And this right over here is 0 degrees Fahrenheit, which really is of no significance. If it was Celsius, we'd be talking about the freezing point. But for Fahrenheit, that happens at 32 degrees. But let's say this is 0 degrees Fahrenheit. And let's plot these two points. temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128."
    },
    {
        "Q": "\nAt about 0:20, Sal says \"We already know that if all three angle, all three of the corresponding angles are congruent to the corresponding angles of ABC, then we're dealing with congruent triangles\". But actually, we're dealing with similar triangles and NOT congruent triangles, right? Because AAA is not a postulate of congruent triangles.",
        "A": "If the angles are congruent then the shapes would be congruent.",
        "video_name": "7bO0TmJ6Ba4",
        "timestamps": [
            20
        ],
        "3min_transcript": "Let's say we have triangle ABC. It looks something like this. I want to think about the minimum amount of information. I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. So for example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees. And we have another triangle that looks like this, it's clearly a smaller triangle, but it's corresponding angles. So this is 30 degrees. This is 90 degrees, and this is 60 degrees, we know that XYZ in this case, is going to be similar to ABC. So we would know from this because corresponding angles are congruent, we would know that triangle ABC And you've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90-degree angle. X corresponds to the 30-degree angle. A corresponds to the 30-degree angle. So A and X are the first two things. B and Y, which are the 90 degrees, are the second two, and then Z is the last one. So that's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough? Well, sure because if you know two angles for a triangle, you know the third. So for example, if I have another triangle that looks like this-- let me draw it like this-- and if I told you that only two of the corresponding angles are congruent. So maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar? Because in a triangle, if you know two of the angles, If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle. So in general, in order to show similarity, you don't have to show three corresponding angles are congruent, you really just have to show two. So this will be the first of our similarity postulates. We call it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. So for example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle, this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. And you can really just go to the third angle in this pretty straightforward way. You say this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity."
    },
    {
        "Q": "SOMEONE HELP ME IM CONFUSED D:\nAt the equation on 8:27, why or how did the 1/2 get in front of the log?\nAfter that I got lost...\n",
        "A": "The half is the exponent of the log result. Now what I am trying to figure out myself is how did the exponent convert to a coefficient!",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            507
        ],
        "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?"
    },
    {
        "Q": "At 8:20 why does the base 2 disappear when changing the exponent to the coefficient? He says this is the 3rd rule we learnt but I don't remember seeing this?\n",
        "A": "At 9:00, he realizes he forgot it and writes it in. As for the third rule, I don t know what you mean.",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            500
        ],
        "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?"
    },
    {
        "Q": "\nI lost him where he appears to have reduced 1/2log2,32 to 5/2, at 9:51. Help.",
        "A": "Dan, log\u00e2\u0082\u0082(32) = 5 because 2*2*2*2*2 = 32 = 2\u00e2\u0081\u00b5 so 1/2 * log\u00e2\u0082\u0082(32) = 1/2 * 5 = 5/2 Also, log\u00e2\u0082\u0082(8) = 3 because 2*2*2 =8 = 2\u00c2\u00b3 so 1/4 * log\u00e2\u0082\u0082(8) = 1/4 * 3 = 3/4 I hope that helps make it click for you.",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            591
        ],
        "3min_transcript": "I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point."
    },
    {
        "Q": "At 4:35, Sal explains a way that he remembers concave and convex... Is there any other way?\n",
        "A": "In Spanish con means with so concave can mean with cave. Another trick to determining concavity is to imagine an elastic band being stretched around the figure, does it touch all sides or does it have to span a gap. If it has to span a gap then the figure must be concave (having a cavity or cave).",
        "video_name": "W9B3VYdC5T8",
        "timestamps": [
            275
        ],
        "3min_transcript": "around the circle. And some of this angle, A+B+C+D+E is just going to be 360 degree And this is work for any convex polygon, and when I say convex polygon I mean it is not that dented words Just to be clear what I'm talking about, it would work for any convex polygon that is kind of I don't want to say regular, regular means it has the same size and angle, but it is not dented, this is a convex polygon. This right here is a concave polygon Let me draw this, right this way, so this would be a concave polygon Let me draw as it having the same number of side, So i just going to dent this two sides and then that's the same side over there, Let me do that and then like that. This has 1,2,3,4,5,6, sides and this has 1,2,3,4,5,6 sides. This is concave, sorry this is a convex polygon, this is concave polygon, All you have to remember is kind of cave in words And so, what we just did is applied to any exterior angle of any convex polygon. I Am a bit confused. This applied to any convex polygon and once again if you take this angle and added to this angle and added to this angle, this angle, that angle and that angle and I'm not applying that all that they are different angles, i could say that one green, and that one some other colour they can all be different but if you shift the angle like this you can see that they just go round the circle. So, once again, I'll just add up to 360 degrees"
    },
    {
        "Q": "\nAt 3:10 Sal said something like clockwise and counterclockwise..\nWhat does he mean by clockwise and counterclockwise ?",
        "A": "Clockwise on the clock rotation goes as top-right-down-left. Counterclockwise on the clock rotation goes as top-left-down-right. Most clocks go Clockwise.",
        "video_name": "W9B3VYdC5T8",
        "timestamps": [
            190
        ],
        "3min_transcript": "right over here. So this going to be a convex angle right over here it's going to have a measure of A, now let me draw angle B, angle B, and i going to draw adjacent to angle A, and what you could do is just to think about it maybe if we draw a line over here, if we draw a line over here that is parallel to this line then the measure over here would also be B,because this is obviously a straight line, it would be like transversal, this of course a responding angles, so if u want to draw adjacent angle, the adjacent to A, do it like that, or whatever angle this is the measure of B and now it is adjacent to A, now let's draw the same thing to C We can draw a parallel line to that right over here. And this angle would also be C and if we want it to be adjacent to that, we could draw it there, so that angle is C we do it in different color, you could do D, right over here or you could shift it over here it'll look like that, or shift over here, it'll look like that If we just kept thinking of parallel, if all of this line were parallel to each other So, let's just draw D like this, so this line is going to parallel to that line Finally, you have E, and again u can draw a line that is parallel to this right over here and this right over here would be angle E or you could draw right over here, right over here And when you see it drawn this way, it's clear that when you add up, the measure, this angle A,B,C,D and E going all the way around the circle, either way around the circle. And some of this angle, A+B+C+D+E is just going to be 360 degree And this is work for any convex polygon, and when I say convex polygon I mean it is not that dented words Just to be clear what I'm talking about, it would work for any convex polygon that is kind of I don't want to say regular, regular means it has the same size and angle, but it is not dented, this is a convex polygon. This right here is a concave polygon Let me draw this, right this way, so this would be a concave polygon Let me draw as it having the same number of side, So i just going to dent this two sides"
    },
    {
        "Q": "At 1:01, you are assuming that the bases of the exponents are the same. How would you solve the expression, if the bases were different numbers? Would you still apply the quotient rule, and add the bases, or what?\n",
        "A": "Sadly, you can t solve it the same way if the bases are different. If you really wanted to solve it, then you d have to actually calculate the exponents and then multiply/divide them. An explanation for this is that if you write out the exponents (3^5 * 4^3 = 3*3*3*3*3*4*4*4), you can t tie them together into one exponent, right? So it s not possible to just add the exponents when the bases are different, because a different number is being multiplied.",
        "video_name": "tvj42WdKlH4",
        "timestamps": [
            61
        ],
        "3min_transcript": "Simplify 3a to the fifth over 9a squared times a to the fourth over a to the third. So before we even worry about the a's, we can actually simplify the 3 and the 9. They're both divisible by 3. So let's divide the numerator and the denominator here by 3. So if we divide the numerator by 3, the 3 becomes a 1. If we divide the denominator by 3, the 9 becomes a 3. So this reduces to, or simplifies to 1a to the fifth times a to the fourth over-- or maybe I should say, a to the fifth over 3a squared times a to the fourth over a to the third. Now this, if we just multiply the two expressions, this would be equal to 1a to the fifth times a to the fourth in the numerator, and we don't have to worry about the one, it doesn't change the value. So it's a to the fifth times a to the fourth in the numerator. And then we have 3a-- let me write the 3 like this-- and then we have 3 times a squared times a to the third in the denominator. can simplify this from here. One sometimes is called the quotient rule. And that's just the idea that if you have a to the x over a to the y, that this is going to be equal to a to the x minus y. And just to understand why that works, let's think about a to the fifth over a squared. So a to the fifth is literally a times a times a times a times a. That right there is a to the fifth. And we have that over a squared. And I'm just thinking about the a squared right over here, which is literally just a times a. That is a squared. Now, clearly, both the numerator and denominator are both divisible by a times a. We can divide them both by a times a. So we can get rid of-- if we divide the numerator And if we divide the denominator by a times a, we just get a 1. So what are we just left with? We are left with just a times a times a over 1, which is just a times a times a. But what is this? This is a to the third power, or a to the 5 minus 2 power. We had 5, we were able to cancel out 2, that gave us 3. So we could do the same thing over here. We can apply the quotient rule. And I'll do two ways of actually doing this. So let's apply the quotient rule with the a to the fifth and the a squared. So let me do it this way. So let's apply with these two guys, and then let's apply it with these two guys. And of course, we have the 1/3 out front. So this can be reduced to 1/3 times-- if we apply the quotient rule with a"
    },
    {
        "Q": "\nAt 4:55 when he was solving a^9-3a^5 .... He got a a4 but it became a^4/3 ??? I kept thinking it would just be 3a^4. How did the a^4 turn into the numerator? and why did 3 stay as the denominator?",
        "A": "Divided by",
        "video_name": "tvj42WdKlH4",
        "timestamps": [
            295
        ],
        "3min_transcript": "did it over here-- that becomes a to the third power. And if we apply it over here with the a to the fourth over a to the third, that'll give us a-- let me do it that same blue color. That'll give us a-- that's not the same blue color. There we go. This will give us a to the 4 minus 3 power, or a to the first power. And of course, we can simplify this as a to the third times a-- well, actually, let me just do it over here. Before I even rewrite it, we know that a to the third times a to the first is going to be a to the 3 plus 1 power. We have the same base, we can add the exponents. We're multiplying a times itself three times and then one more time. So that'll be a to the fourth power. So this right over here becomes a to the fourth power. a to the 3 plus 1 power. And then we have to multiply that by 1/3. So our answer could be 1/3 a to the fourth, or we could equally write it a to the fourth over 3. would have been to apply the product, or to add the exponents in the numerator, and then add the exponents in the denominator. So let's do it that way first. If we add the exponents in the numerator first, we don't apply the quotient rule first. We apply it second. We get in the numerator, a to the fifth times a to the fourth would be a to the ninth power. 5 plus 4. And then in the denominator we have a squared times a to the third. Add the exponents, because we're taking the product with the same base. So it'll be a to the fifth power. And of course, we still have this 3 down here. We have a 1/3, or we could just write a 3 over here. Now, we could apply the quotient property of exponents. We could say, look, we have a to the ninth over a to the fifth. a to the ninth over a to the fifth is equal to a to the 9 minus 5 power, or it's equal to a to the fourth power. And of course, we still have the divided by 3."
    },
    {
        "Q": "\nHow does it work to take away a ten from 10? At 1:27 he takes away one ten from 10, but then has 90.",
        "A": "He is not taking a ten away from 10, he is taking a ten away from 10 tens, or 100. 100-10=90",
        "video_name": "3lHBgFvr3yE",
        "timestamps": [
            87
        ],
        "3min_transcript": "Let's try to subtract 164 from 301, and I encourage you to pause this video and try it on your own first. So let's go place by place, and we can realize where we have to do some borrowing or regrouping. So in the ones place, we have an issue. 4 is larger than 1. How do we subtract a larger number from a smaller number? We also an issue in the tens place. 6 is larger than 0. How do we subtract 6 from 0? So the answer that might be jumping into your head is oh, we've got to do some borrowing or some regrouping. But then you might be facing another problem. You'd say, OK, well, let's try to borrow from the tens place here. So we have a 1. If we could borrow 10 from the tens place, it could be 11. But there's nothing here in the tens place. There's nothing to borrow, so what do we do? So the way I would tackle it is first borrow for the tens place. So we have nothing here, so let's regroup 100 from the hundreds place. So that's equivalent to borrowing a 1 from the hundreds place. So that's now a 2. And now in the tens place, instead of a 0, Now, let's make sure that this still makes sense. This is 200 plus 10 tens. 10 tens is 100, plus 1. 200 plus 100 plus 1 is still 301. So this still makes sense. Now, the reason why this is valuable is now we have something to regroup from the tens place. If we take one of these tens, so now we're left with 9 tens, and we give it to the ones place, so you give 10 plus 1, you're going to be left with 11. And we can verify that we still haven't changed the value of the actual number. 200 plus 90 is 290, plus 11 is still 301. And what was neat about this is now up here, all of these numbers are larger than the corresponding number in the same place. So we're ready to subtract. 11 minus 4 is-- let's see, 10 minus 3 is 7, so 11 minus 4 9 minus 6 is 3, and 2 minus 1 is 1. So we are left with 137."
    },
    {
        "Q": "\nI've recently been learning about this at school. I came on here to study and the beginning of the video seemed right to me, but when it got to around 1:40 , I got a bit confused. I was taught to take the total of everything on the first table, which would be 109, and divide 28/109, then 35/109, then 97/109, then 104/109. I'm very confused on how this creates the same answer, though. I'm thoroughly confused, because my teacher does this concept way differently than this video.",
        "A": "You re using the same total for both your columns.",
        "video_name": "_ETPMszULXc",
        "timestamps": [
            100
        ],
        "3min_transcript": "- [Voiceover] The two-way frequency table below shows data on type of vehicle driven. So, this is type of vehicle driven, and whether there was an accident the last year. So, whether there was an accident in the last year, for customers of All American Auto Insurance. Complete the following two-way table of column relative frequencies. So that's what they're talking here, this is a two-way table of column relative frequencies. If necessary, round your answers to the nearest hundredth. So, let's see what they're saying. They're saying, let's see... Of the accidents within the last year, 28 were the people were driving an SUV, a Sport Utility Vehicle and 35 were in a Sports car. Of the No accidents in the last year, 97 were in SUV and a 104 were in sports cars. Another way you could think of it, of the Sport utility vehicles that were driven and the total, let's see it's 28 plus 97 which is going to be 125. Of that 125, 28 had an accident within the last year Similarly, you could say of the 139 Sports cars 35 had an accident in the last year, 104 did not have an accident in the last year. So what they want us to do is put those relative frequencies in here. So the way we could think about it. One right over here, this represents all the Sport utility vehicles. So one way you could think about, that represents the whole universe of the Sport Utility Vehicles, at least the universe that this table shows. So, that's really representative of the 28 plus the 97. And so, in each of these we want to put the relative frequencies. So this right over here is going to be 28 divided by the total. So over here is 28, but we want this number to be a fraction of the total. Well the fraction of the total is gonna be, 28 over 97 plus 28. Which of course is going to be 125. This one right over here is going to be 97 over 125. And of course, when you add this one and this one, it should add up to one. Likewise, this one's going to be 35 over 139. 35 plus 104. So, 139. And this is going to be 104 over 104 plus 35. Which is 139. So, let me just calculate each of them using this calculator. Let me scroll down a little bit. And so, if I do 28 divided by 125, I get 0.224. They said round your answers to the nearest hundredth. So this is 0.22. No accident within the last year, 97 divided by 125. So 97 divided by 125 is equal to, see here if I rounded"
    },
    {
        "Q": "\nAt 14:28, it mentions using function notation for the answer. Would it be counted wrong of you put \"y\"?",
        "A": "If you write y without saying what y is, then technically it s not right. At 10:57, Sal defines y=f(x), so after that, it s cool to write y wherever you d write f(x).",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            868
        ],
        "3min_transcript": "So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. Negative 1 minus 1.5 is negative 2.5. 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually I'll take a little aside to show you it doesn't matter what order I do this in. If I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3 minus 2 over 1.5 minus negative 1, this should be minus the 2 over 1.5 minus the negative 1. This is equal to what? Negative 3 minus 2 is negative 5 over 1.5 minus negative 1. That's 1.5 plus 1. That's over 2.5. So once again, this is equal the negative 2. So I just wanted to show you, it doesn't matter which one you pick as the starting or the endpoint, as long as If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know the equation is y is equal to negative 2x plus some y-intercept. Let's use one of these coordinates. I'll use this one since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. Of course you have your plus b. So 2 is equal to negative 2 times negative 1 is 2 plus b. 2, minus 2, you're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. Actually if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. Each of these coordinates are the coordinates of x and f of x. So you could even view the definition of slope as change in f of x over change in x. These are all equivalent ways of viewing the same thing."
    },
    {
        "Q": "\nI don't get it. At 3:39 is confusing. Why can't we use the slope equation to get the slope and then the slope would be 2 not -2. Am I missing something? He said if it's switch then it would be be negative?",
        "A": "OMG that is what I meant! Thank you! I forgot that the number would be negative 6. I can t believe I didn t see it!",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            219
        ],
        "3min_transcript": "We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0. So that's a 6-- I want to make it color-coded-- minus 0. So 6 minus 0, that's our change in y. Our change in x is 2 minus 2 minus 5. The reason why I color-coded it is I wanted to show you when I used this y term first, I used the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2 comma 6. This is the coordinate 5 comma 0. I couldn't have swapped the 2 and the 5 then. Then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10."
    },
    {
        "Q": "At 8:05 how do you get 5/2. Please explain.\n",
        "A": "5/6(3) is the same as diving 6 by 3. 6/3=2 5/2",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            485
        ],
        "3min_transcript": "So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6. You get 5/6. So we know that the equation is going to be of the form y is equal to 5/6 x plus b. Now we can substitute one of these coordinates in for b. So let's do. I always like to use the one that has the 0 in it. So y is a zero when x is negative 3 plus b. So all I did is I substituted negative 3 for x, 0 for y. I know I can do that because this is on the line. This must satisfy the equation of the line. Let's solve for b. So we get zero is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5/2 plus b. plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1."
    },
    {
        "Q": "\nAt 4:37 Sal says that x goes into 4 zero times. But x is an unknown - how does he know this?",
        "A": "because it is a un known so he dosen t know but if it is x and x^2 we know that x goes into x x times.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            277
        ],
        "3min_transcript": "So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x? 2 times x is 2x. 2 times 1 is 2. So we get 2 times x plus 1 is 2x plus 2. But we're going to want to subtract this from this up here, so we're going to subtract it. Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them. These guys cancel out. 6 minus 2 is 4. And how many times does x go into 4? We could just say that's zero times, or we could say that 4 is the remainder. So if we wanted to rewrite x squared plus 3x plus 6 over x plus 1-- notice, this is the same thing as x squared plus 3x plus 6 divided by x plus 1, this thing divided by this, we it is equal to x plus 2 plus the remainder divided by x plus 1 plus 4 over x plus 1. This right here and this right here are equivalent. And if you wanted to check that, if you wanted to go from this back to that, what you could do is multiply this by x plus 1 over x plus 1 and it add the two. So this is the same thing as x plus 2. And I'm just going to multiply that times x plus 1 over x plus 1. That's just multiplying it by 1. And then to that, add 4 over x plus 1. I did that so I have the same common denominator. And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here,"
    },
    {
        "Q": "at 3:26, sal puts brackets and says that the equation is in -, how did he know and how did he come to this??\n",
        "A": "The subtraction step of long division requires that you subtract the entire binomial: x^2 + x Subtraction of any polynomial requires that you distribute the minus sign across the polynomial being subtracted. To write the problem another way, you are doing: x^2+3x+6 - (x^2+x) = x^2+3x+6 - x^2 -x = 2x + 6 If you don t distribute the minus, you only subtract the x^2 term. The rest gets added. This would be like saying 57 - 32 = 29 (subtract the 3, but add the 2). Hope this helps.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            206
        ],
        "3min_transcript": "for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x?"
    },
    {
        "Q": "At 8:13 Sal removes x+4 as a factor from numerator and denominator. Why isn't a discontinuity marked at x=-4?\n",
        "A": "When doing this polynomial division, it is automatically assumed that the denominator is not equal to zero, otherwise you cant divide. Yes you are correct that the expression is undefined at x=-4, but we assume that x is not equal to negative 4 so there is no need to write it.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            493
        ],
        "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's"
    },
    {
        "Q": "I do not really understand what Sal means by \"factoring the numerator\" at 7:51. Could someone please explain to me what he is doing?\n",
        "A": "Sal didn t show all the steps. Option 1: (2x+4)/2 can be changed into 2x/2 + 4/2 -- Reduce each fraction and you get x+2 Option 2: Factor the numerator (factoring out GCF=2) (2x+4)/2 becomes 2(x+2)/2 Cancel the common factor of 2, and you get: (x+2)/1 or x+2 Hope this helps.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            471
        ],
        "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's"
    },
    {
        "Q": "\nat \"4:10\" Sal mentions anti derivitive. what is that?",
        "A": "It s another name for the integral (without any bounds).",
        "video_name": "OiNh2DswFt4",
        "timestamps": [
            250
        ],
        "3min_transcript": "of numbers. A transform will take you from one set of functions to another set of functions. So let me just define this. The Laplace Transform for our purposes is defined as the improper integral. I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds. The improper integral from 0 to infinity of e to the minus st times f of t-- so whatever's between the Laplace Transform brackets-- dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is the Laplace Transform? Well let's say that f of t is equal to 1. So what is the Laplace Transform of 1? of time-- well actually, let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. I don't have to rewrite it here, but there's a times 1dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing as the limit. And let's say as A approaches infinity of the integral from 0 to Ae to the minus st. dt. Just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing. could take the limit as something approaches infinity. So anyway, let's take the anti-derivative and evaluate this improper definite integral, or this improper integral. So what's anti-derivative of e to the minus st with respect to dt? Well it's equal to minus 1/s e to the minus st, right? If you don't believe me, take the derivative of this. You'd take minus s times that. That would all cancel out, and you'd just be left with e to the minus st. Fair enough. Let me delete this here, this equal sign. Because I could actually use some of that real estate. We are going to take the limit as A approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper intergrals. So I figured I might as well remind you that we're taking a limit. Now we took the anti-derivative."
    },
    {
        "Q": "Why can you assume that s > 0 at 6:36?\n",
        "A": "S is representing a frequency as in a per second or 1/second which is why you get 1/s for the Laplace of 1. From a physics perspective there is no such thing as a negative frequency because in reality it is the same thing as the frequency, saying something happens -3 times a second does not make a lot of sense. You can also tell that s cannot equal zero from the 1/s.",
        "video_name": "OiNh2DswFt4",
        "timestamps": [
            396
        ],
        "3min_transcript": "valued at 0. And then take the limit of whatever that ends up being as A approaches infinity. So this is equal to the limit as A approaches infinity. If we substitute A in here first, we get minus 1/s. Remember we're, dealing with t. We took the integral with respect to t. e to the minus sA, right? That's what happens when I put A in here. Now what happens when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0. This whole thing becomes 1. And I'm just left with minus 1/s. Fair enough. And then let me scroll down a little bit. to, but that's OK. So this is going to be the limit as A approaches infinity of minus 1/s e to the minus sA minus 1/s. So plus 1/s. So what's the limit as A approaches infinity? Well what's this term going to do? As A approaches infinity, if we assume that s is greater than 0-- and we'll make that assumption for now. Actually, let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as A approaches infinity, what's going to happen? Well this term is going to go to 0, right? e to the minus-- a googol is a very, very small number. And an e to the minus googol is an even smaller number. So then this e to the minus infinity approaches 0, so this This term isn't affected because it has no A in it, so we're just left with 1/s. So there you go. This is a significant to moment in your life. You have just been exposed to your first Laplace Transform. I'll show you in a few videos, there are whole tables of Laplace Transforms, and eventually we'll prove all of them. But for now, we'll just work through some of the more basic ones. But this can be our first entry in our Laplace Transform table. The Laplace Transform of f of t is equal to 1 is equal to 1/s. Notice we went from a function of t-- although obviously this one wasn't really dependent on t-- to a function of s. I have about 3 minutes left, but I don't think that's enough time to do another Laplace Transform. So I will save that for the next video. See you soon."
    },
    {
        "Q": "at 4:02 it really looks hard.\nI don't understand it.\n",
        "A": "He is just setting up a histogram. It looks sort of like a bar graph.",
        "video_name": "4eLJGG2Ad30",
        "timestamps": [
            242
        ],
        "3min_transcript": "So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6. And then on the vertical axis we're going to plot the frequency. So one, two, three, four, five, six, seven. So that's 7, 6, 5, 4, 3, 2, 1. So 0 shows up four times. So we'll draw a little bar graph here. 0 shows up four times. Draw it just like that. 0 shows up four times. That is that information right there. 1 shows up seven times. So I'll do a little bar graph. 1 shows up seven times. Just like that. I want to make it a little bit straighter than that-- 1 shows up seven times. 2-- I'll do it in a different color-- 2 shows up five times."
    },
    {
        "Q": "At 1:35, does Sal says that he groups them into two groups?\n",
        "A": "Yes he does -Andrew",
        "video_name": "2nZsIeaiJUo",
        "timestamps": [
            95
        ],
        "3min_transcript": "You and your friends are making name cards to assign seats for dinner. You each are given an identical strip of blank paper and begin coloring parts of it with crayons. You start with blue, or you start with blue. Your friend Micah starts off with green, Joelle starts with the red, Holden starts with pink, And Kate starts with purple. After one color is used, whose strip has the same part of its area colored as yours? So let's look at my strip right over here. It has how many equal sections? 1, 2, 3, 4, 5, 6 equal sections. And how many of them have been colored blue? Well, let's see. 1, 2, 3, 4 of my equal sections have been colored blue. So I need to figure out which of these other folks have colored exactly 4/6 of their strip blue. Or another way of thinking about it, I want to say, whoever has an equivalent fraction colored, not blue, colored whatever their color might be. So he's got 3 equal sections. He's got 3 equal sections, and he's colored, he's colored 2 of them, 2 of them green. Now is 4/6 and 2/3 the same thing? Well, he has 3 equal sections. I had 6. If I were to group each of these into groups of 2, so if I were to divide by 2, let me do that. If I were divide by 2, so I'm going to group this into groups of 2. Then I would have 1, 2-- so I have 1, 2, 3 equal sections. And then if I were to group the blue strips into groups of 2, then I would have 1, 2 of them that have actually been colored blue. So if I divide both the numerator and the denominator by 2, I get 2/3, which is the exact same fraction as Micah. Another way to think about it, is if you divide each of Micah's sections into 2. Well, then how many total sections does he have? Well, he had 3, now each of those 3 have been divided into 2. So we can multiply by 2 to get 6. So this is multiplying by 2. And each of the 2 in green, there now each of those sections have now been divided into 2. So you can multiply by 2 as well, and you get 4. So 2/3 and 4/6 are the same thing. So Micah has colored in the same amount that I have. Now let's see whether any of these other fractions are the same. So Joelle has colored in 4, has colored in 4 out of 1, 2, 3, 4, 5, 6, 7. 4 out of 7. So it's not clear to me that we can somehow multiply or divide"
    },
    {
        "Q": "At 6:30, Sal talks about the transversal line. What's that?\n",
        "A": "A line that cuts across two or more (usually parallel) lines",
        "video_name": "R0EQg9vgbQw",
        "timestamps": [
            390
        ],
        "3min_transcript": "That is the opposite side of our angle x so we know sine of x is equal to DC over one, or DC over one is just DC. This length right over here is sine of x. Segment AC, same exact logic. Cosine of x is the length of AC over one which is just the length of AC. This length right over here, segment AC its length is cosine of x. That's kind of interesting. Now let's see what we can figure out about this triangle, triangle ACB right over here. How could we figure out CB? Well we know that sine of y, let me write this here. Sine of y is equal to what? It's equal to the length of segment CB over the hypotenuse. The hypotenuse here is the cosine of x, At any point if you get excited pause the video and try to finish the proof on your own. The length of segment CB if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y. Which is neat because we just showed that this thing right over here is equal to this thing right over here. To complete our proof we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that well we already know that the sum of these is equal to the length of DF which is sine of x plus y. Let's see if we can figure out, if we can express DE somehow. What angle would be useful? If somehow we could figure If we could figure out this angle then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle. We know this is angle y over here and we also know that this is a right angle. EC is parallel to AB so you could view AC as a transversal. If this is angle y right over here then we know this is also angle y. These are once again, notice. If AC is a transversal here and EC and AB are parallel then if this is y then that is y. If that's y then this is 90 minus y. If this is 90 degrees and this is 90 minus y then these two angles combined add up to 180 minus y,"
    },
    {
        "Q": "\nDid anyone see Vi Hart draw the dolphin at 1:23 above the 42? Anyone get the reference?",
        "A": "Vi Hart also makes a reference to Douglas Adams and his book the Hitchhiker s Guide to the Galaxy in another one of her videos at her website, Silly Band Fight .",
        "video_name": "a-e8fzqv3CE",
        "timestamps": [
            83
        ],
        "3min_transcript": "While I'm working on some more ambitious projects, I wanted to quickly comment on a couple of mathy things that have been floating around the internet, just so you know I'm still alive. So there's this video that's been floating around about how to multiply visually like this. Pick two numbers, let's say, 12 times 3. And then you draw these lines. 12, 31. Then you start counting the intersections-- 1, 2, 3 on the left; 1, 2, 3, 4, 5, 6, 7 in the middle; 1, 2 on the right, put them together, 3, 7, 2. There's your answer. Magic, right? But one of the delightful things about mathematics is that there's often more than one way to solve a problem. And sometimes these methods look entirely different, but because they do the same thing, they must be connected somehow. And in this case, they're not so different at all. Let me demonstrate this visual method again. This time, let's do 97 times 86. So we draw our nine lines and seven lines time eight lines and six lines. Now, all we have to do is count the intersections-- 1,2, 3, 4, 5, 6, 7, 8, 9, 10. This is boring. How about instead of counting all the dots, we just figure out how many intersections there are. Let's see, there's seven going one way and six Forget everything I ever said about learning a certain amount of memorization in mathematics being useful, at least at an elementary school level. Because apparently, I've been faking my way through being a mathematician without having memorized 6 times 7. And now I'm going to have to figure out 5 times 7, which is half of 10 times 7, which is 70, so that's 35, and then add the sixth 7 to get 42. Wow, I really should have known that one. OK, but the point is that this method breaks down the two-digit multiplication problem into four one-digit multiplication problems. And if you do have your multiplication table memorized, you can easily figure out the answers. And just like these three numbers became the ones, tens, and hundreds place of the answer, these do, too-- ones, tens, hundreds-- and you add them up and voila! Which is exactly the same kind of breaking down into single-digit multiplication and adding that you do during the old boring method. The whole point is just to multiply every pair of digits, make sure you've got the proper number of zeroes on the end, and add them all up. But of course, seeing that what you're actually doing is not something your teachers want you to realize, or else you might remember the every combination concept when you get to multiplying binomials, and it might make it too easy. In the end, all of these methods of multiplication distract from what multiplication really is, which for 12 times 31 is this. All the rest is just breaking it down into well-organized chunks, saying, well, 10 times 30 is this, 10 times 1 this, 30 times 2 is that, and 2 times 1 is that. Add them all up, and you get the total area. Don't let notation get in the way of your understanding. Speaking of notation, this infuriating bit of nonsense has been circulating around recently. And that there has been so much discussion of it is sign that we've been trained to care about notation way too much. Do you multiply here first or divide here first? The answer is that this is a badly formed sentence. It's like saying, I would like some juice or water with ice. Do you mean you'd like either juice with no ice or water with ice? Or do you mean that you'd like either juice with ice or water with ice? You can make claims about conventions and what's right and wrong, but really the burden"
    },
    {
        "Q": "At 3:28, If you watch what Sal does/says, can you conclude that you can do this:\nthe square root of 60 times the square root of 6 = the square root of 60 times 6??\n",
        "A": "Yes, that is correct.",
        "video_name": "74iuGIaBgRc",
        "timestamps": [
            208
        ],
        "3min_transcript": "If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10, plus the square root of nine, times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four, so that is, so let me do this is in another color. so it can be clear. So, this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? Well that's going to be one square root of 10. Actually, let me slow it down a little bit. I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, and that case it's squares of 10s, well, I'm going to be left with just one of that something. I'm just gonna be left with one square root of 10. Which we could just write as the square root of 10. Another way to think about it is, we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is of course, this is just one. So you're just left with the square root of 10. So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside So essentially the same idea. All right, let's see what we can do. So, this is interesting. We have a square root of 1/2. So can I, well actually, what could be interesting is since if I have a square root of something times the square root of something else. So the square root of 180 times the square root of 1/2, this is the same thing as the square root of 180 times 1/2. And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1/2 power, times 1/2 to the 1/2 power, is going to be equal to 180 times 1/2 to the 1/2 power, taking the square root, the principal root is the same thing as raising something to the 1/2 power. And so this is the square root of 80 times 1/2 which is going to be the square root of 90,"
    },
    {
        "Q": "\nAt 5:35, Sal says you can put the answer in either way. On the practice, can you put either option? Thanks!",
        "A": "It would be better to use the second way, because it is more simple. (8/3) sqrt of (2/3)",
        "video_name": "74iuGIaBgRc",
        "timestamps": [
            335
        ],
        "3min_transcript": "times 10, and we just simplified square root of 90 in the last problem, that's equal to the square root of nine times the square root of, principle root of 10, which is equal to three times the square root of 10. Three times the square root of 10. All right, let's keep going. So I have one more of these examples, and like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect square, okay, these are just same directions that we've seen the last few times. And so let's see. If I wanted to do, if I wanted to simplify this, this is equal to the square root of, well, 64 times two is 128, and 64 is a perfect square, so I'm gonna write it as 64 times two, over 27 is nine times three. Nine is a perfect square. So this is going to be the same thing. We could say this is the same thing as the square root of 64 times two, over the square root of nine times three, which is the same thing as the square root of 64 times the square root of two, over square root of nine times the square root of three, which is equal to, this is eight, this is three, so it would be eight times the square root of two, over three times the square root of three. That's one way to say it. Or we could even view the square root of two over the square root of three as a square root of 2/3. So we could say this is eight over three times the square root of 2/3. So these are all possible ways of trying to tackle this. So we could just write it, let's see. Have we removed all factors that are perfect squares? Yes, from inside the radicals and we've combined terms. We weren't doing any adding or subtracting here, from inside the radicals and I think we've done that. So we could say this is going to be 8/3 times the square root of 2/3. And there's other ways that you could express this that would be equivalent but hopefully this makes some sense."
    },
    {
        "Q": "at 15:03 I was not sure if the answer to that was correct\n",
        "A": "it is correct because 9+9+9+9+9+9+9+9=72 or 9x8=72",
        "video_name": "xO_1bYgoQvA",
        "timestamps": [
            903
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt around 3:50 to 4:38, Sal constructs matrix A\nHowever he seems to write \"3Rot\"... I'm not too sure what the 3 part of the Rot is. In the previous video, he doesn't write \"2Rot\"... Is the 3 just notation to represent R3? Since intuitively it feels like a multiplication by 3",
        "A": "@ around 2:40 he says (roughly): ...let me call it 3Rot(theta) to denote that it s in R^3 .",
        "video_name": "gkyuLPzfDV0",
        "timestamps": [
            230,
            278
        ],
        "3min_transcript": "z-component, it looks like that. Then when you rotate it, its z and its y-components will change, but its x-component will stay the same. So then it might look something like this. Let me see if I can give it justice. So then the vector when I rotate it around might look something like that. Anyway, I don't know if I'm giving it proper justice but this was rotated around the x-axis. I think you understand what that means. But just based on the last video, we want to build a transformation. Let me call this rotation 3 theta. Or let me call it 3 rotation theta now that we're dealing in R3. And what we want to do is we want to find some matrix, so I can write my 3 rotation sub theta transformation of x as being some matrix A times the vector x. Since this is a transformation from R3 to R3 this is of course going to be a 3 by 3 matrix. out, you just have to apply the transformation essentially to the identity matrix. So what we do is we start off with the identity matrix in R3, which is just going to be a 3 by 3. It's going to have 1, 1, 1, 0, 0, 0, 0, 0, 0. Each of these columns are the basis vectors for R3. That's e1, e2, e3-- I'm writing it probably too small for you to see-- but each of these are the basis vectors for R3. And what we need to do is just apply the transformation to each of these basis vectors in R3. So our matrix A will look like this. Our matrix A is going to be a 3 by 3 matrix. Where the first column is going to be our transformation, 3 rotation sub theta, applied to that column And then I'm going to apply it to this middle column vector right here. You get the idea, I don't want to write that whole thing again. I'm going to apply 3 rotation sub theta to 0, 1, 0. And then I'm going to apply it-- I'll do it here-- 3 rotation sub theta. I'm going to apply it to this last column vector, so 0, 0, 1. We've seen this multiple times. So let's apply it. Let's rotate each of these basis vectors for R3. Let's rotate them around the x-axis. So the first guy, if I were to draw an R3, what would he look like? He only has directionality in the x direction right? If we call this the x-dimension, if the first entry corresponds to our x-dimension, the second entry corresponds to our y-dimension. And the third entry corresponds to our z-dimension."
    },
    {
        "Q": "\nAt 1:37, how did he define f(x) and g(x)?",
        "A": "Sal simply splitted the root frunction away from the remaining rest. f(x) is the root function, g(x) is the remaining rest.",
        "video_name": "IiBC4ngwH6E",
        "timestamps": [
            97
        ],
        "3min_transcript": "What I want to do in this video is start with the abstract-- actually, let me call it formula for the chain rule, and then learn to apply it in the concrete setting. So let's start off with some function, some expression that could be expressed as the composition of two functions. So it can be expressed as f of g of x. So it's a function that can be expressed as a composition or expression that can be expressed as a composition of two functions. Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function. f prime of g of x times the derivative of the inner function with respect to x. How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1"
    },
    {
        "Q": "At 5:10 would you have to simplify any farther or is that an adequte answer?\n",
        "A": "The answer at 5:10 is in its most simplified form. (3x^2 - x)^(-1/2) can t be multiplied with (6x - 1).",
        "video_name": "IiBC4ngwH6E",
        "timestamps": [
            310
        ],
        "3min_transcript": "And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out So the derivative of f of the outer function with respect to the inner function. So let me write it. It is equal to 1/2 times g of x to the negative 1/2, times 3x squared minus x. This is exactly this based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1/2 power. So the derivative of that whole thing with respect to your something is going to be 1/2 times that something to the negative 1/2 power. That's essentially what we're saying. But now we have to take the derivative of our something with respect to x. And that's more straightforward. g prime of x-- we just use the power rule for each of these terms-- is equal to 6x to the first, or just 6x minus 1. So this part right over here is just going to be 6x minus 1. is this right over here and we're multiplying. And we're done. We have just applied the power rule. So just to review, it's the derivative of the outer function with respect to the inner. So instead of having 1/2x to the negative 1/2, it's 1/2 g of x to the negative 1/2, times the derivative of the inner function with respect to x, times the derivative of g with respect to x, which is right over there."
    },
    {
        "Q": "At 3:00 I don't understand why f'(x) = 1/2x^-1/2. How and why did we get there from f(x) = x^1/2? What's the rule for moving from f(x) to f'(x)?\n",
        "A": "This is a function of the power rule: d/dx( x^n ) \u00e2\u0086\u0092 n\u00e2\u0080\u00a2x^(n - 1) Let: f(x) = sqrt(x) f (x) = d/dx(sqrt(x)) f (x) = d/dx(x^(1/2)) Apply power rule d/dx( x^n ) \u00e2\u0086\u0092 n\u00e2\u0080\u00a2x^(n - 1) where n = 1/2: f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1)\u00e2\u0080\u00a2d/dx(x) f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1)\u00e2\u0080\u00a2dx/dx f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1) f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 2/2) f (x) = 1/2\u00e2\u0080\u00a2x^(-1/2) f (x) = 1/(2\u00e2\u0080\u00a2x^(1/2)) f (x) = [1/(2\u00e2\u0080\u00a2sqrt(x))]",
        "video_name": "IiBC4ngwH6E",
        "timestamps": [
            180
        ],
        "3min_transcript": "How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1 And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out"
    },
    {
        "Q": "\n@3:03 what is does the negative one-half power mean",
        "A": "x^(-0.5) = 1/sqrt(x)",
        "video_name": "IiBC4ngwH6E",
        "timestamps": [
            183
        ],
        "3min_transcript": "How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1 And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out"
    },
    {
        "Q": "In 0:51, did Sal say tenth hundredths or ten hundredths?\n",
        "A": "He said ten hundreds.",
        "video_name": "o31cLUkS23E",
        "timestamps": [
            51
        ],
        "3min_transcript": "Let's try to subtract 9.57 minus 8.09. So try to pause this video and figure this out first before we work through it together. Well, let's just rewrite it. Let's rewrite it. And when I rewrite it, I like to line up the decimals. This one it's a little intuitive. We have 8.09. Just like that, and now we're ready to subtract. And we want to subtract 9 hundredths from 7 hundredths. Well, we don't have enough hundredths up here so let's move over here. Let's see if we can do some regrouping so we always have a higher number on top. So over here we want to subtract 0 tenths from 5 tenths so we have enough tenths over here. So let's regroup. So instead of 5 tenths, I'm going to have 4 tenths, and then I'm going to give that other tenth, which is the same thing as 10 hundredths over here, so this becomes 17 hundredths. 17 minus 9 is 8. 4 minus 0 is 4. So this is going to be 1.48."
    },
    {
        "Q": "\nat 1:54 sal says one people in line",
        "A": "He said it twice, probably on purpose.",
        "video_name": "BIpsQIJUCC8",
        "timestamps": [
            114
        ],
        "3min_transcript": "- [Voiceover] Let's say that you love frozen yogurt. So every day after school you decide to go to the frozen yogurt store at exactly four o'clock, four o'clock PM. Now, because you like frozen yogurt so much, you are not a big fan of having to wait in line when you get there, you're impatient, you want your frozen yogurt immediately. And so you decide to conduct a study. You want to figure out the probability of there being lines of different sizes when you go to the frozen yogurt store after school, exactly at four o'clock PM. So in your study, the next 50 times you observe, you go to the frozen yogurt store at four PM, you make a series of observations. You observe the size of the line. So, let me make two columns here, line size is the left column, and on the right column, let's say this is the number of times observed. So, times observed, observed. O-B-S-E-R-V-E-D, all right, times observed. All right, so let's first think about it. Okay, so you go and you say, hey look, I see no people in line, exactly, or you see no people in line, exactly 24 times. You see one person in line exactly 18 times, and you see two people in line exactly eight times. And, in your 50 visits, you don't see more, you never see more than two people in line. I guess this is a very efficient cashier at this frozen yogurt store. So based on this, based on what you have observed, what would be your estimate of the probabilities of finding no people in line, one people in line, or two people in line, at four PM on the days after school that you visit the frozen yogurt store? So what's the probability of there being no line, a one person line, or a two person line when you visit at four PM on a school day? Well, all you can do is estimate the true probability, the true theoretical probability. We don't know what that is, but you've done 50 observations here right. I know that this adds up to 50, 18 plus eight is 26, 26 plus 24 is 50, so you've done 50 observations here and so you can figure out, well what are the relative frequencies of having zero people? What is the relative frequency of one person, or the relative frequency of two people in line? And then we can use that as the estimates for the probability. So let's do that. So, probability estimate. I'll do it in the next column. So probability, probability estimate, and once again we can do that by looking at the relative frequency. The relative frequency of zero, well we"
    },
    {
        "Q": "\ncan we factor something out from a term which is not even present in every subterm...just a little bit confusing because i've heard we can't factor out something which is not present in each term.. sal just did that at 8:35",
        "A": "You are right in that it is nonsensical to factor something out that is not present in every term undergoing the factoring. Sal didn t do that. At 8:35, each of the 3 terms had involved c_x (c sub x).",
        "video_name": "b7JTVLc_aMk",
        "timestamps": [
            515
        ],
        "3min_transcript": "Remember, this is just the x component of our triple product. Just the x component. But to do this, let me factor out. I'm going to factor out a bx. So let me do this, let me get the bx. So if I were to factor it out-- I'm going to factor it out of this term that has a bx. I'm going to factor it out of this term. And then I'm going to factor it out of this term. So if I take the bx out, I'm going to have an aycy. Actually, let me write it a little bit differently. Let me factor it out of this one first. So then it's going to have an axcx. a sub x, c sub x. So I used this one up. And then I'll do this one now. Plus, if I factor the bx out, I get ay cy. I've used that one now. And now I have this one. I'm going to factor the bx out. So that's all of those. So I've factored that out. And now, from these right over here, let me factor out a negative cx. And so, if I do that-- let me go to this term right over here-- I'm going to have an axbx when I factor it out. So an axbx, cross that out. And then, over here, I'm going to have an ayby. Remember, I'm factoring out a negative cx, so I'm going to have a plus ay, sub by. And then, finally, I'm going to have a plus az, az bz. And what is this? Well, this right here, in green, this is the exact same thing as the dot products of a and c. This is the dot product of the vectors a and c. So that's the dot of a and c times the x component of b minus-- I'll do this in the same-- minus-- once again, this is the dot product of a and b now, minus a dot b times the x component of c. And we can't forget, all of this was multiplied by the unit vector i. We're looking at the x component, or the i component of that whole triple product. So that's going to be all of this. All of this is times the unit vector i. Now, if we do this exact same thing-- and I'm not going to do it, because it's computationally intensive. But I think it won't be a huge leap of faith for you. This is for the x component."
    },
    {
        "Q": "\nQuestion. At 7:03, Sal said \"potential inflection point\" . Just curious, but how could it possibly not be an inflection point if x=0 is a critical number and the second derivative of 0 equals 0.",
        "A": "Sometimes critical points are not inflection points. For example take the graph of y=x^3. At zero, we find that the first derivative is zero [y (0)=3(0)^2], indicating a critical point. However if you look at points slightly to the right of zero and slightly to the left of zero, you ll see that the rate of change (first derivative) is positive on both sides. The graph is not changing concavity, even though zero is an inflection point",
        "video_name": "hIgnece9ins",
        "timestamps": [
            423
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 12:50, do you have to be so close to 0? If it was 1 (which is still greater than zero) 3x-2 would be positive. So that value is not negative for all x>0.\n",
        "A": "You don t. It just has to be between critical numbers. Let s say you had critical numbers of 1, 5, and 7. You could test 0, 2, 6, and 8 to cover all of your bases. Your test numbers don t have to be so close to your critical numbers.",
        "video_name": "hIgnece9ins",
        "timestamps": [
            770
        ],
        "3min_transcript": ""
    },
    {
        "Q": "@ 18:56 for x=0 , if x>0 shouldnt curve be upwards,x<0 be downwards?\n",
        "A": "No. The best way to verify this is to graph the original equation. Perhaps a little more detail would help clarify. When x<0 f (x)>0 therefore the original function is concave up. BUT, when 0<x<2/3, f (x)<0, therefore, f(x) is concave down on that interval. when x>2/3, f (x)>0 again and concave up. Since f has a positive leading coefficient the graph as concave up toward positive and negative infinity.",
        "video_name": "hIgnece9ins",
        "timestamps": [
            1136
        ],
        "3min_transcript": ""
    },
    {
        "Q": "5:45\u00e3\u0080\u0082\u00e3\u0080\u0082\u00e3\u0080\u0082why derivative zai equals 0...he only proved h(y)=0\n",
        "A": "If you remember, the original function was in the form M(x,y)+N(x,y)y = 0. He also assumed u(x)M(x,y) = psi_x and u(x)N(x,y) = psi_y. Using the chain rule for partial derivatives that he showed us that psi = u(x)M(x,y)+u(x)N(x,y)y *. Since *M(x,y)+N(x,y)y = 0 is the same as u(x)M(x,y)+u(x)N(x,y)y = 0, you can say psi = 0.",
        "video_name": "0NyeDUhKwBE",
        "timestamps": [
            345
        ],
        "3min_transcript": "That's the partial of a function purely of y with respect to y. And then that has to equal our new N, or the new expression we got after multiplying by the integrating factor. So that's going to be equal to this right here. This is, hopefully, making sense to you at this point. And that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side. So let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y. And we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant. There's just some constant left over. So for our purposes, we can just say that psi is equal to this. Because this is just a constant, we're going to take the antiderivative anyway, and get a constant on the right hand side. And in the previous videos, the constants all merged together. So we'll just assume that that is our psi. And we know that this differential equation, up here, can be rewritten as, the derivative of psi with respect to x, and that just falls out of the partial derivative chain rule. The derivative of psi with respect to x is equal to 0. If you took the derivative of psi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule. And we know what psi is. So we can write-- or actually we don't even have to. We could use this fact to say, well, if we integrate both sides, that a solution of this differential equation is that I just took the antiderivative of both sides. So, a solution to the differential equation is psi is equal to c. So psi is equal to x to the third y plus 1/2 x squared y squared. And we could have said plus c here, but we know the solution is that psi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here. You have another constant there. And you can just subtract them from both sides. And they just merge into another arbitrary constant. But anyway, there we have it. We had a differential equation that, at least superficially, looked exact. It looked exact, but then, when we tested the exactness of it, it was not exact. But we multiplied it by an integrating factor. And in the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it. And true enough, it was exact."
    },
    {
        "Q": "\nAT 4:30 what happened to the y' how come it did not come down into the new equations, specifically to the right of the equals",
        "A": "Because in the method to solving Exact Equations, you consider what is multiplying your y as your function N(x,y) and the rest as your function M(x,y). The original y is used just to separate those 2 parts, it no longer has any involvement in the rest of the solution.",
        "video_name": "0NyeDUhKwBE",
        "timestamps": [
            270
        ],
        "3min_transcript": "So we get 3x squared plus 2xy. And there we have it. The partial of this with respect to y is equal to the partial of this with respect to N. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation, or that differential equation. So it's exact. Let's solve it. So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function psi where the partial derivative of psi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and we'll get psi is equal to what? 1/2 x squared y squared. And of course, this psi is a function of x and y, so when you take the partial with respect to x, when you go that way, you might have lost some function that's only a So instead of a plus c here, it could've been a whole function of y that we lost. So we'll add that back when we take the antiderivative. But we're not completely done yet, because we have to somehow figure out what this function of y is. And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this. So let's set that up. So what's the partial of this expression with respect to y? So I could write, the partial of psi with respect to y is equal to x to the third plus 2 times 1/2, so it's just x That's the partial of a function purely of y with respect to y. And then that has to equal our new N, or the new expression we got after multiplying by the integrating factor. So that's going to be equal to this right here. This is, hopefully, making sense to you at this point. And that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side. So let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y. And we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant."
    },
    {
        "Q": "\nHow did you get the negative sixteen to become a positive at 0:51? How is it the same?",
        "A": "remember if the problem say is 7- (-9) it makes no sense to solve it. So it s easier if you remember that if there are two minus signs next to each other then change them into a plus sign so there won t have to be any of the minus negative confusion. :) Example: 2+7- (-8)+13 9+15+13 24+13 37",
        "video_name": "03yq7XsErqo",
        "timestamps": [
            51
        ],
        "3min_transcript": "Evaluate 3x squared minus 8x plus 7 when x is equal to negative 2. So to evaluate this expression when x is equal to negative 2, everywhere that we see an x, we just have to substitute it with a negative 2. So it would be 3 times, instead of x squared, it would be 3 times negative 2 squared minus 8 times-- instead of 8 times x, 8 times negative 2 plus 7. And so what does this give us? So we have 3 times negative 2 squared. Well, that's just positive 4. A negative times a negative is a positive. Minus 8 times negative 2 is negative 16 plus 7. So 3 times 4 is 12. From that, you're subtracting negative 16. Subtracting negative 16 is the same thing as adding positive 16. So it's 12 plus 16 plus 7. And so this is equal to 12 plus 16 is 28."
    },
    {
        "Q": "Why is it 4 becquerel at 3:48 ? Is it because of c=8 and you need to half it because cesium are reduced to half after 30 days?\n",
        "A": "Yes, you are right.",
        "video_name": "polop-89aIA",
        "timestamps": [
            228
        ],
        "3min_transcript": "Fair enough? So let's just be clear, this is days since release. In addition, assume that we know that the initial amount of caesium-137 released in the soil is 8 becquerels. Solve for the unknown constants c and r. So the initial in the soil. That's when t is equal to 0. When no days have passed. So we could say that the amount at times 0-- well, that's going to be equal to c times r to the 0 power, which is just going to be equal to c times 1, which is equal to c. And they tell us what A of 0 is. They say A of 0 is 8 becquerels. So this is going to be equal to 8. So our constant here, the c is just going to be equal to 8. What is the value of the constant? We could just write 8 right over there. So the value of the constant c is 8. What is the value of the constant r? So we're starting with 8. So A of 0 is 8. How much are we going to have after 30 days? And the reason why I'm picking 30 days is-- that is the half-life of caesium-137. So A of 30, remember our t is in-- let me just switch colors just for fun. Remember, t is in days. So A of 30. So after 30 days, I am going to-- if I want to use this formula right over here, if I wanted to use the description of this exponential function, we already know that c is 8. It's going to be 8 times r to the 30th power, which is going to be equal to what? Well, if we started with 8, 30 days later, we're going to have half as much. We're going to have 4 becquerels. And now, we can use this to solve for r. So you have 8 times r to the 30th power is equal to 4. Divide both sides by 8. which is the same thing as 1/2. And then, we can raise both sides to the 1/30th power. r to the 30th-- but then, you could think of the 30th root of that or raising that to the 1/30th power-- that's just going to give us r is equal to one half to the 1/30th power. And that is something that's very hard to compute in your head. So I suggest you use a calculator for that. And they hint because we're going to round to the nearest thousandth. So let's get a calculator right out. And so we're talking about one half to the 1/30 power. So we get 0.9771599-- it keeps going. But they tell us to round to the nearest thousandth, so 0.977"
    },
    {
        "Q": "at 2:20 how the heck does (4x^4-x)/(x^4) simplify to 4 - (1/x^3) ?? I'm missing something, that for sure.\n",
        "A": "Recall that in general, (a - b)/c is equivalent to a/c - b/c. So (4x^4-x)/(x^4) = (4x^4)/(x^4) - x/(x^4) = 4 - 1/(x^3). If you are rusty with algebra, you will need to review algebra in order to have a realistic chance of performing well in calculus. Have a blessed, wonderful day!",
        "video_name": "uPksX_O9ARo",
        "timestamps": [
            140
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can find the limit as x approaches negative infinity of the square root of 4x to the fourth minus x over 2x squared plus three. And like always, pause this video and see if you can figure it out. Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at what is the highest degree term in the numerator or in the denominator, or, actually in the numerator and the denominator, and then divide the numerator and the denominator by that highest degree, by x to that degree. Because if we do that, then we're going to end up with some constants and some other things that will go to zero as we approach positive or negative infinity, and we should be able to find this limit. So what I'm talking about, let's divide the numerator by one over x squared and let's divide the denominator by one over x squared. Now, you might be saying, \"Wait, wait, \"I see an x to the fourth here. But remember, it's under the radical here. So if you wanna look at it at a very high level, you're gonna take the square root of this entire expression, so you can really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of a side here. So if I have, if I have one over x squared, all right, let me write it. Let me just, one over x squared times the square root of 4x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of 4x to the fourth minus x. And so this is equal to the square root of 4x to the fourth minus x and all I did is I brought the radical in here. You could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of 4x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator's going to be the square root of four minus one, x to the third power. And then the denominator is going to be equal to, well, you divide 2x squared by x squared. You're just going to be left with two. And then three divided by x squared is gonna be three over x squared. Now, let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero."
    },
    {
        "Q": "\nat 3:00 how can he go from x^2 right to x?",
        "A": "It should be |x| < 1/2.",
        "video_name": "6ynr9N-NQ8E",
        "timestamps": [
            180
        ],
        "3min_transcript": "We can rewrite f of x as being equal to the sum from n equals 0 to infinity of-- let's see, our first term here is 2-- 2 times negative 4x squared to the n-th power. This is a geometric series where our common ratio is negative 4x squared to the n-th power. Now, when will this thing right over here converge? Well, we know that a geometric series will converge if the absolute value of its common ratio is less than 1. So let me write this down. So converge if the absolute value of the common ratio, negative 4x squared, is less than 1. Well, this-- the way it's written right now-- So the absolute value of this is just going to be 4x squared. Right? x squared is going to be non-negative, so 4x squared is going to be non-negative. Negative 4x squared is going to be non-positive. So if you're taking the absolute value of a non-positive thing, that's going to be the same thing as the absolute value of the negative of it. So this just has to be less than 1. And the absolute value of something that is strictly non-negative like this, well, that's just going to be 4x squared-- these two statements are equivalent-- and that has to be less than 1. Can divide both sides by 4, you get x squared is less than 1/4. And so we can say that the absolute value of x has to be less than 1/4, or we could say that negative 1/4 So expressed this way, we're giving the interval of convergence. This thing will converge as long as x is in this interval. Expressed this way, we're really saying the radius of convergence. This will converge as long as x is less than our radius of convergence, as long as the absolute value of x is less than our radius of convergence, as long as x stays less than 1/4 away from 0. To make it a little bit clearer, you could rewrite this as the distance between x and 0, as long as this- this, you could view this as the distance between x and 0-- as long as this stays less than 1/4, this thing is going to converge. So this is the interval of convergence, this, you could view, 1/4, you could view as the radius of convergence. Now with that out of the way, we've thought about where this thing converges, let's think about what it converges to. Well, we've done this multiple times."
    },
    {
        "Q": "\nAt 3:06, can we write the answer as - (m+4)/5\nSince both values have the same denominator, and we can take -1 as a common for the whole numerator, making both m and 4 positive.",
        "A": "Yes, the answer could be written that way. It is equivalent to Sal s version.",
        "video_name": "rtNuo7R3scY",
        "timestamps": [
            186
        ],
        "3min_transcript": "as the negative of 8.55 minus 4.35, and 8.55 minus 4.35, let's see, eight minus four is going to be the negative, eight minus four is four, 55 hundredths minus 35 hundredths is 20 hundredths. So I could write 4.20, which is really just the same thing as 4.2. So all of this can be replaced with a -4.2. So my entire expression has simplified to -5.55, and instead of saying plus -4.2c, I can just write it as minus 4.2c, We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have some more hairy numbers involved. And so, let's see, I have 2/5m minus 4/5 minus 3/5m. So how can I simplify? Well I could add all the m terms together. So let me just change the order. I could rewrite this as 2/5m minus 3/5m minus 4/5. All I did was I changed the order. We can see that I have these two m terms. I can add those two together. So this is going to be 2/5 minus 3/5 times m, and then I have the -4/5 still on the right hand side. Now what's 2/5 minus 3/5? Well that's gonna be -1/5. That's gonna be -1/5. So I have -1/5m minus 4/5. Minus 4/5. I can't simplify it anymore. I can't add this term that involves m somehow So we are done here. Let's do one more. Let's do one more example. So here, and this is interesting, I have a parentheses and all the rest. Like always, pause the video. See if you can simplify this. All right, let's work through it together. Now the first thing that I want to do is let's distribute this two so that we just have three terms that are just being added and subtracted. So if we distribute this two, we're gonna get two times 1/5m is 2/5m. Let me make sure you see that m. M is right here. Two times -2/5 is -4/5, and then I have plus 3/5. Now how can we simplify this more? Well I have these two terms here that don't involve the variable. Those are just numbers. I can add them to each other. So I have -4/5 plus 3/5. So what's negative four plus three? That's going to be negative one. So this is going to be -1/5,"
    },
    {
        "Q": "At 1:03 is the expression like this: -5.55 - 8.55c + 4.35c or is it like this: -5.55 + (-8.55c + 4.35c)?\nWhich expression is the right one, or are they the same?\n",
        "A": "These expressions are equal to each other due to the associative property of addition and the fact that subtraction is the same as adding a negative. This can be seen more clearly by replacing the terms with easier numbers to visualize. For example, 10 - 5 + 2 = 10 + (-5 + 2). You can play around with the groups and order in addition and even subtraction, as long as you remember that subtraction is adding a negative. Be sure to keep your positive and negative signs with their numbers.",
        "video_name": "rtNuo7R3scY",
        "timestamps": [
            63
        ],
        "3min_transcript": "- [Voiceover] What I wanna do in this video is get some practice simplifying expressions and have some hairier numbers involved. These numbers are kind of hairy. Like always, try to pause this video and see if you can simplify this expression before I take a stab at it. All right, I'm assuming you have attempted it. Now let's look at it. We have -5.55 minus 8.55c plus 4.35c. So the first thing I wanna do is can I combine these c terms, and I definitely can. We can add -8.55c to 4.35c first, and then that would be, let's see, that would be -8.55 plus 4.35, I'm just adding the coefficients, times c, and of course, we still have that -5.55 out front. -5.55. I'll just put a plus there. Now how do we calculate -8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. as the negative of 8.55 minus 4.35, and 8.55 minus 4.35, let's see, eight minus four is going to be the negative, eight minus four is four, 55 hundredths minus 35 hundredths is 20 hundredths. So I could write 4.20, which is really just the same thing as 4.2. So all of this can be replaced with a -4.2. So my entire expression has simplified to -5.55, and instead of saying plus -4.2c, I can just write it as minus 4.2c, We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have some more hairy numbers involved. And so, let's see, I have 2/5m minus 4/5 minus 3/5m. So how can I simplify? Well I could add all the m terms together. So let me just change the order. I could rewrite this as 2/5m minus 3/5m minus 4/5. All I did was I changed the order. We can see that I have these two m terms. I can add those two together. So this is going to be 2/5 minus 3/5 times m, and then I have the -4/5 still on the right hand side. Now what's 2/5 minus 3/5? Well that's gonna be -1/5. That's gonna be -1/5. So I have -1/5m minus 4/5. Minus 4/5. I can't simplify it anymore. I can't add this term that involves m somehow"
    },
    {
        "Q": "\nAt 2:17 in this vid of Extraneous solutions of Rad Eq, Sal goes to Quad Formula. However, at this point aren't we able to factor 4x^2 - 3x -7 = 0 to (4x-7)(x+1) to get one of his answers which is a solution, 7/4, and, -1, which is an extraneous solution.",
        "A": "Yes... you can use factoring. It is likely simpler than the quadratic formula.",
        "video_name": "m4eiYHL3PP8",
        "timestamps": [
            137
        ],
        "3min_transcript": "- [Voiceover] So let's say we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side So, we might say, \"Well, let's just get rid of the radical. \"Let's square both sides of this equation.\" So we might say that this is the same thing as two-x minus one squared is equal to the square root of eight minus x, eight minus x squared, and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel, \"Okay we did legitimate operations. \"We did the same thing to both sides. \"That these are equivalent equations.\" But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it, you're losing information. So, for example, this would be true even if the original equation were two-x... Even if the original equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to this right-hand side, and not the yellow one, then we would call that an extraneous solution. So, let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the quadratic formula, here. So the solutions are going to be... X is going to be equal to negative b, so three, plus or minus the square root of b-squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three"
    },
    {
        "Q": "At 0:42, how does he get 4x^2-4x+1=8-x?\nMore specifically, how does he get the second negative four?\n",
        "A": "=(2x-1)^2 =(2x-1)(2x-1) =4x^2-2x-2x+1 =4x^2-4x+1 By using the distributive properties on this perfect square we are able to come up with two 2x and then combine like terms to get -4x.",
        "video_name": "m4eiYHL3PP8",
        "timestamps": [
            42
        ],
        "3min_transcript": "- [Voiceover] So let's say we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side So, we might say, \"Well, let's just get rid of the radical. \"Let's square both sides of this equation.\" So we might say that this is the same thing as two-x minus one squared is equal to the square root of eight minus x, eight minus x squared, and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel, \"Okay we did legitimate operations. \"We did the same thing to both sides. \"That these are equivalent equations.\" But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it, you're losing information. So, for example, this would be true even if the original equation were two-x... Even if the original equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to this right-hand side, and not the yellow one, then we would call that an extraneous solution. So, let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the quadratic formula, here. So the solutions are going to be... X is going to be equal to negative b, so three, plus or minus the square root of b-squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three"
    },
    {
        "Q": "\nAt 0:21, Sal said that when we've got a limit as x approaches infinity of a compilcated function, we can simplify the function to the most important terms, e.g.:\nlim x->\u00e2\u0088\u009e (5x\u00c2\u00b3+3x\u00c2\u00b2-8)/(x\u00c2\u00b3-60x\u00c2\u00b2+18)\u00e2\u0089\u00885x\u00c2\u00b3 /x\u00c2\u00b3 =5.\nMy question is, what's about x as an exponent? What are the most important terms in a function, where some numbers are in the x-th power, for example:\nlim x->\u00e2\u0088\u009e (5x\u00c2\u00b2 +18pi)/(e^x+5x)?",
        "A": "As x goes to infinity, an exponential with a base greater than 1 will eventually outgrow any term with a constant exponent, so the limit in your example is 0 (and it would be 0 even if the first term in the numerator changed to 1000x^1000 and the first term in the denominator changed to 1.001^x). At infinity, exponentials win (unless they re up against something even more powerful, like factorials).",
        "video_name": "KcqO1fX9b_I",
        "timestamps": [
            21
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nwhy at 2:26 did sal say 3x cubed/6x to the fourth is equal to 1/2x?",
        "A": "At 2:03. This is plain algebra; the 3 in the numerator and the 6 in the denominator simplify to 1/2. The x^3 in the numerator and the x^4 in the denominator simplifies to 1/x. Perhaps think of the original fraction as 3(x)(x)(x) / 6(x)(x)(x)(x). Can you see three of the x s cancel out leaving 3/6x which is 1/2x ?",
        "video_name": "KcqO1fX9b_I",
        "timestamps": [
            146
        ],
        "3min_transcript": ""
    },
    {
        "Q": "When we take limits at infinity we basically look if there is a horizontal asymptote. At 03:57 Sal talks about the case where the limit at infinity is equal to infinity. Wouldn't that look like a vertical rather than a horizontal asymptote?\n",
        "A": "If the limit at infinity is infinity, there won t really be an asymptote, it will just look like the line s constantly moving up and to the right.",
        "video_name": "KcqO1fX9b_I",
        "timestamps": [
            237
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhy can he take out the constants at 6:35? I'm guessing it won't change the answer, but I don't get how",
        "A": "the constant rule states that we can take the constant out because a derivative of a constant is 0",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            395
        ],
        "3min_transcript": "let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared, Times dh-- I've already used that pink-- times dh dt. Let's just be very clear. This orange term right over here-- and I'm just using the chain rule-- this is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect This will give us the derivative of h of t to the third power with respect to d with respect"
    },
    {
        "Q": "at 8:27 Saul says \"pi over 12\" but writes 'pi over 2' and carries that mistake through\n",
        "A": "yes, yes it is. but he doesn t carry the mistake through. as he fixes it at 10:28",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            507
        ],
        "3min_transcript": "As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared, Times dh-- I've already used that pink-- times dh dt. Let's just be very clear. This orange term right over here-- and I'm just using the chain rule-- this is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect This will give us the derivative of h of t to the third power with respect to d with respect to do when we apply this operator. How fast is this changing? How is this changing with respect to time? So we can just rewrite this, just so gets a little bit cleaner. Let me rewrite everything I've done. So we've got dV, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3 h of t squared, or I could just write that as 3h squared, times the rate at which the height is changing with respect to time, times dh dt. And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming-- we did express volume as a function of height-- but we're"
    },
    {
        "Q": "\nI don't understand how at 6:30 he separates the function to take the derivative and how you know to do so.",
        "A": "\u00cf\u0080/12 is a constant, so by definition d/dx [3x^2] is equal to 3*d/dx [x^2]. You might know that the derivative of 3x^2 is equal to 3*2x^2-1, 6x^1 or 6x. Likewise, 3 derivative of x^2 is 3*2x or 6x. Same result. So, d/dt [\u00cf\u0080/12*h^3] is the same of \u00cf\u0080/12*d/dx [h^3]. \u00cf\u0080 is not variable.",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            390
        ],
        "3min_transcript": "is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared,"
    },
    {
        "Q": "At 11:29 is the rate of change (Dh/dt) cm/s or cm^3/s\n",
        "A": "cm/s. The unit of the rate depends on the units of the letters on top and bottom. h(cm)/ t (sec) => cm/s",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            689
        ],
        "3min_transcript": "So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h, or the derivative of h of t, because we're assuming that h is a function of time. Now what does this thing right over here get us? Well we're telling us at the exact moment that we set up this problem, we know what dV dt is, we know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment. We were told it is 2 centimeters. So the only unknown we have over here is the rate at which our height is changing with respect to time. Which is exactly what we needed to figure out in the first place. So we just have to solve for that. So we get 1 cubic centimeter-- let me make it clear-- we get 1 cubic centimeter per second-- equal to pi over 2. And I'll write this in a neutral color. Actually, let me write in the same color. Is equal to pi over 2, times 3, times h squared. h is 2 so you're going to get 4 squared centimeters if we kept the units. So 3 times 4. All right let me be careful, that wasn't pi over 2, that was pi over 12. This is a pi over 12 right over here. So you get pi over 12, times 3 times 2 squared, times dh dt. All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to, well 3 times 4 is 12, cancels out with that 12. We get one is equal to pi times dh dt. To solve for dh dt divide both sides by pi. And we get our drum roll now. The rate at which our height is changing with respect to time as we're putting 1 cubic centimeter of water per second the rate at which this height is changing with respect to time is 1 over pi. And I haven't done the dimensional analysis but this is going to be in centimeters per second. You can work through the dimensional analysis if you like by putting in the dimensions right over here. But there you have it. That's how fast our height is going to be changing at exactly that moment."
    },
    {
        "Q": "\nAt 3:15, if Sal had known that the log10 5 = sqrt(0.5), he would've saved time, but it is tough to calculate the square root of 0.5, because square root of 0.5 is sqrt2 / 2, and the square root of 2 is hard to calculate. Am I right?",
        "A": "sqrt(.5)= 0.7071... log 5 = .698... so log 5 =\\= sqrt(.5)",
        "video_name": "OkFdDqW9xxM",
        "timestamps": [
            195
        ],
        "3min_transcript": "is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point-- So this is approximately equal to 2.861. And we can verify it because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power. And this number is closer to 3 than it is to 2. Well, let's verify it. So if I take 5 to that power, and then let me type in-- let me just type in what we did to the nearest thousandth-- 5 to the 2.861. So I'm not putting in all of the digits. What do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good. That this is the power that I have to raise 5 to to get to 100."
    },
    {
        "Q": "\nWhat are combinatorics, mentioned at 1:51?",
        "A": "Combinatorics, often represented by n!, is a field in mathematics dealing with how many ways there are to arrange a certain amount of objects. For example, if I had 4 objects, I could use combinatorics to figure out how many ways I could arrange them. Don t worry about them for now, you will learn about them in later grades.",
        "video_name": "RdehfQJ8i_0",
        "timestamps": [
            111
        ],
        "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season."
    },
    {
        "Q": "At 1:51 you said \"do all the COMBINATORICS\". But you never said what it is a combinatoric?\n",
        "A": "Combinatoric is just a fancy word for math that has to do with different combinations of something.",
        "video_name": "RdehfQJ8i_0",
        "timestamps": [
            111
        ],
        "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season."
    },
    {
        "Q": "So at 1:45 you had to simplify to get 2/3.\n",
        "A": "Yeah, you had to",
        "video_name": "RdehfQJ8i_0",
        "timestamps": [
            105
        ],
        "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season."
    },
    {
        "Q": "What is that symbol he draws at 1:13?\n",
        "A": "greater than or equal to",
        "video_name": "RdehfQJ8i_0",
        "timestamps": [
            73
        ],
        "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season."
    },
    {
        "Q": "7:15 - 7;23 you can't simplify 19/20?\n",
        "A": "no. there is no number that can go into both 19 and 20.",
        "video_name": "bcCLKACsYJ0",
        "timestamps": [
            435
        ],
        "3min_transcript": "it's gonna be six plus 11 over 12, which is equal to, six plus 11 is 17/12. If we wanted to write it as a mixed number, that is what, 12 goes into 17 one time with a remainder of five, so 1 5/12. Let's do one more of these. This is strangely fun. Alright. Let's say that we wanted to add, We're gonna add 3/4 to, we're gonna add 3/4 to 1/5. To one over five. What is this going to be? And once again, pause the video and see if you could work it out. Well we have different denominators here, and we wanna find, we wanna rewrite these so they have the same denominators, so we have to find a common multiple, ideally the least common multiple. So what's the least common multiple of four and five? Well let's start with the larger number, them until we get one that's divisible by four. So five is not divisible by four. 10 is not divisible by four, or perfectly divisible by four is what we care about. 15 is not perfectly divisible by four. 20 is divisible by four, in fact, that is five times four. That is 20. So what we could do is, we could write both of these fractions as having 20 in the denominator, or 20 as the denominator. So we could write 3/4 is something over 20. So to go from four to 20 in the denominator, we multiplied by five. So we also do that to the numerator. We multiply by three times five to get 15. All I did to go from four to 20, multiplied by five. So I have to do the same thing to the numerator, three times five is 15. 3/4 is the same thing as 15/20, and over here. 1/5. What is that over 20? Well to go from five to 20, you have to multiply by four. So we have to do the same thing to the numerator. I have to multiply this numerator times four to get 4/20. it's now written as 15/20 plus 4/20. And what is that going to be? Well that's going to be 15 plus four is 19/20. 19/20, and we're done."
    },
    {
        "Q": "At 7:25pm, How to rewrite a mixed fraction number into a decimal?\n",
        "A": "You turn the mixed number into an improper fraction and the turn it into a decimal. Hope this helps!",
        "video_name": "Gn2pdkvdbGQ",
        "timestamps": [
            445
        ],
        "3min_transcript": "93 goes into 170? Goes into it one time. 1 times 93 is 93. 170 minus 93 is 77. Bring down the 0. 93 goes into 770? Let's see. It will go into it, I think, roughly eight times. 8 times 3 is 24. 8 times 9 is 72. Plus 2 is 74. And then we subtract. 10 and 6. It's equal to 26. Then we bring down another 0. 93 goes into 26-- about two times. 2 times 3 is 6. 18. This is 74. We could keep figuring out the decimal points. You could do this indefinitely. But if you wanted to at least get an approximation, you would say 17 goes into 93 0.-- or 17/93 is equal to 0.182 and then the decimals will keep going. And you can keep doing it if you want. If you actually saw this on exam they'd probably tell you to stop at some point. You know, round it to the nearest hundredths or thousandths place. And just so you know, let's try to convert it the other way, from decimals to fractions. Actually, this is, I think, you'll find a much easier thing to do. If I were to ask you what 0.035 is as a fraction? Well, all you do is you say, well, 0.035, we could write it this way-- we could write that's the same thing as 03-- That's the same thing as 35/1,000. And you're probably saying, Sal, how did you know it's 35/1000? Well because we went to 3-- this is the 10's place. Tenths not 10's. This is hundreths. This is the thousandths place. So we went to 3 decimals of significance. So this is 35 thousandths. If the decimal was let's say, if it was 0.030. There's a couple of ways we could say this. Well, we could say, oh well we got to 3-- we went to the thousandths Place. So this is the same thing as 30/1,000. We could have also said, well, 0.030 is the same thing as 0.03 because this 0 really doesn't add any value."
    },
    {
        "Q": "at 3:00 minutes i got stuck\n",
        "A": "ok. you can complain it. that is not a problem. or you have no internet. that is why you are stuck.",
        "video_name": "Gn2pdkvdbGQ",
        "timestamps": [
            180
        ],
        "3min_transcript": "Let's do a slightly harder one. Let's figure out 1/3. Well, once again, we take the denominator, 3, and we divide it into the numerator. And I'm just going to add a bunch of trailing 0's here. 3 goes into-- well, 3 doesn't go into 1. 3 goes into 10 three times. 3 times 3 is 9. Let's subtract, get a 1, bring down the 0. 3 goes into 10 three times. Actually, this decimal point is right here. 3 times 3 is 9. Do you see a pattern here? We keep getting the same thing. As you see it's actually 0.3333. It goes on forever. And a way to actually represent this, obviously you can't write an infinite number of 3's. Is you could just write 0.-- well, you could write 0.33 repeating, which means that the 0.33 will go on forever. Although I tend to see this more often. Maybe I'm just mistaken. But in general, this line on top of the decimal means that this number pattern repeats indefinitely. So 1/3 is equal to 0.33333 and it goes on forever. Another way of writing that is 0.33 repeating. Let's do a couple of, maybe a little bit harder, but they all follow the same pattern. Let me pick some weird numbers. Let me actually do an improper fraction. Let me say 17/9. So here, it's interesting. The numerator is bigger than the denominator. So actually we're going to get a number larger than 1. But let's work it out. So we take 9 and we divide it into 17. And let's add some trailing 0's for the decimal point here. 1 times 9 is 9. 17 minus 9 is 8. Bring down a 0. 9 goes into 80-- well, we know that 9 times 9 is 81, so it has to go into it only eight times because it can't go into it nine times. 8 times 9 is 72. 80 minus 72 is 8. Bring down another 0. I think we see a pattern forming again. 9 goes into 80 eight times. 8 times 9 is 72. And clearly, I could keep doing this forever and we'd keep getting 8's. So we see 17 divided by 9 is equal to 1.88 where the 0.88 actually repeats forever. Or, if we actually wanted to round this we could say that that is also equal to 1.-- depending where we wanted"
    },
    {
        "Q": "My calc textbook just told me whenever I see this format of integral that I should just apply a formula. It gave the exact formula Sal achieved at 17:35, and just sub what ever x^2 value I get. IE when it is a a^2-x^2, just sub the x^2 into the formula which is exactly the same as what Sal got at 17:35. Given that this video took 20 mins and the formula takes 2 mins, I don't see any reason why I should disagree.\n",
        "A": "Exactly, I have formula for all the problems worked through this chapter. The videos make a nice proof of showing why the formula works but it seems I d never have to do it as Sal does.",
        "video_name": "sw2p2tUIFpc",
        "timestamps": [
            1055,
            1055
        ],
        "3min_transcript": "this times 2 over 2. You might say, Sal, why are you doing that? Because I can rewrite this, let me write my whole thing here. So I have 2 arc sine of x minus 3 over 2, and then I have, I could take this denominator 2 right here. So I say, plus x minus 3 over 2. That 2 is that 2 right there. And then I could write this 2 right here as a square root of 4. Times the square root of 4, times the square root of all of this. 1 minus x minus 3 squared over 4. I think you see where I'm going. I'm kind of reversing everything that I did at the beginning of this problem. And maybe I'm getting a little fixated on making this as simple as possible, but I'm so close, so let me finish. So I get 2 times the arc sine of x minus 3 over 2, which I'm tired of writing, plus x minus 3 over 2, and if we bring this is equal to the square root of 4 times these things. So it's 4 minus x minus 3 squared, all of that that plus c. And we're at the home stretch. This is equal to 2 times the arc sine of x minus 3 over 2 plus x minus 3 over 2 times the radical of 4 minus, let's expand this, x squared minus 6x plus 9. And then this expression right here simplifies to minus minus, it's 6x minus x squared, and then you have a 4 minus 9 minus 5. Which is our original antiderivative. So finally, we're at the very end. plus x minus 3 over 2 times, time the radical of 6x minus x squared minus 5. That right there is the antiderivative of what this thing that we had at the very top of my little chalkboard, which is right there. So that is equal to the antiderivative of the square root of 6x minus x squared minus 5 dx. And I can imagine that you're probably as tired as I am. My hand actually hurts. But hopefully you find that to be vaguely satisfying. Sometimes I get complaints that I only do easy problems. Well, this was quite a hairy and not-so-easy problem."
    },
    {
        "Q": "i need help on 3:19\n",
        "A": "what do you not get?",
        "video_name": "x6xtezhuCZ4",
        "timestamps": [
            199
        ],
        "3min_transcript": "So I'm going to draw 2/3, and I'm going to take 4/5 of it. So 2/3, and I'm going to make it pretty big. Just like this. So this is 1/3. And then this would be 2/3. Which I could do a little bit better job making those equal, or at least closer to looking equal. So there you go. I have thirds. Let me do it one more time. So here I have drawn thirds. 2/3 represents 2 of them. It represents 2 of them. One way to think about this is 2/3 times 4/5 is 4/5 of this 2/3. So how do we divide this 2/3 into fifths? Well, what if we divided each of these sections into 5. So let's do that. So let's divide each into 5. 1, 2, 3, 4, 5. And I could even divide this into 5 if I want. 1, 2, 3, 4, 5. And we want to take 4/5 of this section here. So how many fifths do we have here? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And we've got to be careful. These really aren't fifths. These are actually 15ths, because the whole is this thing over here. So I should really say how many 15ths do we have? And that's where we get this number from. But you see if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Where did that come from? I had 3, I had thirds. And then I took each of those thirds, and I split them into fifths. So then I have five times as many sections. 3 times 5 is 15. But now we want 4/5 of this right over here. This is 10/15 right over here. Notice it's the same thing as 2/3. Now if we want to take 4/5 of that, So we're going to take 8 of them. So 1, 2, 3, 4, 5, 6, 7, 8. We took 8 of the 15, so that is 8/15. You could have thought about it the other way around. You could have started with fifths. So let me draw it that way. So let me draw a whole. So this is a whole. Let me cut it into five equal pieces, or as close as I can draw five equal pieces. 1, 2, 3, 4, 5. 4/5, we're going to shade in 4 of them. 4 of the 5 equal pieces. 3, 4. And now we want to take 2/3 of that. Well, how can we do that?"
    },
    {
        "Q": "I can take the derivative of V to find dv:\n[1/\u00cf\u0080 sin \u00cf\u0080x] => (cos \u00cf\u0080x)(\u00cf\u0080) / \u00cf\u0080\nbut (at 11:25) how did Sal find 1/\u00cf\u0080 sin \u00cf\u0080x from cos \u00cf\u0080x to solve for the anti-derivative?\n",
        "A": "Yes, I understand now... Thank you for your answer! (cos \u00cf\u0080x) = (cos \u00cf\u0080x)(1) = (cos \u00cf\u0080x)(\u00cf\u0080 / \u00cf\u0080); D[sin \u00cf\u0080x] = \u00cf\u0080 cos \u00cf\u0080x; divide both sides by \u00cf\u0080 which yields: D[(sin \u00cf\u0080x) / \u00cf\u0080] = D[(1/\u00cf\u0080)(sin \u00cf\u0080x)] = (\u00cf\u0080 cos \u00cf\u0080x) / \u00cf\u0080 = cos \u00cf\u0080x",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            685
        ],
        "3min_transcript": "Well, we just figured out, from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate this integral right over here. So let's do that. So 1 minus x times cosine of pi x is the same thing as cosine of pi x minus x cosine of pi x. Now, this right here, well, let's just focus on taking the antiderivative. This is pretty easy. But let's try to do this one, because it seems a little bit more complicated. So let's take the antiderivative of x cosine of pi x dx. And what should jump in your mind is, well, this isn't that simple. But if I were able to take the derivative of x, that would simplify. It's very easy to take the antiderivative of cosine of pi x without making it more complicated. And remember, integration by parts tells us that the integral-- I'll write it up here-- the integral of udv is equal to uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos where I prove this and show examples of exactly what that means. But let's apply it right over here. And in general, we're going to take the derivative of whatever the u thing is. So we want u to be something that's simpler when I take the derivative. And then we're going to take the antiderivative of dv. So we want something that does not become more complicated when I take the antiderivative. So the thing that becomes simpler when I take this derivative is x. So if I set u is equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going to be the rest of this. dv is equal to cosine pi x dx. And so v would just be the antiderivative of this with respect to x. v is going to be equal to 1 over pi sine of pi x. If I took the derivative here, derivative of the inside, you get a pi, times 1 over pi, cancels out. Derivative of sine of pi x becomes cosine of pi x. So that's our u, that's our v. This is going to be equal to u times v. So it's equal to x, this x times this. So x over pi sine of pi x minus the integral of v, which is 1 over pi sine of pi x du."
    },
    {
        "Q": "I'm thoroughly confused. At 10:06, Sal gives a formula for integration by parts which he says he's gone through many times, but I just watched the integration by parts videos and didn't find it written it in this form? And I really thought I was getting the hang of these du, dv notations :(\n\nCan someone please explain how \u00e2\u0088\u00abf(x)g'(x)dx=\u00e2\u0088\u00abudv? I mean, I get it's some sort of substitution, but can't understand how f, g, and dx can be reduced to u and dv.\n\nAny help is greatly appreciated! :)\n",
        "A": "Nvm, I was able to derive it: (uv) =uv +u v d(uv)/dx=u*dv/dx+du/dx*v u*dv/dx=d(uv)/dx-du/dx*v \u00e2\u0088\u00ab(u*dv/dx)dx=\u00e2\u0088\u00abd(uv)/dx*dx-\u00e2\u0088\u00ab(du/dx*v)dx \u00e2\u0088\u00abu*dv=\u00e2\u0088\u00abd(uv)-\u00e2\u0088\u00abdu*v \u00e2\u0088\u00abudv=uv-\u00e2\u0088\u00abvdu",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            606
        ],
        "3min_transcript": "So times 20 times the integral from 0 to 1 of f of x cosine of pi x dx. I forgot to write the dx over there. I want to make sure you understand, because this is really the hard part of the problem, just realizing that the integral over this interval is just 1/20 of the whole thing, because over every interval, from 0 to 1, the integral is going to evaluate to the same thing as going from 1 to 2, which will be the same thing as going from 2 to 3, or going from negative 2 to negative 1. So instead of doing the whole interval from negative 10 to 10, we're just doing 20 times the interval from 0 to 1. From negative 10 to 10, there's a difference of 20 here. So we're multiplying by 20. And this simplifies it a good bit. First of all, this part over here simplifies to 20 divided by 10 is 2. So it's 2 pi squared. So it becomes 2 pi squared-- that's just this part over here-- times the integral from 0 to 1. Well, we just figured out, from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate this integral right over here. So let's do that. So 1 minus x times cosine of pi x is the same thing as cosine of pi x minus x cosine of pi x. Now, this right here, well, let's just focus on taking the antiderivative. This is pretty easy. But let's try to do this one, because it seems a little bit more complicated. So let's take the antiderivative of x cosine of pi x dx. And what should jump in your mind is, well, this isn't that simple. But if I were able to take the derivative of x, that would simplify. It's very easy to take the antiderivative of cosine of pi x without making it more complicated. And remember, integration by parts tells us that the integral-- I'll write it up here-- the integral of udv is equal to uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos where I prove this and show examples of exactly what that means. But let's apply it right over here. And in general, we're going to take the derivative of whatever the u thing is. So we want u to be something that's simpler when I take the derivative. And then we're going to take the antiderivative of dv. So we want something that does not become more complicated when I take the antiderivative. So the thing that becomes simpler when I take this derivative is x. So if I set u is equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going to be the rest of this."
    },
    {
        "Q": "1:10\nso there are a infinate number of raidus?\n",
        "A": "yes, a circle has infinite radii, as you can keep on changing the angle by a small amount",
        "video_name": "04N79tItPEA",
        "timestamps": [
            70
        ],
        "3min_transcript": "Draw a circle and label the radius, diameter, center and the circumference. Let me draw a circle, it won't be that well drawn of a circle but I think you get the idea so that is my circle. I'm going to label the center..over here. So I'll do the center I'll call it 'c'. So that is my center....and I'll draw an arrow there that is the center..of the circle and actually the circle itself is the set of all points that are a fixed distance away from that center and that fixed distance away, they're all from that center..that is the radius. So let me draw..the radius. So this distance right over here is the radius That is the radius and..that's going to be the same as this distance! ..which is the same as that distance! I can draw multiple radii. All of these are radii The distance between the center and any point on the circle now a diameter just goes straight across the circle and it's essentially two radii put toghether. So for example this would be a diameter.. that would be a diameter, you have one radii and another radii all on one line going from one side of the circle to another going through the center. So that is a diameter! ^_^........ that is a diameter. and I could have drawn it other ways, I could have drawn it like this that would be another diameter it would have the exact same length and finally, we have to think about the circumference, and the circumference is really just how far you have to go to go around the circle or if you put a string on that circle, how long would the string have to be? so what I'm tracing out in blue right now the length of what im tracing out is the circumference so....right over here..that is the circumference cir-cum-ference. Cir-cum-ference. And we're done!"
    },
    {
        "Q": "at 1:34 what dont undertsand bro\n",
        "A": "The circumference of a circle is basically the distance around a circle. For example, if you had a park or other outdoor area that was shaped in a perfect circle, and you walked all the way around the edge of it, you would have walked along the circumference of the circle. Basically, you can think of the circumference as the perimeter of a circle.",
        "video_name": "04N79tItPEA",
        "timestamps": [
            94
        ],
        "3min_transcript": "Draw a circle and label the radius, diameter, center and the circumference. Let me draw a circle, it won't be that well drawn of a circle but I think you get the idea so that is my circle. I'm going to label the center..over here. So I'll do the center I'll call it 'c'. So that is my center....and I'll draw an arrow there that is the center..of the circle and actually the circle itself is the set of all points that are a fixed distance away from that center and that fixed distance away, they're all from that center..that is the radius. So let me draw..the radius. So this distance right over here is the radius That is the radius and..that's going to be the same as this distance! ..which is the same as that distance! I can draw multiple radii. All of these are radii The distance between the center and any point on the circle now a diameter just goes straight across the circle and it's essentially two radii put toghether. So for example this would be a diameter.. that would be a diameter, you have one radii and another radii all on one line going from one side of the circle to another going through the center. So that is a diameter! ^_^........ that is a diameter. and I could have drawn it other ways, I could have drawn it like this that would be another diameter it would have the exact same length and finally, we have to think about the circumference, and the circumference is really just how far you have to go to go around the circle or if you put a string on that circle, how long would the string have to be? so what I'm tracing out in blue right now the length of what im tracing out is the circumference so....right over here..that is the circumference cir-cum-ference. Cir-cum-ference. And we're done!"
    },
    {
        "Q": "\nFrom 0:44 to 0:46, Sal says \"then we add 3/4 to -10/6.\" But why add? Should the distributive property apply and we multiply? Vote up if this is a question you have too, please.",
        "A": "The distributive property only applies if you re multiplying. Here, you re just adding the three fractions together. It looks confusing because Sal separated two of the fractions and added them together and then he added the third one to those two, but what you re doing to the fractions didn t change -- it s still just three fractions being added.",
        "video_name": "9tmtDBpqq9s",
        "timestamps": [
            44,
            46
        ],
        "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form."
    },
    {
        "Q": "\nat 2:27 it said negative 9/12 and 20/12 but it forgot to write the plus sign.why is thst",
        "A": "Because they are equivalent terms. The rule for subtraction is that you invert/change the sign of the second number and then add. So (-9/12) + (-20/12) is the same as (-9/12) - (20/12). He didn t need to add the plus sign at the end because it didn t change the value of the statement. Hope it helps.",
        "video_name": "9tmtDBpqq9s",
        "timestamps": [
            147
        ],
        "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form."
    },
    {
        "Q": "\nln(sinx)/lnx=sinx/x right?so can i omit the step 6:30-11:00?and concluded that limit of ln(sinx)/lnx=1?\nthanks in anvance",
        "A": "No, ln (sin x) / ln x DOES NOT equal sin x / x For example, when x = e: ln (sin e) / ln e = ln (sin e) \u00e2\u0089\u0088 -0.8897 but sin (e) / e \u00e2\u0089\u0088 0.15112",
        "video_name": "CDf_aE5yg3A",
        "timestamps": [
            390,
            660
        ],
        "3min_transcript": "what's the limit as X approaches zero from the positive direction of Y and once again we don't know. Maybe it's zero to the zero but we don't know what zero to zero actually is. What we could do, what we could do and this is a trick that you see a lot and anytime you get kind of weird things with exponents and whether you're doing limits or derivatives, as you'll see it's often times useful to take the natural log of both sides. Well what happens if you take the natural log of both sides here? On the left-hand side you're going to have the natural log, the natural log, and whenever I think of natural log and E the way I always think about them, the color green for some bizarre reason but we'll say the natural log of Y is equal to. If you take the natural log of this thing, actually let me just, I don't want to skip steps here because this is interesting. This is going to be the natural log of all of this business of sine of X, let me write this way. Sine of X, sine, I want to do this in that orange color. The natural log of over the natural log of X. Well we know from our exponent prior our logarithm properties, the logarithm of something to a power, that's the same thing as the power. One over natural log of X times the logarithm, this case the natural logarithm of whatever taking the sine of X here. Sine of X or we could say the natural log of Y. Want to keep, stay color consistent for at least one more step. The natural log of Y is equal to, if we just rewrite this this is going to be the natural log of sine of X, the natural log of sine of X over the natural log of X. Well this is all interesting but why do we care about this? Well instead of thinking about what is the limit? What is the limit as X approaches zero from the positive direction of Y? Let's think about what the natural log of Y is approaching as we approach, as X approaches zero from the positive direction. Let's figure out what the limit of this expression right over here is as X approaches zero from the positive direction. What is a natural log of Y? What is this whole thing? Not Y, what is the natural log of Y approaching? Let's think about that scenario. Let me write, do this in a new color. We want to figure out what is the limit as X approaches zero from the positive direction of this business and I'll just write it in one color. The natural log of sine of X over the natural log of X. I don't know, I wrote one time in print, one time in cursive. I'll just be consistent right over there. Now why is this interesting? Let's see in the numerator here, this thing's going to approach zero, natural log of zero you're going to approach This thing right over here natural log of,"
    },
    {
        "Q": "At around 10:30 in the video, would it be valid to, instead of splitting up the single limit into two and taking the first derivative of one, just take the second derivative of x*cos(x) / sin(x) to get\ncos(x) + (-x*sin(x)) / cos (x) ? When evaluating it, you still get 1 + 0 / 1, but is that just a coincidence?\n",
        "A": "If I followed what you meant correctly, yes, that is a valid way of doing it. You can continue using multiple iterations of l Hopital s provided the derivatives exist AND you continue to have 0/0 or \u00e2\u0088\u009e/\u00e2\u0088\u009e forms. If you have those forms, you do not necessarily have to split up the limit -- though doing so is often easier. Remember that l Hopital s is NOT valid or true if you don t have 0/0 or \u00e2\u0088\u009e/\u00e2\u0088\u009e . So, you must be careful about that.",
        "video_name": "CDf_aE5yg3A",
        "timestamps": [
            630
        ],
        "3min_transcript": "one over X. This is all going to be equal to, this is equal to the limit as X approaches zero from the positive direction of I could write this as cosine of X, cosine of X. Let's see, if I'm dividing by X. I'm dividing by X, I am going to get, this is going to be X over sine of X, X over sine of X. When I apply and when I try to take the limit here I'm going to get a zero, once again we got a zero over zero. This doesn't feel too satisfying but once again this is where our limit properties might be useful. As you can tell, this is not the most trivial of problems but this is going to be the same thing and this will take a little bit of pattern recognition. because we know that the limit of the product of two functions is equal to the product of their limits, this is the same thing as the limit, the limit as X approaches zero from the positive direction of, if we take this part, let me do this in a different color. If we take this part, that's not a different color. If we take this part right over here, so that's going to be X over sine of X and then times the limit. Let me put parenthesis here times the limit, the limit as X approaches zero from the positive direction of cosine of X, of cosine of X. Now this thing right over here is pretty straightforward. You can just evaluate it at zero, you get one. This thing right over here is equal to one This might ring a bell. You might have seen the limit as X approaches zero, of sine of X over X. This is just the reciprocal of that. This is X over sine of X but when you just superficially try to evaluate it. You get zero over zero so you can then apply L'Hopital's rule to this thing. Once again this is quite an interesting scenario we find ourselves in. This is going to be the same thing as the limit as X approaches zero from the positive direction. Derivative at the top is one, derivative at the bottom, is cosine of X. Well, this is just going to be one over cosine of zero is one, so this is just going to be equal to one. We got to apply L'Hopital's rule again and realize that this limit is going to be equal to one. One times one is one so this thing right over here is equal to one. This thing right over here, this thing right over here is going to be, this thing right over here is going to approach one"
    },
    {
        "Q": "well actually, at 4:40, there is evidence. both of the triangles that were produced by the parallelogram are the exact same size.\n",
        "A": "While the triangles from the parallelogram were congruent, the sides in question were not corresponding sides. DC is congruent to AB Thus, DC/BC equals the cosine, not the sine of \u00e2\u0088\u00a0CBA Thus, none of the options was correct.",
        "video_name": "TugWqiUjOU4",
        "timestamps": [
            280
        ],
        "3min_transcript": "might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see, It's the side opposite the 90 degree side. So this, it's BC. Sine is opposite over hypotenuse, so over BC. Is that what they wrote over here? No. They have DC over BC. Now what is DC equal to? Well, DC is this. And DC is not-- there's no evidence on this drawing right over here that DC is somehow equivalent to AC. So given this information right over here, we can't make this statement, either. So neither of these are true. So let's make sure we got this right. We can go back to our actual exercise, and we get-- oh, that's not the actual exercise. Let me minimize that. And we got it right."
    },
    {
        "Q": "\nAt 3:13, Sal says that tan(\u00e2\u0088\u00a0ABC) is unequal to AC/EF, right? But what if both the sides are in the same figure, like \u00e2\u0088\u00a0EFG? Does that make tan(\u00e2\u0088\u00a0ABC) = EG/EF?",
        "A": "tan(\u00e2\u0088\u00a0ABC) is equal to EG/EF , because triangles ABC and EFG are similar. However, the solution that Sal ruled out said that tan(\u00e2\u0088\u00a0ABC) = AC/EF, which is untrue because there is no way to tell if EG is the same length as AC, because even though the triangles are similar, the size of the triangles can still differ dramatically.",
        "video_name": "TugWqiUjOU4",
        "timestamps": [
            193
        ],
        "3min_transcript": "we're dealing with this right triangle right over here. That's the only right triangle that angle ADC is part of. And so what side is opposite angle ADC? Well, it's side CA, or I guess I say AC, side AC. So that is opposite. And what side is adjacent? Well, this side, CD. CD, or I guess I could call it DC, whatever I want to call it. DC, or CD, is adjacent. Now how did I know that this side is adjacent and not side DA? Because DA is the hypotenuse. They both, together, make up the two sides of this angle. But the adjacent side is one of the sides of the angle that is not the hypotenuse. AD or DA in the sohcahtoa context we would consider to be the hypotenuse. For this angle, this is opposite, this is adjacent, this is hypotenuse. Tangent of this angle is opposite over adjacent-- AC Now is that what they wrote here? No. They wrote AC over EF. Well, where's EF? EF is nowhere to be seen either in this triangle, or even in this figure. EF is this thing right over here. EF is this business right over here. That's EF. It's in a completely different triangle in a completely different figure. We don't even know what scale this is drawn at. There's no way the tangent of this angle is related to this somewhat arbitrary number that's over here. They haven't labelled it. This thing might be a million miles long for all we know. This thing really could be any number. So this isn't the case. We would have to relate it to something within this triangle, or something that's the same length. So if somehow we could prove that EF is the same length as DC, then we could go with that. But there's no way. This is a completely different figure, a completely different diagram. This is a similar triangle to this, but we don't know anything about the lengths. A similar triangle just lets us know that the angles are all might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see,"
    },
    {
        "Q": "at 4:13 why did he put a plus sign and not the minus sign?\n",
        "A": "Sal is doing: -2x^2 + 3x^2 -2 + 3 = +1, not -1. If you aren t sure, use the number line. Go left to -2 on the number line. Then, to do +3, move 3 units to the right. You end up on +1. That is why Sal has +x^2 . Hope this helps.",
        "video_name": "FNnmseBlvaY",
        "timestamps": [
            253
        ],
        "3min_transcript": "Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared squared, so I'm left with the 1x squared. So this right over here simplifies to 1x squared. Or I could literally just write x squared. 1x squared is the same thing as x squared. So plus x squared, and then these there's nothing really left to simplify. So plus 2x plus y squared. And obviously you might have gotten an answer in some other order, but the order in which I write these terms don't matter. It just matters that you were able to simplify it to these four terms."
    },
    {
        "Q": "At 2:56 why does he put -y instead of -1y?\n",
        "A": "-y and -1y are the same thing because -1 times y is -y",
        "video_name": "FNnmseBlvaY",
        "timestamps": [
            176
        ],
        "3min_transcript": "A y is different than a y squared, is different than an xy. Now with that said, let's see if there is anything that we can simplify. So first, let's think about the y terms. So you have a negative 3y there. Do we have any more y term? We have this 2y right over there. So I'll just write it out-- I'll just reorder it. So we have negative 3y plus 2y. Now, let's think about-- and I'm just going in an arbitrary order, but since our next term is an xy term-- let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down-- I'm just reordering the whole expression-- plus 4xy. And then I have minus 4xy right over here. Then let's go to the x squared terms. I have negative 2 times x squared, or minus 2x squared. So let's look at this. Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared"
    },
    {
        "Q": "At around 4:30, it says that by using *(1+1/n)^n* you get e. I understand that, and why, but why do you need the 1 in the beginning? Could you get e just by doing *(1/n)^n* ?\n",
        "A": "No, you could not get e without the 1 in the beginning. Since, as n goes to infinity, 1/n approaches 0 and n grows to infinity, (1/n)^n would approach 0^infinity which is definitely 0. Have a blessed, wonderful day!",
        "video_name": "oQhp3ndj28Y",
        "timestamps": [
            270
        ],
        "3min_transcript": "Now, let's go even higher. Let's take it ... let's do 1+1/ and actually I can now use scientific notation. Let's just say (1+1/1'10^7)^1'10^7, so what do we get here? So, now we went 2.718281692. Let's go even larger. Let's get our last entry here. Let's go, instead of the 7th power, let's go to the eighth power, so now we're (1+1/100,000,000^100,000,000) I don't even know if this calculator can handle this and we get 2.71828181487 and you see that we are quickly approaching, to a very large power, to the number e. The number e in our calculator. You see we've already gotten 1, 2, 3, 4, 5, 6, 7 digits to the right of the decimal point by taking it to the 100 millionth power. So, we are approaching this number. We are approaching, so one way to talk about it is we could say the limit, as n approaches infinity. As n becomes larger and larger, it's not becoming unbounded. It's not going to infinity. It seems to be approaching this number and we will call this number, we will call this magical and mystical number e. We'll call this number e and we see from our calculator that this number and these are kind of, these are almost as famous digits as the digits for Pi, we are getting 2.7182818 and it just Never, never repeating, so it's an infinite string of digits, never, never repeating. Just like Pi. Pi, you remember, is the ratio of the circumference to the diameter of the circle. e is another one of these crazy numbers that shows up in the universe. And in other videos on Khan Academy we go into depth, why this is so magical and mystical. Already this is kind of cool. That I can take an infinite ... If I just add 1 over a number to 1 and take it to that number and I make that number larger and larger and larger, it's approaching this number, but what's even crazier about it is we'll see that this number, which you can view, one way of it, it is coming out of this compound interest. That number, Pi, the imaginary unit which is defined as that imaginary unit squared is a negative 1, that they all fit together in this magical and mystical way and we'll see that again in future videos. But just for the sake of e, what you could imagine what's happening here"
    },
    {
        "Q": "At 11:40, I'm a little confused as to how Sal simplified the equation on the final step. I get that 1/9 becomes 4/9; but what happened to the exponents that were attached to the integers? It doesn't make sense with the exponent rules that are taught here. What am I missing?\n",
        "A": "Because 4 and 1/9 both have the same exponent (1st power) they can be multiplied together.",
        "video_name": "64bH_27Ehoc",
        "timestamps": [
            700
        ],
        "3min_transcript": "And then let me rewrite everything. Because I don't want to do too much on this one step right over here. So this part right over here gives us square root of 3s squared over 16. And then that's going to be times-- I'll open and close the parentheses. So then we have 4 plus. Then in blue, I'll write 3 times 4 to the first power. And then I can rewrite this as times 1/3. We could view this is 1/3 squared. We could view this as 1 over 3 to the first power, squared, or we could view this as 1 over 3 squared to the first power. And I'm going to write it that way. So times 1/9 to the first power. And then plus 3 times 4 squared. And then this we can write as, times 1/9 to the second power. And then this one we could write, plus 3 times 4 to the third times-- But we could also write this based on what we saw over here. We could write this as 1 over 3 squared to the third power, based on this right over here. Let me make this clear. 1 over 3 to the third, to the second power. This is the same thing as 1 over 3 squared to the third power. That's what we showed right over here. So this is the equivalent to 1/9 to the third power. And now we start to see the pattern is starting to clean up a little bit. And let me just do one more step. And then we'll finish this in the next video. So this is equal to square root of 3s squared over 16 times 4 plus 3 times, this is 4/9, plus the next term is three times 4/9 squared. And then we have plus 3 times 4 over 9 to the third power. and on, and on, and on taking the 3 times 4/9 to the successively larger, and larger powers. So this is what we have to find the sum of to find our area. And we're going to do that in the next video. And we're going to use some of the tools we've used to find the sums of infinite geometric series. But we're going to re-do it in this video just so that you don't have to remember that formula or that proof."
    },
    {
        "Q": "\nAt 2:17 in the video, it is shown how the vector is pointed down in the eastwards direction; then if we are given simple numbers in the form of a column vector, how would we know if the arrow should point westwards or otherwise?",
        "A": "If the first number in the column is positive, then it will point partly to the right. If the first number in the column is negative, then it will point partly to the left. If the second number in the column is positive, then it will point partly upward. If the second number in the column is negative, then it will point partly downward.",
        "video_name": "8QihetGj3pg",
        "timestamps": [
            137
        ],
        "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that."
    },
    {
        "Q": "\nAt about 1:30 Sal writes a bunch of stuff I don't understand. What is the meaning?I was also wondering why are there braces around the vectors x and y.",
        "A": "These are 2 vectors in column (vertical) form, and their sum.",
        "video_name": "8QihetGj3pg",
        "timestamps": [
            90
        ],
        "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that."
    },
    {
        "Q": "When Sal is drawing vectors through the video, something came to my mind.\nHow do we know exactly what direction the vectors are going in? Most vectors seem to be drawn going outward from the origin, but couldn't they go inward? Like at 2:09, when he's drawing the purple vector. Couldn't it also be pointing in the opposite direction, but still have the same magnitude?\n\nSorry if the answer is extremely obvious and I just don't see it...\n",
        "A": "The numbers in the video represent the vectors he draws. So when he draws a vector from the origin, it will point to the point (a,b). At 2:09, he s drawing the vector [6,-2], so it points from the origin to the right and down because it s positive (right) 6 and negative (down) 2.",
        "video_name": "8QihetGj3pg",
        "timestamps": [
            129
        ],
        "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that."
    },
    {
        "Q": "So at 3:36 and a little before that, when Sal says that f(-2) is EQUAL TO g(-1) is that in relation to the translation? Like point \"a\" on g(x) has different coordinates than f(x) but they are still equal because they're both have point \"a\" just one is shifted so that's why it appears different?\n",
        "A": "Yes, you are exactly right. This is a big concept in function translation; basically you are trying to find where one function is equal to another, and in this case he means that the function f at -2 is equivalent to function g at -1. Even more specifically, the y-value of x = -2 in f(x) is the same y-value for x= -1 in g(x) (and you can see him drawing horizontal lines across the graphs to illustrate this concept).",
        "video_name": "ENFNyNPYfZU",
        "timestamps": [
            216
        ],
        "3min_transcript": "I'll label it. f of x. And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of negative 3. This is negative 3. This is the point negative 3, f of 3. Now g hits that same value when x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. And we could do that with a bunch of points. We could see that g of 0, which is right there-- let me do it equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of-- well So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. And it's important to realize here. When I get f of x minus 2 here-- and remember the function is being evaluated, this is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counter-intuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2 we would have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see... It gets to about 2 there, then it gets pretty close to 1 right over there."
    },
    {
        "Q": "\nFrom the exponents of i it looks as if i is a negative number,but a negative number times a negative number equals a positive number.But then no number's square is a negative number.But now Sal says -i at 2:56 and that doesn't support that i is negative. Unless he's actually saying (mathematically )(he doesn't know)that -i is positive i.",
        "A": "i is not a negative number, but i^2 is a negative number.....i^2=-1 and i=sqrt. of -1, So -i = -\u00e2\u0088\u009a-1 and hence -i^2=+1",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            176
        ],
        "3min_transcript": "some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power."
    },
    {
        "Q": "\nThis is simple question: @3:40 how does i x i^3, or i x -i, = (-1)(i)(i) or (-1)(-1)?",
        "A": "Do you mean why does i \u00e2\u00a8\u0089 i\u00c2\u00b3 = (-1)(i)(i) = 1? We will use the exponent addition formula for this: i \u00e2\u00a8\u0089 i\u00c2\u00b3 = i\u00e2\u0081\u00b4 = 1. Or we can use this way: i \u00e2\u00a8\u0089 i\u00c2\u00b3 = (-1)(i)(i) = (-1)(-1) = 1 Always remember powers of i: i\u00c2\u00b9 = i, i\u00c2\u00b2 = -1, i\u00c2\u00b3 = -i, i\u00e2\u0081\u00b4 = 1, i\u00e2\u0081\u00b5 = i Hope this helps! \u00e2\u0080\u0094CT-2/002-24",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            220
        ],
        "3min_transcript": "well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power. \"i\" to the fifth power. Well that's just going to be \"i\" to to the fourth times \"i\". And we know what \"i\" to the fourth is. It is one. So its one times \"i\", or it is one times \"i\", or it is just \"i\" again. So once again it is exactly the same thing as \"i\" to the first power. Lets try again just to see the pattern keep going. Lets try \"i\" to the seventh power. Sorry, \"i\" to the sixth power. Well that's \"i\" times \"i\" to the fifth power, that's \"i\" times \"i\" to the fifth, \"i\" to the fifth we already established as just \"i\", so its \"i\" times \"i\", it is equal to, by definition,\"i\" times \"i\" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of \"i\" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of \"i\", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. \"i\" to the seventh power is equal to \"i\" times \"i\" to the sixth power."
    },
    {
        "Q": "At 0:22 Sal says E. What is that number and can i have a link to it?\n",
        "A": "e is a constant that comes up in math and science all the time. It is an irrational and transcendental number the first few digits of which are 2.71828... e is officially defined as: lim h\u00e2\u0086\u00920 (1+h)^(1/h) This same definition can also be expressed as: lim h\u00e2\u0086\u0092 \u00e2\u0088\u009e (1+1/h)^(h)",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            22
        ],
        "3min_transcript": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. \"i\" is defined as the number whose square is equal to negative 1. This is the definition of \"i\", and it leads to all sorts of interesting things. Now some places you will see \"i\" defined this way; \"i\" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power,"
    },
    {
        "Q": "\non 2:56 why did Sal keep it as -1*i instead of make it -i?{OK i know he corrected it at 3:27}",
        "A": "he did that to clarify the the minus is a separate number from i.",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            176,
            207
        ],
        "3min_transcript": "some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power."
    },
    {
        "Q": "at 5:20 Sal said tat i=square root of -1, then how is i=1!\n",
        "A": "i = -1 it doesn t equal 1. I don t know if that answered your question...",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            320
        ],
        "3min_transcript": "Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power. \"i\" to the fifth power. Well that's just going to be \"i\" to to the fourth times \"i\". And we know what \"i\" to the fourth is. It is one. So its one times \"i\", or it is one times \"i\", or it is just \"i\" again. So once again it is exactly the same thing as \"i\" to the first power. Lets try again just to see the pattern keep going. Lets try \"i\" to the seventh power. Sorry, \"i\" to the sixth power. Well that's \"i\" times \"i\" to the fifth power, that's \"i\" times \"i\" to the fifth, \"i\" to the fifth we already established as just \"i\", so its \"i\" times \"i\", it is equal to, by definition,\"i\" times \"i\" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of \"i\" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of \"i\", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. \"i\" to the seventh power is equal to \"i\" times \"i\" to the sixth power. And if you take \"i\" to the eighth, once again it'll be one, \"i\" to the ninth will be \"i\" again, and so on and so forth."
    },
    {
        "Q": "\nIs there any way to check the problem? Like at 2:32, how is he sure that the answer is 1436.8?",
        "A": "He is completely sure since that is the answer the calculator gave him. You can verify by doing the problem again, or if you have a graphing calculator, you can start from the volume and use algebraic manipulation to get back to the radius.",
        "video_name": "IelS2vg7JO8",
        "timestamps": [
            152
        ],
        "3min_transcript": "Find the volume of a sphere with a diameter of 14 centimeters. So if I have a sphere-- so this isn't just a circle, this is a sphere. You could view it as a globe of some kind. So I'm going to shade it a little bit so you can tell that it's three-dimensional. They're giving us the diameter. So if we go from one side of the sphere straight through the center of it. So we're imagining that we can see through the sphere. And we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere-- and we've proved this, or you will see a proof for this later when you learn calculus. But the formula for the volume of a sphere is volume is equal to 4/3 pi r cubed, where r is the radius of the sphere. So they've given us the diameter. And just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. And in fact, the sphere itself is the set the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4/3 pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you could approximate pi with 3.14. Some people even approximate it with 22/7. But we'll actually just get the calculator out to get the exact value for this volume. So this is going to be-- so my volume is going to be 4 divided by 3. And then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi, times 7 to the third power. before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed or cubic centimeters. So we get 1,436. They don't tell us what to round it to. So I'll just round it to the nearest 10th-- 1,436.8. So this is equal to 1,436.8 centimeters cubed. And we're done."
    },
    {
        "Q": "1:21 can you explain why the sphere is radius cubed instead of raidus squared\n",
        "A": "I think that it is because a sphere is 3 dimensional and a circle is 2 dimensional.",
        "video_name": "IelS2vg7JO8",
        "timestamps": [
            81
        ],
        "3min_transcript": "Find the volume of a sphere with a diameter of 14 centimeters. So if I have a sphere-- so this isn't just a circle, this is a sphere. You could view it as a globe of some kind. So I'm going to shade it a little bit so you can tell that it's three-dimensional. They're giving us the diameter. So if we go from one side of the sphere straight through the center of it. So we're imagining that we can see through the sphere. And we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere-- and we've proved this, or you will see a proof for this later when you learn calculus. But the formula for the volume of a sphere is volume is equal to 4/3 pi r cubed, where r is the radius of the sphere. So they've given us the diameter. And just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. And in fact, the sphere itself is the set the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4/3 pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you could approximate pi with 3.14. Some people even approximate it with 22/7. But we'll actually just get the calculator out to get the exact value for this volume. So this is going to be-- so my volume is going to be 4 divided by 3. And then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi, times 7 to the third power. before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed or cubic centimeters. So we get 1,436. They don't tell us what to round it to. So I'll just round it to the nearest 10th-- 1,436.8. So this is equal to 1,436.8 centimeters cubed. And we're done."
    },
    {
        "Q": "At 3:13, why Sal said \"clog\"?\n",
        "A": "It was an accident. He means c Log but pronounced it incorrectly.",
        "video_name": "yEAxG_D1HDw",
        "timestamps": [
            193
        ],
        "3min_transcript": "Fair enough. I think you realize I have not done anything profound just yet. But let's go back. We said that this is the same thing as this. So let's experiment with something. Let's raise this side to the power of C. So I'm going to raise this side to the power of C. That's a kind of caret. And when you type exponents that's what you would use, a caret. So I'm going to raise it to the power of C. So then, this side is x to the B to the C power, is equal to A to the C. All I did is I raised both sides of this equation to the Cth power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent, what happens to the exponents? Well, that's just exponent rule and you just multiply those two exponents. What can we do now? Well, I don't know. Let's take the logarithm of both sides. Or let's just write this-- let's not take the Let's write this as a logarithm expression. We know that x to the BC is equal to A to the C. Well, that's the exact same thing as saying that the logarithm base x of A to the C is equal to BC. Correct? Because all I did is I rewrote this as a logarithm expression. And I think now you realized that something interesting has happened. That BC, well, of course, it's the same thing as this BC. So this expression must be equal to this expression. That if I have some kind of a coefficient in front of the logarithm where I'm multiplying the logarithm, so if I have C log-- Clog base x of A, but that's C times the logarithm base x of A. That equals the log base x of A to the C. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property. So let's review what we know so far about logarithms. We know that if I write-- let me say-- well, let me just with the letters I've been using. C times logarithm base x of A is equal to logarithm"
    },
    {
        "Q": "\nStarting from 6:28, I don't understand what Mr. Khan means when he says that the \"Distance between x and c is less than delta,\" and that the \"distance between f(x) and L is going to be less than epsilon.\" Can someone explain please? Thank you!",
        "A": "You feed numbers called x into a function to get numbers called f(x) out. What he is saying is that as the number going in, i.e. x, gets closer and closer to c .. then the output f(x) is getting closer and closer to L. Now delta is just how far away from c a particular input value x is, and epsilon is how far away from L a particular output value f(x) is. The big idea about limits then, is that even if f(x) is not defined when x = c, you can see where it is going as x gets (arbitrarily) close to c.",
        "video_name": "w70af5Ou70M",
        "timestamps": [
            388
        ],
        "3min_transcript": "defined at C, so we think of ones that maybe aren't C, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range L plus epsilon or L minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon, then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here. But it's the exact same thing. Let me write it this way. Given an epsilon greater than 0-- so that's kind of the first part of the game-- we can find a delta greater than 0, such So what's another way of saying that x is within delta of C? Well, one way you could say, well, what's the distance between x and C is going to be less than delta. This statement is true for any x that's within delta of C. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between C minus delta and C plus delta, and these are the x's that satisfy that right over here, then-- and I'll do this in a new color-- then the distance between your f of x and your limit-- and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is, if the limit truly does exist, it could be super, super small one, we can find a delta. So we can define a range around C so that if we take any x value that is within delta of C, that's all this statement is saying that the distance between x and C is less than delta. So it's within delta of C. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit. The f of x, the difference between f of x, and your limit will be less than epsilon. Your f of x is going to sit some place over there. So that's all the epsilon-delta definition is telling us. In the next video, we will prove that a limit exists by using this definition of limits."
    },
    {
        "Q": "\n7:25 side side side? who came up with that name",
        "A": "The name refers to the method of finding congruence: it is quite logical. The SSS (or side, side, side) Congruence Postulate states that if all 3 sides of a triangle are congruent to that of another triangle, the two triangles are congruent.",
        "video_name": "CJrVOf_3dN0",
        "timestamps": [
            445
        ],
        "3min_transcript": "have the same measure, they're congruent We also know that these two corresponding angles I'll use a double arch to specify that this has the same measure as that So, we also know the measure of angle ABC is equal to the measure of angle XYZ And then finally we know that this angle, if we know that these two characters are congruent, then this angle is gonna have the same measure as this angle as a corresponding angle So, we know that the measure of angle ACB is gonna be equal to the measure of angle XZY Now what we're gonna concern ourselves a lot with is how do we prove congruence? 'Cause it's cool, 'cause if you can prove congruence of 2 triangles then all of the sudden you can make all of these assumptions we're gonna assume it for the sake of introductory geometry course This is an axiom or a postulate or just something you assume So, an axiom, very fancy word Postulate, also a very fancy word It really just means things we are gonna assume are true An axiom is sometimes, there's a little bit of distinction sometimes where someone would say \"an axiom is something that is self-evident\" or it seems like a universal truth that is definitely true and we just take it for granted You can't prove an axiom A postulate kinda has that same role but sometimes let's just assume this is true and see if we assume that it's true what can we derive from it, what we can prove if we assume its true But for the sake of introductory geometry class and really most in mathematics today, these two words are use interchangeably An axiom or a postulate, just very fancy words that things we take as a given Things that we'll just assume, we won't prove them, and then we're just gonna build up from there And one of the core ones that we'll see in geometry is the axiom or the postulate That if all of the sides are congruent, if the length of all the sides of the triangle are congruent, then we are dealing with congruent triangles So, sometimes called side, side, side postulate or axiom We're not gonna prove it here, we're just gonna take it as a given So this literally stands for side, side, side And what it tells is, if we have two triangles and So I say that's another triangle right over there And we know that corresponding sides are equal So, we know that this side right over here is equal into, like, that side right over there"
    },
    {
        "Q": "\nAt the beginning of the video this means that the definition of something being congruent would simply be that it has the same size and shape?\nAnd if if at 3:19, this means that if one side is congruent to another side of another triangle, then all 3 sides of both triangles are both congruent to each other?",
        "A": "Congruent means to be the same size and shape. However, if one side of a triangle is congruent to another side of a triangle, it would not mean all 3 sides are congruent. You would need another side or angle to be congruent to the other triangle to be sure of that.",
        "video_name": "CJrVOf_3dN0",
        "timestamps": [
            199
        ],
        "3min_transcript": "And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing that just means the measure of one line segment is equal to the measure of the other line segment And so we can go thru all the corresponding sides If these two characters are congruent, we also know that BC, we also know that the length BC is gonna be the length of YZ Assuming those are the corresponding sides And we can put these double hash marks right over here to show that these lengths are the same And when we go the third side, we also know that these are going to be has same length or the line segments are going to be congruent So, we also know that the length of AC is going to be equal to the length of XZ Not only do we know that all of the sides, the corresponding sides are gonna have the same length If someone tells that a triangle is congruent We also know that all the corresponding angles are going to have the same measure So, for example: we also know that this angle's measure"
    },
    {
        "Q": "\nAt 1:54, why does Sal put 3-1 over 4 instead of just doing the answer of 2 over four?",
        "A": "because he s showing you the steps",
        "video_name": "0njioQqIxKY",
        "timestamps": [
            114
        ],
        "3min_transcript": "Pedro is supposed to practice piano for 3/4 of an hour every day. Today, he has practiced for 1/4 of an hour. What fraction of an hour does he need to practice? So let's visualize 3/4. So he needs to practice 3/4 of an hour. So if this represents an entire hour, so let's divide it into fourths. So now, divide it into halves. Now, let's make it into four equal sections. Now, it's into fourths. So he needs to practice 3/4 of an hour. That was my attempt at drawing an arrow. So let me shade it in. So he needs to practice 3/4 of an hour. This gets us to 3/4 right over here. So he needs to practice 3 out of the fourths of an hour. So this is 3/4. Now, it says he has already practiced for 1/4 of an hour. So how much more does he need to practice? Well, he needs to practice this much more. And you might already see the answer visually. But let's think about how to represent it as a fraction expression. So let me write it like this. So this is how much he needs to go. How much does he need to practice? So he needs to do a total of 3/4. He's already done 1/4. So if you subtract 1/4 from the 3/4, you're going to get this amount right over here. You're going to get the amount that he needs to practice. Now, this already has the same denominator. So this is going to be equal to 3 minus 1 over 4, which is equal to 2/4. And you see that right over there. He still needs to practice 1 and 2 out of the fourths, 2/4. There's a couple of ways to think about it. You could say, hey, look, this is half of the length of this entire thing. It has a little bit on this end and that end. Or if you make 4 equal blocks, and if you were to shade in 2 of them, you see that you have shaded in 1/2 of the blocks. This is the exact same amount as if I just divide it into 2 sections and I shaded in only 1 of them. 2/4 is the same thing as 1/2. And if you wanted to work it out mathematically, you just have to do the same thing to the numerator and the denominator. So let's divide the numerator and the denominator by 2, because they are both divisible by 2. That's actually their greatest common factor. So 2 divided by 2 is 1. 4 divided by 2 is 2. So what fraction of an hour does he need to practice? He's got to practice 1/2 an hour."
    },
    {
        "Q": "at 6:43 coordinates of vertices are 1,23/4. bt we knw that vertex is the point where the conic section and the axis itersect. how is it possible then?\n",
        "A": "It is not necessary for the focus to always intersect the axes. The focus can be any point in 2D space. It can also obviously be at the origin.",
        "video_name": "w56Vuf9tHfA",
        "timestamps": [
            403
        ],
        "3min_transcript": "We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously, I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually, let me just do part of it, about the parabola just yet. I'm just gonna draw it like that. So we don't know just yet where the directrix and focus is, but we do know a few things. The focus is going to sit on the same, I guess you could say, the same x value as the vertex. So if we draw, this is x equals one, if x equals one, we know from our experience with focuses, foci, (laughs) I guess, that they're going to sit on the same axis as the vertex. So the focus might be right over here and then the directrix is going to be equidistant on the other side, equidistant on the other side. So the directrix might be something like this. Might be right over here. And once again, I haven't figured it out yet, but what we know is that because this point, the vertex, sits on the parabola, by definition has to be equidistant So. This distance has to be the same as this distance right over here and what's another way of thinking about this entire distance? Remember, this coordinate right over here is a, b and this is the line y is equal to k. This is y equals k. So what's this distance in yellow? What's this difference in y going to be? Well, you could call that, in this case, the directrix is above the focus, so you could say that this would be k minus b or you could say it's the absolute value of b minus k. This would actually always work. It'll always give you kind of the positive distance. So if we knew what the absolute value of b minus k is, if we knew this distance, then just split it in half with the directrix is gonna be that distance, half the distance above and then the focus is gonna be half the distance below."
    },
    {
        "Q": "At 17:30 isn't the 99% confidence interval 0.568 +/- 0.08 actually giving 56% to 57.6% ?\n",
        "A": "Because he added the 0 before the . it did seem that way. It got me too :P",
        "video_name": "SeQeYVJZ2gE",
        "timestamps": [
            1050
        ],
        "3min_transcript": "So it is 2.58 times our best estimate of the standard deviation of the sampling distribution, so times 0.031 is equal to 0.0-- well let's just round this up because it's so close to 0.08-- is within 0.08 of the population proportion. Or you could say that you're confident that the population proportion is within 0.08 of your sample mean. That's the exact same statement. So if we want our confidence interval, our actual number that we got for there, our actual sample mean we got was 0.568. I can delete this right here. Let me clear it. I can replace this, because we actually did take a sample. So I can replace this with 0.568. So we could be confident that there's a 99% chance that 0.568 is within 0.08 of the population proportion, which is the same thing as the population mean, which is the same thing as the mean of the sampling distribution of the sample mean, so forth and so on. And just to make it clear we can actually swap these two. It wouldn't change the meaning. If this is within 0.08 of that, then that is within 0.08 of this. So let me switch this up a little bit. So we could put a p is within of-- let me switch this up-- of 0.568. And now linguistically it sounds a little bit more like a confidence interval. We are confident that there's a 99% chance that p is within So what would be our confidence interval? It will be 0.568 plus or minus 0.08. And what would that be? If you add 0.08 to this right over here, at the upper end you're going to have 0.648. And at the lower end of our range, so this is the upper end, the lower end. If we subtract 8 from this we get 0.488. So we are 99% confident that the true population proportion is between these two numbers. Or another way, that the true percentage of teachers who think those computers are good ideas is between-- we're 99% confident-- we're confident that there's a 99% chance that the true percentage of teachers that like the computers is between 48.8% and 64.8%. Now we answered the first part of the question. The second part, how could the survey be changed to narrow"
    },
    {
        "Q": "at around 5:00 you said x 10^4 . where did 4 come from? i thought it was 10^5.\n",
        "A": "The term 0.3979 * 10^5 is not in scientific notation form. He is converting it to that. Scientific notation is a number with a single non-zero digit in front of the decimal times 10 raised to a power. So he is multiplying the 0.3979 by 10 which removes one of the 10s from the 10^5 turning it into 10^4",
        "video_name": "XJBwJjP2_hM",
        "timestamps": [
            300
        ],
        "3min_transcript": ""
    },
    {
        "Q": "1:54 Could we just have calculated both volumes and subtracted the answers together to find the volume of the gold ring\n",
        "A": "Well,if you mean taking the volume of the rectangular glass with the ring and subtract the volume of the rectangular glass without the ring, well you are right. Bang on!",
        "video_name": "ViFLPsLTO1k",
        "timestamps": [
            114
        ],
        "3min_transcript": "Jamie wants to know the volume of his gold ring in cubic inches. He gets a rectangular glass with base 3 inches by 2 inches. So you see that here, the base is 3 inches by 2 inches. And he fills the glass 4 inches high with water. So you see that over here, 4 inches high with water. Jamie drops his gold ring in the glass and measures the new height of the water to be 4.25 inches. So this is after the gold ring is dropped. What is the volume of Jamie's ring in cubic inches? Well when you start with this water right over here and you add his ring, whatever that volume is of his ring is going to displace an equal volume of water and push it up. And so the incremental volume that you now have is essentially going to be the volume of his ring. Well what is the incremental volume here? Well it's going to be the volume. If you think about going from this before volume to the after volume, the difference is the base stays the same. It's 3 inches by 2 inches, the difference is-- to make it a little bit neater-- the base is the same. The height now is 4.25 inches after dropping in the ring So the water went up by 0.25 inches. Let me write that, 0.25 inches is what the water went up by. So we could just think about, what is this incremental volume going to be? So this incremental volume right over here, that I'm shading in with purple. Well to figure that out we just have to measure. We just have to multiply the length times the width times the height times 0.25. So it's just going to be 3 times 2 times 0.25. 3 times 2 is 6, times 0.25, and you could do that either on paper or you might be able do that in your head. 4 times 0.25 is going to be 1, and you have 2 more So this is going to be 1.50. And we multiply it inches times inches times inches. So this is going to be in terms of cubic inches. 1.5 cubic inches is the volume of Jamie's ring, which is actually a pretty sizable volume for a gold ring. Maybe he has a very big finger or he just likes to spend, or I guess is his, whoever bought him the ring likes to spend a lot on gold."
    },
    {
        "Q": "\nAt 2:43 he got the sum, but would he have to simplify it? or would that be the absolute answer?",
        "A": "OMGGGOMGOMG i knew how do to normal division but when he got the reciprocal out of no where that was confusing!",
        "video_name": "Mcm0Q3wGhMo",
        "timestamps": [
            163
        ],
        "3min_transcript": "My wife and I have recently purchased an assortment pack of soap, and we both want to experience all 3 bars. And we are not willing to share soap with each other. So we have a little bit of a conundrum. How do we share these 3 bars of soap so that we each get to experience all of the smells? So this might be a nice rose smell. This might be some type of ivory smell. I don't even know if that's a legitimate smell. This might be some type of sandalwood, which is always very nice. And my wife has an idea. She says, look I'm going to take these 3 bars, so we're starting with 3 bars of soap. And I'm going to divide it into 2 equal groups. And I said, how are you going to do that? And she says, well, I'm just going to take out some type of carving saw or carving knife, or who knows what it is, and I'm just going to cut it right down the middle, so right down this. This is my wife cutting the soap right over here with her saw. So she's cutting the soap right over here. So she has divided the soap into 2 equal groups. So the interesting question here is, how many bars of soap do we each have now? And I encourage you to pause the video and think about that for a second. How many bars of soap do we each have now? Well, let's just visualize my share. Let's say I take this bottom half right over here. So this is Sal's share of the soap. Let me write this out. So this is Sal's share. My wife took this top half. Well, what do I have? Well, I have 1/2, I have another 1/2, and I have another 1/2 of a bar. So I have 3/2 bars of soap. Or you could say that I have 3 times 1/2 bars of soap. Notice, something very interesting happened here. 3 divided by 2 is equal to 3 times 1/2. And we could make it even more interesting, So 3 divided by 2/1 is equal to 3 times 1/2. Notice we went from a division to a multiplication, and we took the reciprocal. And that makes complete sense. We have 3/2 bars now. But what's 3 times 1/2? Well, 3 times 1/2 is equal to 3/2. So just doing this little simple, smelly soap example, we've got a very interesting result. 3 divided by 2 is the same thing as 3 times 1/2, which is the same thing as 3/2."
    },
    {
        "Q": "\nWhat would happen if you did the mobius strip - flake that she showed at 3:56? Would you get the pattern, then the pattern reflected, then the pattern again, and then the pattern reflected and so on?",
        "A": "Yes, you would. That is called a Frieze pattern.",
        "video_name": "8EmhGOQ-DNQ",
        "timestamps": [
            236
        ],
        "3min_transcript": "Stars count as holiday spirit, right? And you can do seven-fold symmetry in a similar way but you're probably going to need your emergency protractor. But you could do nine-fold without a protractor because you can do thirds, and then thirds again. And if you can do fifths without a protractor, you can do tenths too, because it's a fifth times a half. Look, I said happy holidays but I never said which one. Valentine's Day is totally a winter holiday. 11 is prime, though, so time for the protractor again. Look, prime factorization. So now theoretically you can get all sorts of end-fold symmetry, but what about rotational symmetry? There's no mirror lines, which means no folding, so does it even make sense as a question? Cutting a snowflake design efficiently is all about putting the same cut lines on top of each other so you only have to cut them once. So how do you take a rotationally symmetric design like this and put all the layers on top of each other without overlapping anything else? Maybe it's not surprising to see that to get stuff with rotational symmetry to line up, you rotate it. If you make a cut to the center then you can rotate all the way and roll the symmetry up into one unique thing. It's hard to draw accurate rotational symmetry by hand. So to cut out a paper swirl flake, start with a cut, then curl your paper into a cone. You can swirl around once or twice or more. But the important thing is to make sure the cut lines up with itself, because as far as symmetry is concerned, that cut doesn't exist. I like to tape it in place so it doesn't unroll, then cut stuff out. I find that spiraly things work well. Folding the paper is a good way to start a cut, but remember that folding creates symmetry. So I like to use it just to get the scissors in there and then do something asymmetric. Voila, snowflake. For a starflake swirlflake you'll have to curl your paper around five times, or four times. It's funny because I think of this as going around once but really it's going around twice, and a flat sheet of paper goes around once. Anyway, yeah, do that. And then give it a nice spiraly arm or two. You can make a nice fancy starflake swirlflake snowflake, awesome flake. Of course, from snowflakes it's only one small step to folding and cutting freeze patterns, and then wallpaper patterns and, get if you start by folding stuff into a [INAUDIBLE] strip? And then maybe you'll want to start folding and cutting spheres and everything will be a mess,"
    },
    {
        "Q": "At 5:23 , why does 2y' equal -6e^-3x and not positive 6e^-3x. since you multiply the function 2 times itself and negative times negative becomes positive?\n",
        "A": "You actually aren t multiplying the function by itself. The notation is kind of confusing, but 2 is just the coefficient for the first derivative. So 2y = 2 (-3e^-3x) is just -6e^-3x. Because the coefficient (2) is positive, the sign stays the same. I hope this helps.",
        "video_name": "6o7b9yyhH7k",
        "timestamps": [
            323
        ],
        "3min_transcript": "Now let's make that a little more tangible. What would a solution to something like any of these three, which really represent the same thing, what would a solution actually look like? Actually let me move this over a little bit. Move this over a little bit. So we can take a look at what some of these solutions could look like. Let me erase this a little. This little stuff that I have right over here. So I'm just gonna give you examples of solutions here. We'll verify that these indeed are solutions for I guess this is really just one differential equation represented in different ways. But you'll hopefully appreciate what a solution to a differential equation looks like. And that there is often more than one solution. There's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one. So one solution, I'll call it y one. And I could even write it as y one of x to make it explicit that it is a function of x. to the negative three x. And I encourage you to pause this video right now and find the first derivative of y one, and the second derivative of y one, and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it. So let's work through this together. So that's y one. So the first derivative of y one, so we just have to do the chain rule here, the derivative of negative three x with respect to x is just negative three. And the derivative of e to the negative three x with respect to negative three x is just e to the negative three x. And if we take the second derivative of y one, this is equal to the same exact idea, the derivative of this is three times negative three is going to be nine e to the negative three x. And now we could just substitute these values into the differential equation, or these expressions into the differential equation to verify that this So let's verify that. So we first have the second derivative of y. So that's that term right over there. So we have nine e to the negative three x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative three x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative three x. Three e to the negative three x."
    },
    {
        "Q": "\nso\ni knew that (\u00e2\u0088\u009a2)^2 =2 and (-\u00e2\u0088\u009a2)^2=2 why did he chose to (\u00e2\u0088\u009a2)^2 =2\nat 0:48",
        "A": "\u00e2\u0088\u009a2 is less complicated than -\u00e2\u0088\u009a2 Why complicate things!",
        "video_name": "C3QPTCwpIZo",
        "timestamps": [
            48
        ],
        "3min_transcript": "Voiceover:Let's see if we can factor 36a to the eighth, plus 2b to the sixth power and I encourage you to pause the video and try it out on your own. So let's see if we can express this, or re-express this as the difference of squares using imaginary numbers. So we can rewrite 36 as, six squared, and a to the 8th is the same thing as, a to the fourth squared, and so let me actually just rewrite it this way. We can rewrite it as 6a to the fourth squared. That's this first term right over here. And then the second term we can write it as, actually just let me write it this way first. Let me just write it as a square. Plus, the square root of 2, b to the third power, squared. Now we wanted to write it as a difference of squares. So instead of writing it this way, let's get rid of this plus, and let's - so let me clear that out, and I could write it as and negative one, we know, is the same thing as i squared, so we can rewrite this whole thing as 6a to the fourth, squared. And then we have this minus right over here, minus. And so this is i squared. Negative one is i squared. So we can rewrite this in this pink color as i times the square root of two, times b to the third, all of that squared. Notice i squared is negative one. Square root of two squared, is two. B to the third, squared, is b to the sixth power. If I raise something to an exponent, and then raise it to another exponent, I would multiply the two exponents. And so now I've expressed it as a difference of squares, so we're ready to factor. This is going to be equal to 6a to the fourth, minus i times the square root of two, and let me get myself space here, times 6a to the fourth, plus all of this business, i times the square root of 2, times b to the third power, and we are done."
    },
    {
        "Q": "\nAt about 5:47, Sal starts talking about how you can't ever add enough 9s onto 4 to get the exact x value of the maximum. But, if you keep putting on more and more 9s to infinity, you get 4.999... on forever. If I'm not mistaken, 0.999... = 1, so wouldn't the maximum be still at x = (4+0.999...) = (4+1) = 5?",
        "A": "There is no such number as infinity, so you cannot actually add an infinite number of 9s. So, this is a limit, not an actual finite sum.",
        "video_name": "bZYTDst1MOo",
        "timestamps": [
            347
        ],
        "3min_transcript": "Here our maximum point happens right when we hit b. And our minimum point happens at a. For a flat function we could put any point as a maximum or the minimum point. And we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous, and why this needs to be a closed interval. So first let's think about why does f need to be continuous? Well I can easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult or you can't really pick out an absolute minimum or an absolute maximum value over that interval. So let's say that this right over here is my interval. Let's say that's a, that's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here you could say, well look, the function is clearly approaching, as x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maxima because the function never gets to that. So you could say, well let's a little closer here. Maybe this number right over here is 5. So you could say, maybe the maximum is 4.9. Then you could get your x even closer to this value and make your y be 4.99, or 4.999. You could keep adding another 9. So there is no maximum value. Let me draw it a little bit so it looks more like a minimum. There is-- you can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1, or 1.01, or 1.0001. And so you could keep drawing some 0s between the two 1s but there's no absolute minimum value there. Now let's think about why it being a closed interval matters. Why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well let's imagine that it was an open interval. Let's imagine open interval. And sometimes, if we want to be particular, we could make this is the closed interval right of here And if we wanted to do an open interval right over here, that's a and that's b. And let's just pick very simple function,"
    },
    {
        "Q": "at 4:10-4:12 minimum happens at A? I'm so confused, I thought minimum was F(c). What are the white lines for?\n",
        "A": "The white lines are a completely different function superimposed on the illustration, it is not related at all the original function other than they both share the same domain interval.",
        "video_name": "bZYTDst1MOo",
        "timestamps": [
            250,
            252
        ],
        "3min_transcript": "like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well it seems like we hit it right over here, when x is, let's say this is x is c. And this is f of c right over there. And it looks like we had our absolute maximum point over the interval right over there when x is, let's say this is x is equal to d. And this right over here is f of d. So another way to say this statement right over here if f is continuous over the interval, we could say there exists a c and d that are in the interval. are in the interval such that-- and I'm just using the logical notation here. Such that f c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case you're saying, look, we hit our minimum value when x is equal to c. That's that right over here. Our maximum value when x is equal to d. And for all the other Xs in the interval we are between those two values. Now one thing, we could draw other continuous functions. And once again I'm not doing a proof of the extreme value theorem. But just to make you familiar with it and why it's stated the way it is. And you could draw a bunch of functions here Here our maximum point happens right when we hit b. And our minimum point happens at a. For a flat function we could put any point as a maximum or the minimum point. And we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous, and why this needs to be a closed interval. So first let's think about why does f need to be continuous? Well I can easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult or you can't really pick out an absolute minimum or an absolute maximum value over that interval."
    },
    {
        "Q": "3:42 is so confusing\n",
        "A": "Not really. All that means is this: (6x6x6) TIMES (6x6x6x6x6x6). You have to do the parentheses first and then multiply.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            222
        ],
        "3min_transcript": "multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6. green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the sixth power. Well, what's 6 to the sixth power? That's 6 times itself six times. So, it's 6 times 6 times 6 times 6 times 6. Then you get one more, times 6. So what is this whole number going to be? Well, this whole thing-- we're multiplying 6 times itself-- how many times? One, two, three, four, five, six, seven, eight, nine times, right? Three times here and then another six times here. So we're multiplying 6 times itself nine times. 3 plus 6. So this is equal to 6 to the 3 plus 6 power or 6 to the ninth power. And just like that, we/ve stumbled on another exponent property."
    },
    {
        "Q": "At 3:22 which is the simplest form? (3x)^3 , 3^3 *X^3, or 27x^3\n",
        "A": "27x^3 would be simplified.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            202
        ],
        "3min_transcript": "5 times 5 times 5 times 5 times 5 times 5 times 5. One, two, three, four, five, six, seven. This is going to be a really, really, really, really, large number and I'm not going to calculate it right now. If you want to do it by hand, feel free to do so. Or use a calculator, but this is a really, really, really, large number. So one thing that you might appreciate very quickly is that exponents increase very rapidly. 5 to the 17th would be even a way, way more massive number. But anyway, that's a review of exponents. Let's get a little bit steeped in algebra, using exponents. So what would 3x-- let me do this in a different color-- what would 3x times 3x times 3x be? Well, one thing you need to remember about multiplication multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6."
    },
    {
        "Q": "10:18 what happend to the -1? (-1)^2 = just 1? how?\n",
        "A": "A negative times a negative...",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            618
        ],
        "3min_transcript": "So just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the fourth is equal to x to the sixth, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing is x to the a power times y to the a power. We saw that early on in this video. We saw that over here. 3x to the third is the same thing as 3 to the third times x to the third. That's what this is saying right here. 3x to the third is the same thing is 3 to the third times x to the third. is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y"
    },
    {
        "Q": "\nShouldn't you put a paratheses for the x's on 7:04 of the video.",
        "A": "You could, but it s not necessary. Remember our rule for associativity says if we are just multiplying it doesn t matter what order we do it in. This also means that the parentheses aren t necessary.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            424
        ],
        "3min_transcript": "number 6 is the base. We're taking the base to the exponent of 3. When you have the same base, and you're multiplying two exponents with the same base, you can add the exponents. Let me do several more examples of this. Let's do it in magenta. Let's say I had 2 squared times 2 to the fourth times 2 to the sixth. Well, I have the same base in all of these, so I can add the exponents. This is going to be equal to 2 to the 2 plus 4 plus 6, which is equal to 2 to the 12th power. And hopefully that makes sense, because this is going to be 2 times itself two times, 2 times itself four times, 2 times itself six times. When you multiply them all out, it's going to be 2 times itself, 12 times or 2 to the 12th power. Let's do it in a little bit more abstract way, using some What is x to the squared or x squared times x to the fourth? Well, we could use the property we just learned. We have the exact same base, x. So it's going to be x to the 2 plus 4 power. It's going to be x to the sixth power. And if you don't believe me, what is x squared? x squared is equal to x times x. And if you were going to multiply that times x to the fourth, you're multiplying it by x times itself four times. x times x times x times x. So how many times are you now multiplying x by itself? Well, one, two, three, four, five, six times. x to the sixth power. Let's do another one of these. The more examples you see, I figure, the better. mix and match it. Let's say I have a to the third to the fourth power. So I'll tell you the property here, and I'll show you why it makes sense. When you add something to an exponent, and then you raise that to an exponent, you can multiply the exponents. So this is going to be a to the 3 times 4 power or a to the 12th power. And why does that make sense? Well this right here is a to the third times itself four times. So this is equal to a to the third times a to the third times a to the third times a to the third. Well, we have the same base, so we can add the exponents. So there's going to be a to the 3 times 4, right? This is equal to a to the 3 plus 33 plus three plus 3 power, which is the same thing is a the 3 times 4 power or a"
    },
    {
        "Q": "@10:55, if it doesn't matter the order when you multiple, then what is FOIL?\n\nhe didn't need it for this video, i'm confused on when to use it.\n",
        "A": "FOIL is just the standard method of multiplying two binomials (e.g. (a+b)(a-b) ) which you use to make sure you don t miss one of the multiplications necessary (because you need aa, ab, ba and bb) - technically you don t have to do it in that order (because as Sal says the order doesn t matter) but it just helps you make sure you get each of the components.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            655
        ],
        "3min_transcript": "is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2."
    },
    {
        "Q": "\nAt 3:40, if Sal did 6^6*6^6 would that equal (6^6)^6",
        "A": "6^6*6^6 would equal (6^6)^2 (6^6) gets multiplied two times with it self. (6^6)^6 would equal 6^6 * 6^6 * 6^6 * 6^6 * 6^6 * 6^6",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            220
        ],
        "3min_transcript": "multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6. green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the sixth power. Well, what's 6 to the sixth power? That's 6 times itself six times. So, it's 6 times 6 times 6 times 6 times 6. Then you get one more, times 6. So what is this whole number going to be? Well, this whole thing-- we're multiplying 6 times itself-- how many times? One, two, three, four, five, six, seven, eight, nine times, right? Three times here and then another six times here. So we're multiplying 6 times itself nine times. 3 plus 6. So this is equal to 6 to the 3 plus 6 power or 6 to the ninth power. And just like that, we/ve stumbled on another exponent property."
    },
    {
        "Q": "At 9:02 when Sal tells us that the value of (Av) in c(Av) is zero, does that come from 2:57 when he was writing out the set notation for subspace N?\n",
        "A": "Yes. You are assuming that the v is in your subspace and you know that Av=0 for all v in your subspace.",
        "video_name": "jCwRV1QL_Xs",
        "timestamps": [
            542,
            177
        ],
        "3min_transcript": "means they both satisfy this equation, then v1 plus v2 is definitely still a member of n. Because when I multiply a times that, I get the 0 vector again. So let me write that result, as well. So we now know that v1 plus v2 is also a member of n. And the last thing we have to show is that it's closed under multiplication. Let's say that v1 is a member of our space that I defined here, where they satisfy this equation. What about c times v1? Is that a member of n? Well let's think about it. What's our matrix a times the vector-- right? I'm just multiplying this times the scale. I'm just going to get another vector. Lowercase v, so it's a vector. What's this equal to? Well, once again, I haven't prove it to you yet, but it's actually a very straightforward thing to do, to show that when you're dealing with scalars, if you have a scalar here, it doesn't matter if you multiply the scalar times the vector before multiplying it times the matrix or multiplying the matrix times the vector, and then doing the scalar. So it's fairly straightforward to prove that this is equal to c times our matrix a-- I'll make that nice and bold, times our vector v. That these two things are equivalent. Maybe I should just churn out the video that does this, but I'll leave it to you. You, literally, just go through the mechanics component by component. And you show this. But clearly, if there's is true, we already know that v1 is a member of our set, which means that a times v1 is equal to the 0 vector. vector, which is still the 0 vector. So c[v,1] is definitely a member of n. So it's closed under multiplication. And I kind of assumed this right here. But maybe I'll prove that in a different video. But I want to do all this to show that this set n is a valid subspace. This is a valid subspace. It contains a 0 vector. It's close under addition. It's close under multiplication. And we actually have a special name for this. We call this right here, we call n, the null space of a. Or we could write n is equal to-- maybe I shouldn't have written an n. Let me write orange in there. Our orange n is equal to-- the notation is just the null space of a. Or we could write the null space is equal to the orange"
    },
    {
        "Q": "At 1:57, a homogeneous equation is mentioned. What exactly is a homogeneous equation and why is A times vector X =0 a homogeneous equation?\n",
        "A": "Being more precise, A x = 0 is a homogeneous system of equations (a set of homogeneous equations). A linear equation is homogeneous when the constant part is zero. For example, ax + by = c is homogeneous only if c is zero. Ax=b is a way to write a system of equations using matrices.",
        "video_name": "jCwRV1QL_Xs",
        "timestamps": [
            117
        ],
        "3min_transcript": "Let's review our notions of subspaces again. And then let's see if we can define some interesting subspaces dealing with matrices and vectors. So a subspace-- let's say that I have some subspace-- oh, let me just call it some subspace s. This is a subspace if the following are true-- and this is all a review-- that the 0 vector-- I'll just do it like that-- the 0 vector, is a member of s. So it contains the 0 vector. Then if v1 and v2 are both members of my subspace, then v1 plus v2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. You can add any of their two members and you'll get another member of the subspace. And then the last requirement, if you remember, is that subspaces are closed under multiplication. So that if c is real number, and it's just a scalar. And if I multiply, and v1 is a member of my subspace, then if subspace, v1, I'm going to get another member of the subspace. So it's closed under multiplication. These were all of what a subspace is. This is our definition of a subspace. If you call something a subspace, these need to be true. Now let's see if we can do something interesting with what we understand about matrix vector multiplication. Let's say I have the matrix a-- I'll make it nice and bold-- and it's an m by n matrix. And I'm interested in the following situation; I want to set up the homogeneous equation. And we'll talk about why it's homogeneous. Well, I'll tell you in a second. So let's say we set up the equation. My matrix a times vector x is equal to the 0 vector. This is a homogeneous equation, And I want to ask the question-- I talked about subspaces. If I take all of the x's-- if I take the world, the universe, the set of all of the x's that satisfy this equation, do I have a valid subspace? Let's think about this. I want to take all of the x's that are a member of Rn. Remember, if our matrix a has n columns, then I've only defined this matrix vector multiplication. If x is a member of r, and if x has to have exactly n components, only then is it defined. So let me define a set of all the vectors that are a member of Rn where they satisfy the equation a times my vector x is equal to the 0 vector. So my question is, is this a subspace? Is this a valid subspace?"
    },
    {
        "Q": "\nAt 3:34, Sal uses the point (1,2) to find r, but can you use any of the other points? How would you find r by using the next point, (2, 4/3)?",
        "A": "You could use any one of the given points. For (2, 4/3) you would substitute the y-value (so 4/3) into g(x) and the x-value ( so 2) into the x to end up with 4/3=3r^2. You then solve that equation to end up with the r. Just be careful because if the x-value is even, you could have two r s so it is more accurate to use a point with an odd x-value when it is available to eliminate any confusion.",
        "video_name": "Qst1UVtq8pE",
        "timestamps": [
            214
        ],
        "3min_transcript": "And you see that. When x increases by 1, our function increases by 2. So now we know the equation for f of x. f of x is going to be equal to 2 times 2x plus b, or 5. So we figured out what f of x is. Now we need to figure out what g of x is. So g of x is an exponential function. And there's really two things that we need to figure out. We need to figure out what a is, and we need to figure out what r is. And let me just rewrite that. So we know that g of x-- maybe I'll do it down here. g of x is equal to a times r to the x power. And if we know what g of 0 is, that's a pretty useful thing. Because r to the 0th power, regardless of what r is-- or I guess we could assume that r is not equal to 0. People can debate what 0 to the 0 power is. But if r is any non-zero number, we And so that essentially gives us a. So let's just write that down. g of 0 is a times r to the 0 power, which is just going to be equal to a times 1 or a. And they tell us what g of 0 is. g of 0 is equal to 3. So we know that a is equal to 3. So so far, we know that our g of x can be written as 3 times r to the x power. So now we can just use any one of the other values they gave us to solve for r. For example, they tell us that g of 1 is equal to 2. So let's write that down. g of 1, which would be 3 times r to the first power, or just 3-- let me just write it. It could be 3 times r to the first power, or we could just write that as 3 times r. They tell us that g of 1 is equal to 2. Or we get that r is equal to 2/3. Divide both sides of this equation by 3. So r is 2/3. And we're done. g of x is equal to 3 times 2/3. Actually, let me just write it this way. 3 times 2/3 to the x power. You could write it that way if you want, any which way. So 3 times 2/3 to the x power, and f of x is 2x plus 5. So let's actually just type that in. So f of x is 2x plus 5. And we can verify that that's the expression that we want. And g of x is 3 times 2 over 3 to the x power."
    },
    {
        "Q": "At 3:23 minutes, Sal says that the value of |x+10| is going to be greater than or equal to zero. But how can it ever be equal to zero? Shouldn't it just be greater than?\n",
        "A": "when x = -10, then |x+10| = 0",
        "video_name": "15s6B7K9paA",
        "timestamps": [
            203
        ],
        "3min_transcript": "If these two things are equal, and if I want to keep them equal, if I subtract 6 from the right-hand side, I've got to subtract-- or if I subtract 6 times the absolute value of x plus 10 from the right-hand side, I have to subtract the same thing from the left-hand side. So we're going to have minus 6 times the absolute value of x plus 10. And likewise, I want to get all my constant terms, I want to get this 4 out of the left-hand side. So let me subtract 4 from the left, and then I have to also do it on the right, otherwise my equality wouldn't hold. And now let's see what we end up with. So on the left-hand side, the 4 minus 4, that's 0. You have 4 of something minus 6 of something, that means you're going to end up with negative 2 of that something. Negative 2 of the absolute value of x plus 10. Remember, this might seem a little confusing, but remember, if you had 4 apples and you subtract 6 apples, you now have negative 2 apples, Same way, you have 4 of this expression, you take away 6 of this expression, you now have negative 2 of this expression. Let me write it a little bit neater. So it's negative 2 times the absolute value of x plus 10 is equal to, well the whole point of this, of the 6 times the absolute value of x plus 10 minus 6 times the absolute value of x plus 10 is to make those cancel out, and then you have 10 minus 4, which is equal to 6. Now, we want to solve for the absolute value of x plus 10. So let's get rid of this negative 2, and we can do that by dividing both sides by negative 2. You might realize, everything we've done so far is just treating this red expression as almost just like a variable, and we're going to solve for that red expression and then take it from there. So negative 2 divided by negative 2 is 1. 6 divided by negative 2 is negative 3. So we get the absolute value of x plus 10 is equal to negative 3. You might say maybe this could be the positive version or the negative, but remember, absolute value is always non-negative. If you took the absolute value of 0, you would get 0. But the absolute value of anything else is going to be positive. So this thing right over here is definitely going to be greater than or equal to 0. Doesn't matter what x you put in there, when you take its absolute value, you're going to get a value that's greater than or equal to 0. So there's no x that you could find that's somehow-- you put it there, you add 10, you take the absolute value of it, you're actually getting a negative value. So this right over here has absolutely no solution. And I'll put some exclamation marks there for emphasis."
    },
    {
        "Q": "Please can anyone explain how Sal gets from Y/X = -3 to 1/X = -3 x 1/y @ 4:42 and/ or how he gets from X=2/y to 2 x 1/y @ 8:30? Thanks.\n",
        "A": "It s okay, I get it now",
        "video_name": "92U67CUy9Gc",
        "timestamps": [
            282,
            510
        ],
        "3min_transcript": "2 is negative 6. So notice, we multiplied. So if we scaled-- let me do that in that same green color. If we scale up x by 2-- it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor. And if you wanted to go the other way-- let's try, I don't know, let's go to x is 1/3. If x is 1/3, then y is going to be-- negative 3 times 1/3 is negative 1. So notice, to go from 1 to 1/3, we divide by 3. To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n."
    },
    {
        "Q": "\nat 6:53, why is it called \"inverse?\" is it because the greater \"x\" is, the smaller \"y\" gets?",
        "A": "Yes. One variable increases by the same factor that the other decreases. In his example, X multiplies by a factor of 2 while Y divides by a factor of two. If you re graphing the functions, The inverse variation indicates the variables separate from a central point.",
        "video_name": "92U67CUy9Gc",
        "timestamps": [
            413
        ],
        "3min_transcript": "to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n. So if I did it with y's and x's, this would be y is equal to some constant times 1/x. So instead of being some constant times x, it's some constant times 1/x. So let me draw you a bunch of examples. It could be y is equal to 1/x. It could be y is equal to 2 times 1/x, which is clearly the same thing as 2/x. It could be y is equal to 1/3 times 1/x, which is the same thing as 1 over 3x. it could be y is equal to negative 2 over x. And let's explore this, the inverse variation, the same way that we explored the direct variation. So let's pick-- I don't know/ let's pick y is equal to 2/x. And let me do that same table over here. So I have my table. I have my x values and my y values. If x is 1, then y is 2. So if you multiply x by 2, if you scale it up by a factor of 2, what happens to y? y gets scaled down by a factor of 2. You're dividing by 2 now. Notice the difference. Here, however we scaled x, we scaled up y by the same amount. Now, if we scale up x by a factor, when we have inverse variation, we're scaling down y by that same. So that's where the inverse is coming from. And we could go the other way. If we made x is equal to 1/2. So if we were to scale down x, we're going to see that it's going to scale up y. Because 2 divided by 1/2 is 4. So here we are scaling up y. So they're going to do the opposite things. They vary inversely. And you could try it with the negative version of it, as well. So here we're multiplying by 2. And once again, it's not always neatly written for you like this."
    },
    {
        "Q": "At 4:40, how did he divide \"y/x\" by \"y\"? I know that multiplying by the inverse is how you divide a fraction; but I don't understand how \"1/y\" times \"y/x\" would equal \"1/x\" unless you cross multiply; but that wouldn't work because \"1/y\" is already inverted, right?\n",
        "A": "y/x / y = y/x * 1/y = 1y/xy [Simplify by dividing by y] = 1/x I hope this helps!",
        "video_name": "92U67CUy9Gc",
        "timestamps": [
            280
        ],
        "3min_transcript": "2 is negative 6. So notice, we multiplied. So if we scaled-- let me do that in that same green color. If we scale up x by 2-- it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor. And if you wanted to go the other way-- let's try, I don't know, let's go to x is 1/3. If x is 1/3, then y is going to be-- negative 3 times 1/3 is negative 1. So notice, to go from 1 to 1/3, we divide by 3. To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n."
    },
    {
        "Q": "\ncan someone tell me the mathematical reason as for why Sir Khan multiplied 108 by 139 at 5:09 ?",
        "A": "we call also do Cross-Multiplication to get the value of x . which means , multiplying the numerator of LHS with the denominator of RHS and multiplying the numerator of RHS with the denominator of LHS..... provided that there is an equal-to sign between the fractions .",
        "video_name": "Z5EnuVJawmY",
        "timestamps": [
            309
        ],
        "3min_transcript": "And this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? Well, we just write SOHCAHTOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. So we get the cosine of theta is going to be equal to the height that we care about. That's the adjacent side of this right triangle over the length of the hypotenuse, OVER 108. Well, that doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here. So maybe if we can figure out what cosine of theta is based up here, then we can solve for h. So if we look at this data, what is the cosine of theta? And now we're looking at a different right triangle. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters. So it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108. Oh, we already have it labeled here. It's 180. We can assume that this is an isosceles-- that this pyramid is an isosceles triangle. So 180 on that side and 180 on that side. So the cosine is adjacent-- 139-- over the hypotenuse, which is 180, over 180. And these data are the same data. We just showed that. So now we have cosine of theta is h/108. Cosine of theta is 139/180. Or we could say that h/108, which is cosine of theta, Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108/180. So let's get our calculator out and calculate that. So that is going to be 139 times 108, divided by 180, gets us to 83.4 meters. So h is equal to 83.4 meters. The height of the water is 83.4."
    },
    {
        "Q": "\nat around 5:17 why do we multiply by 108 on all sides? Why wouldn't we multiply by 139, 180, or h?",
        "A": "To make h not a fraction, or to remove the 108 from underneath h. Anything else wouldn t have done very well. That is what you need to do to solve for h.",
        "video_name": "Z5EnuVJawmY",
        "timestamps": [
            317
        ],
        "3min_transcript": "And this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? Well, we just write SOHCAHTOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. So we get the cosine of theta is going to be equal to the height that we care about. That's the adjacent side of this right triangle over the length of the hypotenuse, OVER 108. Well, that doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here. So maybe if we can figure out what cosine of theta is based up here, then we can solve for h. So if we look at this data, what is the cosine of theta? And now we're looking at a different right triangle. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters. So it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108. Oh, we already have it labeled here. It's 180. We can assume that this is an isosceles-- that this pyramid is an isosceles triangle. So 180 on that side and 180 on that side. So the cosine is adjacent-- 139-- over the hypotenuse, which is 180, over 180. And these data are the same data. We just showed that. So now we have cosine of theta is h/108. Cosine of theta is 139/180. Or we could say that h/108, which is cosine of theta, Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108/180. So let's get our calculator out and calculate that. So that is going to be 139 times 108, divided by 180, gets us to 83.4 meters. So h is equal to 83.4 meters. The height of the water is 83.4."
    },
    {
        "Q": "at 3:44, how do we know that the second derivative is less than zero? and how do we even know if this function can have a second derivative? what is needed to, what defines a second derivative? thank you!\n",
        "A": "The second derivative is the derivative of the first derivative. e.g. f(x) = x^3 - x^2 f (x) = 3x^2 - 2x f (x) = 6x - 2 So, to know the value of the second derivative at a point (x=c, y=f(c)) you: 1) determine the first and then second derivatives 2) solve for f (c) e.g. for the equation I gave above f (x) = 0 at x = 0, so this is a critical point. f (0) = 6\u00e2\u0080\u00a20 - 2 = -2 Therefore, f(x) is concave downward at x=0 and this critical point is a local maximum. Can you do the same for the other critical point?",
        "video_name": "-cW5hCsc9Yc",
        "timestamps": [
            224
        ],
        "3min_transcript": "So our first derivative should still be equal to zero 'cause our slope of a tangent line right over there is still zero. So F prime of C is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing. We have an upward opening bowl and so here we have a relative minimum value or we could say our second derivative is greater than zero. Visually we see it's a relative minimum value and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says hey look, if we're dealing with some function F, let's say it's a twice differentiable function. So that means that over some interval. and second derivatives are defined and so let's say there's some point, X equals C, where its first derivative is equal to zero, so the slope of the tangent line is equal to zero, and the derivative exists in a neighborhood around C and most of the functions we deal with, if it's differentiable at C, it tends to be differentiable in the neighborhood around C and then we also assume that second derivative exists is twice differentiable. Well then we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with and it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. Then so this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. We have a relative minimum point and if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point. We can't make any strong statement. So with that out of the way, let's just do a quick example just to see if this has gelled. Let's say that I have some twice differentiable function H and let's say that I tell you that H of eight is equal to five, I tell you that H prime of eight is equal to zero, and I tell you that the second derivative at X equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative minimum, relative minimum, maximum point or not enough info? Not enough info or inconclusive?"
    },
    {
        "Q": "\nat 0:55, wouldn't the notation for the arc be ABC then? how would you specify the major arc if there isn't a point on the circle to differentiate between the two?",
        "A": "Hi angela braukman, The notation of the arc would still be two letters because an arc is not a major arc till it is either 180\u00cb\u0099 or more. And to specify it: I would recommend marking your own point if there is no third point. Hope that helps! - Sam",
        "video_name": "GOA9XWEo7QI",
        "timestamps": [
            55
        ],
        "3min_transcript": "- So I have some example questions here from Khan Academy on arc measure. And like always, I encourage you to pause the video after you see each of these questions, and try to solve them before I do. So this first question says what is the arc measure, in degrees, of arc AC on circle P below. So this is point A, that is point C, and when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. When we talk about the minor arc. There's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one, there is the one on the right. And since C isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with the smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as, in degrees, as the measure of the central angle that intercepts the arc. I'll do it in this blue color, that central angle is angle C, P, A. Angle C, P, A, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. One hundred and seventy four degrees, that's the arc measure, in degrees, of arc AC. Let's keep doing these. So let me do another one. So, this next one asks us, in the figure below, in the figure below, segment AD-- so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there, AD and CE are diameters of the circle. So let me draw CE, we're going to connect point C and E. These are diameters. So, let me, so they go straight. Whoops, I'm using the wrong tool, let me... So those are, somehow I should, alright. So, those are di-- whoops, how did that happen? So let me, somehow my pen got really big, alright. That'll be almost there, ok. So CE, there you go. So those are both diameters of the circle P. What is the arc measure of AB, of arc AB in degrees? So arc AB, once again there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to not the major arc they would've said something like A, E, B or A, D, B"
    },
    {
        "Q": "At 1:30, why does Sal give the measure of the minor arc? I thought the question was asking for the major arc on the other side.\n",
        "A": "AC by notation is a minor arc, it just happens to have another letter between them, A major arc is designated by three letters always to distinguish it from a minor arc, so you would need an additional point on the left side (D) to be talking about major arc ADC.",
        "video_name": "GOA9XWEo7QI",
        "timestamps": [
            90
        ],
        "3min_transcript": "- So I have some example questions here from Khan Academy on arc measure. And like always, I encourage you to pause the video after you see each of these questions, and try to solve them before I do. So this first question says what is the arc measure, in degrees, of arc AC on circle P below. So this is point A, that is point C, and when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. When we talk about the minor arc. There's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one, there is the one on the right. And since C isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with the smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as, in degrees, as the measure of the central angle that intercepts the arc. I'll do it in this blue color, that central angle is angle C, P, A. Angle C, P, A, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. One hundred and seventy four degrees, that's the arc measure, in degrees, of arc AC. Let's keep doing these. So let me do another one. So, this next one asks us, in the figure below, in the figure below, segment AD-- so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there, AD and CE are diameters of the circle. So let me draw CE, we're going to connect point C and E. These are diameters. So, let me, so they go straight. Whoops, I'm using the wrong tool, let me... So those are, somehow I should, alright. So, those are di-- whoops, how did that happen? So let me, somehow my pen got really big, alright. That'll be almost there, ok. So CE, there you go. So those are both diameters of the circle P. What is the arc measure of AB, of arc AB in degrees? So arc AB, once again there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to not the major arc they would've said something like A, E, B or A, D, B"
    },
    {
        "Q": "At 1:08, Sal says..\"The actual coordinate in R2 on the Cartesian coordinate..\" what does R2 mean?\n",
        "A": "It means the x-y coordinate plane. R for the real numbers, and 2 for the number of dimensions (loosely speaking)",
        "video_name": "7fYDCUIvZnM",
        "timestamps": [
            68
        ],
        "3min_transcript": "- [Voiceover] Let's say I have some curve C, and it's described, it can be parameterized, I can't say that word, as, let's say x is equal to x of t, y is equal to some function y of t, and let's say that this is valid for t is between a and b. So, t is greater than or equal to a and then less than or equal to b. So, if I were to just draw this on, let me see, I could draw it like this. I'm staying very abstract right now. This is not a very specific example. This is the x axis, this is the y axis. My curve, let's say this is when t is equal to a and then the curve might do something like this. I don't know what it does, let's say it's over there. This is t is equal to b. This actual point right here will be x of b. That would be the x coordinate, And this is, of course, when t is equal to a. The actual coordinate in R2 on the Cartesian coordinates will be x of a, which is this right here, and then y of a, which is that right there. And we've seen that before, that's just a standard way of describing a parametric equation or curve using two parametric equations. What I want to do now is describe this same exact curve using a vector valued function. So, if I define a vector valued function, and if you don't remember what those are we'll have a little bit of review here. Let me say I have a vector valued function r, and I'll put a little vector arrow on top of it. In a lot of textbooks, they'll just bold it and they'll leave scalar valued functions unbolded, but it's hard to draw bold so I'll put a little vector on top. And let's say that r is a function of t. And these are going to be position vectors. Position vectors. And I'm specifying that because in general, when someone talks about a vector, this vector and this vector are considered equivalent as long as they have the same magnitude and direction. No one really cares about what their start and end points are as long as their direction is the same and their length is the same. But, when you talk about position vectors, you're saying, \"No, these vectors are all going to start at zero, at the origin.\" And when you state the position vector you're implicitly saying, \"This is specifying a unique position.\" In this case, it's going to be in two dimensional space, but it could be in three dimensional space, or really four, five, whatever, n dimensional space. So, when you state the position vector, you're literally saying, \"Okay, this vector literally specifies that point in space.\" So, let's see if we can describe this curve as a vector, a position vector valued function. So, we can say r of t, let me switch back to that pink color,"
    },
    {
        "Q": "at 0:57 Sal introduces a new word to me, degenerate, but doesn't explain it thoroughly. Can someone please help me?\n",
        "A": "Imagine a line segment with a point on a side so it s barely a triangle. That s a degenerate triangle in a nutshell",
        "video_name": "KlKYvbigBqs",
        "timestamps": [
            57
        ],
        "3min_transcript": "Let's draw ourselves a triangle. Let's say this side has length 6. Let's say this side right over here has length 10. And let's say that this side right over here has length x. And what I'm going to think about is how large or how small that value x can be. How large or small can this side be? So the first question is how small can it get? Well, if we want to make this small, we would just literally have to look at this angle right over here. So let me take a look at this angle and make it smaller. So let's try to make that angle as small as possible. So we have our 10 side. Actually let me do it down here. So you have your 10 side, the side of length 10, and I'm going to make this angle really, really, really small, approaching 0. If that angle becomes 0, we end up with a degenerate triangle. It essentially becomes one dimension. But as we approach 0, this side starts to coincide or get closer and closer to the 10 side. And you could imagine the case where it actually coincides with it and you actually get the degenerate. So if want this point right over here to get as close as possible to that point over there, essentially minimizing your distance x, the closest way is if you make the angle the way equal to 0, all the way. So let's actually-- let me draw a progression. So now the angle is getting smaller. This is length 6. x is getting smaller. Then we keep making that angle smaller and smaller and smaller all the way until we get a degenerate triangle. So let me draw that pink side. So you have the side of length 10. Now the angle is essentially 0, this angle that we care about. So this side is length 6. And so what is the distance between this point and this point? And that distance is length x. We know that 6 plus x is going to be equal to 10. So in this degenerate case, x is going to be equal to 4. So if you want this to be a real triangle, at x equals 4 you've got these points as close as possible. It's degenerated into a line, into a line segment. If you want this to be a triangle, x has to be greater than 4. Now let's think about it the other way. How large can x be? Well to think about larger and larger x's, we need to make this angle bigger. So let's try to do that. So let's draw my 10 side again. So this is my 10 side. I'm going to make that angle bigger and bigger. So now let me take my 6 side and put it like that. And so now our angle is getting bigger and bigger and bigger. It's approaching 180 degrees. At 180 degrees, our triangle once again"
    },
    {
        "Q": "What does timesish mean? She says it at 5:06.\n",
        "A": "She means taking steps on the number line by multiplying, not adding.",
        "video_name": "N-7tcTIrers",
        "timestamps": [
            306
        ],
        "3min_transcript": "From where you're looking, 5 is plus 1. 3 is also just plus 1 away, but against the flow of time. So it's an anti-1. Minus 4 is twice as far away back in time as it would be if you were looking from a 0 perspective. To get to negative 8, sometimes to make the harder things simple, first, you have to make the simple things harder. Addition is a process of plus 1 plus 1 plus 1. Multiplication is a process of plus n plus n plus n. For multiples of 2, instead of counting plus 1 plus 1 plus 1, you're counting by plus 1 plus 1, plus 1 plus 1, or plus 2, for short. Plus 2 plus 2 plus 2 plus 2. Counting in a plus 2 kind of way. Counting in a plus 5 kind of way. Counting in a plus 10 kind of way. What is 5 times 7? All that means is count in a plus 5 sort of way seven times, which happens to be the same as counting in a plus 7 sort of way 5 times. 1, 2, 3, 4, 5. This is why 8 smells like 2 and 4. 2 sort of way. And 8 is 2 if you're counting in a plus 4 sort of way. Division is fancy counting. A divided by B asks the question, what is A, when counting in a plus B sort of way? 27 divided by 9 means, what is 27, in a plus 9 sort of way? Plus 9 plus 9 plus 9, it's 3. What is seven in a plus 2 sort of way? Just count, 1, 2, 3, and 1/2. Division means counting by plus denominators until you get to the numerator and seeing how long it takes. 7 divided by 1/3 means counting by plus 1/3's until you get to seven. Plus 1/3 plus 1/3 plus 1/3 plus 1/3 plus 1/3, all the way to 21. Exponentiation-- powers-- also fancy counting. This time, instead of counting by plus whatever, you count by times whatever. Powers of 2 are just counting in a times 2 sort of way. The size of the times-ish step is the bottom number. And the number of times-ish steps is a little exponent. Powers of 10 count in a times 10 sort of way. To take 10 to the sixth, you just take six steps of size times 10, 10 times 10 times 10 times 10 times 10 times 10. That's why 8 has a whiff of 3. 8 is 3 when counting in a times 2-ish way. Then, there's roots. To take the fourth root of 81, you're just saying, hey, if 81 is 4 in a timesy sort of way, then how are we counting? It's like going a quarter of the way there, but a timesy quarter, not an addish quarter, times 3 times 3 times 3 times 3. As far as times-ish counting is concerned, 3 is a quarter of 81, which is why 81 to the 1/4 is the same as the fourth root of 81. And personally, I think it's kind of weird that we keep around this root notation, when fractional powers are so much more descriptive. 81 to this 3/4 means you're going 3/4 of the way"
    },
    {
        "Q": "Maybe he said it and I missed it, but how do we know that the zero vector is the only solution to Rx=0, if he only used the particular solution of c1,c2,0,c4,0 around 3:15?\n",
        "A": "It s not, but that s beside the point. We aren t trying to solve c1r1 + c2r2 + c3r3 + c4r4 + c5r5 = 0, only c1r1 + c2r2 + c4r4 = 0. We already know that the other columns are extras .",
        "video_name": "BfVjTOjvI30",
        "timestamps": [
            195
        ],
        "3min_transcript": "combination of my basis vectors? So the first thing that wasn't too much of a stretch of the imagination in the last video, was the idea that these pivot vectors are linearly independent. So r1, r2, and r4. And everything I'm doing, I'm kind of applying to the special case just so that it's easier to understand. But it should be generalizable. In fact, it definitely is generalizable. That all of the pivot columns in reduced row echelon form are linearly independent. And that's because the very nature of reduced row echelon form, is that you are the only pivot column that has a 1 in that respective row. So the only way to construct it is with that vector. You can't construct it with the other pivot columns because they're all going to have 0 in that row. And when I say it's linearly independent, I'm just saying the set of pivot columns. So let me say this in general. The set of pivot columns for any reduced row echelon form And it's just a very straightforward argument. Because each column is going to have a 1 in a very unique place. All of the other pivot columns are going to have a 0 in that same place. And so you can't take any linear combinations to get to that 1 because 0 times anything, minus or plus 0 times anything, can never be equal to 1. So I think you can accept that. Now, that means that the solution to c1 times r1, plus c2 times r2, plus, let me say, c4 times r4. The solution to this equation, because these guys are linearly independent, we know that this only has one solution, and that's c1, c2, and c4 is equal to 0. So another way we could say it is, if we write r times some vector x-- well I'll just write it times this particular x-- where I write it as c1, c2, 0, c4, and 0 is equal to 0. So this will be some special member of your null space. It's a particular solution to the equation. This is equal to one, two, three, four 0's because we have four rows here. Now, if we just expand this out. If we just multiply 1 times c1, plus 0 times c2, minus 1 times 0, plus 4 times 0, you'll get-- or actually a better way to explain it-- this multiplication right here can be written as-- and we've seen this multiple times-- c1"
    },
    {
        "Q": "\nAt 6:58, isn't the formula for finding the area of a circle, pi*diameter?",
        "A": "NO. The circumference is pi x the diameter. Area is pi x r^2.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            418
        ],
        "3min_transcript": "so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles, There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12,"
    },
    {
        "Q": "\nHello! I have a question. In Finding probability example 2 at 9:33 you talk about probability. Why you don't calculated area of the smallest circumference?\nSincerely,Alex.",
        "A": "The area of the smaller circle is already calculated as far as we need it to be in order to construct the ratio of areas. It was given as A=16\u00cf\u0080. The larger circles area was not given, so it needed to be calculated from the given circumference.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            573
        ],
        "3min_transcript": "is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12, So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability-- I'll just write it like this-- the probability that the point also lies in the smaller circle-- so all of that stuff The probability of that is going to be equal to the percentage of this larger circle that is this smaller one, and that's going to be-- that is the smaller circle's area. So it's going to be 16 pi over 324 pi. And the pi's cancel out, and it looks like both of them are divisible by 4. If we divide the numerator by 4, we get 4, if we divide the denominator by 4, what do we get? 4 goes into 320 80 times, it goes into 4 once, so we get an 81. So a probability-- I didn't even draw this to scale, this area is actually much smaller when you do it to scale-- the probability that if you were to randomly select a point from the larger circle, that it also lies in the smaller one is the ratio of their areas, the ratio of the smaller circle to the larger one. And that is 4/81, I guess is the best way to say it."
    },
    {
        "Q": "\n6:05\nWhy is there an infinite number of points contained in the circle ?",
        "A": "There is an infinite number of points because the distance between two points can always be cut in half, whether it is an arc or a line. If we were standing 64 feet apart and every minute we cut the distance in half, within 6 minutes we will be within one foot, then .5ft, .25.ft, .125ft, .0625ft, .03125ft... in theory, and that is what we are talking about, theory, we would never touch. As it should be...",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            365
        ],
        "3min_transcript": "55 is 11 times 5. Not 56, not 28. This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5. So all of these are multiples of 5. 45, that's 9 times 5. 3 is not a multiple of 5. 25, clearly 5 times 5. So I've circled all the multiples of 5. So of all the possibilities, the ones that meet our constraint of being a multiple of 5, there are 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So in this example, the probability of a selecting a number that is a multiple of 5 is 7/12. Let's do another one. the circumference of a circle is 36 pi. Let's draw this circle. so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles, There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle."
    },
    {
        "Q": "\nAt 0:31-- Why does it ask -2+(-7) when you could do -2-(7)?",
        "A": "It is working with negative numbers. This question is showing that adding a negative is the same as subtracting a positive. Either way of writing it is perfectly acceptable.",
        "video_name": "3CKpidALDEg",
        "timestamps": [
            31
        ],
        "3min_transcript": "- [Voiceover] So we've already spent some time introducing ourselves to the idea of adding or subtracting positive and negative numbers. What I now what do in this video is do a bunch of examples using the adding negative numbers on the number line exercise on Khan Academy, so that we can think about different ways to model, or think about, or visualize adding and subtracting positive and negative numbers. So this question asks us, \"Which number line model represents \"the expression negative two \"plus negative seven?\" And a number line model seems like this really fancy thing, but they're just saying, which of these diagrams, which of these drawings are a way to think about what negative two plus negative seven is. So let's think about it. I want to start at negative two and then to that I want to add negative seven. So let's see what's going on here. This number line model I guess we could call it, it's saying this is positive two right over here. Then it subtracts seven. This is the model that would be for positive two plus negative seven, or two, or positive two minus seven. But that's not what we have over here, we have negative two minus seven. This one right over here, let's see, this first... I guess you could say this thing, this arrow, it's going one, two, three, four, five to the left of zero. So starting at negative five, and then it goes another one, two, three, four to the left of that. So you could view this as negative five minus four. Or you could view this as negative five plus negative four, but that's not what we have up here. And then this last one, hopefully this is the answer, otherwise there would be a mistake in the exercise. Let's see, we have this first arrow that takes us two steps to the left of zero. So, one, two. So this is negative two, and then we take another seven steps So one, two, three, four, five, six, seven. So you can think of this arrow as representing negative two, and then we're going to that, so we're starting from this tip of this arrow. We're going to add a negative seven which makes us move seven to the left again to get to negative nine. So that's definitely the model that represents that expression. Let's keep going. All right. Which number line model represents the expression six plus negative two. So here, they've taken our number lines but they've made them go up and down, they've made them vertical. So we wanna really think about it. I want to start at six and I want to add negative two. So on a vertical number line like this, it seems like we're increasing as we go up. I want to start at six above zero. So, I want to start at six above zero, but then I want to add negative two, which should take me two steps back down. This one takes me two steps even higher, so this one right over here is positive six plus two, or positive six plus positive two."
    },
    {
        "Q": "\nAt 1:01, how do you round down?",
        "A": "you round down if the number is 4 or less",
        "video_name": "_MIn3zFkEcc",
        "timestamps": [
            61
        ],
        "3min_transcript": "0.710 Round 9.564, or nine and five hundred sixty-four Thousandths, to the nearest tenth. So lie me write it a bit larger, 9.564 And we need to round to the nearest tenth. So what's the tenth place? The tenths place is right here This represents 5 tenths. This is the ones place, this is the tenths place, this is The hundredths place, and this is the Thousandths place right here So we need to round to the nearest tenth. So if we round up, this will be 9.6 If we round down, this will be 9.5 And just like regular rounding, when we're not Dealing with decimals, you move to one spot, or you look At one place to the right or one place lower, I guess, and You say is that 5 larger If it is, you round if it isn't,you round down So this 9.564 becomes 9.6, or we can call this Nine and sixth tenths. And then we're done!"
    },
    {
        "Q": "\nAt 6:20 ish, how does Sal say that that function meets all three requirements? if c = 0, the lim as x approaches c of either of the functions neither gives us positive or negitive infinity.",
        "A": "Earlier in the video Sal sets up two different ways to meet the requirements for application of l Hospital s rule. One involves the indeterminate form 0/0, and the other involves an indeterminate form with plus or minus infinity in both the numerator and denominator. He s working with an example that meets the first set of conditions, so it doesn t have to meet the other set of conditions.",
        "video_name": "PdSzruR5OeE",
        "timestamps": [
            380
        ],
        "3min_transcript": "Edit, copy, and then let me paste it. So in either of these two situations just to kind of make sure you understand what you're looking at, this is the situation where if you just tried to evaluate this limit right here you're going to get f of c, which is 0. Or the limit as x approaches c of f of x over the limit as x approaches c of g of x. That's going to give you 0/0. And so you say, hey, I don't know what that limit is? But this says, well, look. If this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. And if I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit we get infinity over infinity, or negative infinity or positive infinity over positive or negative infinity. So these are the two indeterminate forms. And to make it all clear let me just show you an example because I think this will make things a lot more clear. do this in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0/0. Sine of 0 is 0. Or the limit as x approaches 0 of sine of x is 0. And obviously, as x approaches 0 of x, that's also going to be 0. So this is our indeterminate form. And if you want to think about it, this is our f of x, that f of x right there is the sine of x. And our g of x, this g of x right there for this first case, is the x. g of x is equal to x and f of x is equal to sine of x. first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. So we get our indeterminate form. So let's see, at least, whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x, and take the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. So I'll do that in the blue. So let me write the derivatives of the two functions. So f prime of x. If f of x is sine of x, what's f prime of x? Well, it's just cosine of x. You've learned that many times. And if g of x is x, what is g prime of x?"
    },
    {
        "Q": "Shouldn't it be lim x->c of f(c)? not f(x)? 2:05\n",
        "A": "No. The limit is of the function f(x), not the limit of f(c) because that is already evaluated at a value.",
        "video_name": "PdSzruR5OeE",
        "timestamps": [
            125
        ],
        "3min_transcript": "Most of what we do early on when we first learn about calculus is to use limits. We use limits to figure out derivatives of functions. In fact, the definition of a derivative uses the notion of a limit. It's a slope around the point as we take the limit of points closer and closer to the point in question. And you've seen that many, many, many times over. In this video I guess we're going to do it in the opposite direction. We're going to use derivatives to figure out limits. And in particular, limits that end up in indeterminate form. And when I say by indeterminate form I mean that when we just take the limit as it is, we end up with something like 0/0, or infinity over infinity, or negative infinity over infinity, or maybe negative infinity over negative infinity, or positive infinity over negative infinity. All of these are indeterminate, undefined forms. And in this video I'm just going to show you what l'Hoptial's rule says and how to apply it because it's fairly straightforward, and it's actually a very useful tool sometimes if you're in some type of a math competition and they ask you to find a difficult limit that when you just plug the numbers in you get something like this. L'Hopital's rule is normally what they are testing you for. And in a future video I might prove it, but that gets a The application is actually reasonably straightforward. So what l'Hopital's rule tells us that if we have-- and I'll do it in abstract form first, but I think when I show you the example it will all be made clear. That if the limit as x roaches c of f of x is equal to 0, and this is another and-- and the limit as x approaches c of f prime of x over g prime of x exists and it equals L. then-- so all of these conditions have to be met. This is the indeterminate form of 0/0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L."
    },
    {
        "Q": "Why at 6:35 does Sal considers that 1/cos(theta) =cos(theta)?\n",
        "A": "He doesn t \u00e2\u0080\u0093 if you go back and listed from 6:30, he is taking the reciprocal of each element in the inequality \u00e2\u0080\u0093 this is also why he switches the direction of the inequalities.",
        "video_name": "5xitzTutKqM",
        "timestamps": [
            395
        ],
        "3min_transcript": "of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta. is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. well, sine of theta is gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value, so I can erase those. If we're in the first or fourth quadrant, our X value is not negative, and so cosine of theta, which is the x-coordinate on our unit circle, is not going to be negative, and so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say for negative pi over two is less than theta is less than pi over two, but over this interval, this is true for any theta over which these functions are defined."
    },
    {
        "Q": "\nWhen Theta was divided by sin theta, shouldn't it be theta/sin theta, why it's sin theta/ theta in the video at 6:25 ?",
        "A": "He took the reciprocal of all 3 terms, which made \u00ce\u00b8/sin\u00ce\u00b8 into sin\u00ce\u00b8/\u00ce\u00b8.",
        "video_name": "5xitzTutKqM",
        "timestamps": [
            385
        ],
        "3min_transcript": "which is the absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now, how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than or equal to the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true a little bit of algebraic manipulation. Let me multiply everything by two so I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta. is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well."
    },
    {
        "Q": "At 3:20, why does Sal take only the positive square root of four, and not both negative and positive two? (I get that eventually it will not matter since you will add or subtract it from 6)\n",
        "A": "Because he is looking at sqrt(4), which means the principal square root of 4 (the positive value you square to get 4). So that is positive 2 only. You ll notice, however, that in the expression he derives using the quadratic formula, there is a +/- symbol in front of the principal square root sign, so you do get both roots coming into play there. Does that help?",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            200
        ],
        "3min_transcript": "So let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus-- so that gives us two roots right over there-- plus or minus square root of b squared minus 4ac over 2a. So let's apply that to this situation. Negative b-- this right here is b. So negative b is negative negative 6. So that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a-- which is 2-- times 2 times c, which is 5. Times 5. All of that over 2 times a. a is 2. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus 36 minus-- so this is 4 times 2 times 5. This is 40 over here. So 36 minus 40. And you already might be wondering what's going to happen here. All of that over 4. Or this is equal to 6 plus or minus the square root of negative 4. 36 minus 40 is negative 4 over 4. And you might say, hey, wait Sal. Negative 4, if I take a square root, I'm going to get an imaginary number. And you would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. So we're essentially going to get two complex numbers when we take the positive and negative version of this root. So let's do that. So the square root of negative 4, that is the same thing as 2i. And we know that's the same thing as 2i, or if you want to think of it this way. Square root of negative 4 is the same thing I could even do it one step-- that's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. And the principal square root of negative 1 is i times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i."
    },
    {
        "Q": "\nWhy did Sal divide (8-6i)/(4) by 2 at 9:17?",
        "A": "Because the numerator can easily be divided by 2, and what s left of the denominator will cancel out when multiplied by the 2 outside the parentheses. 2((8 - 6i) / 4) = 2((4 - 3i) / 2) = 2(4 - 3i) / 2 = 4 - 3i You could, of course, do it the other way around: 2((8 - 6i) / 4) = 2(8 - 6i) / 4 = (8 - 6i) / 2 = 4 - 3i",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            557
        ],
        "3min_transcript": "So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. 3 minus i times 3 minus i, which is-- and you could get practice taking squares of two termed expressions, or complex numbers in this case actually-- it's going to be 9, that's 3 squared, and then 3 times negative i And then you're going to have two of those. So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. 2 divided by 2 is 1. So 3 times 3 is 9. 3 times negative i is negative 3i. is going to be-- I'll do this in blue. 9 minus 1 is going to be 8. We have 8 minus 6i. And then if we divide 8 minus 6i by 2 and 4 by 2, in the numerator, we're going to get 4 minus 3i. And in the denominator over here, we're going to get a 2. We divided the numerator and the denominator by 2. Then we have a 2 out here. And we have a 2 in the denominator. Those two characters will cancel out. And so this expression right over here cancels or simplifies to 4 minus 3i. Then we have a plus 5 needs to be equal to 9 minus 3i. I We have a negative 3i on the left, a negative 3i on the right. We have a 4 plus 5. We could evaluate it. This left hand side is 9 minus 3i, which is the exact same complex number as we have"
    },
    {
        "Q": "At 1:43, Sal takes -6 squared to be positive 36 but if you press -6^2 on a calculator, you get -36. I remember working out a question wrong on Khan Academy. I had at some point in the problem work out g= -1^2 +4. I got 5 as an answer to be wrong. Khan Academy took -1^2 to be -1 and the correct answer was 3. Can somebody please explain?\n",
        "A": "You wrote it like -6\u00c2\u00b2, but you need to write it like (-6)\u00c2\u00b2",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            103
        ],
        "3min_transcript": "We're asked to solve 2x squared plus 5 is equal to 6x. And so we have a quadratic equation here. But just to put it into a form that we're more familiar with, let's try to put it into standard form. And standard form, of course, is the form ax squared plus bx plus c is equal to 0. And to do that, we essentially have to take the 6x and get rid of it from the right hand side. So we just have a 0 on the right hand side. And to do that, let's just subtract 6x from both sides of this equation. And so our left hand side becomes 2x squared minus 6x plus 5 is equal to-- and then on our right hand side, these two characters cancel out, and we just are left with 0. And there's many ways to solve this. We could try to factor it. And if I was trying to factor it, I would divide both sides by 2. If I divide both sides by 2, I would get integer coefficients on the x squared in the x term, but I would get 5/2 for the constant. So it's not one of these easy things to factor. We could complete the square, or we could apply the quadratic formula, which is really So let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus-- so that gives us two roots right over there-- plus or minus square root of b squared minus 4ac over 2a. So let's apply that to this situation. Negative b-- this right here is b. So negative b is negative negative 6. So that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a-- which is 2-- times 2 times c, which is 5. Times 5. All of that over 2 times a. a is 2. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus 36 minus-- so this is 4 times 2 times 5. This is 40 over here. So 36 minus 40. And you already might be wondering what's going to happen here. All of that over 4. Or this is equal to 6 plus or minus the square root of negative 4. 36 minus 40 is negative 4 over 4. And you might say, hey, wait Sal. Negative 4, if I take a square root, I'm going to get an imaginary number. And you would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. So we're essentially going to get two complex numbers when we take the positive and negative version of this root. So let's do that. So the square root of negative 4, that is the same thing as 2i. And we know that's the same thing as 2i, or if you want to think of it this way. Square root of negative 4 is the same thing"
    },
    {
        "Q": "\nAt 5:45 how 9 + 6i is resulted from 2(3+i/2)^2 ? I did the distributive property and came out with (36 + i^4) / 2\nWhy did Sal multiply \"3\" and \"i\" while there is a plus sign between them at 5:35?",
        "A": "You have some errors... To square a binomial, use FOIL. It looks like you only squared the end values. and I have no idea how i became i^4 in your version as i*i = i^2 = -1. It also looks like you never squared the denominator. [(3+i)/2]^2 = (3+i)(3+i)/[2*2] = [9 + 3i + 3i + i^2]/4 = [9 + 6i -1]/4 = [8 + 6i]/4 Multiply it by 2 and it becomes: [8 + 6i]/2 The fraction can then be reduced: 8/2 + 6i/2 = 4 + 3i Hope this helps.",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            345,
            335
        ],
        "3min_transcript": "Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works."
    },
    {
        "Q": "At 8:24 how does Sal get (-i)^2=-1?\n",
        "A": "(-i)^2 = i^2, just like (-2)^2 = 2^2, because when you multiply negative by negative you get positive. i^2 = -1, because that s the definition of i.",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            504
        ],
        "3min_transcript": "you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. 3 minus i times 3 minus i, which is-- and you could get practice taking squares of two termed expressions, or complex numbers in this case actually-- it's going to be 9, that's 3 squared, and then 3 times negative i And then you're going to have two of those. So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. 2 divided by 2 is 1. So 3 times 3 is 9. 3 times negative i is negative 3i."
    },
    {
        "Q": "I don't get why it is 1/a (3:11)? Why not -a?\n",
        "A": "hmm.. let s suppose a was 2 2^4=16 2^3=8 2^2=4 2^1=2 2^0=1 at each step, we re dividing by 2 (that is, a) therefore, 2^-1 should be 1 divided by 2 2^-1= 1/2 replace 2 by a and you get a^-1= 1/a (if you have a question comment below)",
        "video_name": "Tqpcku0hrPU",
        "timestamps": [
            191
        ],
        "3min_transcript": "We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the fourth, what did we do? We multiplied by a again. Or the other way, you could imagine, is when you decrease the exponent, what are we doing? We are multiplying by 1/a, or dividing by a. And similarly, you decrease again, you're dividing by a. And to go from a squared to a to the first, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And, you know, maybe it's 17, maybe it's pi. I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? dividing by a, right? So if you're going from a to the first to a to the zero, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the first, which is just a, and divide by a, right, so we're just going to go-- we're just going to divide it by a, what is a divided by a? Well, it's just 1. So that's where the definition-- or that's one of the intuitions behind why something to the 0-th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable, but now let's go into So what should a to the negative 1 equal? Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a. So let's divide by a again, so 1/a. a to the 0 is one, so what's 1 divided by a? It's 1/a. Now, let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, we want-- you know, it'd be silly now to Every time we decrease the exponent, we're dividing by a, so to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? If you take 1/2 and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully, that gave you a little intuition as to why-- well, first of all, you know, the big mystery is, you know, something to the 0-th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep"
    },
    {
        "Q": "At 3:33, Sal says \"when you take 1/a and divide by a you get 1/a^2\". Why is this?\n",
        "A": "Walk it thru using the rules to divide fractions. 1/a divided by a/1 = 1/a * 1/a = 1/a^2 Hope this helps.",
        "video_name": "Tqpcku0hrPU",
        "timestamps": [
            213
        ],
        "3min_transcript": "dividing by a, right? So if you're going from a to the first to a to the zero, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the first, which is just a, and divide by a, right, so we're just going to go-- we're just going to divide it by a, what is a divided by a? Well, it's just 1. So that's where the definition-- or that's one of the intuitions behind why something to the 0-th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable, but now let's go into So what should a to the negative 1 equal? Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a. So let's divide by a again, so 1/a. a to the 0 is one, so what's 1 divided by a? It's 1/a. Now, let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, we want-- you know, it'd be silly now to Every time we decrease the exponent, we're dividing by a, so to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? If you take 1/2 and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully, that gave you a little intuition as to why-- well, first of all, you know, the big mystery is, you know, something to the 0-th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep And that's the same reason why they defined negative exponents in this way. And what's extra cool about it is not only does it retain this pattern of when you decrease exponents, you're dividing by a, or when you're increasing exponents, you're multiplying by a, but as you'll see in the exponent rules videos, all of the exponent rules hold. All of the exponent rules are consistent with this definition of something to the 0-th power and this definition of something to the negative power. Hopefully, that didn't confuse you and gave you a little bit of intuition and demystified something that, frankly, is quite mystifying the first time you learn it."
    },
    {
        "Q": "at 4:30pm, how do you write a percentage as a fraction or mix number in simplest form? for ex: 146.8 into a fraction or mixed number in simplest form\n",
        "A": "Since the 8 is in the tenths place you create a fraction with 1468 in the numerator and 10 (for the tenths place) in the denominator. This gives you the fraction 1468/10. Now you reduce the fraction by dividing numerator and denominator by 2. This gives you 734/5 as an improper fraction. If you want a mixed number you leave the 146 as is and create a fraction using the 8. This gives you 146 8/10 which reduces to 146 4/5.",
        "video_name": "-gB1y-PMWfs",
        "timestamps": [
            270
        ],
        "3min_transcript": "It's 0.18. You could view this as 1 tenth and 8 hundredths, which is the same thing, or 10 hundredths and 8 hundredths, which is 18 hundredths. So this is written in decimal form. And if we write it as a simplified fraction, we need to see if there is a common factor for 18 and 100. And they're both even numbers, so we know they're both divisible by 2, so let's divide both the numerator and the denominator by 2. So we have 18 divided by 2 over 100 divided by 2. And we're going to get 18 divided by 2 is 9. 100 divided by 2 is 50. And I don't think these guys share any common factors. 50 is not divisible by 3. 9 is only divisible by 3 and 1 and 9. So this is the fraction in simplest form. So we have 18% is the same thing as 0.18, which is the Now, I went through a lot of pain here to show you that this really just comes from the word, from percent, from per 100. But if you ever were to see this in a problem, the fast way to do this is to immediately say, OK, if I have 18%, you should immediately say, anything in front of the percent-- that's that anything, whatever this anything is-- it should be equal to that anything. In this case it's 18/100. And another way to think about it, you could view this as 18.0%. I just added a trailing zero there, just so that you see the decimal, really. But if you want to express this as a decimal without the percent, you just move the decimal to the left two spaces. this becomes 0.18. Or you could immediately say that 18% as a fraction is 18/100. When you put it in simplified form, it's 9/50. But you should also see that 18/100, and we have seen this, is the exact same thing as 18 hundredths, or 0.18. Hopefully, this made some connections for you and didn't confuse you."
    },
    {
        "Q": "At 0:57 why do you devide 36 and 100 by 4?\n",
        "A": "Dividing 36 and 100 by 4 is just simplifying it to its simplest form, which is 9/20.",
        "video_name": "OS1g4PDdNdM",
        "timestamps": [
            57
        ],
        "3min_transcript": "Let's see if we can write 0.36 as a fraction. There are several ways of doing it. The way I like to do it is to say, well, 0.36, this is the same thing as 36 hundredths. Or one way to think about it is this is in the hundredths place. This is in the tens place, or you could view this as 30 hundredths. You could view this as 3 tens, or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36/100. And we've already expressed it as a fraction, but now we could actually simplify it because both 36 and 100 have some common factors. They're both divisible by, well, looks like they're both divisible by 4. So if we could divide the numerator by 4 and the denominator by 4, we're doing the same thing to both. So we're not changing the value of the fraction. 36 divided by 4 is 9, and then 100 divided by 4 is 25. seem to share any other common factors. And so we've written it in simplified form, and we're done."
    },
    {
        "Q": "at 5:31 how is the moon larger enough to block the sun? Isn't it WAY bigger?\n",
        "A": "The sun is bigger! But the Moon is closer to us, so when the Moon is in line with the Earth and Sun, the Sun appears to be blocked out because the Moon is right in it s way. If something is super far away, but you can still see it, it appears really small. But as it moves closer to you, it gets bigger. So because the Moon is closer to us than the Sun, it appears as if the Moon is just as big as the Sun.",
        "video_name": "s8_14yxp1lQ",
        "timestamps": [
            331
        ],
        "3min_transcript": "35 is five times seven and 21 is three times seven. So, you're multiplying by seven up here and here, you have a seven in the denominator, you're dividing by seven, so they're going to cancel out. So, this is going to simplify to five times 66 over 3, and then we could simplify it even more, because 66 is the same thing as three times 22. Three times 22 and so, you have a three in the numerator, you're multiplying by three and three in the denominator, dividing by three. Three divided by three is one, so you're left with five times 22, which is 110. So, it would take her m minutes to eat 35 hot dogs at the same pace. Now, when some of you might have tackled it, you might have had a different equation set up here. Instead of thinking of hot dogs per minute, you might have thought about minutes per hot dog. of minutes per hot dog, you might have said, ok look, it took Mika 66 minutes to eat 20, to eat 21 hot dogs,and it's gonna take her m minutes to eat 35 hot dogs and if it's the same pace, then these two rates are going to be equal. They have to be the same pace. And so, then you can solve for m and actually, this one's easier to solve for m, you just multiply both sides by 35. Multiply both sides by 35 and you're left with, on the right hand side you're left with just an m, and on the left hand side, same, same idea. You're taking 35, you have 35 times 66/21, which we already figured out is 110. So, 110 is equal, is equal to m. So, once again, multiple ways to tackle it, but it's important that we got the same answer."
    },
    {
        "Q": "\nAt 0:09, what does Sal mean by \" one term\"?",
        "A": "a term is anything in a equation for example: x is a term x + 3 are 2 terms remember, terms are only separated by addition and subtraction signs",
        "video_name": "p_61XhXdlxI",
        "timestamps": [
            9
        ],
        "3min_transcript": "Express the area of a rectangle with length 4xy and width 2y as a monomial. Monomial just means just one term. So let's think about a rectangle. So let me draw a rectangle here. And they're telling us that the length is 4xy and they're telling us that the width is 2y. And just as a bit of a refresher, we know the area of a rectangle is just the width times the length, or the height times the width, or however you want to view it-- just the product of its two dimensions. So the area of this rectangle, the area is going to be equal to this length, 4xy times the width times 2y. We can simplify this. We have a 4 times a 2. you can switch the order however you like, as long as it's all multiplication. So 4 times 2, that gives us 8. Then we have this x sitting here. That is the only x we have there, so it's 8 times x. And then we have a y here. We could view that as y to the first. And then we have another y there. We could view that as y to the first. So y times y. You could view that as y squared. Or you could say look, y to the first times y to the first-- same base, add the two exponents-- 1 plus 1 is 2. So it's 8xy squared. This y squared covers both that y and that y over there. And we're done. We've expressed the area of this rectangle as a monomial."
    },
    {
        "Q": "At 6:30\n\nCould you also find the vertex by using the -b/2a that Sal used in the previous video?\n",
        "A": "Yes, but I m pretty sure that you use -b/2a only to find the x-value of the vertex. Well, I guess you do end up finding the vertex, because you could just then plug the x-value into the quadratic to find the y-value of the vertex.",
        "video_name": "FksgVpM_iXs",
        "timestamps": [
            390
        ],
        "3min_transcript": "The discriminant is this part-- b squared minus 4ac. We see the discriminant is negative, there's no solution, which means that these two guys-- these two equations-- never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Let's think a little bit about why that happened. This one is already in kind of our y-intercept form. It's an upward opening parabola, so it looks something like this. I'll do my best to draw it-- just a quick and dirty version of it. Let me draw my axes in a neutral color. Let's say that this right here is my y-axis, that right there is my x-axis. x and y. This vertex-- it's in the vertex form-- occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. It's an upward opening parabola. So this will look something like this. I don't know the exact thing, but that's close enough. Now, what will this thing look like? It's a downward opening parabola and we can actually put this in vertex form. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So, y is equal to-- we could factor in a negative 1-- negative x squared minus 2x plus 2. Actually, let me put the plus 2 further out-- plus 2, all the way up out there. Then we could say, half of negative 2 is negative 1. You square it, so you have a plus 1 and then a minus 1 there. This part right over here, we can rewrite as x minus 1 squared, so it becomes negative x minus 1 squared. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. Here the vertex occurs at x is equal to 1, y is equal to negative 1. The vertex is there, and this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. This is its minimum point. And it's above this guy's maximum point. So they will never intersect, so there is no solution to this system of equations."
    },
    {
        "Q": "at2:15-2:45 i am confused\n",
        "A": "It s easy, just multiply the numerator by the numerator and the denominator by the denominator, but first convert the mixed number to an improper fraction if neccesary",
        "video_name": "XaJQse2u5TQ",
        "timestamps": [
            135,
            165
        ],
        "3min_transcript": "We've already seen that the fraction 2/5, or fractions like the fraction 2/5, can be literally represented as 2 times 1/5, which is the same thing, which is equal to literally having two 1/5s. So 1/5 plus 1/5. And if we wanted to visualize it, let me make a hole here and divide it into five equal sections. And so this represents two of those fifths. This is the first of the fifths, and then this is the second of the fifths, Literally 2/5, 2/5, 2/5. Now let's think about something a little bit more interesting. What would 3 times 2/5 represent? 3 times 2/5. And I encourage you to pause this video and, based on what we just did here, think about what you think this would be equivalent to. as-- so let me just rewrite this as instead of 3 times 2/5 written like this, let me write 2/5 like that-- so this is the same thing as 3 times 2 times 1/5. And multiplication, we can multiply the 2 times the 1/5 first and then multiply by the 3, or we can multiply the 3 times the 2 first and then multiply by the 1/5. So you could view this literally as being equal to 3 times 2 is, of course, 6, so this is the same thing as 6 times 1/5. And if we were to try to visualize that again, so that's a whole. That's another whole. Each of those wholes have been divided into five equal sections. And so we're going to color in six of them. fifth 1/5-- and that gets us to a whole-- and then we have 6/5 just like that. So literally 3 times 2/5 can be viewed as 6/5. And of course, 6 times 1/5, or 6/5, can be written as-- so this is equal to, literally-- let me do the same color-- 6/5, 6 over 5. Now you might have said, well, what if we, instead of viewing 2/5 as this, as we just did in this example, we view 2/5 as 1/5 plus 1/5, what would happen then? Well, let's try it out. So 3 times 2/5-- I'll rewrite it-- 3 times 2/5, 2 over 5, is the same thing as 3 times 1/5 plus 1/5."
    },
    {
        "Q": "Why, at 5:30, did Sal use the word \"Dimension?\" It makes it sound like the rectangle's some sort of galactic unknown universe where extra-terrestial life lurks unseen! Shouln't it be \"Area?\"\n",
        "A": "Dimension is the width, length, or height of an object. At 5:30 he was referring to the width of the square, saying that it s eight units",
        "video_name": "FKJjqEdfB9Y",
        "timestamps": [
            330
        ],
        "3min_transcript": "I can take each of them and divide by two. So what is, what is this, what is this dimension going to be, right over here? Actually let me do that in a different color. What is this dimension right over here going to be? Well two times that is going to be a hundred, so this is going to be 100. And how did we get that? Well we got that by 200 divided by two. 200 divided by two is 100. What's 60 divided by two? Well, 60 divided by two is going to be 30. So this part of the field is going to be 30 in that direction and two in this direction. And once again, not drawn to scale. And then finally, what's this section going to be? It's going to be, it's going to be, eight divided by two, which is four. Notice, hundred times two is 200, 30 times two is 60, four times two is eight. is going to be 100, plus 30, plus four, or 134. Now we've already seen other ways of coming up with this. You say, \"Look, two hundreds divided by two is 100, \"six tens divided by two is three tens, \"eight ones divided by two is four ones.\" And that's exactly what we just did over here, but we visualized it using this kind of a rectangle, breaking it up into chunks that are, maybe easier to imagine dividing by two. We broke it up into two hundreds, two hundreds right over here. We broke it up into six tens, or 60, or 60 right over here. And we broke it up into eight ones, eight ones. We broke up the area, and then we took each of those areas and we divided it by two to find that part of the length. And when you add them all together, you get the entire, you get the entire length. where you take each of the place values and you break up your field or your rectangle like that, but you could do it other ways. You don't always have to break it up that much. For example, let's say, let's say that this... let's say that this area is 856 square units. Square units... Now let's say that this dimension right over here is eight, is eight units. Is eight units. So how could we break this up so it's easier to think about what the other dimension would be, what the length is going to be? is going to be 856 divided by eight. 856 divided by eight is this length right over here. Well you could break it up in to eight hundreds, five tens, and six ones, but you might notice, \"Well five tens, it's not so easy to divide that by eight.\""
    },
    {
        "Q": "where does 0 and 1 come from in 2:01? Is this a formula or rule? Why do we use 0 and 1 in this example and can other numbers be substituted?\n",
        "A": "Sal just plucked them out the air. Any 2 numbers would do, which could be substituted in to the formula to workout what y should equal. I m guessing he chose small numbers so the graph he had to draw wouldn t be to big, but any 2 numbers would do 5, 7, 100 or 1000000 as long as the value of x can be used to calculate the corresponding value of y",
        "video_name": "SSNA9gaAOVc",
        "timestamps": [
            121
        ],
        "3min_transcript": "Is 3 comma negative 4 a solution to the equation 5x plus 2y is equal to 7? So there's two ways to think about it. One, you could just substitute this x and y value into this equation to see if it satisfies-- and then we'll do that way first-- and the other way is if you had a graph of this equation, you could see if this point sits on that graph, which would also mean that it is a solution to this equation. So let's do it the first way. So we have 5x plus 2y is equal to 7, so let's substitute. Instead of x, let us put in 3 for x. So 5 times 3 plus 2 times y-- so y is negative 4-- plus 2 times negative 4 needs to be equal to 7. I'll put a question mark here, because we're not sure yet if it does. So 5 times 3 is 15, and then 2 times negative 4 is negative 8. and this needs to be equal to 7. And of course, 15 minus 8 does equal 7, so this all works out. This is a solution, so we've answered the question. But I also want to show you, this way we just did it by substitution. If we had the graph of this equation, we could also do it graphically. So let's give ourselves the graph of this equation, and I'll do that by setting up a table. There's multiple ways to graph this. You could put it in a slope-intercept form and all of the rest, but I'll just set up a table of x and y values. And I'll graph it, and then given the graph, I want to see if this actually sits on it. And obviously it will, because we've already shown that this works. In fact, we could try the point 3, negative 4, and that actually is on the graph. We could do it on our table, but I won't do that just yet. I'm just going to do this to give ourselves a graph. So let's try it when x is equal to 0. We have 5 times 0 plus 2 times y is equal to 7. to be-- so you're going to have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you have 5 plus 2y is equal to 7. If you subtract 5 from both sides, you get 2y is equal to 2. You get y is equal to 1. So when x is 1, y is 1, and when x is-- well let's try-- well that's actually enough for us to graph. We could keep doing more points. We could even put the point 3, negative 4 there, but let's just try to graph it in this very rough sense right here. So let me draw my x-axis, and then this right over here is my y-axis."
    },
    {
        "Q": "\nAt 1:44, why cannot the answer be -1 to both equations. It should be simple right?",
        "A": "Hey Asish, Sal is trying to demonstrated the fact that one of them must be equal to zero. -1 is not even somewhat related to making the equations zero. I will demonstrate in the equation he is using. (2x - 1) = 0 substitute -` (2 * -1 - 1) = 0 (-2 - 1) = 0 -3 = 0 This is not true, so -1 does not work in the first part. (x + 4) = 0 substitute -1 (-1 + 4) = 0 (3) = 0 3 = 0 That is not possible, so -1 does not work in this part too. Hope that helps! - JK #YouCanLearnAnything",
        "video_name": "-lWVpoPaPBc",
        "timestamps": [
            104
        ],
        "3min_transcript": "- [Instructor] Let's say that we've got the equation two X minus one times X plus four is equal to zero. Pause this video and see if you can figure out the X values that would satisfy this equation, essentially our solutions to this equation. Alright, now let's work through this together. So at first, you might be tempted to multiply these things out, or there's multiple ways that you might have tried to approach it, but the key realization here is that you have two things being multiplied, and it's being equal to zero. So you have the first thing being multiplied is two X minus one. This is expression is being multiplied by X plus four, and to get it to be equal to zero, one or both of these expressions needs to be equal to zero. Let me really reinforce that idea. If I had two variables, let's say A and B, and I told you A times B Well, can you get the product of two numbers to equal zero without at least one of them being equal to zero? And the simple answer is no. If A is seven, the only way that you would get zero is if B is zero, or if B was five, the only way to get zero is if A is zero. So you see from this example, either, let me write this down, either A or B or both, 'cause zero times zero is zero, or both must be zero. The only way that you get the product of two quantities, and you get zero, is if one or both of them is equal to zero. I really wanna reinforce this idea. I'm gonna put a red box around it so that it really gets stuck in your brain, and I want you to think about why that is. Try to come up with two numbers. Try to multiply them so that you get zero, is going to need to be zero. So we're gonna use this idea right over here. Now this might look a little bit different, but you could view two X minus one as our A, and you could view X plus four as our B. So either two X minus one needs to be equal to zero, or X plus four needs to be equal to zero, or both of them needs to be equal to zero. So I could write that as two X minus one needs to be equal to zero, or X plus four, or X, let me do that orange. Actually, let me do the two X minus one in that yellow color. So either two X minus one is equal to zero, or X plus four is equal to zero. X plus four is equal to zero,"
    },
    {
        "Q": "At 10:39, you take the square root of both sides - yet keep the less than/equal sign unchanged.\nCouldn't it be that there might be a case which \"less than\" should turn into \"greater than\" ? In case ||x+y|| <0 ?\n",
        "A": "|x+y| is never < 0.",
        "video_name": "PsNidCBr5II",
        "timestamps": [
            639
        ],
        "3min_transcript": "I said that this thing that I wrote over here, this is the same as that. So this thing up here, which is the same as that, which is less than that also. So we can write that the magnitude of x plus y squared and not the magnitude, the length of the vector x plus y squared is less than this whole thing that I wrote out here. Or less than or equal to. Now, what is this thing? Remember, I mean this might look all fancy with my little double lines around everything. But these are just numbers. This length of x squared, this is just a number. Each of these are numbers and I can just say hey, look, this looks like a perfect square to me. This term on the right-hand side is the exact same thing as the length of x plus the length of y. If you just squared this out you'll get x squared plus 2 times the length of x times the length of y plus y squared. So our length of the vector x plus y squared is less than or equal to this quantity over here. And if we just take the square root of both sides of this, you get the length of our vector x plus y is less than or equal to the length of the vector x by itself plus the length of the vector y. And we call this the triangle inequality, which you might have remembered from geometry. Now why is it called the triangle inequality? Well you could imagine each of these to be separate side of a triangle. We can draw this in R2. Let me turn my graph paper on. Let me see where the graphs show up. If I turn my graph paper on right there, maybe I'll draw it here. So let's draw my vector x. So let's say that my vector x look something like this. Let's say that's my vector x. It's a vector 2, 4. So that's my vector x. And let's say my vector y-- well I'm just going to do it head to tail because I'm going to add the two. So my vector y-- I'm going to do it in nonstandard position. Let's say it's look something like-- let's say my vector y looks something like this. Draw it properly. That's my vector y. What does x plus y look like? And remember, I can't necessarily draw any two vectors on a two-dimensional space like this."
    },
    {
        "Q": "\nAt 1:34, Sal multiplied both numerator and denominator by 10, however he did not multiplied the left hand side of the equation by 10 too! And so, how is the equality still maintained?",
        "A": "When he multiplies the top and bottom by 10, it is equivalent to multiplying by 1, so you don t need to multiply the left hand side to maintain equality. Take a really simple example: 5 = 10/2 I can multiply the RHS by 10/10 to give: 5= 100/20 (which still holds) But I cannot multiply the LHS by just 10, ie: 50 = 100/20 which is the wrong answer But I can do this: 50/10 = 100/20 The key here is as long as you re multiplying by what is essentially 1, you don t need to multiply the other side by anything.",
        "video_name": "a3acutLstF8",
        "timestamps": [
            94
        ],
        "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12"
    },
    {
        "Q": "At 2:21 my answer was 0.5 instead of 1/2 because I chose to divide the decimals. Since that's equivalent would it still be correct to write it in decimal form?\n",
        "A": "I think that it would be fine if you got 0.5 instead of 1/2 because they are equivalent.",
        "video_name": "a3acutLstF8",
        "timestamps": [
            141
        ],
        "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12"
    },
    {
        "Q": "2:07 How can you cross both of those lines when they are not equal to each other and call them parallel?\n",
        "A": "I am not sure what you mean by not equal to each other. Lines, while he does not show it with arrows on the end, go forever.",
        "video_name": "V0xounKGEXs",
        "timestamps": [
            127
        ],
        "3min_transcript": "Let's think a little bit about two terms that you'll see throughout your geometry, and really, mathematical career. One is the idea of things being perpendicular. And usually, people are talking about perpendicular. Actually I'm misspelling it-- perpendicular lines, and the idea of parallel lines. So perpendicular lines are two lines that intersect at a right angle. So what am I talking about? So let's say that this is one line right over here and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here and this line right over here, the way I've drawn them, are parallel lines. They aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other. So these two are parallel. If we have two lines that, let's say, they intersect, but they don't intersect at a right angle, so let's say we have that line and we have this line right over here, and they're clearly not intersecting at a right angle, then we call these neither perpendicular These lines just intersect."
    },
    {
        "Q": "\nat 1:00 doesn't the numbers in A add by itself too?",
        "A": "In the example, the pattern of the numbers in A is a doubling one, that is (1, 2, 4, 8, 16, 32). This means that the increase is one of multiplication (each interval is 2 times larger than the previous one) rather than one of simple addition.",
        "video_name": "Muba9-W2FOQ",
        "timestamps": [
            60
        ],
        "3min_transcript": "Below are ordered pairs that represent the first six terms of two given patterns. The first value in each pair is a term from pattern A. And the second value is a term from pattern B. In the answer box, there are different statements about the two patterns. Choose all correct statements. So let's think about what's going on here. They said the first term is pattern A. So the first term in each of these coordinates is pattern A, or in each pair is pattern A. So pattern A goes from 1, to 2, to 4, to 8, to 16, to 32. So it looks like pattern A, to go from the first term to the second term, we multiplied by 2. And then to go from the second to the third term, we also multiplied by 2. And we just keep multiplying by 2. And we just keep doing that. 8 times 2 is 16. 16 times 2 is 32. Now let's think about what's going on with pattern B. And it's just always 3. So there's a couple of ways you can think about it. You could just say, pattern B's always 3. You could say pattern B starts at 3, and we're just adding 0 every time. Or you could say that pattern B starts at 3, and we are multiplying by 1 every time. Either of those would give you just 3 showing up over and over again. So now that we've looked at these pairs, we show the corresponding terms for pattern A and pattern B, let's look at the choices here and see which of these apply. In pattern A, you can get from any term to the next by multiplying by a constant number. Well, that looks right. We go from the first term to the second term by multiplying by 2. Then we multiply by 2 again to get to the third term. Then we keep multiplying by 2. So that constant number that we're multiplying by to get to the next term is 2. So this looks right. The next pair should be 52 comma 3. If we keep doubling for pattern A-- so this is going to be times 2. 32 times 2 is 64. And then if we'd say that this is 1 times the previous term, we're just going to get a 3 again. So it should be 64 comma 3 should be the next one. They say the next pair should be 52 comma 3. So that's not right. If we graph the pairs, the points will be on the same line. So let's think about that a little bit. Let's think about that. So this is my vertical axis. This is my horizontal axis. On the horizontal axis, I will graph pattern A. And on my vertical axis, I will graph pattern B. And let's see. Pattern A goes all the way up to 32. So I'm going to try my best here. So let's say that this is 32. Then half of that is going to be 16. Half of that is going to be 8."
    },
    {
        "Q": "At 1:17, Sal said something about a \"negative squared is just going to be a 1\". I can't wrap my head around how this actually happened in that particular equation (y= -sqrt(x-3)). Anyone who can clarify this one?\n",
        "A": "What he s talking about is the assumed coefficient in front of the radical. Writing -sqrt(x-3) is the same thing as writing -1 * sqrt(x-3). So if you square -1, you get a positive 1.",
        "video_name": "QWLcNxQ3KvQ",
        "timestamps": [
            77
        ],
        "3min_transcript": "In the relation x is equal to y squared plus 3, can y be represented as a mathematical function of x? So the way they've written it, x is being represented as a mathematical function of y. We could even say that x as a function of y is equal to y squared plus 3. Now, let's see if we can do it the other way around, if we can represent y as a function of x. So one way you could think about it is you could essentially try to solve for y here. So let's do that. So I have x is equal to y squared plus 3. Subtract 3 from both sides, you get x minus 3 is equal to y squared. Now, the next step is going to be tricky, x minus 3 is equal to y squared. So y could be equal to-- and I'm just going to swap the sides. y could be equal to-- if we take the square root of both sides, it could be the positive square root of x minus 3, or it could be the negative square root. If you don't believe me, square both sides of this. You'll get y squared is equal to x minus 3. Square both sides of this, you're going to get y squared is equal to-- well, the negative squared is just going to be a positive 1. And you're going to get y squared is equal to x minus 3. So this is a situation here where for a given x, you could actually have 2 y-values. Let me show you. Let me attempt to sketch this graph. So let's say this is our y-axis. I guess I could call it this relation. This is our x-axis. And this right over here, y is a positive square root of x minus 3. That's going to look like this. So if this is x is equal to 3, it's going to look like this. That's y is equal to the positive square root of x minus 3. And this over here, y is equal to the negative square root of x minus 3, is going to look something like this. because it's going to essentially be the mirror image if you flip over the x-axis. So it's going to look something like this-- y is equal to the negative square root of x minus 3. And this right over here, this relationship cannot be-- this right over here is not a function of x. In order to be a function of x, for a given x it has to map to exactly one value for the function. But here you see it's mapping to two values of the function. So, for example, let's say we take x is equal to 4. So x equals 4 could get us to y is equal to 1. 4 minus 3 is 1. Take the positive square root, it could be 1. Or you could have x equals 4, and y is equal to negative 1. So you can't have this situation. If you were making a table x and y as a function of x,"
    },
    {
        "Q": "\nAt 5:44, where did the three came from?",
        "A": "There was three chairs",
        "video_name": "DROZVHObeko",
        "timestamps": [
            344
        ],
        "3min_transcript": "five different people in five different chairs, and we cared which seat they sit in, we had this five factorial. Factorial is kind of neat little operation there. How can I relate factorial to what we did just now? It looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. So one way to think about what we just did, is we just did five times four times three times two times one, but of course we actually didn't do the two times one, so you could take that and you could divide by two times one. If you did that, this two times one would cancel with that two times one and you'd be left with five times four times three. The whole reason I'm writing this way is that now I can write it in terms of factorial. I could write this as five factorial, five factorial, over two factorial, over two factorial. But then you might have the question I have three seats. Where did this two come from? Well, think about it. I multiplied five times four times three, I kept going until I had that many seats, and then I didn't do the remainder. So the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs, so I was trying to put five things in three places. Five minus three, that gave me two left over. So I could write it like this. I could write it as five ... Let me use the same colors. I could write it as five factorial over, over five minus three, which of course is two, five minus three factorial. Another way of thinking about it, if we wanted to generalize, is if you're trying to figure out the number of permutations and there's a bunch of notations for writing this, if you're trying to figure out the number of permutations or the number of permutations you could put n people in r seats, and there's other notations as well, well, this is just going to be n factorial over n minus r factorial. Here n was five, r was three. Five minus three is two. Now, you'll see this in a probability or a statistics class, and people might memorize this thing. It seems like this kind of daunting thing. I'll just tell you right now, the whole reason why I just showed this to you is so that you could connect it with what you might see in your textbook, or what you might see in a class, or when you see this type of formula, you see that it's not some type of voodoo magic. But I will tell that for me, personally, I never use this formula. I always reason it through, because if you just memorize the formula, you're always going, wait, does this formula apply there? What's n? What's r?"
    },
    {
        "Q": "\nAt 4:51, when Sal writes 5!/2!, couldn't he just write (5/2)! ?",
        "A": "No, Sal could not do this. Factorial does not distribute over division. Example: 4!/2! = (1*2*3*4)/(1*2) = 12, but (4/2)! = 2! = 2*1 = 2. In this particular situation, 5!/2! = (1*2*3*4*5)/(1*2) = 60, but (5/2)! = 2.5! which turns out be approximately 3.32 from using a calculator (the definition of fractional factorials is much more complex than the definition of integer factorials). Have a blessed, wonderful day!",
        "video_name": "DROZVHObeko",
        "timestamps": [
            291
        ],
        "3min_transcript": "so four people could sit in seat two. So we have five times four scenarios where we've seated seats one and seat two. For each of those 20 scenarios, how many people could sit in seat three? Well, we haven't sat, we haven't seaten or sat three of the people yet, so for each of these 20, we could put three different people in seat three, so that gives us five times four times three scenarios. So this is equal to five times four times three scenarios, which is equal to, this is equal to 60. So there's 60 permutations of sitting five people in three chairs. Now this, and my brain, whenever I start to think in terms of permutations, I actually think in these ways. I just literally draw it out because I don't like formulas. I like to actually conceptualize and visualize what I'm doing. five different people in five different chairs, and we cared which seat they sit in, we had this five factorial. Factorial is kind of neat little operation there. How can I relate factorial to what we did just now? It looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. So one way to think about what we just did, is we just did five times four times three times two times one, but of course we actually didn't do the two times one, so you could take that and you could divide by two times one. If you did that, this two times one would cancel with that two times one and you'd be left with five times four times three. The whole reason I'm writing this way is that now I can write it in terms of factorial. I could write this as five factorial, five factorial, over two factorial, over two factorial. But then you might have the question I have three seats. Where did this two come from? Well, think about it. I multiplied five times four times three, I kept going until I had that many seats, and then I didn't do the remainder. So the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs, so I was trying to put five things in three places. Five minus three, that gave me two left over. So I could write it like this. I could write it as five ... Let me use the same colors. I could write it as five factorial over, over five minus three, which of course is two, five minus three factorial. Another way of thinking about it, if we wanted to generalize, is if you're trying to figure out the number of permutations and there's a bunch of notations for writing this, if you're trying to figure out the number of permutations"
    },
    {
        "Q": "At 2:10 , what does Sal mean by \"0 degree terms?\"\n",
        "A": "Basically, the degree refers to the power of the term s variable. So 2x is a 1-degree term and 3x^2 is a second degree term. 0-degree terms simply are constant terms, like the 5 in 4x + 5. They are 0-degree because you can think of the 5 as 5x^0, which is just 5*1.",
        "video_name": "n34dqyVCXs4",
        "timestamps": [
            130
        ],
        "3min_transcript": "f of x is equal to 9 minus x. Actually, it should be 9 minus x squared. And g of x is equal to 5x squared plus 2x plus 1. And then they say find f plus g of x. And this looks a little bit bizarre. What kind of notation is this? And that's really the core of this problem is just to realize that when someone writes f plus g in parentheses like this of x this is just notation for-- so f plus g of x, I just rewrote it. This is the same thing. This is just a kind of shorthand notation for f of x plus g of x. You could view f plus g as a new function that's created by adding the other two functions. But when you view it like this-- so this is really what we have to find. Then, you just have to add these two functions. So f of x, they've given the definition right over there, is 9 minus x squared. And g of x, they've given the definition right over here, So when you add f of x to g of x, this is going to be equal to-- and I'm just rewriting a lot of things just to make it clear. The f of x part is 9 minus x squared. And then you have the plus. I'll do that in yellow. Plus the g of x part, which is 5x squared plus 2x plus 1. And now we can just simplify this a little bit. We can say-- let me just get rid of the parentheses. I'll just rewrite it. So this is equal to 9 minus x squared plus-- since this is a positive, we don't have to worry about the parentheses. So plus 5x squared plus 2x plus 1. And then we have two x squared terms or second degree terms. We have 5x squared. And then we have negative x squared. 5x squared minus x squared is 4x squared. So you get 4x squared when you combine these two terms. And then you have two constant terms, or 0 degree terms. So you have the 9 and you have the 1. And 9 plus 1 gets us to 10. So we're done. f plus g of x is equal to 4x squared plus 2x plus 10. Notice this is a new function that's created by summing the function definitions of f and g."
    },
    {
        "Q": "At 1:10, how did you determine the number range for the buckets?\n",
        "A": "Looking that the time you referenced, do you mean the ranges of ages that Sal put in the t-chart? If so, the reason why is that is it is really easy to group everything in tens. There is no right way to determine the number range. It is just a matter of preference.",
        "video_name": "gSEYtAjuZ-Y",
        "timestamps": [
            70
        ],
        "3min_transcript": "- [Voiceover] So let's say you were to go to a restaurant and just out of curiosity you want to see what the makeup of the ages at the restaurant are. So you go around the restaurant and you write down everyone's age. And so these are the ages of everyone in the restaurant at that moment. And so you're interested in somehow presenting this, somehow visualizing the distribution of the ages, because you want just say, well, are there more young people? Are there more teenagers? Are there more middle-aged people? Are there more seniors here? And so when you just look at these numbers it really doesn't give you a good sense of it. It's just a bunch of numbers. And so how could you do that? Well one way to think about it, is to put these ages into different buckets, and then to think about how many people are there in each of those buckets? Or sometimes someone might say how many in each of those bins? So let's do that. So let's do buckets or categories. So, I like, sometimes it's called a bin. So the bucket, I like to think of it more of as a bucket, the bucket and then the number in the bucket. The number in the bucket. It's the, oops. It's the number (laughing), it's the number in the bucket. Alright. So let's just make buckets. Let's make them 10 year ranges. So let's say the first one is ages zero to nine. So how many people... Why don't we just define all of the buckets here? So the next one is ages 10 to 19, then 20 to 29, then 30 to 39, and 40 to 49, 50 to 59, let me make sure you can read that properly, then you have 60 to 69. And I think that covers everyone. I don't see anyone 70 years old or older here. So then how many people fall into the zero to nine-year-old bucket? Well it's gonna be one, two, three, four, five, six people fall into that bucket. How many people fall into the... How many people fall into the 10 to 19-year-old bucket? One, two, three. Three people. And I think you see where this is going. What about 20 to 29? So that's one, two, three, four, five people. Five people fall into that bucket. Alright, what about 30 to 39? We have one, and that's it. Only one person in that 30 to 39 bin or bucket or category. Alright, what about 40 to 49? We have one, two people. Two people are in that bucket. And then 50 to 59. Let's see, you have one, two people. Two people. And then finally, finally, ages 60-69. Let me do that in a different color. 60 to 69. There is one person, right over there."
    },
    {
        "Q": "\nIsn't it enough to just stop at 4:35, the part where you claim a is a multiple of an integer k times p. Since a/b is said to be 2 integers with no factors in common in the beginning of the video wouldn't a=kp be impossible, since p is prime then this claims p=a/k, and a prime cannot be defined as a integer divided by an integer?",
        "A": "Any prime p can be represented as p/1, both of which are integers.",
        "video_name": "W-Nio466Ek4",
        "timestamps": [
            275
        ],
        "3min_transcript": "I'm just saying that a is some integer right over here. So that's the prime factorization of a. What is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors. pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared times p is equal to a squared. that as some integer k times p. So we can rewrite that as some integer k times p. And so, let's see, if we were to multiply this out. So we get b squared times p-- and you probably see where this is going-- is equal to k squared times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around."
    },
    {
        "Q": "i didn't understand how b\u00c2\u00b2=k\u00c2\u00b2p in 5:45\n",
        "A": "If you understand why a = kp, then it s just simple algebra from the video, starting around 5:14. If you don t know why a = kp, ask about that.",
        "video_name": "W-Nio466Ek4",
        "timestamps": [
            345
        ],
        "3min_transcript": "pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared times p is equal to a squared. that as some integer k times p. So we can rewrite that as some integer k times p. And so, let's see, if we were to multiply this out. So we get b squared times p-- and you probably see where this is going-- is equal to k squared times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around. which is still going to be an integer, times p. So b squared must be a multiple of p. So this lets us know that b squared is a multiple of p. And by the logic that we applied right over here, that lets us know that b is a multiple of p. And that's our contradiction, or this establishes our contradiction that we assumed at the beginning. We assumed that a and b are co-prime, that they share no factors in common other than 1. We assumed that this cannot be reduced. But we've just established, just from this, we have deduced that is a multiple of p and b is a multiple of p. Which means that this fraction can be reduced. We can divide the numerator and the denominator by p. So that is our contradiction. We started assuming it cannot be reduced,"
    },
    {
        "Q": "Just noticed that in this vid about 9:13, when simplifying and rearranging, the \"b\" term was dropped. Just so everyone watching does not get too lost.\nThanks for the vids tho. Actually beginning to understand why I did this the first time.\n",
        "A": "I saw that too, good that you pointed it out.",
        "video_name": "9kW6zFK5E5c",
        "timestamps": [
            553
        ],
        "3min_transcript": "What I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here. So if I multiply this bottom equation times 3-- let me just do-- well, actually, I don't want to make things messier, so this becomes a minus 3 plus a 3, so those cancel out. This becomes a 12 minus a 1. So this becomes 12c3 minus c3, which is 11c3. And then this becomes a-- oh, sorry, I was already done. When I do 3 times this plus that, those canceled out. And then when I multiplied 3 times this, I get 12c3 minus a c3, so that's 11c3. And I multiplied this times 3 plus this, so I get 3c minus this, plus b plus a. So what can I rewrite this by? Actually, I want to make something very clear. This c is different than these c1's, c2's and c3's that I had up here. I think you realize that. But I just realized that I used the letters c twice, and I just didn't want any confusion here. So this c that doesn't have any subscript is a different constant then all of these things over here. Let's see if we can simplify this. We have an a and a minus 6a, so let's just add them. So let's get rid of that a and this becomes minus 5a. If we divide both sides of this equation by 11, what do we get? We get c3 is equal to 1/11 times 3c minus 5a. tell you what c3 is. What is c2? c2 is equal to-- let me simplify this equation right here. Let me do it right there. So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3. And if I divide both sides of this by 3, I get c2 is equal to 1/3 times b plus a plus c3. I'll just leave it like that for now. Then what is c1 equal to? I could just rewrite this top equation as if I subtract 2c2 and add c3 to both sides, I get c1 is equal to a"
    },
    {
        "Q": "\naround 1:54, could you be more specific when you say NO X Value is equal to -14",
        "A": "The question was 24x + 80 = 24x - 14 What he is saying is that it doesn t matter what value you give to 24x, there is no value that you could give to the x and then multiply that number by 24 so that 24x would then make 80 = -14",
        "video_name": "zKotuhQWIRg",
        "timestamps": [
            114
        ],
        "3min_transcript": "Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14."
    },
    {
        "Q": "\nIs there a proper name for the operation Sal calls a \"slash\" at 4:19?",
        "A": "Not like. That is a backslash. Pretty simple character. There can t really be any misinterpreting it, though there could always be additional names for it like anything.",
        "video_name": "2B4EBvVvf9w",
        "timestamps": [
            259
        ],
        "3min_transcript": "relative complement of set B in A. And we're going to talk a lot more about complements in the future. But the complement is the things that are not in B. And so this is saying, look, what are all of the things that are not in B-- so you could say what are all of the things not in B but are in A? So once again, if you said all of the things that aren't in B, then you're thinking about all of the numbers in the whole universe that aren't 17, 19, or 6. And actually, you could even think broader. You're not even just thinking about numbers. It could even be the color orange is not in set B, so that would be in the absolute complement of B. I don't see a zebra here in set B, so that would be its complement. But we're saying, what are the things that are not in B but are in A? And that would be the numbers 5, 3, and 12. Now, when we visualized this as B subtracted from A, OK. I could imagine you took the 17 out. You took the 19 out. But what about taking the 6 out? Shouldn't you have taken a 6 out? Or in traditional subtraction, maybe we would end up with a negative number or something. And when you subtract a set, if the set you're subtracting from does not have that element, then taking that element out of it doesn't change it. If I start with set A, and if I take all the 6s out of set A, it doesn't change it. There was no 6 to begin with. I could take all the zebras out of set A; it will not change it. Now, another way to denote the relative complement of set B in A or B subtracted from A, is the notation that I'm about to write. We could have written it this way. A and then we would have had this little figure like this. That looks eerily like a division sign, but this also means the difference between set A and B where we're talking about-- when we write it this way, we're talking about all the things in set A that are not in Or the relative complement of B in A. Now, with that out of the way, let's think about things the other way around. What would B slash-- I'll just call it a slash right over here. What would B minus A be? So what would be B minus A? Which we could also write it as B minus A. What would this be equal to? Well, just going back, we could view this as all of the things in B with all of the things in A taken out of it. Or all of the things-- the complement of A that happens to be in B. So let's think of it as the set B with all of the things in A taken out of it. So if we start with set B, we have a 17."
    },
    {
        "Q": "At 6:05 the symbol for a null set is written as a 0 with a slash going through it. The slash is with the top to the right ( / ). Can it also be written with the top on the left ( \\ ) ?\n",
        "A": "It s better to donate null set as { } or {0}. However, both slash thing is also correct.",
        "video_name": "2B4EBvVvf9w",
        "timestamps": [
            365
        ],
        "3min_transcript": "Or the relative complement of B in A. Now, with that out of the way, let's think about things the other way around. What would B slash-- I'll just call it a slash right over here. What would B minus A be? So what would be B minus A? Which we could also write it as B minus A. What would this be equal to? Well, just going back, we could view this as all of the things in B with all of the things in A taken out of it. Or all of the things-- the complement of A that happens to be in B. So let's think of it as the set B with all of the things in A taken out of it. So if we start with set B, we have a 17. Then we have a 19. But there's a 19 in set A, so we have to take the 19 out. Then we have a 6. Oh, well, we don't have to take a 6 out of B because the 6 is not in set A. So we're left with just the 6. So this would be just the set with a single element in it, set 6. Now let me ask another question. What would the relative complement of A in A be? Well, this is the same thing as A minus A. And this is literally saying, let's take set A and then take all of the things that are in set A out of it. Well, I start with the 5. Oh, but there's already a 5. There's a 5 in set A. So I have to take the 5 out. Well, there's a 3, but there's a 3 in set A, so I have to take a 3 out. So I'm going to take all of these things out. And so I'm just going to be left with the empty set, often called the null set. will look like this, the null set, the empty set. There's a set that has absolutely no objects in it."
    },
    {
        "Q": "At 3:35, he uses (f(c - h), f(c + h)) to symbolize getting near f(c). However, wouldn't it be more accurate to write:\nlimit as h approaches c of f(h) < f(c) For maximum\nlimit as h approaches c of f(h) > f(c) For minimum\n",
        "A": "I agree with the strictly less than/greater than, otherwise f(c) could be equal to surrounding f(x) values (as implied by the or equal to notation on the inequalities)",
        "video_name": "Hoyv3-BMAGc",
        "timestamps": [
            215
        ],
        "3min_transcript": "there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the-- if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative max, relative maximum value, if f of c x that-- we could say in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in-- and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that is c plus h. That value right over here c minus h. And you see that over that interval, the function at c, f of c is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine-- I encourage you to pause the video, and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write-- let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always"
    },
    {
        "Q": "\nAt 2:46 the lines are not parallel because they have different slope.\nJust find the slope for both lines.",
        "A": "No, they have the same slope - it is 4 for both.",
        "video_name": "BNHLzEv6Mjg",
        "timestamps": [
            166
        ],
        "3min_transcript": "Or I guess another way to think about it, it represents an x and y value that satisfy both of these constraints. So one system that has one solution is the system that has y is equal to 0.1x plus 1, and then this blue line right here, which is y is equal to 4x plus 10. Now, they only want us to identify one system of two lines that has a single solution. We've already done that. But just so you see it, there's actually another system here. So this is one system right here, or another system would be the green line and this red line. This point of intersection right here, once again, that represents an x and y value that satisfies both the equation y is equal to 0.1x plus 1, and this point right here satisfies the equation y is equal to 4x minus 6. there's one point of intersection of these two equations or these two lines, and this system also has one solution because it has one point of intersection. Now, the second part of the problem, they say identify one system of two lines that does not have a single solution or does not have a solution, so no solution. So in order for there to be no solution, that means that the two constraints don't overlap, that there's no point that is common to both equations or there's no pair of x, y values that's common to both equations. And that's the case of the two parallel lines here, this blue line and this green line. Because they never intersect, there's no coordinate on the coordinate plane that satisfies both equations. So there's no x and y that satisfy both. So the second part of the question, a system that has no solution is y is equal to 4x plus 10, and then the other And notice, they have the exact same slope, and they're two different lines, they have different intercepts, so they never, ever intersect, and that's why they have no solutions."
    },
    {
        "Q": "At 0:48, Sal divides it by 8 but why does he multiply it by 8 also?\n",
        "A": "He is just factoring out the 8; it is still the same answer. -> 8+4+2 is the same as 2(4+2+1).",
        "video_name": "GMoqg_s4Dl4",
        "timestamps": [
            48
        ],
        "3min_transcript": "Factor 8k squared minus 24k minus 144. Now the first thing we can do here, just eyeballing each of these terms, if we want to simplify it a good bit is all of these terms are divisible by 8. Clearly, 8k squared is divisible by 8, 24 is divisible by 8, and 144-- it might not be as obvious is divisible by 8-- but it looks like it is. 8 goes into one and 144, 8 goes into 14 one time. 1 times 8 is 8. Subtract, you get a 6. 14 minus 8 is 6. Bring down the 4. 8 goes into 64 eight times. So it goes into 144 18 times. So let's just factor out an 8 of this. And then that will simplify our expression. It will actually give us a leading 1 coefficient. So this will become 8 times k squared minus 24 divided by 8 is 3k minus 18. And remember if anything has the form x squared plus bx plus c, where you have a leading 1 coefficient-- this is implicitly a one-- we have that here in this expression in parentheses. Then we literally just need to-- and we can do this multiple ways-- but we need to find two numbers whose sum is equal to the coefficient on x. So two numbers whose sum is equal to negative 3 and whose product is equal to the constant term. And whose product is equal to negative 18. So let's just think about the factors of negative 18 here. Let's see if we can do something interesting. So it could be 1. And since it's negative, one of the numbers has to be positive, one has to be negative 1 and 18 is if it was positive. And then one of these could be positive and then one of these could be negative. But no matter what if this is negative If you switch them, then they add up to negative 17. So those won't work. So either we could write it this way, positive or negative 1, and then negative or positive 18 to show that they have to be different signs. So those don't work. Then you have positive or negative 3. And then negative or positive 6, just to know that they are different signs. So if you have positive 3 and negative 6, they add up to negative 3 which is what we need them to add up to. And clearly, positive 3 and negative 6, their product is negative 18. So it works. So we're going to go with positive 3 and negative 6 as our two numbers. Now, for this example-- just for the sake of this example-- We'll do this by grouping. So what we can do is we can separate this middle term right here as the sum of 3k negative 6k. So I could write the negative 3k as plus 3k minus 6k. And then let me write the rest of it."
    },
    {
        "Q": "\nAre the charts mentioned at the 9:00 marker something that is given or would I have to fill it in myself while solving the problem. If so, how would i go about doing that?",
        "A": "You do not memorize the values. If you are doing homework/classwork, the tables are provided at the end of the book. On a test, the table should be attached to the test or the teacher should allow you to look up the critical value for a specified level of significance and degrees of freedom using the text. I hope this helps.",
        "video_name": "2QeDRsxSF9M",
        "timestamps": [
            540
        ],
        "3min_transcript": "Plus 45 minus 40 is 5 squared is 25. So plus 25 over 40. Plus the difference here is 3 squared is 9, so it's 9 over 60. Plus we have a difference of 10 squared is plus 100 over 30. And this is equal to-- and I'll just get the calculator out for this-- this is equal to, we have 100 divided by 20 plus 36 divided by 20 plus 16 divided by 30 plus 25 divided by 40 plus 9 divided by 60 plus 100 divided by 30 gives us 11.44. So this right here is going to be 11.44. This is my chi-square statistic, or we could call it a big capital X squared. Sometimes you'll have it written as a chi-square, but this statistic is going to have approximately a chi-square distribution. Anyway, with that said, let's figure out, if we assume that it has roughly a chi-square distribution, what is the probability of getting a result this extreme or at least this extreme, I guess is another way of thinking about it. Or another way of saying, is this a more extreme result than the critical chi-square value that there's a 5% chance of getting a result that extreme? So let's do it that way. Let's figure out the critical chi-square value. And if this is more extreme than that, then we will reject our null hypothesis. So let's figure out our critical chi-square values. So we have an alpha of 5%. And actually the other thing we have to figure out is the degrees of freedom. five, six sums, so you might be tempted to say the degrees of freedom are six. But one thing to realize is that if you had all of this information over here, you could actually figure out this last piece of information, so you actually have five degrees of freedom. When you have just kind of n data points like this, and you're measuring kind of the observed versus expected, your degrees of freedom are going to be n minus 1, because you could figure out that nth data point just based on everything else that you have, all of the other information. So our degrees of freedom here are going to be 5. It's n minus 1. So our significance level is 5%. And our degrees of freedom is also going to be equal to 5. So let's look at our chi-square distribution. We have a degree of freedom of 5. We have a significance level of 5%. And so the critical chi-square value is 11.07."
    },
    {
        "Q": "\nAt 5:31, Sal decides to use the line as a transversal. If the point of this video is to prove that a line has a constant slope, wouldn't it be bad science to use an assumption of a line's constant slope to prove it has a constant slope?",
        "A": "Using the line as a transversal is making no assumptions about the slope. All he is doing here is using the fact that if you have a line that crosses two parallel lines (in other words, if you have a transversal) that the corresponding angles are congruent. This is something we know from geometry. From here he is able to prove that the triangles are similar and thus the equation for the slope is the same for both segments.",
        "video_name": "24WMbh1BBKc",
        "timestamps": [
            331
        ],
        "3min_transcript": "if you can establish that 2 triangles are similar, then the ratio between corresponding sides is going to be the same. So if these 2 are similar, then the ratio of this side to this side is going to be the same as the ratio of-- let me do that pink color-- this side to this side. And so you can see why that will be useful in proving that the slope is constant here, because all we have to do is look. If these 2 triangles are similar, then the ratio between corresponding sides is always going to be the same. We've picked 2 arbitrary sets of points. Then this would be true, really, for any 2 arbitrary set of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical because the green lines literally go in the horizontal direction. The purple lines go in the vertical direction. So we know that these are both right angles. So we have 1 corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these 2 green lines. So I'll continue them. These are line segments, but if we view them as lines and we just continue them, on and on and on. So let me do that, just like here. So this line is clearly parallel to that 1. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now, we make a very similar argument for this angle, but now we use the 2 vertical lines. We know that this segment, we could continue it as a line. So we could continue it, if we wanted, as a line, And we could continue this one as a vertical line. We know that these are both vertical. They're just measuring-- they're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it. Corresponding angles of the transversal of 2 parallel lines are congruent. We learned that in geometry class. And there you have it. All of the corresponding-- this angle is congruent to this angle. This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. Just let me write that down so we know that these are both similar triangles. And now we can use the common ratio of both sides. So for example, if we called this side length a."
    },
    {
        "Q": "at 4:30 why does Sal move 4,-1 clock wise for the rotation example? Is negative clockwise and positive counter?\n\nThanks\n",
        "A": "Just to answer your question, when you rotate it counter-clockwise it is positive. when it is rotated clockwise it is negative. I hope this answers your question :)",
        "video_name": "_eAWDuLYVfg",
        "timestamps": [
            270
        ],
        "3min_transcript": "So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3/2. So this one looks right in between the two. Or actually it could be someplace in between. But either way, we just have to think about it qualitatively. If you had a line that looked something like that, and if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over. So it's pretty clear that this could be a reflection. Now rotation actually makes even more sense, or at least in my brain makes a little more sense. If you were to rotate around to this point right over here, this point would map to that point, and that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation Actually maybe I'll put that in magenta, as well. To 7, negative 3, just like that. And we want to go from negative 5, 5 to negative 2, 3. So I could definitely imagine a translation right over here. This point went 3 to the right and 2 down. This point went 3 to the right and 2 down. So a translation definitely makes sense. Now let's think about a reflection. So it would be tempting to-- let's I could reflect around that, but that won't help this one over here. And to get from that point to that point, I could reflect around that, but once again, that's not going to help that point over there. So a reflection really doesn't seem to do the trick. And what about a rotation? Well to go from this point to this point, we could rotate around this point. We could go there, but that won't help this point right While this is rotating there, this point is going to rotate around like that and it's going to end up someplace out here. So that's not going to help. So it looks like this one can only be a translation."
    },
    {
        "Q": "\nIn 1:15, why can't it be a translation? I didn't understand.",
        "A": "Because one point would have to move 6 to the right and 4 down. While the other would move 12 to the right and 8 down. In a translation, all points should be moved the same distance.",
        "video_name": "_eAWDuLYVfg",
        "timestamps": [
            75
        ],
        "3min_transcript": "Transformation C maps negative 2, 3 to 4, negative 1. So let me do negative 2 comma 3, and it maps that to 4, negative 1. And point negative 5 comma 5, it maps that to 7, negative 3. And so let's think about this a little bit. How could we get from this point to this point, and that point to that point? Now it's tempting to view this that maybe a translation is possible. Because if you imagined a line like that, you could say, hey, let's just shift this whole thing down and then to the right. These two things happen to have the same slope. They both have a slope of negative 2/3, and that point would map to that point. But that's not what we want. We don't want negative 2, 3 to map to 7, negative 3. We want negative 2, 3 to map to 4, negative 1. So you could get this line over this line, but we won't map the points that we want to map. So this can't be, at least I can't think of a way, that this could actually be a translation. Now let's think about whether our transformation could be a reflection. Well, if we imagine a line that has-- let's see, these both have a slope of negative 3. These both have a slope of negative 2/3. So if you imagined a line that had a slope of positive 3/2 that was equidistant from both-- and I don't know if this is. Let's see, is this equidistant? Is this equidistant from both of them? It's either going to be that line or this line right over-- So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3/2. So this one looks right in between the two. Or actually it could be someplace in between. But either way, we just have to think about it qualitatively. If you had a line that looked something like that, and if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over. So it's pretty clear that this could be a reflection. Now rotation actually makes even more sense, or at least in my brain makes a little more sense. If you were to rotate around to this point right over here, this point would map to that point, and that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation"
    },
    {
        "Q": "\nI got lost at around 6:00\nCan someone please help me?",
        "A": "Do you have a more specific question? I can help if I know how : )",
        "video_name": "dvoHB9djouc",
        "timestamps": [
            360
        ],
        "3min_transcript": "And to that, I'm going to add the squared distance between 5 and my mean. And since I'm squaring, it doesn't matter if I do 5 minus 6, or 6 minus 5. When I square it, I'm going to get a positive result regardless. And then, to that I'm going to add the squared distance between 7 and my mean. So 7 minus 6 squared. All of this, this is my population mean that I'm finding the difference between. And then, finally, the squared difference between 14 and my mean. And then, I'm going to find, essentially, the mean of these squared distances. So I have five squared distances right over here. So let me divide by 5. So what will I get when I make this calculation, right over here? This is going to be equal to 1 minus 6 is negative 5, negative 5 squared is 25. 3 minus 6 is negative 3, now if I square that, I get 9. 5 minus 6 is negative 1, if I square it, I get positive 1. 7 minus 6 is 1, if I square it, I get positive 1. And 14 minus 6 is 8, if I square it, I get 64. And then, I'm going to divide all of that by 5. And I don't need to use a calculator, but I tend to make a lot of careless mistakes when I do things while making a video. So I get 25 plus 9 plus 1 plus 1 plus 64 divided by 5. So I get 20. So the average squared distance, or the mean squared distance, from our population mean is equal to 20. Remember, it's the squared distance away from my population mean. So I squared each of these things. I liked it, because it made it positive. And we'll see later it has other nice properties about it. Now the last thing is, how can we represent this mathematically? We already saw that we know how to represent a population mean, and a sample mean, mathematically like this, and hopefully, we don't find it that daunting anymore. But how would we do the exact same thing? How would we denote what we did, right over here? Well, let's just think it through. We're just saying that the population variance, we're taking the sum of each-- so we're going to take each item, we'll start with the first item. And we're going to go to the n-th item in our population. We're talking about a population here. And we're going to take-- we're not"
    },
    {
        "Q": "\nAround 6:10 he's talking about the 20 being the squared distance away from the population mean - is there a time when you would take the square root?",
        "A": "Yep. The square root of the variance is called the standard deviation, which will be another crucial concept that you ll get to pretty soon.",
        "video_name": "dvoHB9djouc",
        "timestamps": [
            370
        ],
        "3min_transcript": "And to that, I'm going to add the squared distance between 5 and my mean. And since I'm squaring, it doesn't matter if I do 5 minus 6, or 6 minus 5. When I square it, I'm going to get a positive result regardless. And then, to that I'm going to add the squared distance between 7 and my mean. So 7 minus 6 squared. All of this, this is my population mean that I'm finding the difference between. And then, finally, the squared difference between 14 and my mean. And then, I'm going to find, essentially, the mean of these squared distances. So I have five squared distances right over here. So let me divide by 5. So what will I get when I make this calculation, right over here? This is going to be equal to 1 minus 6 is negative 5, negative 5 squared is 25. 3 minus 6 is negative 3, now if I square that, I get 9. 5 minus 6 is negative 1, if I square it, I get positive 1. 7 minus 6 is 1, if I square it, I get positive 1. And 14 minus 6 is 8, if I square it, I get 64. And then, I'm going to divide all of that by 5. And I don't need to use a calculator, but I tend to make a lot of careless mistakes when I do things while making a video. So I get 25 plus 9 plus 1 plus 1 plus 64 divided by 5. So I get 20. So the average squared distance, or the mean squared distance, from our population mean is equal to 20. Remember, it's the squared distance away from my population mean. So I squared each of these things. I liked it, because it made it positive. And we'll see later it has other nice properties about it. Now the last thing is, how can we represent this mathematically? We already saw that we know how to represent a population mean, and a sample mean, mathematically like this, and hopefully, we don't find it that daunting anymore. But how would we do the exact same thing? How would we denote what we did, right over here? Well, let's just think it through. We're just saying that the population variance, we're taking the sum of each-- so we're going to take each item, we'll start with the first item. And we're going to go to the n-th item in our population. We're talking about a population here. And we're going to take-- we're not"
    },
    {
        "Q": "At 3:20, how did Sal go from cos(2*theta) to cos^2(theta) - sin^2(theta)? I didn't get that part, can someone please help me? Thanks!\n",
        "A": "That s a double angle formula. It s a known trig identity.",
        "video_name": "lXShNH1G6Pk",
        "timestamps": [
            200
        ],
        "3min_transcript": "this expression evaluated at negative pi over 4, so 1 plus square root of 2 times sine of negative pi over 4, over cosine of 2 times negative pi over 4. Now, negative pi over 4, sine of negative pi over 4 is going to be negative square root of 2 over 2, so this is negative square root of 2 over 2, we're assuming this is in radians, if we're thinking in degrees, this would be a negative 45-degree angle, so this is one of the, one of the trig values that it's good to know and so if you have, if you have 1, so let's see, actually, let me just rewrite it, so this is going to be equal to 1 plus square root of 2 times that is going to be negative 2 over 2, so this is going to be minus 1, that's the numerator over here. All of this stuff simplifies to negative 1 this is going to be cosine of negative pi over 2, right? This is negative pi over 2, cosine of negative pi over 2, if you thought in degrees, that's going to be negative 90 degrees. Well, cosine of that is just going to be zero, so what we end up with is equal to zero over zero, and as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up, but we have this indeterminate form, it does not mean the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or does not, is not an indeterminate form, that theta is equal to pi over 4 and we'll see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have 1 plus the square root of 2, sine theta, over cosine 2 theta, the things that might be useful here are our trig identities and in particular, cosine of 2 theta seems interesting. Let me write some trig identities involving cosine of 2 theta. I'll write it over here. So we know that cosine of 2 theta is equal to cosine squared of theta minus sine squared of theta which is equal to 1 minus 2 sine squared of theta which is equal to 2 cosine squared theta minus 1, and you can go from this one to this one to this one just using the Pythagorean identity. We proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are going to be differences of squares, so we can factor them in interesting ways,"
    },
    {
        "Q": "At 10:05, why is f(x) continuous at a when a = -\u00cf\u0080/4? Isn't f(-\u00cf\u0080/4) undefined?\n",
        "A": "Sal is just making a general statement about limit equality between functions. E.g if f(x)=a for all x, and g(x)=a for all x!=10 and g(10)=1234 then g(x) -> 4 when x -> 10. The functions have exactly the same limits, regardless of x-value.",
        "video_name": "lXShNH1G6Pk",
        "timestamps": [
            605
        ],
        "3min_transcript": "1 minus square root of 2 times the negative square root of 2 over 2, so negative, negative, you get a positive, square root of 2 times square root of 2 is 2, over 2 is going to be 1. So this is going to be equal to 1/2. And so, I want to be very clear. This expression is not the same thing as this expression. They are the same thing at all values of theta, especially if we're dealing in this open interval except at theta equals negative pi over 4. This one is not defined and this one is defined, but as we've seen multiple times before, if we find a function that is equal to our original or an expression, is equal to our original expression, and all values of theta except, except where the original one was not defined at a certain point, but this new one is defined and is continuous there, well then these two limits are going to be equal, then this limit is going to be 1/2, and I've said this in previous videos. It might be very tempting to say, well, I'm just going to algebraically simplify this in some way to get this, and I'm not going to worry about too much about these constraints, and then I'm just going to substitute negative pi over 4, and you will get this answer which is the correct answer but it's really important to recognize that this expression and this expression are not the same thing and what allows you to do this is, is the truth that if you have two functions, if you have f and g, two functions equal, let me write it this way, equal, equal for all x, except for all, wait, let me just write this this way, for all x except for a, then the limit, then, and let me write it this way, equal for all except, for all x except a continuous at a, then, then the limit of f of x as x approaches a is going to be equal to the limit of g of x as x approaches a, and I said this in multiple videos and that's what we are doing right here, but just so you can make sure you got it right, the answer here is 1/2."
    },
    {
        "Q": "\nAt 1:21 what does it mean by past the . that the numbers are smaller than the 1's?",
        "A": "The numbers to the right of the decimal point represent fractions, anything less than 1. For instance, 0.1 is one tenth and is the same as 1/10.",
        "video_name": "BItpeFXC4vA",
        "timestamps": [
            81
        ],
        "3min_transcript": "So I have a number written here. It's a 2, a 3, and a 5. And we already have some experience with numbers like this. We can think about 'what does it represent'. And to think about that we just have to look at the actual place values. So this right-most place right over here. This is the ones place. So this 5 represents five ones, or I guess you could say that's just going to be 5. This 3, this is in the tens place. This is the tens place, so we have three tens. So that's just going to be 30. And the 2 is in the hundreds place. So putting a 2 there means that we have two hundreds. So this number we can view as two hundred, thirty, five. Or you could view it as two hundred plus thirty plus five. Now what I want to do in this video is think about place values to the right of the ones place. And you might say 'wait, wait, I always thought that the ones place was the place furthest to the right.' Well everything that we've done so far, it has been. But to show that you can go even further to the right We call that a 'decimal point'. And that dot means that anything to the right of this is going to be place values that are smaller, I guess you could say, than the ones place. So right to the left you have the ones place and the tens place and the hundreds place, and if you were to keep going you'd go to the thousands place and the ten thousands place. But then if you go to the right of the decimal point now you're going to divide by 10. So what am I talking about? Well, right to the right of the decimal point you are going to have-- find a new color-- this is going to be the tenths place. Well what does that mean? Well whatever number I write here that tells us how many tenths we're dealing with. So if I were to write the number 4 right over here, now my number is 2 hundreds plus 3 tens plus 5 ones plus 4 tenths. Or you could write this as 4 tenths. Not tens, 4 tenths. Or 4 tenths is the same thing as this right over here. So this is a super important idea in mathematics. I can now use our place values to represent fractions. So this right over here, this 'point 4', this is 4/10. So another way to write this number-- I could write it this way, I could write it as two hundred, thirty-- let me do the thirty in blue-- two hundred and thirty five and four tenths. So I could write it like this, as a mixed number. So this up here would be a decimal representation: 235.4 And this right over here would be a mixed number representation: 235 and 4/10 but they all represent 200 plus 30 plus 5 plus 4/10."
    },
    {
        "Q": "at 2:51 it says multiply both by 2/3 cant you just divide both by 3 and then multiply them by 2?\n",
        "A": "Yes you can but that s the same thing as multiplying both sides by 2/3. Because 2*1/3 = 2 divided 3 = 2/3",
        "video_name": "wo7DSaPP8hQ",
        "timestamps": [
            171
        ],
        "3min_transcript": "So this first sentence, they say-- let me do this in a different color-- they say for the past few years, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. So we could say, hey, Maple is approximately Old River, or M is approximately River plus 1,000. Or since we don't know the exact amount-- it says it's about 1,000 more, so we don't know it's exactly 1,000 more-- we can just say that in a normal year, Old Maple Farms, which we denote by M, has a larger amount of apples than River Orchard. So in a normal year, M is greater than R, right? It has about 1,000 more apples than Old Maple Farms. Now, they say due to cold weather this year-- so let's talk about this year now-- the harvests at both farms were So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by 1/3. Now if I go down by 1/3, that's the same thing as being 2/3 of what I was before. Let me do an example. If I'm at x, and I take away 1/3x, I'm left with 2/3x. So going down by 1/3 is the same thing as multiplying the quantity by 2/3. So if we multiply each of these quantities by 2/3, we can still hold this inequality, because we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by 2/3. So 2/3 of M is still going to be greater than 2/3 of R. And you could even draw that in a number line if you like. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's 0 on our number line. So in a normal year, M is has 1,000 more than R. So in a normal year, M might be over here and maybe R is over here. I don't know, let's say R is over there. Now, if we take 2/3 of M, that's going to stick us some place around, oh, I don't know, 2/3 is right about there. So this is M-- let me write this-- this is 2/3 M. And what's 2/3 of R going to be? Well, if you take 2/3 of this, you get to right about there, that is 2/3R. So you can see, 2/3R is still less than 2/3M, or 2/3M is greater than 2/3R. Now, they say both farms made up for some of the shortfall"
    },
    {
        "Q": "\nI don't know if this is a mistake, but at 3:23 in the video, Sal says Maple Farms lost 2/3 of their farm, not 1/3.Is this a mistake, or did i miss something?",
        "A": "At 2.22, Sal explains why it is 2/3 : 1/3 less than an amount x is x-1/3x which equals 2/3x. Hope this helps! :)",
        "video_name": "wo7DSaPP8hQ",
        "timestamps": [
            203
        ],
        "3min_transcript": "So this first sentence, they say-- let me do this in a different color-- they say for the past few years, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. So we could say, hey, Maple is approximately Old River, or M is approximately River plus 1,000. Or since we don't know the exact amount-- it says it's about 1,000 more, so we don't know it's exactly 1,000 more-- we can just say that in a normal year, Old Maple Farms, which we denote by M, has a larger amount of apples than River Orchard. So in a normal year, M is greater than R, right? It has about 1,000 more apples than Old Maple Farms. Now, they say due to cold weather this year-- so let's talk about this year now-- the harvests at both farms were So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by 1/3. Now if I go down by 1/3, that's the same thing as being 2/3 of what I was before. Let me do an example. If I'm at x, and I take away 1/3x, I'm left with 2/3x. So going down by 1/3 is the same thing as multiplying the quantity by 2/3. So if we multiply each of these quantities by 2/3, we can still hold this inequality, because we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by 2/3. So 2/3 of M is still going to be greater than 2/3 of R. And you could even draw that in a number line if you like. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's 0 on our number line. So in a normal year, M is has 1,000 more than R. So in a normal year, M might be over here and maybe R is over here. I don't know, let's say R is over there. Now, if we take 2/3 of M, that's going to stick us some place around, oh, I don't know, 2/3 is right about there. So this is M-- let me write this-- this is 2/3 M. And what's 2/3 of R going to be? Well, if you take 2/3 of this, you get to right about there, that is 2/3R. So you can see, 2/3R is still less than 2/3M, or 2/3M is greater than 2/3R. Now, they say both farms made up for some of the shortfall"
    },
    {
        "Q": "At 1:07 to 2:59, couldn't he find the area and then multiply it by 3/4 and you get an answer?\n",
        "A": "Sure, if what you needed was the area. But the exercise is asking for the arc length which is part of the circumference and not the area. Just remember that arc length, and circumference for that matter, is a measurement of length as the name implies while area is not.",
        "video_name": "YjWCDdNlXxc",
        "timestamps": [
            67,
            179
        ],
        "3min_transcript": "- [Instructor] Find the area of the semicircle. So pause this video and see if you can figure it out. So let's see. We know that the area of a circle is equal to pi times our radius squared. So, if we think about the entire circle, what is the area going to be? Well, they tell us what our radius is. Our radius is equal to two, so the area, if we're talking about the whole circle, it would be equal to pi times two squared. Pi times two squared. Two squared is of course two times two, which is equal to four, so our area is going to be equal to four times pi. Now, I wouldn't put four pi here, because that would be the entire circle. They want the area of just the semicircle, of just this region right over here. Well, the semicircle is half of the circle, so if I want the area of the semicircle, this is gonna be half this. So instead of four pi, it is going to be two pi square units. Let's do another example. So here, instead of area, we're asked to find the arc length of the partial circle, and that's we have here in this bluish color right over here, find this arc length. And you can see this is going three fourths of the way around the circle, so this arc length is going to be three fourths of the circumference. So what is the circumference? Well, we know the circumference is equal to two pi times the radius. They tell us what the radius is. It's equal to four, so our circumference is equal to two pi times four. Let's see, we can just change the order in which we multiply so it's two times four times pi. This is going to be equal to eight pi. This is going to be equal to eight pi. Now, that is the circumference of the entire circle. it's going to be three fourths times the circumference of the entire circle. So three over four times eight pi. What is that going to be? Well, what's three fourths times eight? Well, three times eight is 24 divided by four is six. So this is going to be equal to six pi. Another way to think about it, one fourth of eight is two, so three fourths is going to be six. Or another way to think about it is, one fourth of eight pi is two pi, and so three of those is going to be equal to six pi. So the arc length of the partial circle is six pi, and once again we knew that because it was three fourths of the way around. The way that I knew it was three fourths is that this is a 90 degree angle. This is 90 degrees, which is one fourth of the way around a circle, so the arc length that we care about is the three fourths of our circumference."
    },
    {
        "Q": "At 2:03 to 2:13, I wish he would have wrote out the 10^5-10^7 to show what he did in his head.\n",
        "A": "That would certainly be helpful!",
        "video_name": "497oIjqRPco",
        "timestamps": [
            123,
            133
        ],
        "3min_transcript": "We have 7 times 10 to the fifth over 2 times 10 to the negative 2 times 2.5 times 10 to the ninth. So let's try to simplify this a little bit. And I'll start off by trying to simplify this denominator here. So the numerator's just 7 times 10 to the fifth. And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do over 2 times 2.5 times 10 to the negative 2 times 10 to the ninth. And this is going to be equal to-- so the numerator I haven't changed yet-- 7 times 10 to the fifth over-- and here in the denominator, 2 times-- let me do this in a new color now. 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the ninth, when you multiply two numbers that are being raised to exponents and have the exact same base-- we can add the exponents. So this is going to be 10 to the 9 minus 2, or 10 to the seventh. So times 10 to the seventh. And now we can view this as being equal to 7 over 5 times 10 to the fifth over 10 to the seventh. Let me do that in that orange color to keep track of the colors. 10 to the seventh. Now, what is 7 divided by 5? 7 divided by 5 is equal to-- let's see, it's 1 and 2/5, or 1.4. So I'll just write it as 1.4. And then 10 to the fifth divided by 10 to the seventh. So that's going to be the same thing as-- and there's two ways to view this. You could view this as 10 to the fifth times 10 You get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now, are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9. It's a digit between 1 and 9, including 1 and 9. And it's being multiplied by 10 to some power. So it looks like we're done. This simplified to 1.4 times 10 to the negative 2."
    },
    {
        "Q": "\nAt 1:40 Sal says that (7 * 10^5) / (5 * 10^7) = (7 / 5) * (10^5 / 10^7). Why can we detach 7 / 5 like this? Shouldn't we get a different answer? I'm sure Sal says the right thing, I just don't get why it works.",
        "A": "How many sides does your polygon have?",
        "video_name": "497oIjqRPco",
        "timestamps": [
            100
        ],
        "3min_transcript": "We have 7 times 10 to the fifth over 2 times 10 to the negative 2 times 2.5 times 10 to the ninth. So let's try to simplify this a little bit. And I'll start off by trying to simplify this denominator here. So the numerator's just 7 times 10 to the fifth. And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do over 2 times 2.5 times 10 to the negative 2 times 10 to the ninth. And this is going to be equal to-- so the numerator I haven't changed yet-- 7 times 10 to the fifth over-- and here in the denominator, 2 times-- let me do this in a new color now. 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the ninth, when you multiply two numbers that are being raised to exponents and have the exact same base-- we can add the exponents. So this is going to be 10 to the 9 minus 2, or 10 to the seventh. So times 10 to the seventh. And now we can view this as being equal to 7 over 5 times 10 to the fifth over 10 to the seventh. Let me do that in that orange color to keep track of the colors. 10 to the seventh. Now, what is 7 divided by 5? 7 divided by 5 is equal to-- let's see, it's 1 and 2/5, or 1.4. So I'll just write it as 1.4. And then 10 to the fifth divided by 10 to the seventh. So that's going to be the same thing as-- and there's two ways to view this. You could view this as 10 to the fifth times 10 You get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now, are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9. It's a digit between 1 and 9, including 1 and 9. And it's being multiplied by 10 to some power. So it looks like we're done. This simplified to 1.4 times 10 to the negative 2."
    },
    {
        "Q": "\nAround 2:25 he mentions \"f of x\", this mightt sound like a stupid question but what does this mean? Function?",
        "A": "yup yup yups!",
        "video_name": "W0VWO4asgmk",
        "timestamps": [
            145
        ],
        "3min_transcript": "Welcome to the presentation on limits. Let's get started with some-- well, first an explanation before I do any problems. So let's say I had-- let me make sure I have the right color and my pen works. OK, let's say I had the limit, and I'll explain what a limit is in a second. But the way you write it is you say the limit-- oh, my color is on the wrong-- OK, let me use the pen and yellow. OK, the limit as x approaches 2 of x squared. Now, all this is saying is what value does the expression x squared approach as x approaches 2? Well, this is pretty easy. If we look at-- let me at least draw a graph. I'll stay in this yellow color. So let me draw. x squared looks something like-- let me use x square looks something like this, right? And when x is equal to 2, y, or the expression-- because we don't say what this is equal to. It's just the expression-- x squared is equal to 4, right? So a limit is saying, as x approaches 2, as x approaches 2 from both sides, from numbers left than 2 and from numbers right than 2, what does the expression approach? And you might, I think, already see where this is going and be wondering why we're even going to the trouble of learning this new concept because it seems pretty obvious, but as x-- as we get to x closer and closer to 2 from this direction, and as we get to x closer and closer to 2 to this direction, what does this expression equal? Well, it essentially equals 4, right? The way I think about it is as you move on the curve closer and closer to the expression's value, what does the expression equal? In this case, it equals 4. You're probably saying, Sal, this seems like a useless concept because I could have just stuck 2 in there, and I know that if this is-- say this is f of x, that if f of x is equal to x squared, that f of 2 is equal to 4, and that would have been a no-brainer. Well, let me maybe give you one wrinkle on that, and hopefully now you'll start to see what the use of a limit is. Let me to define-- let me say f of x is equal to x squared when, if x does not equal 2, and let's say it equals 3 when x equals 2."
    },
    {
        "Q": "So around the time 4:43 in the video, he is talking about simplifying your answer. The answer I got that I think I need to simplify is -3x+9y=-6. My question is would it be possible to divide the positive 9 by a negative 3 or the only way you would be able to divide that positive is by a positive vice versa.\n",
        "A": "Yes it would be possible. I would divide it by -3. So you would have -3x/-3=x and 9y/-3=-3y and -6/-3=2 . The simplified version would be x+3y+2 :)",
        "video_name": "XOIhNVeLfWs",
        "timestamps": [
            283
        ],
        "3min_transcript": "And then 21 divided by seven is three times four is 12. So just like that I was able to get rid of the fractions. And now I wanna get all the X's and Y's on one side. So I wanna get this 14x onto the left side. So let's see if I can do that. So I'm gonna do that by. To get rid of this I would want to subtract 14x. I can't just do it on the right hand side I have to do it on the left hand side as well. So I wanna subtract 14x, and then what am I left with? Let me give myself a little bit more space. So on the left hand side I have negative 14x plus 21y. Plus 21y is equal to. Let's see and I subtracted 14x to get rid of this. And then I have this is equal to 12. Now let's see, am I done? Do these share any, do 14, 21, and 12 Let's see, 14 is divisible by two and seven. 21 is divisible by three and seven. 12 is divisible by two, six, three, four. But all of these aren't divisible by the same number. 14, let's see. 14 is divisible two, so is 12, but 21 isn't. 14 is divisible by seven, so is 21, but 12 isn't. And 21 and 12 are divisible by three but 14 isn't. So I think this is about as simplified as I could get. If there was a common factor for all three of these numbers then I would divide all of them the way I did in that previous example. But that's not the case right over here. So it's negative 14x plus 21y is equal to 12. So let me see if I can remember that and type that in. So it is, negative 14x plus 21y is equal to 12. It worked out."
    },
    {
        "Q": "At 2:44 , SAl said the slope is 7. How is it 7 ? is he assuming ?\n",
        "A": "In the video there s a graphic clip with given information: Let f be a differentiable function for all x. If f(-2) = 3 and f (x) \u00e2\u0089\u00a4 7 for all x, then what is the largest possible value of f(10). f (x) is the slope of f(x).",
        "video_name": "EXLVMGSDQbI",
        "timestamps": [
            164
        ],
        "3min_transcript": "So really, the way to get the largest possible value of f-- we don't have to necessarily invoke the mean value theorem, although the mean value theorem will help us know for sure-- is to say well, look, the largest possible value of f of 10 is essentially if we max this thing out. If we assume that the instantaneous rate of change just stays at the ceiling right at 7. So if we assumed that our function, the fastest growing function here would be a line that has a slope exactly equal to 7. So the slope of 7 would look-- and obviously, I'm not drawing this to scale. Visually, this looks more like a slope of 1, but we'll just assume this is a slope of 7 because it's not at the same-- the x and y are not at the same scale. So slope is equal to 7. where do we get to when x is equal to 10? When x is equal to 10, which is right over here, well what's our change in x? So what's our change in x? Let's just think about it this way. Our change in y over change in x is going to be what? Well our change in y is going to be f of 10 minus f of 2. f of 2 is 3, so minus 3, over our change in x. Our change in x is 10 minus negative 2. 10 minus negative 2 is going to be equal to 7. This is the way to max out what our value of f of 10 might be. If at any point the slope were anything less than that, because remember, the instantaneous rate of change can never be more than that. So if we start off even a little bit lower, than the best we can do is get to that. That would get us too steep. So it has to be like that. And then we would get to a lower f of 10. Every time you have a slightly lower rate of change, then it kind of limits what happens to you. So remember, our slope can never be more than 7. So this part should be parallel. So this should be parallel to that right over there. This should be parallel. But we can never have a higher slope than that. So the way to max it out is to actually have a slope of 7. And so what is f of 10 going to be? So let's see, 10 minus negative 2, that is 12. Multiply both sides by 12, you get 84. So f of 10 minus 3 is going to be equal to 84. Or f of 10 is going to be equal to 87. So if you have a slope of 7, the whole way, you travel 12."
    },
    {
        "Q": "at 3:30 sal multiplies vector a with a 'scalar quantity'-1.....but how come -1 is scalar....1 represents magnitude and the negative sign represents direction so it has to be a vector. and if it is a vector how do we plot it on a graph\n",
        "A": "That s creative! But not exactly how that works. You could think of the number 2 as having a + in front of it, that we just don t write. The negative symbol is a part of the number. It says we are going negative one in some direction, it does not say whether we are going negative 1 up, or left or right or down or diagonally, just that we are going negative 1. Does that make sense?",
        "video_name": "ZN7YaSbY3-w",
        "timestamps": [
            210
        ],
        "3min_transcript": "But how do we define multiplying 3 times this vector? Well one reasonable thing that might jump out at you is, why don't we just multiply the 3 times each of these components? So this could be equal to.. we have 2 and 1.. And we're going to multiply each of these with 3. So 3 times 2 and 3 times 1. And then the resulting vector is still going to be a 2-dimensional vector. And it's going to be the 2-dimensional vector (6,3). Now I encourage you to get some graph paper out and to actually plot this vector, and think about how it relates to this vector right over here. So let me do that.. So the vector (6,3), if we started at the origin.. We would move 6 in the horizontal direction.. 1, 2, 3, 4, 5, 6.. And 3 in the vertical.. 1, 2, 3.. So it gets us right over there, so it would look like this. Well notice, one way to think about it is what has changed, and what has not changed about this vector? Well what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it.. did not change the direction that my vector is going in. Or at least in this case it didn't.. But it did change its magnitude. Its magnitude is now 3 times longer, which makes sense! Because we multiplied it by 3. One way to think about it is we scaled it up by 3. The scalar scaled up the vector. That might make sense. Or it might make an intuition of where that word scalar came from. The scalar, when you multiply it, it scales up a vector. It Increased its magnitude by 3 without changing its direction. Well let's do something interesting.. Let's multiply our vector a by a negative number. So let's just multiply -1 times a. Well using the convention that we just came up with.. We would multiply each of the components by -1. So 2 times -1 is -2, and 1 times -1 is -1. So now -1 times a is going to be (-2,-1) So if we started at the origin, we would move in the horizontal direction -2, and in the vertical -1 So now what happened to the vector? When I did that? Well now it flipped its direction! Multiplying it by this -1, it flipped it's direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction. Which makes sense, that multiplying by a negative number would do that."
    },
    {
        "Q": "\nAt 4:03 Sal said the magnitude didn't change. But he added, essentially, 1 more of the original vector in the negative direction. Am Imissing something?",
        "A": "The magnitude remained the same only for the -1 scalar.",
        "video_name": "ZN7YaSbY3-w",
        "timestamps": [
            243
        ],
        "3min_transcript": "Well notice, one way to think about it is what has changed, and what has not changed about this vector? Well what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it.. did not change the direction that my vector is going in. Or at least in this case it didn't.. But it did change its magnitude. Its magnitude is now 3 times longer, which makes sense! Because we multiplied it by 3. One way to think about it is we scaled it up by 3. The scalar scaled up the vector. That might make sense. Or it might make an intuition of where that word scalar came from. The scalar, when you multiply it, it scales up a vector. It Increased its magnitude by 3 without changing its direction. Well let's do something interesting.. Let's multiply our vector a by a negative number. So let's just multiply -1 times a. Well using the convention that we just came up with.. We would multiply each of the components by -1. So 2 times -1 is -2, and 1 times -1 is -1. So now -1 times a is going to be (-2,-1) So if we started at the origin, we would move in the horizontal direction -2, and in the vertical -1 So now what happened to the vector? When I did that? Well now it flipped its direction! Multiplying it by this -1, it flipped it's direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction. Which makes sense, that multiplying by a negative number would do that. If you took 5 times -1, well now you're going in the other direction you're at -5, you're 5 to the left of zero. So it makes sense that this would flip its direction. So you could imagine, if you were to take something like -2 times your vector a, -2 times your vector a.. And I encourage you to pause this video and try this on your own.. What would this give? And what would be the resulting visualization of the vector? Well let's see, this would be equal to -2 times 2 is -4, -2 times 1 is -2, so this vector.. if you were to start at the origin! remember you don't have to start at the origin.. but if you were.. it would go 0, 1, 2, 3, 4.. 1, 2.. It looks just like this.. And so just to remind ourselves.. our original vector a looked like this.."
    },
    {
        "Q": "\n10:50 what are pictograms?",
        "A": "Pictograms are the pictures/symbols themselves. A pictograph is the entire graph.",
        "video_name": "qrVvpYt3Vl0",
        "timestamps": [
            650
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhy at 1:58 is 8x-7x= to x ?",
        "A": "8-7=1. 1x is the same thing as x. It is easier to say x than 1x, but you could actually say 1x, and it wouldn t be wrong.",
        "video_name": "E0TNh9uWesw",
        "timestamps": [
            118
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:57 in the video, he says that 8x - 7x = x, which I think I understand. But what if my number was larger than just x? What would I do from there, divide both sides?\n\nThanks\n",
        "A": "Yes, you would. It s nice that you can know (or take a good guess at) what to do.",
        "video_name": "E0TNh9uWesw",
        "timestamps": [
            117
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAround 2:00, 8x - 7x = x? I assume the missing 1 is just a place holder?",
        "A": "Yeah, if there is no coefficient in front of a variable, that means that there is a coefficient of 1: 1x = x",
        "video_name": "E0TNh9uWesw",
        "timestamps": [
            120
        ],
        "3min_transcript": ""
    },
    {
        "Q": "On that last one (5:31) shouldn't we have multiplied 2/1 by three to get the denominators to be the same before we multiplied it... to get our final answer?\n",
        "A": "if you add 2/1 to -(1/3), then yes. But this is a multiplication problem. So just multiply the numerators and multiply the denominators respectively.",
        "video_name": "a_Wi-6SRBTc",
        "timestamps": [
            331
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nx < -1 (-infinity,-1) since the -1 is in the parenthesis, shouldn't it be less than or equal to? @1:29",
        "A": "The parentheses mean only less than . To have x <= -1, the interval would be written as (-infinity, -1] The square bracket indicates that x can also be equal to -1.",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            89
        ],
        "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by"
    },
    {
        "Q": "\nat 5:40, why do you subtract -8x instead of adding 12x to both sides or adding 5 or 7 to both first. Is there a reason you pick 8x to subtract first?",
        "A": "That is what I would have advised my students to do so that the coefficient of the variable stays positive. There is no reason for subtracting 8 except to possibly remind us of the need to flip the inequality sign when dividing by a negative. You should have gotten the same answer.",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            340
        ],
        "3min_transcript": "Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we that's negative 7, and then we have this plus 8x left over. Now, I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides, or adding and subtracting from them. The right-hand side becomes-- this thing cancels out, 8x minus 8x, that's 0. So you're just left with a negative 7. And now I want to get rid of this negative 5. So let's add 5 to both sides of this equation. These 5's cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now, we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation, or really any type of an inequality, we want to divide both sides by negative 20. But we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality. So let's remember that. So if we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides divided by negative 20, we have to swap the inequality. The greater than or equal to has to become a less than or equal sign. And, of course, these cancel out, and you get x is less"
    },
    {
        "Q": "\nwhy do you flip the sign at 2:35???",
        "A": "oh ok thank you for explaining this to me",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            155
        ],
        "3min_transcript": "So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater"
    },
    {
        "Q": "\nAt 4:00, while setting up the problem, verbally \"8x minus 5\", then at 4:24 while simplifying the inequality, verbally says \"distribute the negative 5\".\nWhile trying to respect the order of operations, this interchangeable \"subtraction\" and \"negative\" concept is very confusing. Any help would be appreciated, thank you in advance!",
        "A": "Do you already know that subtraction is adding the opposite? For example, having $10 and then spending $4 could be written 10 - 4 (ten take away four) or it could be written 10 + -4 (having ten and then losing 4). Think of that for this distribution problem: 8x - 5(4x + 1) could be written 8x + -5(4x + 1). Now can you see how it is a negative 5 that needs to be distributed?",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            240,
            264
        ],
        "3min_transcript": "So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we"
    },
    {
        "Q": "At 1:13, I thought that multiplying or dividing in an inequality you were supposed to flip the sign?\n",
        "A": "For an inequality, you only flip the sign if you re multiplying or dividing by a negative number. At that point in the video, he s dividing by positive 4, so there s no need to flip the inequality sign.",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            73
        ],
        "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by"
    },
    {
        "Q": "At 4:34 Sal calculates the probability of small to be (1/26-1/2600). Why isn't this probability equal to (9/10*9/10*1/26)?\nSince, Probability of losing number = 9/10 and Probability of winning alphabet = 1/26.\n",
        "A": "I did the problem like you say. Once you buy a ticket, the expected values are as follows: Expected Value of $5 payout = probability*value = 1 * (-$5) EV(grand prize) = P(x)*x = (1/10*1/10*1/26) * (10405) = 4.0019 EV(small prize) = (1/10*9/10*1/26) * 100 + (9/10*1/10*1/26) * 100 + (9/10*9/10*1/26) * 100 [ there are 2 ways to get 1 number, 1 way to get no numbers] = 0.34615 + 0.34615 + 3.1154 Total is 7.81 - 5 = 2.81",
        "video_name": "6vlBOHckmzU",
        "timestamps": [
            274
        ],
        "3min_transcript": "What's the probability of the grand prize? I'll do that over here, probability of grand prize. Well the probability that he gets the first letter right is one in 10, there's 10 digits there. Probability he gets the second letter right is one in 10, these are all independent and probability he gets the letter right, there's 26 equally likely letters that might be in the actual one so he has a one in 26 chance of that one as well. The probability of the grand prize is one in 2600. This is one in 2600. Now what's the probability of getting the small price? Well let's see, he has a one in 26 chance. The small prize is getting the letter right but not getting both of the numbers right. but we're not done here just with the one in 26 because this one in 26, this includes all the scenarios where he gets the letter right, including the scenarios where he wins the grand prize, where he gets the letter and he gets the two numbers right. We need to do is we need to subtract out the situation, the probability of getting the two numbers, getting the letter and the two numbers right and we already know what that is, it's one in 2600. It's one and 26 minus one and 2600. The reason why I have to subtract out at this 2600 is he has one in 26 chance of getting this letter right. That includes the scenario where he gets everything right but the small prize is only where you get the letter and one or none of these. If you get both of these then you're at the grand prize case. You essentially have to subtract out the probability that you won the grand prize, to figure out the probability of the small prize. Now what's the probability of essentially losing? The probability of neither. Well it's just kind of that's everything else. It would be one minus these probabilities right over here. It would be one minus the probability of the small prize. The probability of the small minus the probability of the grand, these are the possible outcomes so they have to add up to one or a 100%. This is one less probability small minus probability of large or I'll say grand prize. Let's fill this in. The probability of the small one, this right over ... I'm using that red too much. This right over here is one in 26 minus one in 2600"
    },
    {
        "Q": "at 1-1:20, how did you choose the bounds? The upper limit can be anything right? I was confused because 3 is the value you get when you plug X= -2 into the y=x^2-1 formula. I thought it had extra significance. As for the lower bound, I understand y= -1 is the minimum of the parabola. Is that the reason for it being the bound? Thank you.\n",
        "A": "Since the only part of the function that is being rotated is the one for positive x values, it is a mere coincidence that 3 is the intersection of the whole function and x = -2. The part of the function for negative x values was never considered and the number 3 was chosen at random.",
        "video_name": "jxf7XqvZWWg",
        "timestamps": [
            80
        ],
        "3min_transcript": "Let's do another example, and this time we're going to rotate our function around a vertical line that And if we do that-- so we're going to rotate y is equal to x squared minus 1-- or at least this part of it-- we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disc method. So what I want to do is construct some discs. So that's one of the discs right over here. It's going to have some depth, and that depth is going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disc is going to be the area as a function of y times the depth of the disc times dy. And then we just have to integrate it over the interval that we care about, and we're doing it all in terms of y. And in this case, we're going to integrate going to hit-- this y-intercept right over here is y is equal to negative 1. And let's go all the way to y is equal to, let's say y is equal to 3 right over here. So from y equal negative 1 to y equals 3. And that's going to give us the volume of our upside-down gumdrop-type-looking thing. So the key here, so that we can start evaluating the double integral, is to just figure out what the area of each of these discs are as a function of y. And we know that area as a function of y is just going to be pi times radius as a function of y squared. So the real key is, what is the radius as a function of y for any one of these y's? So what is the radius as a function of y? So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. so you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And you get x is equal to the principal root of the square root of y plus 1. So this we can write as x-- or we can even write it as f of y if we want-- f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance here at any point? Well, this distance-- let me make it very clear. So it's going to be our total distance in the horizontal direction. So this first part as we're-- and I'm going to do it in a different color so we can see. So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius as a function of y"
    },
    {
        "Q": "At 7:57, Sal mentions that dA is used as a shorthand. Isn't it also used to generalize dxdy so that the order of integration (and the coordinate system) is not specified? I feel like this would make it more applicable in proofs and theorems and such.\n",
        "A": "dA is often used to indicate integration over an area without specifying how the integration will be performed. dA can even indicate integration over a curved surface in 3-D.",
        "video_name": "twT-WZChfZ8",
        "timestamps": [
            477
        ],
        "3min_transcript": "So we could take, we're summing in the y-direction first. We would get a sheet that's parallel to the y-axis, now. So the top of the sheet would look something like that. So if we're coming the dy's first, we would take the sum, we would take the integral with respect to y, and it would be, the lower bound would be y is equal to c, and the upper bound is y is equal to d. And then we would have that sheet with a little depth, the depth is dx, and then we could take the sum of all of those, sorry, my throat is dry. I just had a bunch of almonds to get power to be able to record these videos. But once I have one of these sheets, and if I want to sum up all of the x's, then I could take the infinite sum of infinitely small columns, or in this view, sheets, infinitely the upper bound is x is equal to b. And once again, I would have the volume of the figure. And all I showed you here is that there's two ways of doing the order of integration. Now, another way of saying this, if this little original square was da, and this is a shorthand that you'll see all the time, especially in physics textbooks, is that we are integrating along the domain, right? Because the x-y plane here is our domain. So we're going to do a double integral, a two-dimensional integral, we're saying that the domain here is two-dimensional, and we're going to take that over f of x and y times da. And the reason why I want to show you this, is you see this in physics books all the time. I don't think it's a great thing to do. Because it is a shorthand, and maybe it looks simpler, but for me, whenever I see something that I don't know how to compute or that's not obvious for me to know how to compute, So I wanted to just show you that what you see in this physics book, when someone writes this, it's the exact same thing as this or this. The da could either be dx times dy, or it could either be dy times dx, and when they do this double integral over domain, that's the same thing is just adding up all of these squares. Where we do it here, we're very ordered about it, right? We go in the x-direction, and then we add all of those up in the y-direction, and we get the entire volume. Or we could go the other way around. When we say that we're just taking the double integral, first of all, that tells us we're doing it in two dimensions, over a domain, that leaves it a little bit ambiguous in terms of how we're going to sum up all of the da's. And they do it intentionally in physics books, because you don't have to do it using Cartesian coordinates, using x's and y's. You can do it in polar coordinates, you could do it a ton of different ways. But I just wanted to show you, this is another way to having an intuition of the volume under a surface. And these are the exact same things as this type of"
    },
    {
        "Q": "\nAt 8:52 Sal says we can do this in polar coordinates. Can someone explain to me how this would work?",
        "A": "3D polar could be either with an extra angle coordinate or another radius coordinate , spherical or cylindrical respectively. You might try 2D polar first, I would go back and watch integration videos in 2D and do a parallel situation in polar. But basically, the idea is to split the area into sectors of a circle where dTHETA becomes infinitely small. The area of a sector is 1/2*pi*r^2. Be mindful of your endpoints - integrating from pi/4 to 3pi/2 is much different than -pi/4 to pi/4. Hope this helps.",
        "video_name": "twT-WZChfZ8",
        "timestamps": [
            532
        ],
        "3min_transcript": "the upper bound is x is equal to b. And once again, I would have the volume of the figure. And all I showed you here is that there's two ways of doing the order of integration. Now, another way of saying this, if this little original square was da, and this is a shorthand that you'll see all the time, especially in physics textbooks, is that we are integrating along the domain, right? Because the x-y plane here is our domain. So we're going to do a double integral, a two-dimensional integral, we're saying that the domain here is two-dimensional, and we're going to take that over f of x and y times da. And the reason why I want to show you this, is you see this in physics books all the time. I don't think it's a great thing to do. Because it is a shorthand, and maybe it looks simpler, but for me, whenever I see something that I don't know how to compute or that's not obvious for me to know how to compute, So I wanted to just show you that what you see in this physics book, when someone writes this, it's the exact same thing as this or this. The da could either be dx times dy, or it could either be dy times dx, and when they do this double integral over domain, that's the same thing is just adding up all of these squares. Where we do it here, we're very ordered about it, right? We go in the x-direction, and then we add all of those up in the y-direction, and we get the entire volume. Or we could go the other way around. When we say that we're just taking the double integral, first of all, that tells us we're doing it in two dimensions, over a domain, that leaves it a little bit ambiguous in terms of how we're going to sum up all of the da's. And they do it intentionally in physics books, because you don't have to do it using Cartesian coordinates, using x's and y's. You can do it in polar coordinates, you could do it a ton of different ways. But I just wanted to show you, this is another way to having an intuition of the volume under a surface. And these are the exact same things as this type of Sometimes they won't write a domain, sometimes they'd write over a surface. And we'll later do those integrals. Here the surface is easy, it's a flat plane, but sometimes it'll end up being a curve or something like that. But anyway, I'm almost out of time. I will see you in the next video."
    },
    {
        "Q": "at 0:10, Why did you call the space \"blank\"?\n",
        "A": "Blank represents any unknown or missing value. It is commonly used in language and math lessons. The goal is to find the/an answer that fits in the blank and completes the sentence or problem.",
        "video_name": "OPpmp-kAuE4",
        "timestamps": [
            10
        ],
        "3min_transcript": "- [Voiceover] So, we have blank plus one-16th is equal to three halves. So, I encourage you to pause the video and just figure out what blank is. What plus one-16th is equal to three-halves? All right. Now, let's work through this together. One thing that might make it a little bit simpler for our brains is if we were to express one-16th and three-halves with a common denominator. When we think about a common denominator, we will look at a common multiple of the denominators here. And lucky for us, 16 is already divisible by 2. It's divisible by 16 as well, so it is the common multiple, or it is the least common multiple of 16 and 2. There are other common multiples, but the smallest one is going to be 16 times 1, which is also divisible by 2. So, let's try both of these, let's try both of these fractions. Let's rewrite this equation where both of these fractions have 16 as their denominators. This one obviously already has it. So, let's write that. So, we're going to have, we're going to have blank plus one-16th, is equal to, is equal to, let's see, let's write three-halves as something over 16. Well, to get our denominator from 2 to 16, we have to multiply by 8. So, we have to multiply the numerator by 8 as well. So, 3 times 8 is going to be 24. Now, at this point, you might be able to do it in your head. Blank plus one-16th is equal to 24-16ths. We could say, okay, this could, this could... We could do this as a certain number of 16ths. So, how many 16ths plus one-16th is going to be 24-16ths? Well, 23-16ths. That's 23-16ths, and I add one more 16th, I'm going to have 24-16ths. Another way that you could have thought about it, Actually, you could have even thought about it from the first step, is you could say, look, if blank plus one-16th is equal to three-halves, then you could say that blank, three-halves, three-halves minus one-16th, minus one-16th. This is another way that you could have tackled it. If blank plus one-16th is three-halves, then blank is going to be equal to three-halves minus one-16th. And we know that this is going to be equal to... Three-halves, we already know, it's the same thing as 24-16ths. So, 24 over 16 minus one-16th, minus one-16th, which we figured out is 23/16ths, which is equal to... I'll just rewrite it again, 23 over 16. Let's do another example. So, here, this is a little bit different. I have blank minus three-fourths is equal to two-thirds. And there's a couple of ways to think about it. If blank minus three-fourths is equal to two-thirds,"
    },
    {
        "Q": "shouldnt the x+5 over 100 @ 16:12 be a 10 because of ([x--5]/10)^2 ?!\n",
        "A": "Yes, in the exponential term of the yellow integral either the denominator should be 10 or the square should be limited to the numerator",
        "video_name": "hgtMWR3TFnY",
        "timestamps": [
            972
        ],
        "3min_transcript": "too much-- because the area under the curve just under 0-- there's no area. It's just a line. You have to say between a range. So you have to say the probability between, let's say, minus-- and actually, I can type it in here on our-- I can say, the probability between, let's say, minus 0.005 and plus 0.05 is-- well, it rounded, so it says there, close to 0. Let me do it-- between minus 1 and between 1, all right? It calculated at 7%, and I'll show you how I calculated this in a second. So let me get the Screen Draw tool. This between minus 1 and 1-- and I'll show you the behind the scenes what Excel is doing-- we're going from minus 1, which is roughly right here, to 1. And we're calculating the area under the curve, all right? Or, for those of you who know calculus, we're calculating the integral from minus 1 to 1 of this function, where the standard deviation is right here, is 10, and the mean is minus 5. And actually, let me put that in. So we're calculating for this example, the way it's drawn right here, the normal distribution Let's see. Our standard deviation is 10 times the square root of 2 pi, times e to the minus 1/2, times x minus our mean. Our mean is negative right now, right? Our mean is minus 5. So it's x plus 5 over the standard deviation squared, which is the variance. So that's 100 squared dx. This is what this number is right here. This 7%, or actually 0.07, is the area right under there. this isn't an easy integral to evaluate analytically, even for those of us who know our calculus. So this tends to be done numerically. And kind of an easy way to do this-- well, not an easy way-- but a function has been defined, called the cumulative distribution function, that is a useful tool for figuring out this area. So what the cumulative distribution function is, is essentially-- let me call it the cumulative distribution function-- it's a function of x. It gives us the area under this curve. So let's say that this is x right here. That's our x. It tells you the area under the curve up to x. Or so another way to think about it-- it tells you, what is the probability that you land at some value less than your x value? So it's the area from minus infinity to x of our probability density function, dx. When you actually use the Excel normal distribution function,"
    },
    {
        "Q": "why is the middle one horizontal at 1:35 in the video?\n",
        "A": "As Sal explain from 1:36 onward, the reason it was initially horizontal, is to remind us, how we have to put the numbers vertically, with the ones aligned, and the tens aligned. It s just a reminder that an addition might be presented in a test in this (horizontal) notation, and that we have to arrange them in the specific way to carry out the addition procedure.",
        "video_name": "9hM32lsQ4aI",
        "timestamps": [
            95
        ],
        "3min_transcript": "We have three different addition problems right over here. And what I want you to do so you get the hang of things is to pause the video and try them on your own. But as you do them, I want you to really keep in mind and think about what the carrying actually means. So I assume you've tried on your own. Now I'll work through them with you. So we have 9 + 6. 9 ones + 6 ones. Well 9 + 6 is 15. Well, we could write the 5 in the ones place, and then we carry the 1. But what did we just do? What does this 1 represent? Well, we put it in the tens place. 1 ten represents 10. So all we said is that 9 + 6 is equal to 10 + 5, is equal to 1 ten + 5, which is equal to 15. Now, in the tens place, we have 1 + 0 + 9, which is 10. And so we can write 0 and carry the 1. 1 + 0 + 9 is 10. Now what does that really mean? Well this is 1 ten + 0 tens + 9 tens, which is 10 tens. Or another way of thinking about 100, it's 1 hundred, and 0 tens. So that's all that carrying represents. So now we have 1 + 7 + 9. That is going to be 17. Now, we have to remind ourselves, this is in the hundreds place. So this is actually 1 hundred, plus 7 hundreds, plus 9 hundreds, or 17 hundreds. Or 1 thousand, 7 hundreds. And of course we have this 5 here. And we are done. Now let's try to tackle this one. Now the reason why I wrote it that way is to make sure that we remind ourselves to align the proper places under the appropriate places. So this problem we could rewrite as 373 + 88 (stacked vertically). We want to write the ones under the ones place, and the tens under the tens place, so that we're adding the appropriate place values. The 1 in the ones place, and the 1 in the tens place. 10 + 1 is 11. 1 + 7 is 8. 8 + 8 is 16. But this is actually 16 tens. So let me just write down the 16. What is 16 tens? Well, that's 160. So this 6 is a sixty. And then I have 100. 1 + 3 is 4. But these are actually hundreds, so it's 4 hundreds. So we get 461. Now finally, 9 + 3 is 12. 2 ones, and 1 ten. 12 is the same thing as 10 + 2. Now in the tens place: 1 + 4 + 9 is 14. So we write the 4, and carry the 1. But remind ourselves, this is actually 10 + 40 + 90, which is 140. Which is the same thing as 40 + 100. And then 1 + 1 + 2 is 4. But this is the hundreds place. So this is actually 1 hundred, plus 1 hundred, plus 2 hundreds or 4 hundreds."
    },
    {
        "Q": "At 2:00 what if it remained as dt? Could we still do it with respect to another variable, like in differentiation?\n",
        "A": "Yes! You can still do the problem but treating all the other variables as constants. So you re answer would be: (e^a + 1/a)t + c Everything within the parenthesis can be thought of as a single constant. So the problem would be the same as say integral(3) = 3t + c Hope that helps!",
        "video_name": "hXg-6YgAARk",
        "timestamps": [
            120
        ],
        "3min_transcript": "I thought I would do a few more examples of taking antiderivatives, just so we feel comfortable taking antiderivatives of all of the basic functions that we know how to take the derivatives of. And on top of that, I just want to make it clear that it doesn't always have to be functions of x. Here we have a function of t, and we're taking the antiderivative with respect to t. And so you would not write a dx here. That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative, of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. The derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t-- so plus sine of t. We've found the antiderivative. Now we don't have a t. We're taking the indefinite integral with respect to-- actually, this is a mistake. This should be with respect to a. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be da. That's what we are integrating or taking the antiderivative with respect to. So what is this going to be equal to? Well once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a da, so this one right over here-- a d I'll do it in green-- plus the indefinite integral, or the antiderivative, of 1/a da. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e in the exponential function in general is so amazing. And if we just replaced a with x or x with a, you get the derivative with respect to a of e to the a"
    },
    {
        "Q": "\nAt 3:45 couldn't the indefinite integral of (1/a) also be ln(a)+c or does the (a) part have to be its absolute value like Sal put it, ln(IaI)+c?\nThanks in advance",
        "A": "Sometimes people don t bother with the absolute value for this integral, but in that case you re stating the integral only for positive values of x, because ln(x) is undefined for x<=0. When you include the absolute value, you state the integral for all x in the domain of 1/x (that is, all x except 0).",
        "video_name": "hXg-6YgAARk",
        "timestamps": [
            225
        ],
        "3min_transcript": "would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. The derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t-- so plus sine of t. We've found the antiderivative. Now we don't have a t. We're taking the indefinite integral with respect to-- actually, this is a mistake. This should be with respect to a. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be da. That's what we are integrating or taking the antiderivative with respect to. So what is this going to be equal to? Well once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a da, so this one right over here-- a d I'll do it in green-- plus the indefinite integral, or the antiderivative, of 1/a da. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e in the exponential function in general is so amazing. And if we just replaced a with x or x with a, you get the derivative with respect to a of e to the a So the antiderivative here, the derivative of e to the a, the antiderivative is going to be e to the a. And maybe you can shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me-- always important. Remember the constant. So you have a constant factor right over here. Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1/a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative, so there could be a constant factor out here as well. And we are done. We found the antiderivative of both of these expressions."
    },
    {
        "Q": "At 5:57 Vi mentions an equilateral right triangle. How can an equilateral triangle also be a right triangle? Doesn't each angle have to be 60 degrees?\n",
        "A": "Whenever Vi says equilateral right triangle, she means isosceles right triangle. She actually makes that mistake a lot.",
        "video_name": "Oc8sWN_jNF4",
        "timestamps": [
            357
        ],
        "3min_transcript": "And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd square root 2 squared as much stuff. And square root 2 squared is, of course, two, which is the amount of stuff you got. Odd how dragons turn out to be exactly two dimensional. But in the end, you get a fill-up thing with a fractal edge. And that's a lot like a filled in dragon dungeon. 2D area of pink steel on the inside, infinite fractal patina on the outside. Except the dragon curve still gets weirdness points for getting its 2D-ness from an infinitely squiggled up line rather than triangles that are 2D to begin with. And now you're plagued by another thought. What about an infinitely long line, a true line, rather than the line segments you've been dealing with? In a way, an infinitely long line doesn't have a length. Not a defined one. Like the infinitely short line, there's no real number capable of describing it. And if there's no number for it, how can you multiply that number by 2? If, when you make a line twice as long, you don't get a line with exactly twice the length, is a line really one dimensional? And if the Koch curve is made out"
    },
    {
        "Q": "\nAt 2:50what would be an example of what he's explaining?",
        "A": "Sal is saying that given two functions, if each is substituted into the other, different answers will (usually) result. Say we had f(x) = x - 3 and g(x) = - 4x + 7. Substituting f(x) into g will give us - 4 ( x - 3 ) +7 which simplifies to - 4x +19. But substituting g(x) into f will give us ( - 4x + 7 ) - 3 which simplifies to - 4x +4. Hope this helps!",
        "video_name": "_b-2rZpX5z4",
        "timestamps": [
            170
        ],
        "3min_transcript": "Voiceover:When we first got introduced to function composition, we looked at actually evaluating functions at a point, or compositions of functions at a point. What I wanna do in this video is come up with expressions that define a function composition. So, for example, I wanna figure out, what is, f of, g of x? f of, g of x. And I encourage you to pause the video, and try to think about it on your own. Well, g of x in this case, is the input to f of x. So, wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So, f of g of x is going to be equal to the square root of- Well instead of an x, we would write a g of x. g of x, g of x squared. g of x squared, minus one. Now what is g of x equal to? So this is going to be equal to the square root of, g of x, is x over 1 plus x. We're going to square that. We're going to square that, minus 1. So f of g of x, is also a function of x. So f of g of x is a square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x, squared, minus one. Now let's go the other way round. What is g of f of x? What is g of f of x? And once again, I encourage you to pause the video, and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to, this is going to be equal to, f of x, over- so you can appreciate it better. f of x over, one plus f of x. One plus f of x. And what's that equal to? Well, f of x is equal to the square root, of x squared minus one. x squared minus one. So it's gonna be that over 1, plus the square root. One plus the square root of x squared minus one. So this is a composition f of g of x, you get this thing. This is g of f of x, where you get this thing. And to be clear, these are very different expressions. So typically, you want the composition one way. This isn't gonna be the same as the composition the other way, unless the functions are designed in a fairly special way."
    },
    {
        "Q": "\nAt 2:35, can you cancel out the square root of x - 1 in the numerator and the denominator? What would the answer be if you could?",
        "A": "First, it is sqrt(x^2-1), not x-1 in both numerator & denominator. Second, you can t cancel them out because the sqrt(x^2-1) in the denominator is being added to 1. You can only cancel items being multiplied (factors), not terms (items being added/subtracted).",
        "video_name": "_b-2rZpX5z4",
        "timestamps": [
            155
        ],
        "3min_transcript": "Voiceover:When we first got introduced to function composition, we looked at actually evaluating functions at a point, or compositions of functions at a point. What I wanna do in this video is come up with expressions that define a function composition. So, for example, I wanna figure out, what is, f of, g of x? f of, g of x. And I encourage you to pause the video, and try to think about it on your own. Well, g of x in this case, is the input to f of x. So, wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So, f of g of x is going to be equal to the square root of- Well instead of an x, we would write a g of x. g of x, g of x squared. g of x squared, minus one. Now what is g of x equal to? So this is going to be equal to the square root of, g of x, is x over 1 plus x. We're going to square that. We're going to square that, minus 1. So f of g of x, is also a function of x. So f of g of x is a square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x, squared, minus one. Now let's go the other way round. What is g of f of x? What is g of f of x? And once again, I encourage you to pause the video, and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to, this is going to be equal to, f of x, over- so you can appreciate it better. f of x over, one plus f of x. One plus f of x. And what's that equal to? Well, f of x is equal to the square root, of x squared minus one. x squared minus one. So it's gonna be that over 1, plus the square root. One plus the square root of x squared minus one. So this is a composition f of g of x, you get this thing. This is g of f of x, where you get this thing. And to be clear, these are very different expressions. So typically, you want the composition one way. This isn't gonna be the same as the composition the other way, unless the functions are designed in a fairly special way."
    },
    {
        "Q": "\nat 11:20, I dont understand, like why does the y axis have to have the price, and the x axis is months why cant the y axis be months, and x axis be price ?",
        "A": "X is usually used for representing an independent variable. Y is usually for dependent variables. Since time is independent (time goes on no matter what) the months are put on the X axis. I don t expect text books to do it the other way around.",
        "video_name": "36v2EXZRzUE",
        "timestamps": [
            680
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 5:19, Why didn't he use positive 4 instead of \"I minus x\"? which is from what I can understand here is, 1 minus 4? And how did he get the answer 4 instead?",
        "A": "Using the lower value of the domain, -3. 1-(-3) = 4",
        "video_name": "rpI-X9Gn5a4",
        "timestamps": [
            319
        ],
        "3min_transcript": "So when does this thing hit its low point? o this thing hits, hits its low point when x is as small as possible. An x is going to be as small as possible when x is approaching negative 6. So if x were equal to negative 6, it can't equal negative 6 herer but if x is equal to negative 6, then this thing over here would be equal to negative 6 plus 7, would be, would be 1. So if x is greater than negative 6, g of x is going to be greater than 1, or another way to think about it is if negative 6 is less than x, then 1 is going to be less than g of x. And the reason I said that is if I put negative 6 into this, negative 6 plus 7 is equal to 1. Now this gonna hit a high end when it as large as possible. The largest value in this interval that we can take on is x being equal to negative 3. So when x is equal to negative 3, negative 3 plus 7 is equal to equal to, so we can actually take on x equals negative 3 in which case g of x actually will take on positive 4. So, let' do that for each of these. Now here we have 1 minus x, so this is going to take on its smallest value when x is as large as possible. So the largest value x can approach for, it can't quite take on for, but it's going to approach for. So if x, let's see, if we said x was 4, although that's not this clause here, 1 minus x, 1 minus 4 is negative 3. So as long as x is less than 4, then negative 3 is going to be less than g of x. I wanna make sure that makes sense to you because it can be little bit confusing because this takes on its minimum value when x is approaching, or it's approaching its minimum value when x is approaching its, when x is approaching its maximum because So if you take the upper end, even though this doesn't actually include 4, but as we approach for, we could say, OK, 1 negative 4 is negative 3 so that's, so g of x is always going to be greater than that, as well it's going to be going to be a less than. Well what happens as we approach x being equal to negative 3? So, 1 minus negative 3 is going to be positive 4. So this is going to be positive 4 right over here. And these are both less than, not less than or equal to because these are both less than right over here. And now let's think about this right over here. So 2x minus 11 is gonna hit its maximum value when x is as large as possible. So its maximum value's going to be hit when x is equal to 6 So 2 times 6 is 12, minus 11. Well that's going to be 1. So its maximum value's going to be 1. It's actually going to be able to hit because x can be equal to 6. Its minimum value is going to be when x is equal to 4, and actually can be equal to"
    },
    {
        "Q": "When Salman does range past 3:00, why are x values inputted?\n",
        "A": "Since this is a piecewise function, you have to figure out the range for each piece . And, each piece is used depending upon the value of X. So, if you are looking for the range of the 1st piece, you look at the X-values that would use that piece and what they create for output values. Hope this helps.",
        "video_name": "rpI-X9Gn5a4",
        "timestamps": [
            180
        ],
        "3min_transcript": "is less than x and I'm leaving -- so let's write it here. All real numbers -- actually let me write this way x, I could write it more math-y. I could say x is a member of the real numbers such that, such that negative 6 is less than x. Negative 6 is less than x and I also think about the upper bound. So as x goes, I just wanna make sure that we fill in all the gaps between x being a greater than negative 6 and x is less than or equal to 6. So let's see. As we go up to and including negative 3, we're in this clause. As soon as we cross negative 3, we fall into this clause up to 4, but as soon as we get 4, we're in this clause up to and including 6. So x at the high end is said to be less than or equal to 6, less then or equal to 6. Now another way to say this and kind of less math-y notation is x, x real number, any the real number such that, such that negative 6 is less than x is less than or equal to 6. These two statements are equivalent. So now let's think about the range of this function. Let's think about the range, and the range is, this is the set of all inputs , oh sorry, this is the set of all outputs that this function can take on, or all the values that this function can take on. And to do that, let's just think about as x goes, but x varies or x can be any values in this interval. What are the different values that g of x could take on? Let's think about that. g of x is going to be between what and what? g of x is going to be between what and what? g of x is going to be between what and what? And it might actually, this might be some equal signs there but I'm gonna worry about So when does this thing hit its low point? o this thing hits, hits its low point when x is as small as possible. An x is going to be as small as possible when x is approaching negative 6. So if x were equal to negative 6, it can't equal negative 6 herer but if x is equal to negative 6, then this thing over here would be equal to negative 6 plus 7, would be, would be 1. So if x is greater than negative 6, g of x is going to be greater than 1, or another way to think about it is if negative 6 is less than x, then 1 is going to be less than g of x. And the reason I said that is if I put negative 6 into this, negative 6 plus 7 is equal to 1. Now this gonna hit a high end when it as large as possible. The largest value in this interval that we can take on is x being equal to negative 3. So when x is equal to negative 3, negative 3 plus 7 is equal to"
    },
    {
        "Q": "\nat 1:55, you crossed out the 9 and made it a 3, and you made the 3 into a 1. when you did the simplified equation, instead of writing 3/100+1/1*, you wrote 3/100+1/3* shown at 2:22. when i did the problem 3/100+1/1, the answer was 103/100. when you did the problem you answered 109/300.\ndid you make a mistake? or did i miss something in a previous video?",
        "A": "Why do you have the extra * at the end of two numbers? Sal simplified 1/3 + 9/100 * 1/3 into 1/3 + 3/100 * 1/1. Because *1/1 is the same as *1 and doesn t change anything, he wrote it as 1/3 + 3/100 instead, leaving out the *1/1.",
        "video_name": "MZpULgKhaEU",
        "timestamps": [
            115,
            142
        ],
        "3min_transcript": "Most liquids, when cooled, will simply shrink. Water, on the other hand, actually expands when it is frozen. Its volume will increase by about 9%. Suppose you have 1/3 of a gallon of water that gets frozen. What is the volume of the ice that you now have? So you're starting with 1/3 of a gallon of water. They tell us that when it gets frozen, when it turns into ice, its volume is going to expand by 9%. So the new volume is going to be your existing volume. So this is the original volume, 1/3 of a gallon, and it's going to expand by 9%. So your frozen volume is going to be your original volume plus 9% of your original volume. So you could say it's 9% times 1/3. So this right over here is going to be the expanded volume. We could turn things to decimals or whatever else, but they tell us to express your answer as a fraction. So let's make sure that everything here is a fraction, and then we'll just try to simplify. So the one thing that's sitting here that is not a fraction is our 9%. Well, what does 9% actually represent? Well, 9% literally means 9 per 100. So we could rewrite this as-- so this is going to be equal to 1/3 plus, instead of writing 9%, I'll write that as 9 per 100, and then once again times 1/3. And we can simplify this expression right over here. We have a 9 in the numerator, a 3 in the denominator. If we divide both of them by 3, we get a 3 and a 1. And so we're left with 1/3 plus 300 times 1/1. Well, that's just going to be 3/100. write this in orange still, or maybe I'll do it in a new color-- plus 3/100. And now we have to add something, or two numbers that have different denominators. So let's find a common denominator. So this is going to be equal to, well, the least common multiple of 3 and 100. And they share no common factor, so it's really just going to be the product of 3 and 100-- the least common multiple is So it's going to be something over 300 plus something over 300. Now to go from 3 to 300, in the denominator you multiply by 100, so you have to multiply the numerator by 100 as well. So 1/3 is the same thing as 100/300. And to go from 100 to 300, we have to multiply by 3 in the denominator, so we have to multiply by 3 in the numerator as well. So 3/100 is the same thing as 9/300."
    },
    {
        "Q": "\nhow can the temperature be same on a particular day every year. At 2:37 you can observe Sal saying in future January 7 also you will have the highest average temperature as 29C .",
        "A": "It isn t the same on a particular day every year, but if you averaged all of the temperatures that it had previously been on that day of the year, it would be the temperature that Sal used. Therefore, the temperature could really have been 30C or 26C, but it is most likely to be 29C.",
        "video_name": "mVlCXkht6hg",
        "timestamps": [
            157
        ],
        "3min_transcript": "pretty apparent why they are suggesting that we use a trigonometric function to model this. Because our seasonal variations they're cyclical. They go up and down. Actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. This axis right over here. This is the passage of the days. Let's do d for days and that's going to be in days after January seventh. So this right over here would be January seventh. And the vertical axis, this is the horizontal axis. The vertical axis is going to be in terms of Celsius degrees. The high is 29 and I could write 29 degrees Celsius. The highest average day. If this is zero then 14, which is the lowest average day. 14 degrees Celsius. The highest average day, which they already told us, is January seventh we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day and the reason why a trigonometric function is a good idea is because it's cyclical. If this is January seventh, if you go 365 days in the future, you're back at January seventh. If the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function so we're going to hit our low point exactly halfway in between. So we're going to hit our low point exactly halfway in between. Something right like that. So our function is going to look like this. I'm going to draw the low point right over there and this is a high point. That's a high point right over there. That looks pretty good. Then, I have the high point right over here and then, I just need to connect them and there you go. I've drawn one period of our trigonometric function and our period is 365 days. If we go through 365 days later we're at the same point in the cycle, we are at the same point in the year. We're at the same point in the year. So, what I want you to do right now is, given what I've just drawn, try to model this right. So, this right over here, let's write this as T as a function of d. Try to figure out an expression for T as a function of d and remember it's going to be some trigonometric function. So, I'm assuming you've given a go at it and you might say, \"Well this looks like a cosine curve, maybe it could be \"a sine curve, which one should I use?\" You could actually use either one, but I always like to go with the simpler one."
    },
    {
        "Q": "At 2:48, why did he take 25^2 and not -25^2 like he did with the -b part of the abc formula?\n",
        "A": "He made a mistake saying it, it is supposed to be (-25)\u00c2\u00b2 but because when you square a number or its negative, it will be the same thing, so he got away with it. Good job spotting it. You should report so they can add an annotation to youtube that he meant (-25)\u00c2\u00b2 to clear the confusion for others.",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            168
        ],
        "3min_transcript": "So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8. this is over here. So let's say so we have 625 minus-- let's see., this is going to be 16 times 36. 16 times 36 is equal to 49. It's a nice perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7, all of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32, 32/8, which is equal to 4. And then our other solution, let me do that in a different color."
    },
    {
        "Q": "1:27 is really confusing. Need some clarification\n",
        "A": "He expanded (2x-6)^2, making it now 4x^2-24x+36. When you have the square of a binomial, you take square of the first monomial=4x^2; multiply (2)(the first)(the second): (2)(2x)(-6): -24x. Then you square the second=36. 4x^-24x+36. If you didn t get that you can multiply (2x-6)(2x-6). You will get the same.",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            87
        ],
        "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8."
    },
    {
        "Q": "at 2:17 after he subtracted x from both sides why did he right -25x when in the equation above it he had written -24x, and where did he subtract the x on the right hand side from\n",
        "A": "becuz hes subtracting a -24 by a -x and the x on the right hand side hes combining with the -24. so a -24 minus -x=-25",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            137
        ],
        "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8."
    },
    {
        "Q": "\nAt 0:55, isn't PEMDAS the same as Please Excuse My Dear Aunt Sally?",
        "A": "Yes. It is the same thing. :) some people find it easier to remember PEMDAS, others say GEMDAS (where G stands for Grouping symbols), and yet others memorize the mnemonic Please Excuse My Dear Aunt Sally. Another version of it is Please Excuse My Dear Aunt Sally, she limps from left to right. This helps the student remember which direction do do these symbols in. :) Hope this helps! Sylvia.",
        "video_name": "0uCslW40VHQ",
        "timestamps": [
            55
        ],
        "3min_transcript": "Evaluate the expression 5y to the fourth minus y squared when y is equal to 3. So every place we see a y here, we could just replace it with a 3 to evaluate it. So it becomes 5 times 3 to the fourth power minus 3 squared. All I did is every time we saw a y here, I put a 3 there. Every time we saw a y, I put a 3. So what does this evaluate to? And we have to remember our order of operations. Remember, parentheses comes first. Sometimes it's referred to as PEMDAS. Let me write that down. PEMDAS, PEMDAS. P is for parentheses. E is for exponents. M and D are for Multiplication and Division. They're really at the same level of priority. And then addition and subtraction If you really want to do it properly, it should be P-E, and then multiplication and division are really at the same level. And addition and subtraction are at the same level. But what this tells us is that we do parentheses first. But then after that, exponentiation takes priority over everything else here. before we multiply anything or before we subtract anything. So the one exponent we'd have to evaluate is 3 squared. So let's remember. 3 to the first is just 3. It's just 3 times itself once. 3 squared is equal to 3 times 3, 3 multiplied by itself twice. That's equal to 9. 3 to the third power is equal to 3 times 3 times 3. Or you could view it as 3 squared times 3. So it'll be 9. 3 times 3 is 9. 9 times 3 is equal to 27. 3 to the fourth is equal to 3 times 3 times 3 times 3. So 3 times 3 is 9. 3 times 3 is 9. So it's going to be the same thing as 9 times 9. So this is going to be equal to 81. So we now know what 3 to the fourth is. We know what 3 squared is. Let's just put it in the expression. So this is going to be equal to 5 times 3 to the fourth. 3 to the fourth is 81. And we have 3 squared right over here. It is equal to 9. 5 times 81 minus 9. Let's figure out what 5 times 81 is. So 81 times 5. 1 times 5 is 5. 8 times 5 is 40. So this right over here is 405. So it becomes 405 minus 9. So that is going to be equal to-- if we were subtracting 10, it would be 395. But we're subtracting one less than that. So it's 396. And we're done."
    },
    {
        "Q": "\nIn the video on calculating the square footage of a house, at 2:50 how did you get the width of the green rectangle to be 25ft? Shouldn't it be 26ft, the same as the width of the blue one?",
        "A": "It isn t drawn to scale.",
        "video_name": "xo4VpX2IIMk",
        "timestamps": [
            170
        ],
        "3min_transcript": "It has a width of 20 feet, and it has a length of 20 feet. So that would be a rectangle right over there. We should be able to figure out its area. Then I could set up another rectangle that has a width of 26 feet and has a length. That's its length right over there. And we could think about in a second what that length actually is. Actually, let's think about that. How would we figure out what this length actually is? Well, this length plus 5 feet is going to be the same thing as this length over here. It's the same as the opposite wall of this rectangle. So this length plus 5 feet is 20 feet. Well, this must be 15 feet. So this blue rectangle is 15 feet long and 26 feet wide. Now let's add another rectangle. We could have one that's 18 feet long and then goes the entire length of the house. Goes the entire length of the house like that. How do we figure out the width of this rectangle? Well, we know that this is 8 feet. We know that this is 20 feet, and we know that this is 26 feet. So the entire width is going to be 26 feet plus 20 feet. So 26 plus 20 gets us to 46. Plus 8 gets us to 54 feet. So this is 54 feet right over here. Did I do that right? 8 plus 36 would be 34, plus 20 is 54 feet. And then finally we have one last rectangle to deal with, this rectangle right over here, which is 15 feet long and 25 feet wide. And so now we can calculate the areas of the different rectangles. So the total area is going to be the 20 feet by the 20 feet. So let's just multiply them. So it's going to be 20 times 20. Plus 15 times 26. That's this area right over there. Plus 18 times 54, which is this area right over there. And then finally, plus 15 times 25, which is this area right over here. So we just have to now evaluate these things. So what is 20 times 20? Well, this is going to be 400. What's 15 times 26? Well, let's multiply it out. 26 times 15. 6 times 5 is 30. 2 times 5 is 10, plus 3 is 13."
    },
    {
        "Q": "At 1:00, couldn't there be another line that bisects angle BCA starting from point C and ending at a point that is between A and B?\n",
        "A": "Yes. There could be another line.",
        "video_name": "21vbBiCVijE",
        "timestamps": [
            60
        ],
        "3min_transcript": "I have triangle ABC here. And in the last video, we started to explore some of the properties of points that are on angle bisectors. And now, what I want to do in this video is just see what happens when we apply some of those ideas to triangles or the angles in triangles. So let's bisect this angle right over here-- angle BAC. And let me draw an angle bisector. So the angle bisector might look something-- I want to make sure I get that angle right in two. Pretty close. So that looks pretty close. So that's the angle bisector. Let me call this point right over here-- I don't know-- I could call this point D. And then, let me draw another angle bisector, the one that bisects angle ABC. So let me just draw this one. It might look something like that right over there. And I could maybe call this point E. So AD bisects angle BAC, and BE bisects angle ABC. So the fact that this green line-- AD bisects this angle right over here-- be equal to that angle right over there. They must have the same measures. And the fact that this bisects this angle-- angle ABC-- tells us that the measure of this angle-- angle ABE-- must be equal to the measure of angle EBC. Now, we see clearly that they have intersected at a point inside of the triangle right over there. So let's call that point I just for fun. I'm skipping a few letters, but it's a useful letter based on what we are going to call this in very short order. And there's some interesting things that we know about I. I sits on both of these angle bisectors. And we saw in the previous video that any point that sits on an angle bisector is equidistant from the two sides of that angle. So for example, I sits on AD. So it's going to be equidistant from the two sides of angle BAC. So this is one side right over here. And then this is the other side right over there. So because I sits on AD, we know that these two distances are going to be the same, assuming that this is the shortest distance between I and the sides. And then, we've also shown in that previous video that, when we talk about the distance between a point and a line, we're talking about the shortest distance, which is the distance you get if you drop a perpendicular. So that's why I drew the perpendiculars right And let's label these. This could be point F. This could be point G right over here. So because I sits on AD, sits on this angle bisector, we know that IF is going to be equal to IG. Fair enough. Now, I also sits on this angle bisector. It also sits on BE, which says that it must be equidistant. The distance to AB must be the same as I's distance to BC. I's distance to AB we already just said is this right over here."
    },
    {
        "Q": "ok, be sure to watch all the way to the end ! sal makes an important correction after minute 5:26. for a minute there i thought i still didn't get how to simplify fractions! whew... :))\n",
        "A": "This is a duplicate - see the question above",
        "video_name": "CLrImGKeuEI",
        "timestamps": [
            326
        ],
        "3min_transcript": "All of that over negative 2. And let's see if we can simplify the radical expression here, the square root of 60. Let's see, 60 is equal to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect square here. We do have a 2 times 2 in there. It is 2 times 2 times 15, or 4 times 15. So we could write, the square root of 60 is equal to the square root of 4 times the square root of 15, right? The square root of 4 times the square root of 15, that's what 60 is. 4 times 15. And so this is equal to-- square root of 4 is 2 times the square of 15. So we can rewrite this expression, right here, as being equal to negative 8 plus or minus 2 times the square Now both of these terms right here are divisible by either 2 or negative 2. So let's divide it. So we have negative 8 divided by negative 2, which is positive 4. So let me write it over here. Negative 8 divided by negative 2 is positive 4. And then you have this weird thing. Plus or minus 2 divided by negative 2. And really what we have here is 2 expressions. But if we're plus 2 and we divide by negative 2, it will be negative 1. And if we take negative 2 and divide by negative 2, we're going to have positive 1. So instead of plus or minus, you could imagine it is going to be minus or plus. But it's really the same thing. It's really now minus or plus. If it was plus, it's now going to be a minus. If it was a minus, it's now going to be a plus. Minus or plus 2 times the square root of 15. Or another way to view it is that the two solutions here These are both values of x that'll satisfy this equation. And if this confuses you, what I did, turning a plus or minus into minus plus. Let me just take a little bit of an aside there. I could write this expression up here as two expressions. That's what the plus or minus really is. There's a negative 8 plus 2 roots of 15 over negative 2. And then there's a negative 8 minus 2 roots of 15 over negative 2. This one simplifies to-- negative 8 divided by negative 2 is 4. 2 divided by negative 2 is negative 1. 2 times a 4 minus the square root of 15. And then over here you have negative 8 divided by negative 2, which is 4. And then negative 2 divided by negative 2, which is plus the square of 15."
    },
    {
        "Q": "\nAt 2:06 how did he get the four in the equation? (8^2 -4 (-1)(-1)...",
        "A": "It is part of the quadratic formula: x = [-B +/= sqrt(B^2 - 4AC)] / 2A",
        "video_name": "CLrImGKeuEI",
        "timestamps": [
            126
        ],
        "3min_transcript": "Use the quadratic formula to solve the equation, negative x squared plus 8x is equal to 1. Now, in order to really use the quadratic equation, or to figure out what our a's, b's and c's are, we have to have our equation in the form, ax squared plus bx plus c is equal to 0. And then, if we know our a's, b's, and c's, we will say that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. So the first thing we have to do for this equation right here is to put it in this form. And on one side of this equation, we have a negative x squared plus 8x, so that looks like the first two terms. But our constant is on the other side. So let's get the constant on the left hand side and get a 0 here on the right hand side. So let's subtract 1 from both sides of this equation. The left hand side of the equation will become negative x squared plus 8x minus 1. Now we have it in that form. We have ax squared a is negative 1. So let me write this down. a is equal to negative 1. a is equal to negative 1. It's implicit there, you could put a 1 here if you like. A negative 1. Negative x squared is the same thing as negative 1x squared. b is equal to 8. So b is equal to 8, that's the 8 right there. And c is equal to negative 1. That's the negative 1 right there. So now we can just apply the quadratic formula. The solutions to this equation are x is equal to negative b. Plus or minus the square root of b squared, of 8 squared, the green is the part of the formula. The colored parts are the things that we're substituting into the formula. Minus 4 times a, which is negative 1, times negative 1, times c, which is also negative 1. And then all of that-- let me extend the square root sign a little bit further --all of that is going to be over 2 times a. In this case a is negative 1. So let's simplify this. So this becomes negative 8, this is negative 8, plus or minus the square root of 8 squared is 64. And then you have a negative 1 times a negative 1, these just cancel out just to be a 1. So it's 64 minus is 4. That's just that 4 over there. All of that over negative 2. So this is equal to negative 8 plus or minus the"
    },
    {
        "Q": "\nAt 0:35 what would be the way to solve for x?",
        "A": "3x + 5 = 17 To solve for x, we need to be on one side of the equation by itself. To do this we could first subtract 5 from both sides of the equation so that we would get 3x = 12. Then we can divide both sides of the equation by 3 so that the equation then becomes x = 4 Did that help?",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            35
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:34 Arnt you supposed to do minus 3 on both sides? 3x=17 im new to this thing",
        "A": "For 3x=17, we need to divide by 3 because x is being multiplied by 3 and the opposite of multiplication is division, not subtraction. Does that make sense?",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            34
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 8:21 can you make the fraction -8/7 a decimal?",
        "A": "Yes! It is also equal to -(8/7). So you can make it a negative mixed number -(1 and 1/7). You ignore the 1 for a moment, and you do 1 divided by 7, which is 0.14285714285, and 0.14285714285 plus the ignored 1 is equal to 1.14285714285.",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            501
        ],
        "3min_transcript": ""
    },
    {
        "Q": "xD is that all you simplify it to!? seems like you alreddy got the awnser at 8:21\n",
        "A": "yes you want to make sure u simplify everything if u want it to be correct",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            501
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:58 What if the outcomes werent all equally likely? How would I solve it?\n",
        "A": "As long as the other holly wands have a probability not all zero, the answer would be the same. This is because the probability of a set of events happening must be greater than the probability of a proper subset of them happening.",
        "video_name": "0uHhk7P9SNo",
        "timestamps": [
            118
        ],
        "3min_transcript": "- So, this right over here is a screenshot of the Describing Subsets of Sample Spaces exercise on Khan Academy, and I thought I would do a couple of examples, just because it's good practice just thinking about how do we describe sets and subsets. So it reads, Harry Potter is at Ollivanders Wand Shop. As we all know, the wand must choose the wizard, so Harry cannot make the choice himself. He interprets the wand selection as a random process so he can compare the probabilities of different outcomes. The wood types available are holly, elm, maple, and wenge. The core materials on offer are phoenix feather, unicorn hair, dragon scale, raven feather and thestral tail. All right! Based on the sample space of possible outcomes listed below, what is more likely? And so, we see here, we have four different types of woods for the wand, and then each of those could be combined with five different types of core, Phoenix Feather, Unicorn Hair, Dragon Scale, And so, that gives us four different woods and each of those can be combined with five different cores. 20 possible outcomes. And they don't say it here but they way they're talking I guess we can, I'm going to go with the assumption that they're equally likely outcomes, although it would have been nice if they said that, \"These are all equally likely\" but these are the 20 outcomes. And so, which of these are more likely? The wand that selects Harry will be made of holly or unicorn hair. So, how many of those outcomes involve this? So if, Holly are these five outcomes and if you said, \"Holly or Unicorn Hair\" it's going to be these five outcomes plus, well this one involves Unicorn Hair but we've already included this one, but the other ones that's not included for the Holly, that involve Unicorn Hair, are the Elm Unicorn, the Maple Unicorn and the Wenge Unicorn. So it's these five, plus these three, right over here. So eight of these 20 outcomes. And if these are all equally likely outcomes, that means there is an 8/20 probability of a wand So this is 8/20 or, that's the same thing as 4/10 or 40% chance. Now, the wand that selects Harry will be made of Holly and Unicorn Hair. Well, Holly and Unicorn Hair, that's only one out of the 20 outcomes. So this, or course, is going to be a higher probability. It actually includes this outcome and then seven other outcomes. So, the first choice includes the outcome for the second choice plus seven other outcomes. So this is definitely going to be a higher probability. Let's do a couple more of these or at least one more of these. You and a friend are playing \"Fire-Water-Sponge\". I've never played that game. In this game, each of the two players chooses fire, water or sponge. Both players reveal their choice at the same time and the winner is determined based on the choices. I guess this is like Rock, Paper, Scissors. Fire beats sponge by burning it."
    },
    {
        "Q": "\nAt 1:34, why did he switched around y-x=5 to y=5+x?",
        "A": "if you add x to both sides of the equal sign, then you can eliminate the negative x on the left side, and add a positive x on the right side (-x + x = 0, and 0 + x = x)",
        "video_name": "0BgUKHTW37E",
        "timestamps": [
            94
        ],
        "3min_transcript": "We're asked to solve and graph this system of equations here. And just as a bit of a review, solving a system of equations really just means figuring out the x and y value that will satisfy both of these equations. And one way to do it is to use one of the equations to solve for either the x or the y, and then substitute for that value in the other one. That makes sure that you're making use of both constraints. So let's start with this bottom equation right here. So we have y minus x is equal to 5. It's pretty straightforward to solve for y here. We just have to add x to both sides of this equation. So add x. And so the left-hand side, these x's cancel out, the negative x and the positive x, and we're left with y is equal to 5 plus x. Now, the whole point of me doing that, is now any time we see a y in the other equation, we can replace it with a 5 plus x. So the other equation was-- let me do it in this orange This second equation told us, if we just rearranged it, that y is equal to 5 plus x, so we can replace y in the second equation with 5 plus x. That makes sure we're making use of both constraints. So let's do that. We're going to replace y with 5 plus x. So this 9x plus 3y equals 15 becomes 9x plus 3 times y. The second equation says y is 5 plus x. So we're going to put 5 plus x there instead of a y. 3 times 5 plus x is equal to 15. And now we can just solve for x. We get 9x plus 3 times 5 is 15 plus 3 times x is 3x is equal to 15. So we can add the 9x and the 3x, so we get 12x, plus 15, is Now we can subtract 15 from both sides, just so you get only x terms on the left-hand side. These guys cancel each other out, and you're left with 12x is equal to 0. Now you divide both sides by 12, and you get x is equal to 0/12, or x is equal to 0. So let me scroll down a little bit. So x is equal to 0. Now if x equals 0, what is y? Well, we could substitute into either one of these equations up here. If we substitute x equals 0 in this first equation, you get 9 times 0 plus 3y is equal to 15. Or that's just a 0, so you get 3y is equal to 15. Divide both sides by 3, you get y is equal to 15/3 or 5. y is equal to 5."
    },
    {
        "Q": "\nat 1:30 of the video why do we need to subtract 5 or add -5 to the problem 3 times.",
        "A": "Because 3x has an x in it, and x is a variable meaning it can change. so if: x=2 3x = 6 there are too many possibilites, so we would call that another term But -5 is just -5! So it would not change. That is why we do that. Hope that helped!",
        "video_name": "cvB8b4AACyE",
        "timestamps": [
            90
        ],
        "3min_transcript": "We're asked to solve for x. And we have this compound inequality here, negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And really, there's two ways to approach it, which are really the same way. And I'll do both of them. And I'll actually do both of them simultaneously. So one is to just solve this compound inequality And I'll just rewrite it. Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And the other way is to think of it as two separate inequalities, but both of them need to be true. So you could also view it as negative 16 has to be less than or equal to 3x plus 5. And 3x plus 5 needs to be less than or equal to 20. This statement and this statement are equivalent. This one may seem a little bit more familiar because we can independently solve each of these inequalities and just remember the \"and.\" This one might seem a little less traditional because now we have three sides to the statement. We have three parts of this compound inequality. But what we can see is that we're actually In any situation, we really just want to isolate the x on one side of the inequality, or in this case, one part of the compound inequality. Well, the best way to isolate this x right here is to first get rid of this positive 5 that's sitting in the middle. So let's subtract 5 from every part of this compound inequality. So I'm going to subtract 5 there, subtract 5 there, and subtract 5 over there. And so we get negative 16 minus 5 is negative 21, is less than or equal to 3x plus 5 minus 5 is 3x, which is less than or equal to 20 minus 5, which is 15. And we could essentially do the same thing here. If we want to isolate the 3x, we can subtract 5 from both sides. We get negative 21. Negative 21 is less than or equal to 3x. And we get, subtracting 5 from both sides. And notice, we're just subtracting 5 from every part of this compound inequality. So this statement and this statement, once again, are the exact same thing. Now, going back here, if we want to isolate the x, we can divide by 3. And we have to do it to every part of the inequality. And since 3 is positive, we don't have to change the sign. So let's divide every part of this compound inequality by 3. You divide every part by 3. This is equivalent to dividing every part of each of these inequalities by 3. And then we get negative 21 divided by 3 is negative 7, is less than or equal to x, which is less than or equal to 15 divided by 3 is 5. You get negative 7 is less than or equal to x, and x is less than or equal to 15/3, which is 5. This statement and this statement are completely equal. And we've solved for x. We've given you the solution set. And if we want to graph it on a number line,"
    },
    {
        "Q": "\nAt 7:41 Sal says that an ellipse is the locus of all points. What does Sal mean when he says locus?",
        "A": "No, he does not mean focus. A locus can be defined as a set of points satisfying a given condition. Sal defines a locus as a set of points which are such that the sum of the distance between the point and two fixed points (f1 and f2) is constant.",
        "video_name": "QR2vxfwiHAU",
        "timestamps": [
            461
        ],
        "3min_transcript": "that focus, is equal to g plus h, or this big green part, which is the same thing as the major diameter of this ellipse, which is the same thing as 2a. Hopefully that that is good enough for you. Now, the next thing, now that we've realized that, is how do we figure out where these foci stand. Or, if we have this equation, how can we figure out what these two points are? Let's figure that out. So, the first thing we realize, all of a sudden is that no matter where we go, it was easy to do it with these points. But even if we take this point right here and we say, OK, what's this distance, and then sum it to that distance, that should also be equal to 2a. And we could use that information to actually figure out where the foci lie. So, let's say I have -- let me draw another one. And then we want to draw the axes. For clarity. Let me write down the equation again. Just so we don't lose it. x squared over a squared plus y squared over b squared is equal to 1. Let's take this point right here. These extreme points are always useful when you're trying to prove something. Or they can be, I don't want to say always. Now, we said that we have these two foci that are symmetric around the center of the ellipse. This is f1, this is f2. And we've already said that an ellipse is the locus of all points, or the set of all points, that if you take each of these points' distance from each of the focuses, and add them up, you get a constant number. And we've figured out that that constant number is 2a. So we've figured out that if you take this distance right here and add it to this distance right here, So we could say that if we call this d, d1, this is d2. We know that d1 plus d2 is equal to 2a. And an interesting thing here is that this is all symmetric, right? This length is going to be the same, d1 is is going to be the same, as d2, because everything we're doing is symmetric. These two focal lengths are symmetric. This distance is the same distance as this distance right there. So, d1 and d2 have to be the same. There's no way that you could -- this is the exact center point the ellipse. The ellipse is symmetric around the y-axis. So if d1 is equal to d2, and that equals 2a, then we know that this has to be equal to a. And this has to be equal to a. I think we're making progress. And the other thing to think about, and we already did that in the previous drawing of the ellipse is, what"
    },
    {
        "Q": "\nhow can it be reduced to a scalar multiple of the first vector?\n2:18",
        "A": "the first vector is 2 3 and second vector 4 6 let C1 be the scalar multiple of first vector and C2 be the scalar multiple of second vector but, second vector = [4 6] = 2 [2 3] = 2 first vector therefore, Linear combination = (C1 + 2C2) [2 3] by substituting *(C1 + 2C2)* as C3, C3 becomes a new scalar multiple for the first vector",
        "video_name": "CrV1xCWdY-g",
        "timestamps": [
            138
        ],
        "3min_transcript": "Let's say I had the set of vectors-- I don't want to do it that thick. Let's say one of the vectors is the vector 2, 3, and then the other vector is the vector 4, 6. And I just want to answer the question: what is the span of And let's assume that these are position vectors. What are all of the vectors that these two vectors can represent? Well, if you just look at it, and remember, the span is just all of the vectors that can be represented by linear combinations of these. So it's the set of all the vectors that if I have some constant times 2 times that vector plus some other constant times this vector, it's all the possibilities that I can represent when I just put a bunch of different real numbers for c1 and c2. Now, the first thing you might realize is that, look, this vector 2, this is just the same thing as So I could just rewrite it like this. I could just rewrite it as c1 times the vector 2, 3 plus c2 times the vector-- and here, instead of writing the vector 4, 6, I'm going to write 2 times the vector 2, 3, because this vector is just a multiple of that vector. So I could write c2 times 2 times 2, 3. I think you see that this is equivalent to the 4, 6. 2 times 2 is 4. 2 times 3 is 6. Well, then we can simplify this a little bit. We can rewrite this as just c1 plus 2c2, all of that, times 2, 3, times our vector 2, 3. And this is just some arbitrary constant. It's some arbitrary constant plus 2 times some other arbitrary constant. So we can just call this c3 times my vector 2, 3. So in this situation, even though we started with two vectors is equal to all of the vectors that can be constructed with some linear combination of these, any linear combination of these, if I just use this substitution right here, can be reduced to just a scalar multiple of my first vector. And I could have gone the other way around. I could have substituted this vector as being 1/2 times this, and just made any combination of scalar multiple of the second vector. But the fact is, that instead of talking about linear combinations of two vectors, I can reduce this to just a scalar combination of one vector. And we've seen in R2 a scalar combination of one vector, especially if they're position vectors. For example, this vector 2, 3. It's 2, 3. It looks like this. All the scalar combinations of that vector are just going to lie along this line. So 2, 3, it's going to be right there. They're all just going to lie along that line right there, so along this line going in both directions forever."
    },
    {
        "Q": "at 5:23, whats place values?\n",
        "A": "Place values are which place a specific digit is in a number. It goes ones, tens, hundreds, thousands, hundred thousands, and so on. So, for a number 7,932, the 2 is in the ones place, the 3 is in the tens place, the 9 is in the hundreds place, and the 7 is in the thousands place.",
        "video_name": "UzXoxglkr98",
        "timestamps": [
            323
        ],
        "3min_transcript": "is eight 10s, or 80. One other way we could've thought about this is 240, as we've already said, is 24 10s. If we divide 24 10s by three, we end up with eight 10s, and eight 10s is equal to 80. If we have eight 10s, that equals 80. So this is one other way that we could've thought about it, both ways using the zero or our knowledge of 10s to break this division problem down so we didn't have to deal with a large three-digit number but could deal with simpler, smaller numbers. Let's try another one, this time let's do thousands, let's make this one trickier. What about something like 42 hundred, or 4,200, divided by seven. So here, again we can break this number down. 42 hundred, this number 42 hundred, or 4,200, can be written as 42 times 100, because our pattern tells us when we have a number, like a whole number like 42, and we multiply by 100, we keep our whole number of 42 and we add two zeroes now. So 42 times 100, and then we still need to have our divided by seven at the end. Reverse these numbers so that 42 and seven can be next to each other, 42 divided by seven, because that's a division problem, a division factor we might already know. 42 divided into groups of seven is six, 100 times six is 600. So our solution, going back up here to 4,200 divided by seven, is 600. Or we could've thought about it again, still thinking about place value but using words here instead of digits, 4,200 is 42 hundreds, 42, I can write that out, 42 hundreds, and if you divide 42 hundreds into groups of seven or into seven groups, each group will have six hundreds, or 600 in it. So either way, 42 hundred, 4,200 divided by seven is 600. So here again, we were able to solve a tricky problem,"
    },
    {
        "Q": "\nAt 1:15, why does Sal say \"this is the probability of 5 coin flips, NOT of the outcomes of X\"? I thought X was defined as the outcome of 5 coin flips.\nPlease answer ASAP and thanks in advance.",
        "A": "He isn t finding the probability of X at that moment. He says that he needs to FIRST find out what the probability is of 5 coin flips is so that LATER he can find the probability of X. X is defined as the # of heads out of five flips.",
        "video_name": "WWv0RUxDfbs",
        "timestamps": [
            75
        ],
        "3min_transcript": "- [Voiceover] Let's define a random variable x as being equal to the number of heads, I'll just write capital H for short, the number of heads from flipping coin, from flipping a fair coin, we're gonna assume it's a fair coin, from flipping coin five times. Five times. Like all random variables this is taking particular outcomes and converting them into numbers. And this random variable, it could take on the value x equals zero, one, two, three, four or five. And I what want to do is figure out what's the probability that this random variable takes on zero, can be one, can be two, can be three, can be four, can be five. To do that, first let's think about how many possible outcomes are there from flipping a fair coin five times. Let's think about this. Let's write possible outcomes. Possible outcomes from five flips. These aren't the possible outcomes for the random variable, this is literally the number of possible outcomes from flipping a coin five times. For example, one possible outcome could be tails, heads, tails, heads, tails. Another possible outcome could be heads, heads, heads, tails, tails. That is one of the equally likely outcomes, that's another one of the equally likely outcomes. How many of these are there? For each flip you have two possibilities. Let's write this down. Let me... The first flip, the first flip there's two possibilities, times two for the second flip, times two for the third flip. Actually maybe we'll not use the time notation, you might get confused with the random variable. Two possibilities for the first flip, two possibilities for the second flip, two possibilities for the third flip, two possibilities for the fourth flip, and then two possibilities for the fifth flip, or two to the fifth equally likely possibilities which is, of course, equal to 32. This is going to be helpful because for each of the values that the random variable can take on, we just have to think about how many of these equally likely possibilities would result in the random variable taking on that value. Let's just delve into it to see what we're actually talking about. I'll do it in this light, let me do it in... I'll start in blue. Let's think about the probability that our random variable x is equal to one. Well actually, let me start with zero. The probability that our random variable x is equal to zero. That would mean that you got no heads out of the five flips. Well there's only one way, one out of the 32 equally likely possibilities, that you get no heads. That's the one where you just get five tails. So this is just going to be, this is going to be equal to"
    },
    {
        "Q": "why At 2:27, Sal puts +10 below the 15 when the 15 is positive? shouldn't have been negative the 10?\n",
        "A": "Sal wanted to eliminate the - 10 from the left-hand side of the equation so that only the variable term would remain there. That s why he added + 10 to both sides of the equation.",
        "video_name": "Z7C69xP08d8",
        "timestamps": [
            147
        ],
        "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5."
    },
    {
        "Q": "At 4:30 he just \"sees\", that \u00ce\u00bb\u00c2\u00b2-4\u00ce\u00bb-5 is equal to (\u00ce\u00bb-5)(\u00ce\u00bb+1). How does one just see something like this? Did I miss something?\n",
        "A": "-5 and 1 are two factors of -5 that sum to -4",
        "video_name": "pZ6mMVEE89g",
        "timestamps": [
            270
        ],
        "3min_transcript": "The second term is 0 minus 2, so it's just minus 2. The third term is 0 minus 4, so it's just minus 4. And then the fourth term is lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well everything became a negative, right? We negated everything. And then the terms around the diagonal, we've got a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well the determinant of this is just this times that, minus this times that. So it's lambda minus 1, times lambda minus 3, minus these two guys multiplied by each other. So minus 2 times minus 4 is plus eight, minus 8. matrix right here, which simplified to that matrix. And this has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible and its determinant has to be equal to 0. So now we have an interesting polynomial We can multiply it out. We get what? We get lambda squared, right, minus 3 lambda, minus lambda, plus 3, minus 8, is equal to 0. Or lambda squared, minus 4 lambda, minus 5, is equal to 0. expression right here is known as the characteristic polynomial. Just a little terminology, polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product is minus 5, when you add them you get minus 4. It's minus 5 and plus 1, so you get lambda minus 5, times lambda plus 1, is equal to 0, right? Minus 5 times 1 is minus 5, and then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1."
    },
    {
        "Q": "\nWhat does the symbol that Sal draws at 2:34 mean?",
        "A": "If you are referring to the \u00e2\u0088\u00a9 symbol, that means intersection. For instance, let A and B be to possible events. If we want to talk about the probability of event A taking place, we would write P(A), and similarly P(B) for B. What if we want to talk about the probability of both A and B happening at the same time? We would write that as P(A \u00e2\u0088\u00a9 B) - the probability of A and B. C = A \u00e2\u0088\u00a9 B is a new set containing all the elements common to both sets A and B - the intersection of the two.",
        "video_name": "VjLEoo3hIoM",
        "timestamps": [
            154
        ],
        "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--"
    },
    {
        "Q": "at 6:30 , why did Sal write the \"0.30*$1=0.30\"??\n",
        "A": "Because he had just worked out that you had a 30% chance of winning the game which when multiplied by what your prize will give you your average gain (30 cents as opposed to playing 35 cents to play the game).",
        "video_name": "VjLEoo3hIoM",
        "timestamps": [
            390
        ],
        "3min_transcript": "right over here, just a straight up vertical line, just means given-- given that the first was green. Now, what is the probability that the second marble is green given that the first marble was green? Well, we drew the scenario right over here. If the first marble is green, there are 4 possible outcomes, not 5 anymore, and 2 of them satisfy your criteria. So 2 of them satisfy your criteria. So the probability of the first marble green being green and the second marble being green, is going to be the probability that your first is green, so it's going to be 3/5, times the probability that the second is green, given that the first was green. Now you have 1 less marble in the bag, and we're assuming that the first pick was green, so you'll only have 2 green marbles left. Let's see, 3/5 times 2/4. Well, 2/4 is the same thing as 1/2. This is going to be equal to 3/5 times 1/2, which is equal to 3/10. Or we could write that as 0.30, or we could say there's a 30% chance of picking 2 green marbles, when we are not replacing. So given that, let me ask you the question again. Would you want to play this game? Well, if you played this game many, many, many, many, many times, on average you have a 30% chance of winning $1. And we haven't covered this yet, but so your expected value is really going to be 30% times $1-- this gives you a little bit of a preview-- which is going to be $0.30 30% chance of winning $1. many, many times, that playing the game is going to give you $0.30. Now, would you want to give someone $0.35 to get, on average, $0.30? No, you would not want to play this game. Now, one thing I will let you think about is, would you want to play this game if you could replace the green marble, the first pick. After the first pick, if you could replace the green marble, would you want to play the game in that scenario?"
    },
    {
        "Q": "What does the upside down U symbol at 2:32-2:35 mean?\n",
        "A": "In this example, Sal is asking what is the probability of both the first AND second being green . The upside down U symbol in this case stands for the AND. The symbol typically stands for intersection and is used in set theory to refer to common numbers or letters in sets. For example, the intersection of {1, 3, 5, 7} and {4, 5, 6, 7} is {5, 7} because those are the numbers that you can find in both sets.",
        "video_name": "VjLEoo3hIoM",
        "timestamps": [
            152,
            155
        ],
        "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--"
    },
    {
        "Q": "\nAt 0:40, does it matter what kind of brackets to draw? Or can any be used?",
        "A": "Set notation traditionally uses the curly brackets { }",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            40
        ],
        "3min_transcript": "What I want to do in this video is familiarize ourselves with the notion of a sequence. And all a sequence is is an ordered list of numbers. So for example, I could have a finite sequence-- that means I don't have an infinite number of numbers in it-- where, let's say, I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence. So an example of an infinite sequence-- let's say we start at 3, and we keep adding 4. So we go to 3, to 7, to 11, 15. And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount, we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1"
    },
    {
        "Q": "Why did he add the K-1 thing at about 3:11\n",
        "A": "Around 3:11 Sal put in the k - 1 in to show the term before K. hope it helps",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            191
        ],
        "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1."
    },
    {
        "Q": "at 4:28 Sal writes {a sub k} k = 1, is it correct for him to write that k = 1 because if you look at the sequence then you can see that (a sub 1) is 3?\n",
        "A": "Be careful here. k is the number of the term (first term, second term, third term, fourth term), not the value of the term ( 1, 4, 7, and 10). In Sal s first example, k went from 1 to 4 because there are 4 terms in the sequence., but the values of those terms were 1 , 4, 7, and 10. The number of the first term ( 1 ) just happened to be the same as the value of the first term ( 1 ).",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            268
        ],
        "3min_transcript": "I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1. an allowable input, the domain, is restricted to positive integers. Now, how would I denote this business right over here? Well, I could say that this is equal to-- and people tend to use a. But I could use the notation b sub k or anything else. But I'll do a again-- a sub k. And here, we're going from our first term-- so this is a sub 1, this is a sub 2-- all the way to infinity. Or we could define it-- if we wanted to define it explicitly as a function-- we could write this sequence as a sub k, where k starts at the first term and goes to infinity, with a sub k is equaling-- so we're starting at 3. And we are adding 4 one less time. For the second term, we added 4 once. For the third term, we add 4 twice."
    },
    {
        "Q": "\nAt 3:09, why does he put \"k-1\" into that function?",
        "A": "Starting on the first term, we need to move (\u00f0\u009d\u0091\u0098 \u00e2\u0088\u0092 1) spaces in the sequence in order to get to the \u00f0\u009d\u0091\u0098:th term. (Moving 1 space gets us to the 2nd term, moving 2 spaces gets us to the 3rd term, and so on...) And each time we move a space, we add 3, so by the time we get to the \u00f0\u009d\u0091\u0098:th term we have added 3(\u00f0\u009d\u0091\u0098 \u00e2\u0088\u0092 1).",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            189
        ],
        "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1."
    },
    {
        "Q": "at time 3:42, why the radius of the surface is radical 2?\n",
        "A": "The radius isn t set at the square root of 2: it is the square root of y. The radius is variable as the function increases in either direction, and is equal to the distance between the y axis and f(x). The radius is only the square root of 2 when f(x) is 2.",
        "video_name": "43AS7bPUORc",
        "timestamps": [
            222
        ],
        "3min_transcript": "And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing? Well, what we can do, instead of creating discs where the depth is in little dx's, what if we created discs where the depth is in dy? So let's think about that a little bit. So let's create-- let's think about constructing a disc at a certain y-value. So let's think about a certain y-value, and we're going to construct a disc right over there that has the same radius of the shape at that point. So that's our disc. That's our disc right over here. of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disc in terms of y? And as you could imagine, we're going to do this definite integral, and it is a definite integral, with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these discs. Or I guess you could say the face of this coin. Well, to find that area it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, over here. So we can also call of this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y. It's going to be the square root of y. It's going to be the square root of y is our radius. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x, and actually this is f of y. No, it would be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius, radius squared."
    },
    {
        "Q": "did he graph it in a wrong way at 6:03\n",
        "A": "He did not. Since 1/2x was there, he started the grAph at (0, -6) , before going down by 1/2 from there.",
        "video_name": "unSBFwK881s",
        "timestamps": [
            363
        ],
        "3min_transcript": "So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1/2 x minus 6. Now, our inequality is not greater than or equal, it's just greater than negative x over 2 minus 6, or greater than negative 1/2 x minus 6. So using the same logic as before, for any x-- so if you take any x, let's say that's our particular x we want to pick-- if you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point-- in fact, you draw an open circle there-- because you can't include the point of negative 1/2 x minus 6. But it's going to be all the y's greater than that."
    },
    {
        "Q": "\ni'm confused,at1:25 do you graph the whole equation",
        "A": "You do graph the line just as you would if you were given the equality: y = 4x + 3 Once you have graphed the line, you look at the inequality to see if you should shade the area above the line or below it. For y > 4x + 3, you would shade above, and make the line a dotted line For y >= 4x+3, you would shade above, and make the line a solid line For y < 4x + 3, you would shade below, and make the line a dotted line For y <= 4x+3, you would shade below, and make the line a solid line",
        "video_name": "unSBFwK881s",
        "timestamps": [
            85
        ],
        "3min_transcript": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is"
    },
    {
        "Q": "@ around 4:15 Sal talks about:\n\n\"...the set of all of the positions or all of the position vectors that specify the triangle that is essentially formed by connecting these dots.\"\n\nI'm thinking that yes, the \"...set of all of the positions...\"; but not \"...all of the position vectors...\" which seem to span the interior of the (transformed or untransformed) triangle (if their tails are at the origin). Anyone disagree?\n",
        "A": "When it comes to vectors in standard position (starting at the origin) (i.e. position vectors), we only care about the position of the head. The line which composes the body of the vector is immaterial, merely a graphical representation which is not preserved in the numerical representation.",
        "video_name": "qkfODKmZ-x4",
        "timestamps": [
            255
        ],
        "3min_transcript": "very easy to operate any transformation on each of these basis vectors that only have a 1 in its corresponding dimension, or with respect to the corresponding variable, and everything else is 0. So all of this is review. Let's actually use this information to construct some interesting transformations. So let's start with some set in our Rn. And actually everything I'm going to do is going to be in R2, but you can extend a lot of this into just general dimensions. But we're dealing with R2 right here. Obviously, it's only 2 dimensions right here. Let's say we have a triangle formed by the points, let's say the first point is right here. Let's say it's the point 3, 2. And then you have the point, let's say that your next point in your triangle, is the point, let's just make it the I shouldn't have written that as a fraction. I don't know why I did that. 3, 2. Then you have the point minus 3, 2. And that's this point right here. Minus 3, 2. And then let's say, just for fun, let's say you have the point, or the vector-- the position vector, right? 3, minus 2. Which is right here. Now each of these are position vectors, and I can draw them. I could draw this 3, 2 as in the standard position by drawing an arrow like that. I could do the minus 3, 2 in its standard position like that. And 3, minus 2 I could draw like that. But more than the actual position vectors, I'm more concerned with the positions that they specify. And we know that if we take the set of all of the the triangle that is essentially formed by connecting these dots. The transformation of this set-- so we're going to apply some transformation of that-- is essentially, you can take the transformation of each of these endpoints and then you connect the dots in the same order. And we saw that several videos ago. But let's actually design a transformation here. So let's say we want to-- let's just write down and words what we want to do with whatever we start in our domain. Let's say we want to reflect around the x-axis. Reflect around-- well actually let's reflect around the y-axis. We essentially want to flip it over. We want to flip it over that way. So I'm kind of envisioning something that'll look something like that when we flip it over."
    },
    {
        "Q": "\n3:25 Isn't this essentially another logarithmic property?",
        "A": "Yes. \u00f0\u009d\u0091\u008f^(log[\u00f0\u009d\u0091\u008f] (4)) = 4. 4^(log[4] (\u00f0\u009d\u0091\u008f)) = \u00f0\u009d\u0091\u008f \u00e2\u0087\u0094 4 = \u00f0\u009d\u0091\u008f^(1/(log[4] (\u00f0\u009d\u0091\u008f))). Thereby log[\u00f0\u009d\u0091\u008f] (4) = 1/(log[4] (\u00f0\u009d\u0091\u008f))",
        "video_name": "qtsMgdZ98Yg",
        "timestamps": [
            205
        ],
        "3min_transcript": "write what the base is. Well this is going to be true regardless of which base you choose as long as you pick the same base. This could be base nine, base nine in either case. Now typically, people choose base 10. So 10 is the most typical one to use and that's because most peoples calculators or they might be logarithmic tables for base 10. So here you're saying the exponent that I have to raise A to to get to B is equal to the exponent I have to raise 10 to to get to B, divided by the exponent I have to raise 10 to to get to A. This is a really really useful thing to know if you are dealing with logarithms. And we prove it in another video. But now we'll see if we can apply it. So now going back to this yellow expression, this once again, is the same thing as one divided by this right over here. So let me write it that way actually. This is one divided by log base B of four. to re-write it. So this is going to be equal to, this is going to be equal to one, divided by, instead of writing it log base B of four, we could write it as log of four, and if I just, if I don't write the base there we can assume that it's base 10, log of four over log of B. Now if I divide by some fraction, or some rational expression, it's the same thing as multiplying by the reciprocal. So this is going to be equal one times the reciprocal of this. Log of B over log of four, which of course is just going to be log of B over log of four, I just multiplied it by one, and so we can go in the other direction now, using this little tool we established at the beginning of the video. This is the same thing as log base four of B. So we have a pretty neat result that actually came out here, we didn't prove it for any values, although we have a pretty general B here. If I take the, If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases. This is log base B, what exponent do I have to raise B to to get to four? And then here I have what exponent do I have to raise four to to get to B? Now it might seem a little bit magical until you actually put some tangible numbers here. Then it starts to make sense, especially relative to fractional exponents. For example, we know that four to the third power is equal to 64. So if I had log base four of 64, that's going to be equal to three. And if I were to say log base 64 of four,"
    },
    {
        "Q": "At 0:58, Sal mentions \"mu\". Is that the symbol for the arithmetic mean?\n",
        "A": "In the context of statistics, the letter mu (\u00ce\u00bc) often indeed refers to the arithmetic mean. In mathematics in general, it is used for many things in mathematics - sometimes even just as a variable (much like n for a natural number). The context should make it clear which \u00ce\u00bc is meant.",
        "video_name": "PWiWkqHmum0",
        "timestamps": [
            58
        ],
        "3min_transcript": "Let's say that you're curious about studying the dimensions of the cars that happen to sit in the parking lot. And so you measure their lengths. Let's just make the computation simple. Let's say that there are five cars in the parking lot. The entire size of the population that we care about is 5. And you go and measure their lengths-- one car is 4 meters long, another car is 4.2 meters long, another car is 5 meters long, the fourth car is 4.3 meters long, and then, let's say the fifth car is 5.5 meters long. So let's come up with some parameters for this population. So the first one that you might want to figure out is a measure of central tendency. And probably the most popular one is the arithmetic mean. So let's calculate that first. So we're going to do that for the population. So we're going to use mu. Well, we just have to add all of these data points up and divide by 5. And I'll just get the calculator out just so it's a little bit quicker. This is going to be for 4 plus 4.2 plus 5 plus 4.3 plus 5.5. And then, I'm going to take that sum and then divide by 5. And I get an arithmetic mean for my population of 4.6. So that's fine. And if we want to put some units there, it's 4.6 meters. Now, that's the central tendency or measure of central tendency. We also might be curious about how dispersed is the data, especially from that central tendency. So what would we use? the population variance. And the population variance is one of many ways of measuring dispersion. It has some very neat properties the way we've defined it as the mean of the squared distances from the mean. It tends to be a useful way of doing it. So let's just a bit. Let's actually calculate the population variance for this population right over here. Well, all we need to do is find the distance from each of these points to our mean right over here. And then, square them. And then, take the mean of those two squared distances. So let's do that. So it's going to be 4 minus 4.6 squared plus 4.2 minus 4.6 squared plus 5 minus 4.6 squared"
    },
    {
        "Q": "How is \u00cf\u0086 - 1 = 1/\u00cf\u0086 ? How is \u00cf\u0086 ^2-\u00cf\u0086 -1=0?? (At 1:38) and (4:17)\n",
        "A": "\u00cf\u0086 = 1 + (1/ \u00cf\u0086) He subtracts 1 from both sides. \u00cf\u0086 - 1 = 1 + (1/ \u00cf\u0086) - 1. On the right side, 1 - 1 is just 0. You are left with \u00cf\u0086 - 1 = 1/ \u00cf\u0086 \u00cf\u0086 = 1 + (1/ \u00cf\u0086) He multiplies both sides by \u00cf\u0086, like this: (\u00cf\u0086)\u00cf\u0086 = (1 + 1/\u00cf\u0086) (\u00cf\u0086). You get \u00cf\u0086^2 = \u00cf\u0086 + 1 To make a quadratic equation, subtract \u00cf\u0086 and 1 from the right side. You get \u00cf\u0086^2 - \u00cf\u0086 - 1 = 0",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            98,
            257
        ],
        "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let"
    },
    {
        "Q": "At 11:30 Sal says b/(a-b) = 1/((a-b)/b)\nHe says that he took the reciprocal, I guess I don't understand the mechanics behind taking reciprocals of fractures? Can someone explain?\n",
        "A": "Reciprocals are basically when you divided 1 by a certain fraction. What this does is it swaps both the top and the bottom parts of the fraction. So, the reciprocal of, to use this example, b/(a-b) is 1/b/(a-b), or (a-b)/b. However, you might note that the two are not equal. Therefore, in order to use it in an equation, you need to have 1/(a-b)/b, which is equal to the original fraction.",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            690
        ],
        "3min_transcript": "It just keeps on showing up in a ton of different ways when you look at a pentagram like this. If you look at something like a pentagon, a regular pentagon where all the angles are the same and all the sides are the same, a regular pentagon. If you take any of the diagonals of a regular pentagon, so right over here, if you take this diagonal right over here, the ratio of this green side to--and when I'm talking about the diagonals, ones that actually aren't one of the edges-- the ratio of any of the diagonals to any of the sides is once again, this golden ratio. So it keeps showing up on and on and on. And we can do interesting things with the golden ratio. Let's say that we had a rectangle, where the ratio of the width to the height is the golden ratio. So let's try that out. So let's say that this is its height. This is its width. And that the ratio-- So let's call this a. Let's call this b. That 1.61 so on and so forth. Let me scroll down a little bit. So that is going to be equal to phi. So that's something interesting to do. Maybe that's a nice looking rectangle of some sort. But let me put out a square here. So let me separate this into a b by b square. So this is a b by b square right over here. And then-- actually let me do it a little bit, let me draw it a little bit differently, this rectangle actually isn't exactly the way I would want to draw it-- so the ratio might look a little bit like this. So the ratio of the width to the length, or the width to the height, is going to be the golden ratio. So a over b is going to be that golden ratio. And let me separate out a little b by b square over here. So this has width b as well. And so this distance right over here is going to be a minus b. Actually, I should say, we have a b by b square, right over here. This is b by b. And then we're left with a b by a minus b rectangle. Now wouldn't it be cool if this was also the golden ratio? And so let's try it out. Let's find the ratio of b to a minus b. So the ratio of b to a minus b. Well, that's going to be equal to 1 over the ratio of a minus b to b. I just took the reciprocal of this right over here. And this is just going to be equal to 1 over a over b. Let me write this, a over b minus 1. I just rewrote this right there. And that's just going to be equal to 1 over phi. The ratio of a to b, we said, by definition was phi minus 1. But what is phi minus 1? Well phi minus 1 is 1 over phi. It's this cool number. So it's equal to 1 over 1 over 1 over phi, which is once again,"
    },
    {
        "Q": "\n@ 4:57, how come a is = to it's coefficient, and b = to it's coefficient?",
        "A": "He is using the quadratic formula to solve this quadratic equation (if you changed it to x s to make it more clear: x^2 - x - 1 = 0 The letters in the video are phi s, but it really makes no difference.",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            297
        ],
        "3min_transcript": "equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it, It can also be expressed in these kind of recursive square roots underneath each other. So this is already starting to get very, very, very intriguing, but let's get back to business. Let's actually solve for this magic number, this magic ratio that we started thinking about. And really from a very simple idea, that the ratio of the longer side to the shorter side is equal to the ratio of the sum of the two to the longer side. So let's just solve this as a traditional quadratic. Let's get everything on the left-hand side. So we're going to subtract phi plus 1 from both sides. And we get phi squared minus phi minus 1 is equal to 0. And we can solve for phi now using the quadratic formula, which we've proven in other videos. You can prove using completing the square. But the quadratic formula you say, negative b. Negative b is the coefficient on this term right here. So let me just write it down, a is equal to 1, that's the coefficient on this term. b is equal to negative 1, that's the coefficient on this term. or it's really the constant term right over there. So the solutions to this, phi-- and we're actually only going to care about the positive solution because we're thinking about a positive-- when we go to our original problem here, we're assuming that these are both positive distances, so we care about a positive value right over here. We get phi is equal to-- do it in orange-- negative b. Well negative negative 1 is 1 plus or minus the square root of b squared. b squared is going to be 1 minus 4ac. a is 1, c is negative 1. So negative 4 times negative 1 is positive 4. So 1 plus 4, all of that over 2a. So a is 1, so all of that over 2. So phi is equal to 1. And once again, we only care about the positive solution This is going to be the square root of 5. If you have 1 minus the square root of 5, you're going to get a negative in the numerator. So we only care about the positive solution. 1 plus the square root of 5 over 2."
    },
    {
        "Q": "\nHow does Phi=1+1/Phi = Phi^2=Phi+1\nDoes 1/(x^2)=x\n(3:14)",
        "A": "If 1 + 1/phi = phi then you can multiply both sides by phi: phi(1 + 1/phi) = phi x phi (now use the distributive property on the left side: phi + 1 = phi^2 I hope my answer is clear enough",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            194
        ],
        "3min_transcript": "And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it,"
    },
    {
        "Q": "Hey Guys, and Khan, I just have one question at aaround 3:10 3:09, just say that you times both side by \u00c3\u0098 (\u00c3\u0098=phi) , but I don't understand how u get \u00c3\u0098^2= \u00c3\u0098+1, because the origional equation is \u00c3\u0098= 1+1/\u00c3\u0098, and then when you times both side by \u00c3\u0098, \u00c3\u0098^2= 1+ 1/\u00c3\u0098*\u00c3\u0098 = \u00c3\u0098^2 = 1+1, because 1/\u00c3\u0098*\u00c3\u0098 = 1/1 ... or have I got a brain malfunction ??...\n",
        "A": "I believe you re missing one step here. You are forgetting to fully distribute phi on the right side of the equation, (1+ 1/phi), for I don t believe you are multiplying the constant 1 by phi. If you multiply the whole equation by phi , then you ll first have (phi = 1 + 1/phi) * phi; then after distribution (phi * phi) = (1 * phi) + (phi * 1/phi), which of course leads to the equation phi^2 = phi +1. I this still isn t making sense I would suggest watching some of Sal s Algebra videos. Cheers!",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            190,
            189
        ],
        "3min_transcript": "And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it,"
    },
    {
        "Q": "At 1:33 is that the only symbol for phi?\n",
        "A": "yes like this:\u00cf\u0086",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            93
        ],
        "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let"
    },
    {
        "Q": "\nAt about 11:35 Sal determines that the reciprocal of b/a-b= 1/a-b/b. Why isn't it just a-b/b?",
        "A": "b/(a-b)=1/((a-b)/b) This is equivalent to b/(a-b). The second expression is the reciprocal of the reciprocal, not just the reciprocal, of b/(a-b), so it is equivalent to it. I hope this helps!",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            695
        ],
        "3min_transcript": "That 1.61 so on and so forth. Let me scroll down a little bit. So that is going to be equal to phi. So that's something interesting to do. Maybe that's a nice looking rectangle of some sort. But let me put out a square here. So let me separate this into a b by b square. So this is a b by b square right over here. And then-- actually let me do it a little bit, let me draw it a little bit differently, this rectangle actually isn't exactly the way I would want to draw it-- so the ratio might look a little bit like this. So the ratio of the width to the length, or the width to the height, is going to be the golden ratio. So a over b is going to be that golden ratio. And let me separate out a little b by b square over here. So this has width b as well. And so this distance right over here is going to be a minus b. Actually, I should say, we have a b by b square, right over here. This is b by b. And then we're left with a b by a minus b rectangle. Now wouldn't it be cool if this was also the golden ratio? And so let's try it out. Let's find the ratio of b to a minus b. So the ratio of b to a minus b. Well, that's going to be equal to 1 over the ratio of a minus b to b. I just took the reciprocal of this right over here. And this is just going to be equal to 1 over a over b. Let me write this, a over b minus 1. I just rewrote this right there. And that's just going to be equal to 1 over phi. The ratio of a to b, we said, by definition was phi minus 1. But what is phi minus 1? Well phi minus 1 is 1 over phi. It's this cool number. So it's equal to 1 over 1 over 1 over phi, which is once again, So once again, the ratio of this smaller rectangle, of its height to its width, is once again this golden ratio, this number that keeps showing up. And then we could do the same thing again. We could separate this into an a minus b by a minus b square. Just like that. And then we'll have another golden rectangle, sometimes it's called, right over there. And then we could separate that into a square and another golden rectangle. Then we could separate that into a square and then another golden rectangle. Then another golden rectangle. Actually let me do it like this. This would be better. So let me separate. Let me do the square up here. So this is an a minus b by a minus b square and then we have another golden rectangle right over here. I could put a square right in there. Then we'll have another golden rectangle. Then we could put another square right over there, you have another golden rectangle. I think you see where this is going."
    },
    {
        "Q": "at 1:41, when would we use the golden ratio?\n",
        "A": "Music, art, and nature according to Sal. Check out the pics on the end!",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            101
        ],
        "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let"
    },
    {
        "Q": "At 1:50, sal said 8-24-3= negative 19. 8-24 is 16, not 22.\n",
        "A": "just listen to the guy",
        "video_name": "-rxUip6Ulnw",
        "timestamps": [
            110
        ],
        "3min_transcript": "What I want to do is think about whether this expression right over here would evaluate the same way whether or not we had parentheses. So to think about that, let's first think about how it would evaluate if we add the parentheses. So if we add the parentheses, we want to do what's ever in the parentheses first. And so here we have 8 minus 3, which is equal to 5. So this simplifies to 5 times 5 times 8 minus 3. And now we want to do the multiplication before we do subtraction. This goes back to order of operations. You do your multiplication and division first. Well, you do your parentheses first. Then if you have multiplication, division, addition, and subtraction all in a row, you want to do your multiplication and your division first. So here we're going to multiply 5 times 8 to get 40, and then we're going to subtract 3 to get 37. Now, let's think about what this would evaluate to if we did not have the parentheses. So it would be 8 minus 3 times 8 minus 3. about the order of operations. The convention is to do your multiplication first. So you're actually going to multiply the 3 times the 8 before you subtract it from this 8 and then before you subtract this 3. So we took away the parentheses, but the order of operations say, hey, do this multiplication first. We could even put a parentheses here to emphasize that. So this will become 8 minus 8 minus 24. Let me write it this way. 8 minus 24 minus 3. 8 minus 24 minus 3. Now, 8 minus 24 is negative 16. You subtract another 3, you're going to get to negative 19. So clearly, you get very, very different values"
    },
    {
        "Q": "\nat 0:18 why do we put a sqwiggle under (8-3)",
        "A": "That means we will solve that part first. (8-3) Also, you have to solve that part first, because it is in parentheses.",
        "video_name": "-rxUip6Ulnw",
        "timestamps": [
            18
        ],
        "3min_transcript": "What I want to do is think about whether this expression right over here would evaluate the same way whether or not we had parentheses. So to think about that, let's first think about how it would evaluate if we add the parentheses. So if we add the parentheses, we want to do what's ever in the parentheses first. And so here we have 8 minus 3, which is equal to 5. So this simplifies to 5 times 5 times 8 minus 3. And now we want to do the multiplication before we do subtraction. This goes back to order of operations. You do your multiplication and division first. Well, you do your parentheses first. Then if you have multiplication, division, addition, and subtraction all in a row, you want to do your multiplication and your division first. So here we're going to multiply 5 times 8 to get 40, and then we're going to subtract 3 to get 37. Now, let's think about what this would evaluate to if we did not have the parentheses. So it would be 8 minus 3 times 8 minus 3. about the order of operations. The convention is to do your multiplication first. So you're actually going to multiply the 3 times the 8 before you subtract it from this 8 and then before you subtract this 3. So we took away the parentheses, but the order of operations say, hey, do this multiplication first. We could even put a parentheses here to emphasize that. So this will become 8 minus 8 minus 24. Let me write it this way. 8 minus 24 minus 3. 8 minus 24 minus 3. Now, 8 minus 24 is negative 16. You subtract another 3, you're going to get to negative 19. So clearly, you get very, very different values"
    },
    {
        "Q": "At 4:30 Sal says that all bi's should be members of R3. Shouldn't be all bi's members of R4 as B has 4 column components ? Or my understanding is wrong ?\n",
        "A": "How many components do B s column vectors have? What vector space are they in? How many components do B s row vectors have? What vector space are they in? (3, R3. 4, R4.) (Rn is defined as the set of all the vectors with n real number components.) This matrix B itself is 4x3 , and it s not an Rn or an Rm vector - although parts of it (e.g., its rows and columns) are. An mx1 matrix is an Rm column vector. A 1xn matrix is an Rn row vector.",
        "video_name": "x1z0hOyjapU",
        "timestamps": [
            270
        ],
        "3min_transcript": "And I showed you that in the last video. With that said, let's actually compute some matrix-matrix products just so you get the hang of it. So let's say that I have the matrix A. Let's say that A is equal to the matrix 1, minus 1, 2, 0, minus 2, and 1. I keep the numbers low to keep our arithmetic fairly straightforward. And let's say that I have the matrix B, and let's say that it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and then 3, 1, 0, 2. So A is a 2 by 3 matrix, 2 rows, 3 columns. So by our definition, what is the product AB going to be equal to? Well, we know it's well-defined because the number of columns here is equal to the number of rows, so we can actually take these matrix vector products-- you'll see that in a second-- so AB is equal to the matrix A times the column vector, 1, 2, 3. That's going to be the first column in our product matrix. And the second one is going to be the matrix A times the column 0, 0, 1. The third column is going to be the matrix A times the column vector 1, 1, 0. And then the fourth column in our product vector is going to be the matrix A times the column vector 1, minus 1, 2. And this, when we write it like this, it should be clear be the number of rows in B, because the column vectors in B are going to have the same number of components as the number of rows in B, so all of the column vectors in B-- so if we call this B1, B2, B3, B4, all of my bi's-- let me write it this way-- all of my bi's where this i could be 1, 2, or 3, or 4, are all members of R3. So we only have matrix vector products well-defined when the number of columns in your matrix are equivalent to essentially the dimensionality of your vectors. That's why that number and that number has to be the same. Well, now we've reduced our matrix-matrix product problem to just four different matrix vector product problems, so we can just multiply these. This is nothing new to us, so let's do it."
    },
    {
        "Q": "\ni did not clearly understand how Sal undistributed s(s+5) -7(s+5)=0\nat 2:15 .. could someone help me out please ?",
        "A": "so simple see that video once again seeit was s^2 +5s -7s -35 then he (s^2 +5s)_ _ _ _ _ _ (i) -(7s +35) _ _ _ _ _ _(ii) in (i) s is common so, s(s+5) in (ii) -7 is common so, -7(s+5) done this is the easiest way",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            135
        ],
        "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers"
    },
    {
        "Q": "\nDo you come out with two answers when you do these kinds of equations? If S=S is a true statement, how could it be possible that S could equal both -5 or 7 as shown at 4:00? I'm so confused! HELP! :-O",
        "A": "S can t equal both values at the same time. But, both values (-5 and 7) will make the equation be true. This means each value on its own is a valid solution to the equation. Hope this helps.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            240
        ],
        "3min_transcript": "So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx."
    },
    {
        "Q": "\nAt 2:39 isn't -2*8 equals -16?",
        "A": "-2*8 is -16 but he added the -2, making it -18",
        "video_name": "jlID_mIJXi4",
        "timestamps": [
            159
        ],
        "3min_transcript": "And what we will have will be h of what we inputted, h of g of 8. So let's just do it one step at a time. Let's figure out what g of 8 is. And I'm going to color code it, so we can keep track of things. g of 8 is equal to-- well, g of t, we have our definition here. So our input now, 8 is going to be our t, so our input is 8. So every place where we see a t in this function definition, we replace it with an 8. So it's going to be negative 2 times 8 minus 2 minus-- and this might be a little daunting, but let's just replace this t with an 8 and then see if we can make sense of it. h minus-- and let me do it in the right color-- minus 2 minus h of 8. we saw a t, we replaced it with the input 8. Now let's see if we can calculate this. This is going to be equal to negative 2 times 8 is negative 16. Minus 2 is negative 18. Let me do that the same. So this is going to be equals negative 18 minus-- what is h of 8 going to be equal to? So let's do that over here. So h of 8-- this thing, h of 8. Now we go to the definition of h. Don't worry about later we're going to input all this business into h again. Just let's worry about it one step at a time. We need to calculate h of 8. So h of 8 is just going to be-- well, every time we see an x, we replace it with an 8-- it's going to be 3 times 8, which is equal to 24. So this value right over here is 24. We are subtracting it, so we have minus 24. That's negative 42. So all of this business is going to be equal to-- did I do that right? Yeah-- negative 42. So we figured out what g of 8 is. It is negative 42. So this right over here is negative 42. And now we can input negative 42 into h. Let me do it right over here. h of negative 42-- remember, negative 42 is the same thing as g of 8. So this is h of g of 8 is the same thing as h of negative 42. Let me do that in the same color. This is going to be equal to 3 times negative 42, which is equal to-- this is negative 126. And we are done. So it seemed convoluted at first,"
    },
    {
        "Q": "At 8:20 when x=1, why is that y is incremented by 'half' of dy/dx( 1.5 ) and not 1.5 itself?\n",
        "A": "Because our \u00ce\u0094x = 0.5, so we only advance half a unit. dy/dx = 1.5 means that an increment of 1 in x would carry an increment of 1.5 in y, but since we are only incrementing x by 0.5, we only increment y by 0.75.",
        "video_name": "q87L9R9v274",
        "timestamps": [
            500
        ],
        "3min_transcript": "which is right over here. And so, for this next stretch, the next stretch is going to look like that. And as you can see just by doing this, we haven't been able to approximate what the particular solution looks like and you might say, \"Hey, so how do we know \"that's not so good of an approximation?\" And my reply to you is well, yeah I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand I took fairly large delta X steps. If I wanted a better approximation I could have lowered the delta X and let's do that. So let's take another scenario. So let's do another scenario where instead of delta X equal one, let's say delta X equals 1/2. So once again, X, Y and the derivative of Y with respect to X. So we know this first point. We're given this initial condition. When the X is zero, Y is one and so the slope of the tangent line is going to be one. But then if we're incrementing by 1/2 so then when X is, I'll just write it as 0.5. 0.5. What is our new Y going to be? Well we're gonna assume that our slope from this to this is this slope right over here. So our slope is one, so if we increase X by 0.5 we're gonna increase Y by 0.5 and we're going to get to 1.5. So, we can get 0.5, 1.5. We get to that point right over there. Actually you're having trouble seeing that. This stuff right over here is this point right over here and now our new slope is going to be 1.5. Which is going to look like. Which is going to look like actually not quite that steep. and it's starting to get a little bit messy but it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment X by another 0.5 where you get to one. So now if you increment by 0.5 and your slope is 1.5, your Y is going to increment by half of that by 0.75 and so, you're gonna get to 2.25. So now you get to one, 2.25 which is this point right over here. Once again, this is a better approximation. Remember, in the original one Y of one you know should be equal to E. Y of one in the actual solution should be equal to E. 2.7 on and on and on and on and on. Now in this one, Y of one got us to two. In this one Y of one got us to 2.25. Once again, closer to the actual reality, closer to E. Instead of stepping by 0.5, if we stepped by 0.1 we would get even closer."
    },
    {
        "Q": "\nAt 8:29, how did Sal get y=2.5 when x=1? I thought it would be 1+(1.5*.5)",
        "A": "Sal got y=2.25 when x=1. 1.5 + (1.5*.5) =2.25",
        "video_name": "q87L9R9v274",
        "timestamps": [
            509
        ],
        "3min_transcript": "which is right over here. And so, for this next stretch, the next stretch is going to look like that. And as you can see just by doing this, we haven't been able to approximate what the particular solution looks like and you might say, \"Hey, so how do we know \"that's not so good of an approximation?\" And my reply to you is well, yeah I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand I took fairly large delta X steps. If I wanted a better approximation I could have lowered the delta X and let's do that. So let's take another scenario. So let's do another scenario where instead of delta X equal one, let's say delta X equals 1/2. So once again, X, Y and the derivative of Y with respect to X. So we know this first point. We're given this initial condition. When the X is zero, Y is one and so the slope of the tangent line is going to be one. But then if we're incrementing by 1/2 so then when X is, I'll just write it as 0.5. 0.5. What is our new Y going to be? Well we're gonna assume that our slope from this to this is this slope right over here. So our slope is one, so if we increase X by 0.5 we're gonna increase Y by 0.5 and we're going to get to 1.5. So, we can get 0.5, 1.5. We get to that point right over there. Actually you're having trouble seeing that. This stuff right over here is this point right over here and now our new slope is going to be 1.5. Which is going to look like. Which is going to look like actually not quite that steep. and it's starting to get a little bit messy but it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment X by another 0.5 where you get to one. So now if you increment by 0.5 and your slope is 1.5, your Y is going to increment by half of that by 0.75 and so, you're gonna get to 2.25. So now you get to one, 2.25 which is this point right over here. Once again, this is a better approximation. Remember, in the original one Y of one you know should be equal to E. Y of one in the actual solution should be equal to E. 2.7 on and on and on and on and on. Now in this one, Y of one got us to two. In this one Y of one got us to 2.25. Once again, closer to the actual reality, closer to E. Instead of stepping by 0.5, if we stepped by 0.1 we would get even closer."
    },
    {
        "Q": "\nat 3:35, how does he know that it's x squared and not anything else",
        "A": "Okay, well he knows that he has to find the greatest common factor (GCF) among x^2, x^3, and x^4. So what terms can he evenly divide each of the above terms by, so that there is no remainder? First of all, there s x. Another option would be x-squared (x^2). x^2 divided by x^2 = 1 x^3 divided by x^2 = x x^4 divided by x^2 = x^2 But nothing larger than x^2 can fit evenly into x-squared. The only options are x and x^2. x^2 is the GREATEST, so that is what he will use to factor out of all the numbers.",
        "video_name": "tYknkDjp-bQ",
        "timestamps": [
            215
        ],
        "3min_transcript": "Three is divisible into all of them. And that's it 'cause we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five 'cause five isn't divisible into 12 or six; so the greatest common factor is going to be three. Another way we could have done this is we could have said where are the non-prime factors of each of these numbers. 12 you could have said, OK I can get 12 by saying one times 12 or two times six or three times four. Six you could have said, let's see, that could be one times six or two times three. So those are the factors of six. And then 15 you could have said well one times 15 or three times five. And so you say the greatest common factor? that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor the coefficient is going to be three. And then we look at these powers of X. We have X to the fourth. I'm using a different color. We have X to the fourth, X to the third, and X squared. Well what's the largest power of X that's divisible into all of those? Well it's going to be X squared. X squared is divisible into X to the fourth and X to the third and of course X squared itself. So the greatest common monomial factor is 3x squared. This length right over here, this is 3x squared. So if this is 3x squared, we can then figure out what the width is. If we were to divide 12x to the fourth by 3x squared, what do we get? Well 12 divided by three is four is X squared. Notice 3x squared times 4x squared is 12x to the fourth. And then we move over to this purple section. If we take 6x to the third divided by 3x squared, six divided by three is two. And then X to the third divided by X squared is just going to be X. And then last but not least, we have 15 divided by three is going to be five. X squared divided by X squared is just one, so it's just gonna be five. So the width is going to be 4x squared, plus 2x, plus five. So once again, the length, we figure that out, is the greatest common monomial factor of these terms. It's 3x squared. And the width is 4x squared plus 2x, plus five. And one way to think about it is we just factored this expression over here. We could write that."
    },
    {
        "Q": "At 1:50 why bother multiplying by 1/2? You could have multiplied the right by 2; it wouldn't have been easier than adding the extra step?\n",
        "A": "They are not a bother. Both ways are correct.",
        "video_name": "STcsaKuW-24",
        "timestamps": [
            110
        ],
        "3min_transcript": "The height of a triangle is four inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals one half base times height for the area of a triangle. OK. So let's think about it a little bit. We have the-- let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base, let's call that b. And then this is the height. This is the height right over here. And then the area is equal to one half base times height. Now in this first sentence they tell us at the height of a triangle is four inch is less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take one half the base times the height Or we could say that 30 inches squared is equal to one half times the base, times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get-- let me do this in yellow. We get 30 is equal to one half times-- let's distribute the b-- times b, let me make it clear. So let's do it this way. Times b over 2, times b, minus 4. I just multiplied the one half times the b. Now let's distribute the b over 2. So 30 is equal to b squared over 2, be careful. b over 2 times b is just b squared over 2. And then b over 2, times negative 4 is negative 2b. here let's multiply both sides of this equation by 2. So let's multiply that side by 2. And let's multiply that side by 2. On the left hand side you get 60. On the right hand side 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic-- we have a second degree term right here-- is to get all of the terms on one side of the equation, having them equal 0. So let's subtract 60 from both sides of this equation. And we get 0 equal to b squared, minus 4b, minus 60. And so what we need to do here is just factor this thing right now, or factor it. And then, no-- if I have the product of some things, and that equals 0, that means that either one or both"
    },
    {
        "Q": "At 1:06, why wouldn't it be 30^2 instead of 30 since they directions tell you that area equals 30in^2?\n",
        "A": "in^2 is a term for telling you the area of something. That is virtually all it is, just like you would use My room has 50 feet of flooring you could easily say My room s floor is 50 ft^2 and it would mean the same thing. You only have to worry about the in^2 thing when you get into physics and such. I hope that helps!",
        "video_name": "STcsaKuW-24",
        "timestamps": [
            66
        ],
        "3min_transcript": "The height of a triangle is four inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals one half base times height for the area of a triangle. OK. So let's think about it a little bit. We have the-- let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base, let's call that b. And then this is the height. This is the height right over here. And then the area is equal to one half base times height. Now in this first sentence they tell us at the height of a triangle is four inch is less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take one half the base times the height Or we could say that 30 inches squared is equal to one half times the base, times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get-- let me do this in yellow. We get 30 is equal to one half times-- let's distribute the b-- times b, let me make it clear. So let's do it this way. Times b over 2, times b, minus 4. I just multiplied the one half times the b. Now let's distribute the b over 2. So 30 is equal to b squared over 2, be careful. b over 2 times b is just b squared over 2. And then b over 2, times negative 4 is negative 2b. here let's multiply both sides of this equation by 2. So let's multiply that side by 2. And let's multiply that side by 2. On the left hand side you get 60. On the right hand side 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic-- we have a second degree term right here-- is to get all of the terms on one side of the equation, having them equal 0. So let's subtract 60 from both sides of this equation. And we get 0 equal to b squared, minus 4b, minus 60. And so what we need to do here is just factor this thing right now, or factor it. And then, no-- if I have the product of some things, and that equals 0, that means that either one or both"
    },
    {
        "Q": "This question has been asked before, I think but it hasn't been answered yet so I'll just bring it up again. At 3:42 Sal mentions that he'll explain why he's picking the numbers he's picking. My question is that could you have made the height of the rectangle any fraction with the tau in the denominator as long as the limit to it went to infinity? Are we trying to force the area to 1 perhaps?\n",
        "A": "Yes, we are trying to get the area to be 1, so that it will approach the dirac delta function as \u00cf\u0084 approaches 0.",
        "video_name": "4qfdCwys2ew",
        "timestamps": [
            222
        ],
        "3min_transcript": "This is part of the definition of the function. I'm going to tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it to be equal to 1. I'm defining this. Now, you might say, Sal, you didn't prove it to me. No, I'm defining it. I'm telling you that this delta of x is a function such that its integral is 1. So it has this infinitely narrow base that goes infinitely high, and the area under this-- I'm telling you-- is of area 1. And you're like, hey, Sal, that's a crazy function. I want a little bit better understanding of how someone can construct a function like this. So let's see if we can satisfy that a little bit more. But then once that's satisfied, then we're going to start taking the Laplace transform of this, and then we'll start manipulating it and whatnot. Let's say that I constructed another function. Let's call it d sub tau And this is all just to satisfy this craving for maybe a better intuition for how this Dirac delta function can be constructed. And let's say my d sub tau of-- well, let me put it as a function of t because everything we're doing in the Laplace transform world, everything's been a function of t. So let's say that it equals 1 over 2 tau, and you'll see why I'm picking these numbers the way I am. 1 over 2 tau when t is less then tau and greater than minus tau. And let's say it's 0 everywhere else. So this type of equation, this is more reasonable. functions, and we can actually define it as a combination of unit step functions. So if I draw, that's my x-axis. And then if I put my y-axis right here. That's my y-axis. Sorry, this is a t-axis. I have to get out of that habit. This is the t-axis, and, I mean, we could call it the y-axis or the f of t-axis, or whatever we want to call it. That's the dependent variable. So what's going to happen here? It's going to be zero everywhere until we get to minus t, and then at minus t, we're going to jump up to some level. Just let me put that point here. So this is minus tau, and this is plus tau. So it's going to be zero everywhere, and then at minus tau, we jump to this level, and then we stay constant at that level until we get to plus tau."
    },
    {
        "Q": "\nwhat is an abaquabapattern 2:05",
        "A": "It s abacaba or abacabadabacaba. You can learn about it in Vi s other video, Fractal Fractions. Hope this helps :)",
        "video_name": "pjrI91J6jOw",
        "timestamps": [
            125
        ],
        "3min_transcript": "You may have heard of turkducken, a turkey stuffed with a duck, stuffed with a chicken. Yeah, it's a cute idea, but mathematically uninspiring. Much more interesting would be some sort of fractal-fowl arrangement. Say the turkey were stuffed with two ducks, and say each duck were stuffed with two hens. You'd get a turduckduckenenenen, or turduckenenduckenen. I'm not sure yet. Then each hen could be stuffed with two quails, or whatever. And by the time you get down to sparrows, you have a whole flock of them in there. That sounds fun and practical. Let's do it! But being an extremely practical person who realizes that exponential quails will not fit inside of one turkey, I am using tiny unborn quails. Eight of them inside four hens, inside two ducks, inside one turkey. This structure is much more interesting because birds belong in trees, binary trees, that is. Obviously this is superior to the old linear bird-stuffing paradigm, and while you're deboning seven birds, you have lots of time to consider the question of binary bird stuffing nomenclature. How do you traverse this tree of syllables? you'll have to say the syllables in a linear order. You could go layer by layer, biggest birds first, like turduckenenenenailailailailailailailail. Or, you could go down through the layers, like turduckenailailenailailduckenailailenailail. The second one is certainly more complicated to say, but I like the way the structure of it suggests the structure of the whole. And say you've made two of these and put them in a goose. In the first scheme, the new bird names get inserted into the name to get gooturturduckduckduckenenenenenenenenailailailailailailailailailailailailailailailail, while for the second, you start with goose and then just repeat the old word twice to get gooturduckenailailenailailduckenailailenailailturduckenailailenailailduckenailailenailail. So it's nice that part of it stays the same. Of course, those aren't the only possible naming schemes. Maybe you go from left to right on the tree. So this would be, quailenquailduckquailenquailkeyquailenquailduckquailenquail OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs"
    },
    {
        "Q": "\nAt 3:54, it shows the egg with the whites still surrounding the yolk. How did she get it to stay together instead of all smushing around? Did she leave it inside of the shells? If so, then how do you eat it?",
        "A": "They were hard-boiled and peeled.",
        "video_name": "pjrI91J6jOw",
        "timestamps": [
            234
        ],
        "3min_transcript": "OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs two ducks in this turkey-- if I can close it. Eventually I had to go for sewing up the turkey most of the way with the one duckenailailenailail, and then stuffing the other duckenailailenailail in. Now you can arrange it nicely with the original legs, and wings, and stuffing to make it look perfectly natural. There. Anyway, once you've got that, you're pretty much good to go as far as Thanksgiving is concerned. You should already have your gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes with organic hyperbolic plain, string bean vector field with Borromean onion rings on top, double helix cut ham, pi, tau, and so on. And now, finally, you've got your turduckenailailenailailduckenailailenailail, or quailenquailduckquailenquailkeyquailenquailduckquailenquail, or turduckduckenenenenailailailailailailailail whatever. Each slice gives a different cross section of this binary bird. Here, you can see a quail egg wrapped in the light meat of the hen, wrapped in the darker meat of the duck, wrapped in the light meat of the turkey. Another cross section shows two eggs. Now you can sit down, eat your mathematically-inspired food at least mathematics is always there for you, making sense, and truth, and beauty, and birds. Lots of birds."
    },
    {
        "Q": "At 3:07, when she sows the turkey to hold everything, I wondered if she was actually going to eat that, considering the fact that there is now string somewhere in there. Or will she just take the string out when she gets to it?\n",
        "A": "She probably took the string out once it was cooked.",
        "video_name": "pjrI91J6jOw",
        "timestamps": [
            187
        ],
        "3min_transcript": "you'll have to say the syllables in a linear order. You could go layer by layer, biggest birds first, like turduckenenenenailailailailailailailail. Or, you could go down through the layers, like turduckenailailenailailduckenailailenailail. The second one is certainly more complicated to say, but I like the way the structure of it suggests the structure of the whole. And say you've made two of these and put them in a goose. In the first scheme, the new bird names get inserted into the name to get gooturturduckduckduckenenenenenenenenailailailailailailailailailailailailailailailail, while for the second, you start with goose and then just repeat the old word twice to get gooturduckenailailenailailduckenailailenailailturduckenailailenailailduckenailailenailail. So it's nice that part of it stays the same. Of course, those aren't the only possible naming schemes. Maybe you go from left to right on the tree. So this would be, quailenquailduckquailenquailkeyquailenquailduckquailenquail OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs two ducks in this turkey-- if I can close it. Eventually I had to go for sewing up the turkey most of the way with the one duckenailailenailail, and then stuffing the other duckenailailenailail in. Now you can arrange it nicely with the original legs, and wings, and stuffing to make it look perfectly natural. There. Anyway, once you've got that, you're pretty much good to go as far as Thanksgiving is concerned. You should already have your gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes with organic hyperbolic plain, string bean vector field with Borromean onion rings on top, double helix cut ham, pi, tau, and so on. And now, finally, you've got your turduckenailailenailailduckenailailenailail, or quailenquailduckquailenquailkeyquailenquailduckquailenquail, or turduckduckenenenenailailailailailailailail whatever. Each slice gives a different cross section of this binary bird. Here, you can see a quail egg wrapped in the light meat of the hen, wrapped in the darker meat of the duck, wrapped in the light meat of the turkey. Another cross section shows two eggs. Now you can sit down, eat your mathematically-inspired food"
    },
    {
        "Q": "\nI am extremely confused as to how he got s=1.04 at time 1:52... He says that s=sqrt(sum(x-xbar)^2/(n-1)), however, when I do this, I get s=2.8074. I've tried looking at other sites and they say the same thing about the formula that I wrote above. I've used the formula many times, and I still don't understand how he gets 1.04. Any help?",
        "A": "I m getting 1.04 the same as Sal. In Excel it would be: =SQRT(((1.5-2.34)^2+(2.9-2.34)^2+(0.9-2.34)^2+(3.9-2.34)^2+(3.2-2.34)^2+(2.1-2.34)^2+(1.9-2.34)^2)/6) and this gives you 1.042209",
        "video_name": "K4KDLWENXm0",
        "timestamps": [
            112
        ],
        "3min_transcript": "7 patients blood pressures have been measured after having been given a new drug for 3 months. They had blood pressure increases of, and they give us seven data points right here-- who knows, that's in some blood pressure units. Construct a 95% confidence interval for the true expected blood pressure increase for all patients in a population. So there's some population distribution here. It's a reasonable assumption to think that it is normal. It's a biological process. So if you gave this drug to every person who has ever lived, that will result in some mean increase in blood pressure, or who knows, maybe it actually will decrease. And there's also going to be some standard deviation here. It is a normal distribution. And the reason why it's reasonable to assume that it's a normal distribution is because it's a biological process. It's going to be the sum of many thousands and millions of random events. And things that are sums of millions and thousands of random events tend to be normal distribution. And we don't know anything really about it outside of the sample that we have here. Now, what we can do is, and this tends to be a good thing to do, when you do have a sample just figure out everything that you can figure out about that sample from the get-go. So we have our seven data points. And you could add them up and divide by 7 and get your sample mean. So our sample mean here is 2.34. And then you can also calculate your sample standard deviation. Find the square distance from each of these points to your sample mean, add them up, divide by n minus 1, because it's a sample, then take the square root, and you get your sample standard deviation. I did this ahead of time just to save time. Sample standard deviation is 1.04. And when you don't know anything about the population distribution, the thing that we've been doing from the get-go is estimating that character with our sample So we've been estimating the true standard deviation of the population with our sample standard deviation. Now in this problem, this exact problem, we're going to run into a problem. We're estimating our standard deviation with an n of only 7. So this is probably going to be a not so good estimate because-- let me just write-- because n is small. In general, this is considered a bad estimate if n is less than 30. Above 30 you're dealing in the realm of pretty good estimates. So the whole focus of this video is when we think about the sampling distribution, which is what we're going to use to generate our interval, instead of assuming that the sampling distribution is normal like we did in many"
    },
    {
        "Q": "\n@3:32 why are we movie 1 step toward x and 7 towards y?? or 1 step back towards negative x and 7 steps towards positive y??",
        "A": "There are 2 different ways to write the slope 7, you can just say 7, or you can say 7/1 because they are the same thing, so when you move 7 up you move 1 to the right, the reason you can move 1 to the left and 7 down is because the line will be in the same spot either way.",
        "video_name": "MxiqyE2uMCo",
        "timestamps": [
            212
        ],
        "3min_transcript": "So that tells us that when x is equal to negative 4, then y is equal to negative 11. So we can use this information in what we have or the part of our equation that we've been able to figure out so far. We know that when x is equal to negative 4, y is going to be equal to negative 11. So what b do we need to make that happen? So y is negative 11 when x is equal to negative 4. So negative 11 is equal to 7 times x-- and in this case x is negative 4-- plus b. And now we can just solve for b. A b that makes this equation, or that satisfies the constraint that when x is equal to negative 4, y is equal to negative 11. So let's see, we get negative 11 is equal to 7 times negative 4 can add a 28 to both sides of this equation. So let's add a 28. I'm just trying to isolate the b on the right-hand side. And so on the left-hand side, negative 11 plus 28, that is just positive 17. These guys cancel out on purpose. And I just have a b on the right-hand side. So I get b is equal to 17. Let me write it in green. That's not green. We get b is equal to 17. So we know m is 7, they told us that right at the beginning. And now we know b is 17. So the equation of our line is y is equal to 7x, that's our slope. 7 times x plus b, and b here is 17. And if we wanted to graph it, it would look something like this. I'll just do a real rough graph. and this is my y-axis. The y-intercept is 17. So that means that the point 0, 17 is on this line. So this point right over here is going to be 0, 17. And our slope is 7. So that means if we move to the right one, we move up t seven. So it's a high slope. So if we move to the right one, we move up seven. Or if we move back one, will move down seven. So we'll move down seven, so the line will look roughly like this. Obviously, haven't done it very exactly, but our line is going to look like. That's going to be a pretty steep upward-sloping line. It has a very high slope, slope of seven. If you move one in the x direction, you have to move up seven. And its y-intercept is at y is 17. When x is 0, y is 17."
    },
    {
        "Q": "\nAt 5:19, he says ' x - -3, is the same thing as x + 3, I don't understand... I always have problems trying to simplify the point-slope form...",
        "A": "First you must understand that the opposite of the opposite of 3 is 3. Think of it this way: First you find the opposite of 3 (-3) and then find the opposite of that (3). I hope that answers your question.",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            319
        ],
        "3min_transcript": "there, or that 6 right there-- and we want to subtract from that our starting x value. Well, our starting x value is that right over there, that's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3, right there. I'm just saying, if we go from that point to that point, our y went down by 6, right? We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. Y went down by 6. And, if we went from that point to that point, what happened to x? We went from negative 3 to 6, it should go up by 9. And if you calculate this, take your 6 minus negative 3, that's the same thing as 6 plus 3, that is 9. And what is negative 6/9? Well, if you simplify it, it is negative 2/3. So that is our slope, negative 2/3. So we're pretty much ready to use point slope form. We have a point, we could pick one of these points, I'll just go with the negative 3, 6. And we have our slope. So let's put it in point slope form. All we have to do is we say y minus-- now we could have taken either of these points, I'll take this one-- so y minus the y value over here, so y minus 6 is equal to our slope, which is negative 2/3 times x minus our Well, our x-coordinate, so x minus our x-coordinate is negative 3, x minus negative 3, and we're done. We can simplify it a little bit. x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both"
    },
    {
        "Q": "Do you have to do what Sal does at about 7:39? He simplified the fraction. Is this necessary?\n",
        "A": "Doesn t seem necessary",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            459
        ],
        "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form."
    },
    {
        "Q": "At 0:08 what does the line over the letter stand 4?\n",
        "A": "That line indicates that you are talking about a segment defined by the points A and B. Those are the two ends of the line you are talking about. You may also see other similar forms of notation, like a small triangle next to the letters ABC describing triangle ABC, or something that looks like a small angle next to some letters indicating that the angle is defined by whatever is after it, like XYZ.",
        "video_name": "bJF9R8_-0O0",
        "timestamps": [
            8
        ],
        "3min_transcript": "Are line segment AB and line segment CD congruent? So let's look at these right over here. AB is this line segment right over here, and CD is this one right over here. AB has length of 1. It goes from 2 to 3. And CD has length of 1. It goes from 4 to 5. So they have the exact same length. These are just line segments with the exact same length. So yes, they are congruent. Yes. Let's do a couple more. Are AB and CD congruent? And you can just eyeball this. AB is much longer than CD. AB is of length 4. It goes from negative 5 to negative 1. CD is only of length 1. It goes from 1 to 2. So no, they are not congruent. They have different lengths. Let's do one more. Are AB and CD congruent? Let's see-- AB has length 2. It goes from negative 1 to 1. CD is of length 3. So they're not congruent. No."
    },
    {
        "Q": "At 1:12, why is the derivative of sqrt(3)x^2/36 = root 3x/18? I tried using chain rule and quotient rule on that term and I ended up getting 4 root 3x as my final answer. Was there an easier way to do that, or a correct way rather?\n",
        "A": "No need for the chain rule or the quotient rule here. All you have is x^2 times a constant (which happens to be sqrt(3)/36). The derivative of x^2 is 2x, and you multiply that by the constant to get the result indicated.",
        "video_name": "eS-_ZFzHjYA",
        "timestamps": [
            72
        ],
        "3min_transcript": "Where we left off in the last video, we had come up with an expression as a function of x of our combined area based on where we make the cut. And now we just need to figure out where this hits a minimum value. And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over-- let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So A prime, the derivative of our combined area equal to-- well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x, well, it's the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2/16, which is just over 8. And then times-- we're just doing the chain rule-- times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1, so times negative 1. So we'll multiply negative 1 right over here. And so we can rewrite all of that as-- this is going to be equal to the square root of 3/18 x plus-- let's see, I could write this as positive x/8. So I could write this as 1/8 x, right? And then minus 100/8, which is negative 12.5-- minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x-values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to-- if you add the x terms, you get square root of 3/18 plus 1/8 x. To solve for x, divide both sides by this business. You get x is equal to 12.5 over square root of 3 over 18"
    },
    {
        "Q": "\nStarting at 1:05, why did you just divide by 2 to get the derivative?",
        "A": "There s 2 things going on here. First, to get from sqrt(3)x/36 to sqrt(3)x/18, you multiply by 2, not divide. He does this due to the Power Rule for derivatives. Suppose that you have a function a*x^n, where a and n are both constants. The derivative of this function is (a*n)*x^(n-1). You multiply the constant by the exponent and then reduce the exponent by 1. As you see in the video, we go from having an x^2 to an x^1. There s a video on this website that describes this rule in more detail.",
        "video_name": "eS-_ZFzHjYA",
        "timestamps": [
            65
        ],
        "3min_transcript": "Where we left off in the last video, we had come up with an expression as a function of x of our combined area based on where we make the cut. And now we just need to figure out where this hits a minimum value. And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over-- let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So A prime, the derivative of our combined area equal to-- well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x, well, it's the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2/16, which is just over 8. And then times-- we're just doing the chain rule-- times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1, so times negative 1. So we'll multiply negative 1 right over here. And so we can rewrite all of that as-- this is going to be equal to the square root of 3/18 x plus-- let's see, I could write this as positive x/8. So I could write this as 1/8 x, right? And then minus 100/8, which is negative 12.5-- minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x-values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to-- if you add the x terms, you get square root of 3/18 plus 1/8 x. To solve for x, divide both sides by this business. You get x is equal to 12.5 over square root of 3 over 18"
    },
    {
        "Q": "At 3:42, the second missing number that you were explaining with the help of the number line.\nI am someone who gets very confused with Math, and the second 'number line' explanation, got me very confused. Rather I'd like to get this clarified.\nWhen trying to find a missing number i.e. ___ -(-2) = -7, as mentioned above. If minus of (minus some number) is nothing but a positive number, can we start with +2 on the number line and find out how much is needed to get to -7?\n",
        "A": "Yes, -(-2) becomes +2 and you can start with +2 on the number line to get to -7. Hopefully, I didn t confuse you any further.",
        "video_name": "KNGa11O2uLE",
        "timestamps": [
            222
        ],
        "3min_transcript": "four is equal to, let me just write it without my boxes there, is equal to negative eight minus negative 12. Well what happens when you subtract a negative? Well that ends up being adding the positive. So this is gonna be the same thing as negative eight plus positive 12, or four is equal to negative 8 plus 12, which we know is true. You add eight you get to zero, negative eight plus eight is zero, then you add another four, you get to positive four. So that one was pretty interesting. Let's do another one, we can't get enough practice, this is too much fun. All right. So let's say we had... let's say we had something blank minus negative two is equal to negative seven. How do we think about this? Well the first thing my brain wants to do is to simplify subtracting a negative. that's the same thing as adding positive two. So this is an equivalent statement to say, something subtracting a negative two is the same thing as adding a positive two. Something plus two is equal to negative seven. And now we can get the number line out to think about it. So, get my number line out. So, my goal, my goal is I wanna get to negative seven. So there's negative five, negative six, negative seven right there. And I wanna add two to something to get to negative seven. So if I'm adding two that means I'm jumping two, I'm jumping two to the right, that's what adding two looks like. So what number do I need to start at that if I jump two to the right, if I add two I end up at negative seven? Well I would have to start two to the left of negative seven. If I start two to the left, one, two, if I started... Let me do this in a different color. If I start at this number right here I end up at negative seven. Well what number is this? This number that is two to the left of negative seven. Well it's negative nine. So this right over here is negative nine. Negative nine plus two, so if you start at negative nine and you move two to the right you end up at negative seven. Or, you could say, negative nine minus negative two, 'cause if you subtract negative two you're going to move to the right. If you subtract positive two, you move to the left. But then if you subtract negative two, you move two to the right, to negative seven. Another fascinating one. Let's do one more of these, I'm having maybe a little bit too much fun. All right. So let's say we have negative four is equal to blank minus nine. So let's get the number line out and always pause the video, see if you can try it on your own. But let's get my number line out"
    },
    {
        "Q": "what is si in 0:25 and deta in 2:04\n",
        "A": "Psi and theta are Greek letters. Greek letters are often used to denote angles.",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            25,
            124
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 0:25, is there a special meaning si has? Oh, and theta too. (2:01.)",
        "A": "psi and theta are just like using x in algebra they are just variables",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            25,
            121
        ],
        "3min_transcript": ""
    },
    {
        "Q": "how does sal gets 1/2 at 5:50 - 5:56\n",
        "A": "He divided both sides of the equation by 2. i.e. if 2 pens are worth 1 dollar, then 1 pen is worth 1/2 a dollar.",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            350,
            356
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:56, what does base angle mean?\n",
        "A": "Let s back up for a second. Sal has a triangle with an unknown side (a chord) and two radii. Because the two radii are congruent, then we know the triangle is isosceles. A base angle in an isosceles triangle is one of the two angles formed by the base (non congruent side) and one congruent side. A vertex angle is the angle formed by the two congruent sides of the triangle, in this case two radii.",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            236
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 1: 40 ish what's an alternate interior angle it means whats in alternate interior angle it said it at around 1:40 in the video",
        "A": "An Alternate interior angle are angles that are on opposite sides of the transversal, and are between the lines cut by the ttransversal",
        "video_name": "LhrGS4-Dd9I",
        "timestamps": [
            100
        ],
        "3min_transcript": "What we're going to prove in this video is a couple of fairly straightforward parallelogram-related proofs. And this first one, we're going to say, hey, if we have this parallelogram ABCD, let's prove that the opposite sides have the same length. So prove that AB is equal to DC and that AD is equal to BC. So let me draw a diagonal here. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. Nope. That's not any better. That is about as good as I can do. So if we view DB, this diagonal DB-- we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent-- so angle ABD. That's that angle right there-- is going to be congruent to angle BDC, because they are alternate interior angles. So we know that angle ABD is going to be congruent to angle BDC. Now, you could also view this diagonal, DB-- you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC right over here is going to be congruent to angle ADB for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles are congruent when you have a transversal intersecting And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle. Then they have this side in common. And then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle-side-angle that these two triangles are congruent. So let me write this down. We have shown that triangle-- I'll go from non-labeled to pink to green-- ADB is congruent to triangle-- non-labeled to pink to green-- CBD. And this comes out of angle-side-angle congruency."
    },
    {
        "Q": "\nWhy do we have to go down by one and a half ?\nduring 3:45",
        "A": "To make a right triangle between point N and the line of reflection",
        "video_name": "kj3ZfOQGKdE",
        "timestamps": [
            225
        ],
        "3min_transcript": "I want to drop it to, I want to drop it to the line that I'm going to reflect on, and then I'm going to go the same distance onto the other side to find to find the corresponding point in the image. So how do I do that? Well if this line, if this purple line has a slope of negative one, a line that is perpendicular to it a line that is perpendicular to it so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one. But we want the negative of that. So the slope here needs to be one. And luckily, that's how I drew it. The slope here needs to be equal to one which is however much I change in the X direction I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four and we decrease X by four. Now, if we want to stay on this line to find the reflection, we just do the same thing. so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four. So this is, this point corresponds to this point right over there. Now, let's do the same thing. Let's do the same thing for point N. For point N, we already know, if we drop a perpendicular, and this is perpendicular, it's going to have a slope of one because this purple line has a slope of negative one. The negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we want to do that on the other side. We want to stay on this perpendicular line. So we want to go left one and a half, and down one and a half. And we get to this point right over here, which is the point So we are now equidistant. We're on this perpendicular line, still, but we're equidistant on the other side. So the image of IN is going to go through negative six, negative four, and three, negative eight. So let me draw that. Let me see if I can remember, negative six, negative four, and three, negative eight. So I have a bad memory. So negative six, negative four, and three, negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And actually, we can do the exact same thing with points T and point O. Let me do that. So point T, to point T, To get from point T to the line in the shortest distance, once again, we drop a perpendicular. This line is going to have a slope of one, because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by"
    },
    {
        "Q": "At about 1:40 Sal shows that you can multiply the numerator by the reciprocal (8 x 36/8, or simplified which is 8 x 9/2) to get the denominator. But why? I mean how does he know he can do that? Where does this idea come from? I feel like there's something that I am missing here.\n",
        "A": "they have top find out what to multiply the next ratio by so you flip the numerator and the denominator (pardon my spelling) which you can sliplify to 9/2 . WHich means you multiply thenext ratio by 9 over 2 that is how they know this.",
        "video_name": "GO5ajwbFqVQ",
        "timestamps": [
            100
        ],
        "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this,"
    },
    {
        "Q": "\nCouldn't Sal have just converted 5/4 into 1.25 instead of going through all the motions at 0:60?",
        "A": "I like the other ways because it gives me more than one POV and more ways to solve it.",
        "video_name": "GO5ajwbFqVQ",
        "timestamps": [
            60
        ],
        "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this,"
    },
    {
        "Q": "\nI'm doing proportions right now, and I'm trying the first method that he used at 0:25 , but when I use the method on the problem, 9/8 = 11/r, and I try to divide 11 by 9 to see what times 9 is equal to 11, I'm getting a repeating decimal! What should I do? Thanks.",
        "A": "Thanks this is super helpful!",
        "video_name": "GO5ajwbFqVQ",
        "timestamps": [
            25
        ],
        "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this,"
    },
    {
        "Q": "At 4:55, I'm not sure where Sal gets the extra ten. Is it one of the two 10s left over from the beginning?\n",
        "A": "He got an extra ten at that time because Sal rewrote 29.12 into 2.912. Since 29.12 equals 2.912 x 10.",
        "video_name": "xxAFh-qHPPA",
        "timestamps": [
            295
        ],
        "3min_transcript": "1 +1 = 2 and 2 + 7 = 9 and we have a 2 here. So 91 x 32 is 2912 But I didn't have 91 by 32 I had 9.1 x 3.2 So what I am going to do is count the number of digits behind the decimal point. I have 2 digits behind the point, so that's how many I need in the answer. I will stick the decimal right over there. So this part here comes out as 29.12 So you might say we are done as this looks like sientific notation As we have a number times by a power of ten. But remember this number has to be greater than or equal to one but this is not less than ten, so what we can do is write this number in scientific notation. and use the power of ten part to multiply this power of ten part. so 29.12 is the same thing as 2.912 x 10 notice what I had to do to go from there to there. I just moved the decimal to the left, so what can I do to this to get back I could multiply this (2.912) by 10 or move the decimal to the right. So I want to write this so 2.912 x 10 is the same as 29.12 so 2.912 is in scientific notation but I still have to multiply it by another part times another ten so to finnish up this problem I have Whats that, well that is ten Well that is going to be this part over here That's just ten squared. So it is 2.912 times ten to the second power. And we are done."
    },
    {
        "Q": "at 1:23, he said the oval had no sides, but doesnt it have 1?\n",
        "A": "it has infinite tangents, but no sides, just like a circle.",
        "video_name": "8xbIS2UkQxI",
        "timestamps": [
            83
        ],
        "3min_transcript": "- [Voiceover] Oh look, it's my cousin Fal! Cousin Fal, what can I do for you? - [Voiceover] Hey Sal, what's up? - [Voiceover] Oh, well not much, I'm just sitting here thinking about what video I should do next. - [Voiceover] Well then maybe you can help me. I'm starting a shape collection. - [Voiceover] Okay, so how do you need help? - [Voiceover] Well look at these shapes over here! I have no way to classify them! - [Voiceover] Okay, well there's a bunch of ways that you could classify them. The first way, you could think about how many sides each of these shapes have, so that's one way to think about it, how many sides they have. - [Voiceover] How do I figure that out? - [Voiceover] Okay, well let's look at each of these and think about how many sides they have. So this first shape right over here this is one side. We see this kind of straight right over here, it's the straight line between these two corners, so that's one side. Two sides, three sides, and four sides. So this shape right over here has four sides. - [Voiceover] Okay, I think I get it. - [Voiceover] Okay,well let's see. This one, well same idea, this one also looks like it has one, two, three, and four. Four sides. - [Voiceover] What about this one? - [Voiceover] Okay, well this one, actually this is interesting. This one actually had no corners, and it has no sides, there's no straight edges over here, this thing is all curved. So since it has no corners, it also has no sides, or at least the way that I'm thinking about it. So I would say that this one over here has zero sides. Let me write that a little bit neater for you. - [Voiceover] Okay yeah, I like to write things really neat. - [Voiceover] Okay, zero sides. And it has zero corners. Actually let me write them for all of them. This one has four sides, and we could also count the corners. We have one, that's where the two sides meet. One, two, three, four corners. it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three,"
    },
    {
        "Q": "\nAt the end of the video he messed up. It was around 3:34. It confused me",
        "A": "he said so all the side are not the same length when ther are 2 small and 2 long sides",
        "video_name": "8xbIS2UkQxI",
        "timestamps": [
            214
        ],
        "3min_transcript": "it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three, - [Voiceover] Oh, thanks a bunch Uncle-- Cousin Sal. But I always get confused whether you're my cousin or my uncle. But I'll call you Cousin Sal. But I read in a shape magazine that there was other ways that I can classify. I can classify based on whether the sides have the same size. So which of these have all the sides are the same size? - [Voiceover] Okay, well calm down. Let's look at this. So if you look at this one, it has four sides. But we see, that this side right over here, this green side, is clearly shorter than this purple side. So all the four sides here are not the same length. But if we look at this shape right over here, this side at least looks like the same length as this side. And that looks like the same length as this side, and that looks like the same length as that side. So it looks over here like all four sides are the same length. And that's also true for this shape. This side looks the same length as that side, looks the same length as that side, looks the same length as that side. This side looks the same length as that side, This is the same length as that one. - [Voiceover] Oh, thank you so much Cousin Sal! I'm ready to go classify my shape collection. This is awesome! - [Voiceover] Well I'm glad that I could help!"
    },
    {
        "Q": "\nCan a circle or an oval be considered to have 1 side? At 1:32 Sal said, \"Or at least the way that I am thinking about it.\" It makes me wonder if there is a rule about this.",
        "A": "no because it must have a starting and ending point to be a side but if we flatten the perimete of the circle (for example put a thread around a circle then measure the length ) to be a side then that s what we call a circumference",
        "video_name": "8xbIS2UkQxI",
        "timestamps": [
            92
        ],
        "3min_transcript": "- [Voiceover] Oh look, it's my cousin Fal! Cousin Fal, what can I do for you? - [Voiceover] Hey Sal, what's up? - [Voiceover] Oh, well not much, I'm just sitting here thinking about what video I should do next. - [Voiceover] Well then maybe you can help me. I'm starting a shape collection. - [Voiceover] Okay, so how do you need help? - [Voiceover] Well look at these shapes over here! I have no way to classify them! - [Voiceover] Okay, well there's a bunch of ways that you could classify them. The first way, you could think about how many sides each of these shapes have, so that's one way to think about it, how many sides they have. - [Voiceover] How do I figure that out? - [Voiceover] Okay, well let's look at each of these and think about how many sides they have. So this first shape right over here this is one side. We see this kind of straight right over here, it's the straight line between these two corners, so that's one side. Two sides, three sides, and four sides. So this shape right over here has four sides. - [Voiceover] Okay, I think I get it. - [Voiceover] Okay,well let's see. This one, well same idea, this one also looks like it has one, two, three, and four. Four sides. - [Voiceover] What about this one? - [Voiceover] Okay, well this one, actually this is interesting. This one actually had no corners, and it has no sides, there's no straight edges over here, this thing is all curved. So since it has no corners, it also has no sides, or at least the way that I'm thinking about it. So I would say that this one over here has zero sides. Let me write that a little bit neater for you. - [Voiceover] Okay yeah, I like to write things really neat. - [Voiceover] Okay, zero sides. And it has zero corners. Actually let me write them for all of them. This one has four sides, and we could also count the corners. We have one, that's where the two sides meet. One, two, three, four corners. it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three,"
    },
    {
        "Q": "What does Sal mean when he says \"we can take the scalar out\" at 2:49? What does a scalar mean in this case? Thank you!\n",
        "A": "coefficient on the h",
        "video_name": "t4GfuftHH5c",
        "timestamps": [
            169
        ],
        "3min_transcript": "we're approaching negative one, as we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero, it looks like it is equal to one, and you could, and the limit is also equal to one, we could see that as we approach it from the left, we are approaching one, as we approach from the right, we are approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, and is defined at x equals zero, and the limit as x approaches zero is equal to the same as, is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to, negative one. Let's do a few more of these. So we have the limit of negative two times f of x plus three times h of x, as x approaches negative three. Well, once again, we can use our limit properties. We know that this is the same thing as the limit, as x approaches negative three of negative two f of x, plus the limit as x approaches negative three of three times h of x. And this is the same thing as the limit, or I should say this is equal to negative two times the limit as x approaches negative three, the limit of f of x, as x approaches negative three, plus, we can take this scaler out, three times the limit of h of x, as x approaches negative three. And so we just have to figure out what the limit and the limit of h of x as x approaches negative three. So I'll first do f of x, so f of x right over here, limit is x approaches negative three, when we approach negative three from the left-hand side, it seems the value of the function is approaching three, and as we approach it from the right-hand side, it seems like the value of the function is zero. So our left-handed and right-handed limits are approaching different things, so this limit actually does not exist. Does not exist. This limit actually here does exist. But since this limit doesn't exist, and we need to figure out this limit in order to figure out this entire limit, this whole thing does not exist. Does not, does not exist. The limits that it's made up of need to exist, this is a combined limit, so each of the pieces need to exist in order for the, in this case, the scaled up sum, to actually exist."
    },
    {
        "Q": "Am I the only one who was blown away at how quickly Sal calculated the height of the container in his head at 12:22 when he said, \"5 divided by 1.65 squared ... is going to be roughly a little under 2 meters tall\" ? (actual answer is h=1.84 meters). Once again, Sal is AMAZING !\n",
        "A": "It was kind of easy to estimate that. 5/(1.65)^2 is approximately 5/(2)^2. 5/(2)^2=5/4=1.25 Since we made the denominator bigger, our answer is smaller than the actual, so it was probably more around (1.75, 2). This wasn t meant to insult or downgrade Sal s awesomeness, though. smile",
        "video_name": "tSMuKcN-RKM",
        "timestamps": [
            742
        ],
        "3min_transcript": "is exactly this right over here. So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down. c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65. Concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now-- We know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest So we just have to figure out what our cost is. And we already know what our cost is as a function of x, so we just have to put 1.65 into this equation. Evaluate the function at 1.65. So let's do that. Our cost is going to be equal to 20 times 1.65. because I'm using an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to 163. I'll just say $163.5. So it's approximately. So the cost-- let me do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54. So $163.54, which is quite an expensive box. This is fairly expensive material here. Although it's a fairly large box. 1.65 meters in width, and it's going to be twice that in length. And then you could figure out what its height is going to be. Although it's not going to be too tall. 5 divided by 1.65 squared. I don't know, it'll be roughly a little under two meters tall. So it actually is quite a large box made out of quite expensive material. The minimum cost to make this box is going to be $163.54."
    },
    {
        "Q": "At 3:47 can someone explain why he set the radian measurement sin(theta+pi/2) as the y coordinate?\n",
        "A": "He s comparing cos(theta) with sin(theta+pi/2). For example, what s sin(pi/6)? What s cos(pi/6+pi/2)?",
        "video_name": "h-TPSylHrvE",
        "timestamps": [
            227
        ],
        "3min_transcript": "the opposite side, and what's the hypotenuse? This is a unit circle, so it's going to be one. In this case, sine of theta is equal to the length of the opposite side. The length of the opposite side is equal to sine theta. And same exact logic. The cosine of theta is equal to adjacent over hypotenuse, is equal to adjacent over hypotenuse. And so that's... since the hypotenuse is equal to one, it's just the length of the adjacent side, so cosine of theta is the length of the adjacent side. So this is all a little bit of review, just showing how the unit circle definition is an extension of the Soh Cah Toa definition. But now let's do something interesting. This is the angle theta. Let's think about the angle theta plus pi over two. So the angle theta plus pi over two. So if I were to essentially add pi over two to this, I'm going to get a ray that is perpendicular to the first ray, pi over two. If we think in degrees, pi over two radians, I'm talking in radians. Pi over two radians is equivalent to 90 degrees. So we're essentially adding 90 degrees to it. So this angle right over here, that angle right over here is theta plus pi over two. Now, what I want to explore in this video, and I guess this is the interesting part of the video, is can we relate sin of theta plus pi over two to somehow sine of theta or cosine of theta? I encourage you to pause this video and try to think this through on your own before I work it out. Well let's think about what sine of theta plus pi over two is. We know from the unit circle definition, the sine of this angle, which is theta plus pi over two, is the Y coordinate. It's that, it's this value right over here. Or another way of thinking about it, it's the length of this line in magenta. This right over here is the sine of theta, So that right over there. Now how does that relate to what we have over here? Well when you look at it, it looks like we just took this triangle, and we just kind of... we rotated it. We rotated it counter clockwise by 90 degrees, which essentially what we did do. Because we took this terminal side, and we added 90 degrees to it, or pi over two radians. a little bit more rigorous about it, if this whole white angle here is theta plus pi over two, and the part that's in the first quadrant is pi over two, then this part right over here, that must be equal to theta. And if we think about it, if we try to relate the side this side that I've put in magenta relative to this angle theta using the Soh Cah Toa definition, here, relative to this angle theta in yellow, this is the adjacent side. So let's think about it a little bit. So if we were.. so what deals with the adjacent and the hypotenuse,"
    },
    {
        "Q": "does anyone else sees a realistic calculator at 3:43 to 4:27?\n",
        "A": "No I think that you are the only one.",
        "video_name": "ZElOxG7_m3c",
        "timestamps": [
            223,
            267
        ],
        "3min_transcript": "If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a, because we know what bc and theta actually are. So a squared is going to be equal to b squared... so it's going to be equal to 144, plus c squared which is 81, so plus 81, minus two times b times c. So, it's minus two, I'll just write it out. Minus two times 12 times nine, times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times nine is 108. Minus 216 times the cosine of 87 degrees. Now, let's get our calculator out in order to approximate this. And remember, this is a squared. Actually, before I get my calculator out, let's just solve for a. So a is just going to be the square root of this. So a is going to be equal to the square root... of all of this business, which I can just copy and paste. It's going to be equal to the square root of that. So let me copy and paste it. So a is going to be equal to the square root of that, which we can now use the calculator to figure out. Let me increase this radical a little bit, so that we make sure we're taking the square root of this whole thing. So let me get my calculator out. So I want to find that square root of 220. Actually, before I do that, let me just make sure I'm in degree mode, and I am in degree mode. Because we're evaluating a trig function in degrees here. So that's fine, so let me exit. So it's going to be 225 minus 216, of 87 degrees. Not 88 degrees, 87 degrees. And we deserve a drumroll now. This is going to be 14.61, or 14.618. If, say, we wanted to round to the nearest tenth, just to get an approximation, it would be approximately 14.6. So a is approximately equal to 14.6, whatever units we're using long."
    },
    {
        "Q": "\nstarting at 0:17 seconds within the video....\n\nThere are situations where two sides are known but there are no labels for which side is side a or b or c. Should I always assume the largest-looking side is side c?",
        "A": "The side that says a = ? is the A side, the side that says b = 12 is the B side, and the side that says c = 9 is the C side. If you are trying to solve a side on a triangle that is not labeled for use of the law of cosines, then A is automatically the side you are trying to solve, and B and C are the other sides. It doesn t matter which side is B and which side is C, as they are interchangeable. I hope this helps!",
        "video_name": "ZElOxG7_m3c",
        "timestamps": [
            17
        ],
        "3min_transcript": "- [Voiceover] Let's say that I've got a triangle, and this side has length b, which is equal to 12, 12 units or whatever units of measurement we're using. Let's say that this side right over here, this side right over here, has length c, and that happens to be equal to nine. And that we want to figure out the length of this side, and this side has length a, so we need to figure out what a is going to be equal to. Now, we won't be able to figure this out unless we also know the angle here, because you could bring the blue side and the green side close together, and then a would be small, but if this angle was larger than a would be larger. So we need to know what this angle is as well. So let's say that we know that this angle, which we will call theta, is equal to 87 degrees. So how can we figure out a? Well, lucky for us, we have the Law of Cosines, which gives us a way for determining a third side if we know two of the sides and the angle between them. The Law of Cosines tells us that a squared is going to be equal b squared plus c squared. Now, if we were dealing with a pure right triangle, if this was 90 degrees, then a would be the hypotenuse, and we would be done, this would be the Pythagorean Theorem. But the Law of Cosines gives us an adjustment to the Pythagorean Theorem, so that we can do this for any arbitrary angle. So Law of Cosines tell us a squared is going to be b squared plus c squared, minus two times bc, times the cosine of theta. And this theta is the angle that opens up to the side that we care about. If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a, because we know what bc and theta actually are. So a squared is going to be equal to b squared... so it's going to be equal to 144, plus c squared which is 81, so plus 81, minus two times b times c. So, it's minus two, I'll just write it out. Minus two times 12 times nine, times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times nine is 108."
    },
    {
        "Q": "At 7:25, he says \"X sub i-1.\" What does \"sub\" mean?\n",
        "A": "Sub just means that it is in the subscript, the smaller text following the bottom half of the X.",
        "video_name": "ViqrHGae7FA",
        "timestamps": [
            445
        ],
        "3min_transcript": "the conventions in order to actually calculate the area, or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle-- so let me write this down. So it's going to be rectangle one-- so the area of rectangle one-- so rectangle one plus the area of rectangle two plus the area of rectangle three-- I think you get the point here-- plus all the way to the area of rectangle n. And so what are these going to be? Rectangle one is going to be its height, which is f of x0 or f of a. Either way. x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta-- actually I can write as f of x0, What is our height of rectangle two? It's f of x1 times delta x. What's our area of rectangle three? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums, all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. I'll use that same shade. It is f of x sub n minus 1 times delta x. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating the areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this, as the sum, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle one. So it's going to be f of-- well, if we're in the i-th rectangle, then the left boundary is going to be x sub i minus 1 times delta x. And so here, right over here, is a general way of thinking about approximating the area under a curve using rectangles, where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here"
    },
    {
        "Q": "\nIn the Video at mark 2:36 I do not understand how it is true when f(x) when x=0 is equal to 0. wouldn't f(x) be equal to 0?",
        "A": "No, he does not say that f(x) is equal to 0, when x=0. The question was, what is the limit as x--->0- (coming from the left of 0), which is 1. This is why he says it is true. lim x--->0- f(x) = 1 => True.",
        "video_name": "_WOr9-_HbAM",
        "timestamps": [
            156
        ],
        "3min_transcript": "This would be true if instead of saying from the positive direction, we said from the negative direction. From the negative direction, the value of the function really does look like it is approaching 0. For approaching 1 from the negative direction, when x is right over here, this is f of x. When x is right over here, this is f of x. When x is right over here, this is f of x. And we see that the value of f of x seems to get closer and closer to 0. So this would only be true if they were approaching from the negative direction. Next question. Limit of f of x, as x approaches 0 from the negative direction, is the same as limit of f of x as x approaches 0 from the positive direction. Is this statement true? Well, let's look. Our function, f of x, as we approach 0 from the negative direction-- I'm using a new color-- as we approach 0 from the negative direction, so right over here, this is our value of f of x. Then as we get closer, this is our value of f of x. As we get even closer, this is our value of f of x. like it is approaching positive 1. From the positive direction, when x is greater than 0, let's try it out. So if, say, x is 1/2, this is our f of x. If x is, let's say, 1/4, this is our f of x. If x is just barely larger than 0, this is our f of x. So it also seems to be approaching f of x is equal to 1. So this looks true. They both seem to be approaching the limit of 1. The limit here is 1. So this is absolutely true. Now let's look at this statement. The limit of f of x, as x approaches 0 from the negative direction, is equal to 1. Well, we've already thought about that. The limit of f of x, as x approaches 0 from the negative direction, we see that we're getting closer and closer to 1. As x gets closer and closer to 0, f of x gets closer and closer to 1. So this is also true. Well, it definitely exists. We've already established that it's equal to 1. So that's true. Now the limit of f of x as x approaches 1 exists, is that true? Well, we already saw that when we were approaching 1 from the positive direction, the limit seems to be approaching 1. We get when x is 1 and 1/2, f of x is 1. When x is a little bit more than 1, it's 1. So it seems like we're getting closer and closer to 1. So let me write that down. The limit of f of x, as x approaches 1 from the positive direction, is equal to 1. And now what's the limit of f of x as x approaches 1 from the negative direction? Well, here, this is our f of x. Here, this is our f of x. It seems like our f of x is getting closer and closer to 0, when we approach 1 from values less than 1. So over here it equals 0. So if the limit from the right-hand side is a different value than the limit from the left-hand side,"
    },
    {
        "Q": "6:45 Sal's referring to HL (Hypotenuse-Leg) postulate right?\n",
        "A": "He doesn t mention that at 6:45",
        "video_name": "f8svAm237xM",
        "timestamps": [
            405
        ],
        "3min_transcript": "So this side could pivot over here. We can kind of rotate it over there. But there's only one way, now, that this orange side can reach this green side. Now the only way is this way over here. And we were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one. And so then it constrains what the triangle can become. So I don't want to make you say, in general, SSA, you do not want to use it as a postulate. I just wanted to make it clear that there is the special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. And then finally, there's a circumstance that this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle. So where you have the A in SSA is a right angle. So if you had it like this. If you have a right angle and you have some base of unknown length but you fix this length right saying it's congruent to some other triangle, and if you know that the next length is fixed-- and if you think about it, this next side is going to be the side opposite the right angle. It's going to have to be the hypotenuse of the right angle. Then you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you could do it is to bring it down over here. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. And here, they wrote the angle first. You could view this as angle-side-side. And they were able to do it because now they can write \"right angle,\" and so it doesn't form that embarrassing acronym. And this would also be a little bit common sense. Because if you know two sides of a right triangle-- and we haven't gone into depth on this in the geometry by Pythagorean theorem, you can always figure out the third side. So if you have this information about any triangle, you can always figure out the third side. And then you can use side-side-side. So I just wanted to show you this little special case. But in general, the important thing is that you can't just use SSA unless you have more information."
    },
    {
        "Q": "\nAt 0:45, Mr. Sal has gotten me confused.\nHe says \" twenty per hundred.\"\nTwenty WHAT?",
        "A": "What he means when he said that: -cent means hundred . He was explaining what percent means per and cent = per and hundred. Get what I m saying? what s he s basically saying is twenty per-cent. Hope this cleared it up for you!",
        "video_name": "Lvr2YsxG10o",
        "timestamps": [
            45
        ],
        "3min_transcript": "We're asked to shade 20% of the square below. Before doing that, let's just even think about what percent means. Let me just rewrite it. 20% is equal to-- I'm just writing it out as a word-- 20 percent, which literally means 20 per cent. And if you're familiar with the word century, you might already know that cent comes from the Latin for the word hundred. This literally means you can take cent, and that literally means 100. So this is the same thing as 20 per 100. If you want to shade 20%, that means, if you break up the square into 100 pieces, we want to shade 20 of them. 20 per 100. So how many squares have they drawn here? four, five, six, seven, eight, nine, ten squares. If we go vertically, we have one, two, three, four, five, six, seven, eight, nine, ten. So this is a 10 by 10 square. So it has 100 squares here. Another way to say it is that this larger square-- I guess that's the square that they're talking about. This larger square is a broken up into 100 smaller squares, so it's already broken up into the 100. So if we want to shade 20% of that, we need to shade 20 of every 100 squares that it is broken into. So with this, we'll just literally shade in 20 squares. So let me just do one. So if I just do one square, just like that, I have just shaded 1 per 100 of the squares. 100 out of 100 would be the whole. I've shaded one of them. That one square by itself would be 1% If I were to shade another one, if I were to shade that and that, then those two combined, that's 2% of the It's literally 2 per 100, where 100 would be the entire square. If we wanted to do 20, we do one, two, three, four-- if we shade this entire row, that will be 10%, right? One, two, three, four, five, six, seven, eight, nine, ten. And we want to do 20, so that'll be one more row. So I can shade in this whole other row right here. And then I would have shaded in 20 of the 100 squares. Or another way of thinking about it, if you take this larger square, divide it into 100 equal pieces, I've shaded in 20 per 100, or 20%, of the entire larger square. Hopefully, that makes sense."
    },
    {
        "Q": "\nCouldn't you have substituted 180 = pi into -pi/3 at 2:39?",
        "A": "Yes. And that is essentially what Sal did. He multiplied by 180 and divided by pi; thus, he cancelled out the pi, and put in a 180, so he basically just substituted it. But your way is a nice shortcut; so if you can do it that way in your head, feel free to.",
        "video_name": "z0-1gBy1ykE",
        "timestamps": [
            159
        ],
        "3min_transcript": "and on the righthand side here, 360 divided by two is 180. And we have still the units which are degrees. So we get pi radians are equal to 180 degrees. Which actually answers the first part of our question. We wanted to convert pi radians, well we just figured out! Pi radians are equal to 180 degrees. If you want to think about it, pi radians are halfway around the circle Halfway around the circle like that, and it is the same thing as 180 degrees. So now lets think about the second part. We want to convertnegative pi over three radians. --Switch to a new color-- so negative pi over three, so how do we convert that? Well, to figure this out we need to know how many degrees there are per radian. We need to multiply this by degrees -- I'm going to write the word out instead of the circle here -- It would be really hard to visualize that, degrees per radian So how many degrees are there per radian? well we know that for 180 degrees we have pi radians. Or you can say there are 180 over pi degrees per radian. This is going to work out: We have however manyradians we have times the number of degrees per radian. So of course the units are going to work out. The radians cancel out, the pi also cancels. And you are left with 180 divided by 3, leaving us with Negative 60, and we don't want to forget the units We could write them out, the only unit left is degrees. WE could write out the word degrees or just put that symbol there."
    },
    {
        "Q": "At 3:09, why is the answer -60? I thought degree measurements could not be negative, like length measurements.\n",
        "A": "Angle measurement can be negative. Starting on the positive x-axis, it is negative when you draw it clockwise, and positive when you draw it counter-clockwise.",
        "video_name": "z0-1gBy1ykE",
        "timestamps": [
            189
        ],
        "3min_transcript": "and on the righthand side here, 360 divided by two is 180. And we have still the units which are degrees. So we get pi radians are equal to 180 degrees. Which actually answers the first part of our question. We wanted to convert pi radians, well we just figured out! Pi radians are equal to 180 degrees. If you want to think about it, pi radians are halfway around the circle Halfway around the circle like that, and it is the same thing as 180 degrees. So now lets think about the second part. We want to convertnegative pi over three radians. --Switch to a new color-- so negative pi over three, so how do we convert that? Well, to figure this out we need to know how many degrees there are per radian. We need to multiply this by degrees -- I'm going to write the word out instead of the circle here -- It would be really hard to visualize that, degrees per radian So how many degrees are there per radian? well we know that for 180 degrees we have pi radians. Or you can say there are 180 over pi degrees per radian. This is going to work out: We have however manyradians we have times the number of degrees per radian. So of course the units are going to work out. The radians cancel out, the pi also cancels. And you are left with 180 divided by 3, leaving us with Negative 60, and we don't want to forget the units We could write them out, the only unit left is degrees. WE could write out the word degrees or just put that symbol there."
    },
    {
        "Q": "\nFor the formula [Sn=a/(1-r)], the beginning value doesn't have to necessarily be k=1, for example at 5:02? Does that imply that any value for \"k\" can be used, such as starting at 2 or 1000?",
        "A": "That formula only works for k=0 to infinity.",
        "video_name": "2BgWWsypzLA",
        "timestamps": [
            302
        ],
        "3min_transcript": "And that gives us that value right over there. Plus 0.4008 times 10 to the negative 4 to the second power. And we keep on going. And so in this form, it looks a little bit clearer, like a geometric series, an infinite geometric series. And if we wanted to write that out with sigma notation, we could write this as the sum from k equals 0 to infinity of, well, what's our first term going to be? It's going to be 0.4008 times our common ratio, which we could write out as either 10 to the negative fourth or 0.0001. I'll just write it as 10 to the negative fourth. 10 to the negative fourth to the k-th power, to the k-th power. this clearly can be represented as a geometric series-- is, well, what is the sum? You might say, well, that's just going to be 4008 repeating over and over. But I want to express it as a fraction. And so I want you to pause the video. Use what would you already know about finding the sum of an infinite geometric series to try to express this thing right over here as a fraction. So I'm assuming you've had a go at it. So let's think about it. We've already seen, we've already derived in previous videos, that the sum of an infinite geometric series-- let me do this in a neutral color. If I have a series like this, k equals 0 to infinity of ar to the k power, that this sum is going to be equal to a over 1 minus r. We've derived this actually in several other videos. So in this case, this is going to be-- well, our a here is 0.4008. And it's going to be that over 1 minus our common ratio, minus-- So what's this going to be? Well, this is going to be the same thing as 0.4008. If you take 1 minus 1 ten thousandths, or you could do this as 10,000 ten thousandths minus 1 ten thousandth, you're going to have 9,999 ten thousandths. Once again, you could view-- let me write this out just so this doesn't look confusing. 1 is the same thing as 10,000/10,000. And you're subtracting 1/10,000. And so you're going to get 9,999/10,000. And so this is going to be the same thing"
    },
    {
        "Q": "\nAt 6:05, Sal said that the equation of a circle is x^2 + y^2 = r^2, but I remember in the previous lessons that the standard form is (x \u00e2\u0080\u0093 a)^2 + (y \u00e2\u0080\u0093 b)^2 = r^2. Can someone please clarify the differences?",
        "A": "a and b (a,b) are used to determine the center of the circle. In the equation x^2 + y^2 = r^2, we have a and b equal to zero, so the center of the circle would be (0,0). If the equation of the circle were to be (x-3)^2 + (y-4)^2 = r^2, the center of the circle would be (3,4). So basically, the 2 equations that Sal gave mean the same thing, but the a and b are used to define the center of the circle. Comment if you have any further questions.",
        "video_name": "lvAYFUIEpFI",
        "timestamps": [
            365
        ],
        "3min_transcript": "I think that makes more sense. And you can call this the minor radius. So let's just do an example. And I think when I've done an example with actual numbers, it'll make it all a little bit clearer. So let's say I were to show up at your door with the following: If I were to say x squared over 9 plus y squared over 25 is equal to 1. So what is your radius in the x-direction? This is your radius in the x-direction squared. So your radius in the x-direction if we just map it, we would say that a is equal to 3. Because this is a squared. And if we were just map it we'd say this is b squared than this tells us that b is equal to 5. is centered at the origin. Let me draw the ellipse first. So, first of all, we have our radius in the y-direction is larger than our radius in the x-direction. The ellipse is going to be taller and skinnier. It's going to look something like that. Draw some axes, so that could be your x-axis, your y-axis. This is your radius in the y-direction. So this distance right here is going to be 5, and so will this distance. And this is your radius in the x-direction. So this will be 3, and this will be 3. You have now plotted this ellipse. Nothing too fancy about it. And actually just to kind of hit the point home that the circle is a special case of an ellipse. We learned in the last video that the equation of a circle x squared plus y squared is equal to r squared. So if we were to divide both sides of this by r squared, we would get-- and this is just little algebraic manipulation-- x squared over r squared plus y squared over r squared is equal to 1. Now in this case, your a is r and so is your b. So your semi-minor axis is r and so is your semi-major axis of r. Or, in other words, this distance is the same as that distance, and so it will neither be short and fat nor tall and skinny. It'll be perfectly round. And so that's why the circle is a special case of an ellipse. So let me give you a slightly-- It'll look a lot more complicated, and this is something you might see on exam. But I just want to show you that this is just a shifting. Let's say we wanted to shift this ellipse."
    },
    {
        "Q": "at 9:15 Sal says that. if we add 2 to any function then we shift it down by two. but in reality that is not true.\nwe should subtract 2 in function to shift it down.\n",
        "A": "I would usually make the same mistake. You see, what helped me was writing the equation like this, for the standard form: (x-(h))^2/a^2 + (y-(k))^2/b^2.... So if either h or k is negative, it would then be a double negative, which in turn makes it positive, like so: (x-(-h))/a^2 + (y-(-k))/b^2=1 or: (x+h)^2/a^2 + (y+k)^2/b^2. Really hope this helps you like it did me!!",
        "video_name": "lvAYFUIEpFI",
        "timestamps": [
            555
        ],
        "3min_transcript": "So instead of the origin being at x is equal to 0, the origin will now be at x is equal 5. So a way to think about that is what does this term have to be so that at 5 this term ends up being 0. Well I'll actually draw it for you, because I think that might be confusing. So if we shift that over the right by 5, the new equation of this ellipse will be x minus 5 squared over 9 plus y squared over 25 is equal to 1. So if I were to just draw this ellipse right now, it would look like this. I want to make it look fairly similar to the ellipse I had before. Just shifted it over by five. And the intuition we learned a little bit in the circle video where I said, oh well, you know, if you have x minus something that means that the new origin is now at positive 5. And you could memorize that. You could always say, oh, if I have a minus here, that the origin is at the negative of whatever this number is, so it would be a positive five. You know, if you had a positive it would be the opposite that. But the way to really think about it is now if you go to x is equal to 5, when x is equal to 5, this whole term, x minus 5, will behave just like this x term will here. When x is equal to 5 this term is 0, just like when x was 0 here. So when x is equal to 5, this term is 0, and then y squared over 25 is equal 1, so y has to be equal five. Just like over here when x is equal is 0, y squared over 25 had to be equal to 1, y is equal to either And I really want to give you that intuition. And then, let's say we wanted to shift this equation down by two. So our new ellipse looks something like this. A lot of times you learned this in conic sections. But this is true any function. When you shift things, you shift it this way. If you shift this graph to the right by five, you replace all of the x's with x minus 5. And if you were to shift it down by two, you would replace all the y's with y plus 2. So let me draw our new ellipse first, just to show you what I'm doing. So our new ellipse is going to look something like that. I'm shifting the yellow ellipse down by two."
    },
    {
        "Q": "So when Sal says at 1:49 that the area is 1/2 b\u00c2\u00b7h it is the answer of b\u00c2\u00b7h cut in half?\n",
        "A": "You can think of it as half of a rectangle",
        "video_name": "YTRimTJ5nX4",
        "timestamps": [
            109
        ],
        "3min_transcript": "Let's say we've got a rectangle and we have two diagonals across the rectangle-- that's one of them, and then we have the other diagonal --and this rectangle has a height of h-- so that distance right there is h --and it has a width of w. What we're going to show in this video is that all of these four triangles have the same area. Now right when you look at it, it might be reasonably obvious that this bottom triangle will have the same area as the top triangle, as this top kind of upside down triangle. That these to have the same area, that might be reasonably obvious. they have the same dimension for their base, this width, and they have the same height because this distance right here is exactly half of the height of the rectangle. They have the same proportions. Now it's probably equally obvious that this triangle on the left has the same area as this triangle on the right. That's probably equally obvious. What is not obvious is that these orange triangles angles have the same area as these green, blue triangles. And that's what we're going to show right here. So all we have to do is really calculate the areas of the different triangles. So let's do the orange triangles first. and before doing that let's just remind ourselves what the area of a triangle is. Area of a triangle is equal to 1/2 times the base of the triangle times the height of the triangle. That's just basic geometry. Not with that said, let's figure out the area of the orange triangle. It's going to be 1/2 times the base. right here: it is w. So 1/2 times w. I want to do that in a different color; the color I wrote the w in. Now what's the height here? Well we already talked about it: it's exactly half way up the height of the rectangle. So times 1/2 times the height of the rectangle. So what's that going to be? You have 1/2 times 1/2 is 1/4 times width times height. So the area of that triangle is 1/4 width height. So is that one. Same exact argument; they have equal area. Now what's the area of these green or these green/blue triangles? Well once again-- we'll do this in a green color --area"
    },
    {
        "Q": "At 2:40, you begin to explain that you want to change dy / dx = 1 / cos y into something in terms of x by using the Pythagorean Identity and setting that (the Pythagorean Identity) equal to cos y . Could you just have inserted your original definition of y (y = sin^-1 x) instead? You would end up with:\n\ndy / dx = 1 / (cos (sin^-1 x)).\n\nDoes that work ? tried it with a few arbitrary numbers for x and was getting the same answer for both equations, but wasn't sure if it technically was correct.\n",
        "A": "Yes. If you want to visualize, try to type cos(arcsin(x)) and sqrt(1-x^2) into Google or is cos(arcsin(x)) equal to sqrt(1-x^2) into WolframAlpha.",
        "video_name": "yIQUhXa-n-M",
        "timestamps": [
            160
        ],
        "3min_transcript": "So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We can take the derivative of both sides with respect to X. So, derivative of the left-hand side with respect to X and the derivative of the right-hand side with respect to X. But what's the derivative of the left-hand side with respect to X going to be? And here we just apply the chain rule. It's going to be the derivative of sine of Y with respect to Y. Which is going to be, which is going to be cosine of Y times the derivative of Y with respect to X. So times dy, dx, times dy,dx. And the right-hand side, what's the derivate of X with respect to X, well that's obviously just going to be equal to one. And so we could solve for dy,dx, divide both sides by cosine of Y. to X is equal to one over cosine of Y. Now this still isn't that satisfying cuz I have the derivative in terms of Y. So let's see if we can re-express it in terms of X. So, how could we do that? Well, we already know that X is equal to sine of Y. We already know that X is equal to sine of Y. So, if we could rewrite this bottom expression in terms, instead of cosine of Y. If we could use our trigonometric identities to rewrite it in terms of sine of Y, then we'll be in good shape because X is equal to sine of Y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of Y plus cosine squared of Y is equal to one. Or, if we want to solve for cosine of Y, subtract sine squared of Y from both sides. We know that cosine squared of Y is equal to one minus sine squared of Y. is equal to the principal root of one minus sine squared of Y. So, we could rewrite this as being equal to one over, one over, instead of cosine of Y, we could rewrite it as one minus sine squared of Y. Now why is this useful? Well, sine of Y is just X. So this is the same if we just substitute back in, let me just write it that way so it's a little bit clear. I could write it as sine Y squared. We know that this thing right over here is X. So this is going to be equal to, and we deserve a little bit of a drumroll. One over the square root of one minus, instead of sin of Y, we know that X is equal to sin of Y. So, one minus X squared. And so, there you have it. The derivative with respect to X of the inverse sine of X is equal to one over"
    },
    {
        "Q": "I'm confused...\n\n1) In the last 2 videos \"Proof: product of rational & irrational is irrational\" and \"Proof: sum of rational and irrational is irrational\" state in the title that the sum or product of rational and irrational is irrational.\n\n2) BUT, @5:31 in this video Sal says, \"You don't know whether the product is going to be rational or irrational unless someone tells you the specific numbers\".\n\nSo which is correct, 1 or 2?\n",
        "A": "There is no conflict. The last 2 videos were about sum/products of a rational and an irrational number. At 5:31, Sal is talking about the product of 2 irrational numbers. So, this is a whole different scenario. Hope this helps.",
        "video_name": "16-GZWi66CI",
        "timestamps": [
            331
        ],
        "3min_transcript": "and people don't tell you anything else, they don't tell you which specific irrational numbers they are, you don't know whether their sum is going to be rational or irrational. Now let's think about products. Similar exercise, let's say we have a times b is equal to c, ab is equal to c, a times b is equal to c. And once again, let's say someone tells you that both a and b are irrational. Pause this video and think about whether c must be rational, irrational, or whether we just don't know. Try to figure out some examples like we just did when we looked at sums. Alright, so let's think about, let's see if we can construct examples where c ends up being rational. Well one thing, as you can tell I like to use pi, pi might be my favorite irrational number. If a was one over pi well, what's their product going to be? Well, their product is going to be one over pi times pi, that's just going to be pi over pi, which is equal to one. Here we got a situation where the product of two irrationals became, or is, rational. But what if I were to multiply, and in general you could this with a lot of irrational numbers, one over square root of two times the square of two, that would be one. What if instead I had pi times pi? Pi times pi, that you could just write as pi squared, and pi squared is still going to be irrational. This is irrational, irrational. It isn't even always the case that if you multiply the same irrational number, that it's always going to be irrational. For example, if I have square root of two times, I think you see where this is going, times the square root of two, I'm taking the product of two irrational numbers. In fact, they're the same irrational number, but the square root of two times the square root of two, well, that's just going to be equal to two, which is clearly a rational number. So once again, when you're taking the product of two irrational numbers, you don't know whether the product is going to be rational or irrational unless someone tells you the specific numbers. Whether you're taking the product or the sum of irrational numbers, in order to know whether the resulting number is irrational or rational, you need to know something about what you're taking the sum or the product of."
    },
    {
        "Q": "\nI don' understand what happened at 6:57 - 7:42. I tried to follow by reading what you said, and I couldn't follow",
        "A": "He said: -2+2*1=0, 4+2(-2)=0, and -10+2(5)=0. This came from the row operation: 2R2+R3 replace R3.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            417,
            462
        ],
        "3min_transcript": "I'm going to replace this row with that. 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. 4 minus 2 times 7, is 4 minus 14, which is minus 10. Now what can I do next. You can kind of see that this row, well talk more about what this row means. When all of a sudden it's all been zeroed out, there's nothing here. If I had non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. I'm just going to move over to this row. The first thing I want to do is I want to make this leading coefficient here a 1. row by minus 1. If I multiply this entire row times minus 1. I don't even have to rewrite the matrix. This becomes plus 1, minus 2, plus 5. I think you can accept that. Now what can we do? Well, let's turn this right here into a 0. Let me rewrite my augmented matrix in the new form that I have. I'm going to keep the middle row the same this time. My middle row is 0, 0, 1, minus 2, and then it's augmented, and I get a 5 there. What I want to do is I want to eliminate this minus 2 here. Why don't I add this row to 2 times that row. Then I would have minus 2, plus 2, and that'll work out. What do I get. Well, these are just leading 0's. That's just 0. 4 plus 2 times minus 2, that is minus 4. That's 4 plus minus 4, that's 0 as well. Then you have minus 10 plus 2 times 5. Well, that's just minus 10 plus 10, which is 0. That one just got zeroed out. Normally, when I just did regular elimination, I was happy just having the situation where I had these Everything below it were 0's. I wasn't too concerned about what was above our 1's. What I want to do is, I want to make those into a 0 as well. I want to make this guy a 0 as well. What I can do is, I can replace this first row with that first row minus this second row."
    },
    {
        "Q": "\nAt 4:21, why is Khan allowed to subtract the second row from the first row and put it in the second row? I thought he could only do that vice versa: subtracting first row from the second row and making that the new second row.",
        "A": "I agree with you and I do it the way you thought is correct, because it s less confusing for me; however: Both operations are equivalent, because if you multiply both sides of row 1 - row 2 by -1, you get row 2 - row 1 (try it), and these are just scalar multiples of each other. a-c = b-d <=> c-a = d-b",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            261
        ],
        "3min_transcript": "coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry Let me do that. The first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get. 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I want to turn it into a 0. Let's replace this row with this row minus 2 times that row. What I'm going to do is, this row minus 2"
    },
    {
        "Q": "Where did Sal get x2 = 0 1 0 and x4= 0 0 1 at 14:30?\n",
        "A": "I m still a little confused about this also. As far as I can tell the solution is correct with or without x2 and x4 components.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            870
        ],
        "3min_transcript": "I just subtracted these from both sides of the equation. This right here is essentially as far as we can go to the solution of this system of equations. I can pick, really, any values for my free variables. I can pick any values for my x2's and my x4's and I can solve for x3. What I want to do right now is write this in a slightly different form so we can visualize a little bit better. Of course, it's always hard to visualize things in four dimensions. So we can visualize things a little bit better, as to the set of this solution. Let's write it this way. If I were to write it in vector form, our solution is the vector x1, x3, x3, x4. Well it's equal to-- let me write it like this. It's equal to-- I'm just rewriting, I'm just essentially rewriting this solution set in vector form. So x1 is equal to 2-- let me write a little column there-- plus x2. Let me write it this way. Plus x2 times something plus x4 times something. x1 is equal to 2 minus 2 times x2, or plus x2 minus 2. I put a minus 2 there. I can say plus x4 times minus 3. I can put a minus 3 there. This right here, the first entries of these vectors literally represent that equation right there. x1 is equal to 2 plus x2 times minus 2 plus x4 times minus 3. x3 is equal to 5. Put that 5 right there. Plus x4 times 2. x2 doesn't apply to it. We can just put a 0. 0 times x2 plus 2 times x4. Now what does x2 equal? You could say, x2 is equal to 0 plus 1 times x2 plus 0 times x4. x2 is just equal to x2. It's a free variable. Similarly, what does x4 equal to? x4 is equal to 0 plus 0 times x2 plus 1 times x4. What does this do for us? Well, all of a sudden here, we've expressed our solution set as essentially the linear combination of the linear combination of three vectors. This is a vector."
    },
    {
        "Q": "At 2:58, Sal mentions \"previous videos where I was trying to figure out whether things were linearly independent or not.\" Where are these videos?\n",
        "A": "Videos about linear independence within, I suspect, the linear algebra subjects.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            178
        ],
        "3min_transcript": "The coefficient there is 1. The coefficient there is 1. The coefficient there is 2. You have 2, 2, 4. 2, 2, 4. 1, 2, 0. 1, 2, there is no coefficient the x3 term here, because there is no x3 term there. We'll say the coefficient on the x3 term there is 0. And then we have 1, minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. What I want to do is I want to augment it, I want to augment it with what these equations need to be equal to. Let me augment it. What I am going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. I think you can see that this is just another way of writing this. coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry"
    },
    {
        "Q": "What determines the row you will use to minus with another row to replace a particular row? for example at 3:00 you replaced row 1 with (1st row - 2nd row)...why didn't you use the (3rd row - 2 times 1st row)? since it will still give you the same answer...?\n",
        "A": "You can replace any row with: 1) its multiple 2) its difference from any other row or from any multiple of a row.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            180
        ],
        "3min_transcript": "The coefficient there is 1. The coefficient there is 1. The coefficient there is 2. You have 2, 2, 4. 2, 2, 4. 1, 2, 0. 1, 2, there is no coefficient the x3 term here, because there is no x3 term there. We'll say the coefficient on the x3 term there is 0. And then we have 1, minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. What I want to do is I want to augment it, I want to augment it with what these equations need to be equal to. Let me augment it. What I am going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. I think you can see that this is just another way of writing this. coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry"
    },
    {
        "Q": "\nWhy did Sal say 1 minus minus 1 at 4:37?",
        "A": "It s just another way of saying 1 minus negative 1. Or 1 - (-1). He s subtracting row 2 from row 1. The 4th column in row 1 is 1, the 4th column in row to is -1. So 1 - (-1) = 2.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            277
        ],
        "3min_transcript": "Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry Let me do that. The first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get. 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I want to turn it into a 0. Let's replace this row with this row minus 2 times that row. What I'm going to do is, this row minus 2 I'm going to replace this row with that. 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. 4 minus 2 times 7, is 4 minus 14, which is minus 10. Now what can I do next. You can kind of see that this row, well talk more about what this row means. When all of a sudden it's all been zeroed out, there's nothing here. If I had non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. I'm just going to move over to this row. The first thing I want to do is I want to make this leading coefficient here a 1."
    },
    {
        "Q": "At 11:16, can't you solve for the free variables if you switch the order of the variables and make them your pivot variables by changing how you change the matrix?\n",
        "A": "If you use the matrix method in this video, you have to swap entire column vectors (but not the last, constant one) with each other. It seems to me that then the equations are the same, and the variables remain in their own columns, and the result should be equivalent, but as you suggest expressed as x2 and x4 in terms of x1 and x3. I put the column vectors in the order x2, x4, x1, x3 , and got: x2 = 19/4 - (x1)/2 - 3(x3)/4, x4 = -5/2 + (x3)/2. What do you get?",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            676
        ],
        "3min_transcript": "If I have any zeroed out rows, and I do have a zeroed out row, it's right there. This is zeroed out row. Just the style, or just the convention, is that for reduced row echelon form, that has to be your last row. We have the leading entries are the only -- they're all 1. That's one case. You'd want to divide that equation by 5 if this was a 5. So your leading entries in each row are a 1. That the leading entry in each successive row is to the right of the leading entry of the row before it. This guy right here is to the right of that guy. This is just the style, the convention, of reduced row echelon form. If you have any zeroed out rows, it's in the last row. And finally, of course, and I think I've said this multiple times, this is the only non-zero entry in the row. What does this do for me? Now I can go back from this world, back to my linear equations. We remember that these were the coefficients on x1, these were the coefficients on x2. my constants out here. I can rewrite this system of equations using my reduced row echelon form as x1, x1 plus 2x2. There's no x3 there. So plus 3x4 is equal to 2. This equation, no x1, no x2, I have an x3. I have x3 minus 2x4 is equal to 5. I have no other equation here. This one got completely zeroed out. I was able to reduce this system of equations to this system of equations. The variables that you associate with your pivot entries, we call these pivot variables. x1 and x3 are pivot variables. The variables that aren't associated with the pivot x2 and x4 are free variables. Now let's solve for, essentially you can only solve for your pivot variables. The free variables we can set to any variable. I said that in the beginning of this equation. We have fewer equations than unknowns. This is going to be a not well constrained solution. You're not going to have just one point in R4 that solves You're going to have multiple points. Let's solve for our pivot variables, because that's all we can solve for. This equation tells us, right here, it tells us x3, let me do it in a good color, x3 is equal to 5 plus 2x4. Then we get x1 is equal to 2 minus x2, 2 minus 2x2."
    },
    {
        "Q": "\nAt 2:25, why is the exponent on 7 zero? Wouldn't it be one? I know that everything to the first power is just itself, but why does that x^0 have to go there instead of 7^1?",
        "A": "He writes it to show you that there is an x with a degree on the 7, even though we usually don t write it.",
        "video_name": "REiDXCN0lGU",
        "timestamps": [
            145
        ],
        "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done."
    },
    {
        "Q": "\nAround the 1:45 mark, it's mentioned that the 7 is a coefficient of the x^0. I've always understood that the lonely number at the end is a constant, not a coefficient. Which is it and why?",
        "A": "x^0 = 1 and anything multiplied by 1 is itself. Therefore, it s perfectly logical to say that the constant is the coefficient of x^0 power. It helps in solving certain problems as you go further in your mathematical career. For now, you should just understand the logic behind it and why it makes sense.",
        "video_name": "REiDXCN0lGU",
        "timestamps": [
            105
        ],
        "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done."
    },
    {
        "Q": "At 1:05, can we say that the exponent on x always counts as part of x, and has nothing to do with the coefficient?\n",
        "A": "Yes, exactly.",
        "video_name": "REiDXCN0lGU",
        "timestamps": [
            65
        ],
        "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done."
    },
    {
        "Q": "at 2:56 he adds an h(y). I don't understand his explanation\n",
        "A": "If you were taking the indefinite anti-derivative of some function, you would add a +C on the end, to account for any constants that may have been lost. In this case, we took the indefinite anti-derivative of a partial function. So we add the h(y) on the end to account for any function of y that may have been lost. If you think about it backwards, if you took the partial derivative of Psi with respect to x, the function of y goes to zero.",
        "video_name": "eu_GFuU7tLI",
        "timestamps": [
            176
        ],
        "3min_transcript": "We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx, so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx, so that becomes dy dx, is equal to-- what's 0 divided by dx? Well it's just 0. And there we have it. We have written this in the form that we need, in this form. And now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of M? This is the M function, right? This was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x, and then just a 2. So the partial of this with respect to y is minus 2x. This would be 0, this would be minus 2x. So there you have it. The partial of M with respect to y is equal to the partial of N with respect to x. My is equal to Nx. So we are dealing with an exact equation. So now we have to find psi. The partial of psi with respect to x is equal to M, which is equal to 3x squared minus 2xy plus 2. Take the anti-derivative with respect to x on both sides, and you get psi is equal to x to the third minus x squared y-- because y is just a constant-- plus 2x, plus some function of y. Because we know psi is a function of x and y. So when you take a derivative, when you take a partial with lost. So it's like the constant, when we first learned taking anti-derivatives. And now, to figure out psi, we just have to solve for h of y. And how do we do that? Well let's take the partial of psi with respect to y. That's going to be equal to this right here. So The partial of psi with respect to y, this is 0, this is minus x squared. So it's minus x squared-- this is o-- plus h prime of y, is going to be able to what? That's going to be equal to our n of x, y. It's going to be able to this. And then we can solve for this. So that's going to be equal to 6y squared minus x squared plus 3. You can add x squared to both sides to get rid of this and this. And then we're left with h prime of y is equal to 6y squared plus 3. Anti-derivative-- so h of y is equal to what is this-- 2y"
    },
    {
        "Q": "At 0:56, I'm confused about the form. Some places I see have the from M(x,y)dx + N(x,y)dy = 0, while others (I have the same textbook as Khan) have the form like he has it, M(x,y) +N(x,y)dy/dx. Anyone know which way is correct? I've been using the dx and dy at the end to figure out which function is M and which function is N too. How do you pick your Ms and Ns without it?\n",
        "A": "M(x,y)dx + N(x,y)dy = 0 divide by dx, and you get: M(x,y) +N(x,y)dy/dx = 0.",
        "video_name": "eu_GFuU7tLI",
        "timestamps": [
            56
        ],
        "3min_transcript": "Welcome back. I'm just trying to show you as many examples as possible of solving exact differential equations. One, trying to figure out whether the equations are exact. And then if you know they're exact, how do you figure out the psi and figure out the solution of the differential equation? So the next one in my book is 3x squared minus 2xy plus 2 times dx, plus 6y squared minus x squared plus 3 times dy is equal to 0. So just the way it was written, this isn't superficially in that form that we want, right? What's the form that we want? We want some function of x and y plus another function of x and y, times y prime, or dy dx, is equal to 0. We're close. We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx, so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx, so that becomes dy dx, is equal to-- what's 0 divided by dx? Well it's just 0. And there we have it. We have written this in the form that we need, in this form. And now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of M? This is the M function, right? This was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x, and then just a 2. So the partial of this with respect to y is minus 2x. This would be 0, this would be minus 2x. So there you have it. The partial of M with respect to y is equal to the partial of N with respect to x. My is equal to Nx. So we are dealing with an exact equation. So now we have to find psi. The partial of psi with respect to x is equal to M, which is equal to 3x squared minus 2xy plus 2. Take the anti-derivative with respect to x on both sides, and you get psi is equal to x to the third minus x squared y-- because y is just a constant-- plus 2x, plus some function of y. Because we know psi is a function of x and y. So when you take a derivative, when you take a partial with"
    },
    {
        "Q": "\nHow can you understand this at 6:38 with out more help?",
        "A": "i am asuming you mean at 1:38 becasue the video is less than 2 minutes long. At that time he is placing the 4.1 on the number line and you would find 4 and then realise that youneed to go .1 positive (to the right) from that point.",
        "video_name": "uC09taczvOo",
        "timestamps": [
            398
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhy do we want x and y values that give zero in perfect squares? #3:35\nThanks for help.",
        "A": "Technically, we could have the x and y values be anything, and that would probably give us a random point along the circle we graph. However, the zeroes of a graph are generally a good place to start as they tend to be easier to find than any other point.",
        "video_name": "XyDMsotfJhE",
        "timestamps": [
            215
        ],
        "3min_transcript": "We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared a, b, essentially, the point that makes both of these equal to 0. And that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2, and the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0, and the y value that makes this 0. So the center is negative 2, 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2, 2. So our center is negative 2, 2. So that's right over there. X is negative 2, y is positive 2. So let's see, this would be 1, 2, 3, 4, 5. So you have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5. Let's check our answer. We got it right."
    },
    {
        "Q": "At 1:00, Sal says that it has to be plus y squared minus 4y, but isn't the y squared a minus?\n",
        "A": "The original problem has a + y^2. It s only when Sal circles it that the upstroke on the + disappears so that it looks like it s a minus sign.",
        "video_name": "XyDMsotfJhE",
        "timestamps": [
            60
        ],
        "3min_transcript": "We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x, and complete the square in terms of y, to put it into a form that we can recognize. So first let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here, because I'm going to complete the square. And then I have my y terms. I'll circle those in-- well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need to review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2, square it to get 4. And that comes straight out of the idea if you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x, plus 2 squared. We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared"
    },
    {
        "Q": "\n3:15, how did he know that it 6.5 equals to 3.5 and 3?",
        "A": "Because, 3.5 plus 3 equals 6.5.",
        "video_name": "loAA3TCNAvU",
        "timestamps": [
            195
        ],
        "3min_transcript": "especially a triangle like this, a right triangle, is just going to be half of a rectangle like this. We just care about half of its area. So this area is going to be 1/2 times 2 times 3.5. 1/2 times 2 is equal to 1. 1 times 3.5 is 3.5 square units. So the area of that part is going to be 3.5 square units. Let's think about the area of this triangle right over here. Well, once again we have its height is 3.5. Its base is 7. So its area is going to be 1/2 times 7 times 3.5. 1/2 times 7 is 3.5 times 3.5. So this part is 3.5, and I'm going to multiply that times 3.5 again. Let's figure out what that product is equal to. 3.5 times 3.5. 5 times 5 is 25. Let's cross that out. Move one place over to the left. 3 times 5 is 15. 3 times 3 is 9, plus 1 is 10. So that gets us to 5 plus 0 is 5. 7 plus 5 is 12, carry the 1. 1 plus 1 is 2. And we have a 1. We have two digits to the right of the decimal, one, two. So we're going to have two digits to the right of the decimal in the answer. The area here is 12.25 square units. Now this region may be a little bit more difficult, because it's kind of us this weird trapezoid looking thing. But one thing that might pop out at you is that you can divide it very easily into a rectangle and a triangle. And we can actually figure out the dimensions that we need to figure out the areas of each of these. We know what the width of this rectangle you want to call it. It's going to be 2 units plus 7 units. So this is going to be 9. We know that this distance is 3.5. If this distance right over here is 3.5, then this distance down here has to add up with 3.5 to 6.5, so this must be 3. Now we can actually figure out the area. The area of this rectangle is just going to be its height times its length, or 9 times 3.5. 9 times 3.5. And one way you could do it-- we could even try to do this in our head-- this is going to be 9 times 3 plus 9 times 0.5. 9 times 3 is 27. 9 times 0.5, that's just half of nine, so it's going to be 4.5. 27 plus 4 will get us to 31, so that's going to be equal to 31.5."
    },
    {
        "Q": "\nAt 3:39,why sai said because of SAS so those triangles are similar?isn't SAS is for congruent?",
        "A": "SAS applies both to similarity and congruence. In this case, Sal was using SAS to show similarity.",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            219
        ],
        "3min_transcript": "Let's look at the shorter side on either side of this angle. So the shorter side is two, and let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be XT. So what we want to compare is the ratio between-- let me write it this way. We want to see, is XY over XT equal to the ratio of the longer side? Or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well. Is that equal to the ratio of XZ over the longer of the two sides-- when you're looking at this angle right here, on either side of that angle, for the larger triangle-- over XS? And it's a little confusing, because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between, and then the longer side on either side of this angle. and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see XY. This is two. XT is 3 plus 1 is 4. XZ is 3, and XS is 6. So you have 2 over 4, which is 1/2, which is the same thing as 3/6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle, for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides. And I'm running out of space here. Let me write it right above here. We can write that triangle XYZ is similar to triangle-- and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. There are some shared angles. This guy-- they both share that angle, the larger triangle"
    },
    {
        "Q": "\nAt 5:47, why is it written as a square root, it makes no sense. is it representing a long line of decimals. Someone tell me if I'm having a stupid moment.",
        "A": "The square root is used because it s more accurate and easier to write than a list of decimals.",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            347
        ],
        "3min_transcript": "and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. There are some shared angles. This guy-- they both share that angle, the larger triangle So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity, but right now, we can't really do anything as is. Let's try this one out, this pair right over here. So these are the first ones that we have actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here, is 3 square roots of 3 over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6, and then So this is going to give us-- let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here-- if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So 1 over 3 root 3 needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3"
    },
    {
        "Q": "At 0:57, what is mx?\n",
        "A": "mx is the slope of the line, or the steepness (is that a word??)",
        "video_name": "YBYu5aZPLeg",
        "timestamps": [
            57
        ],
        "3min_transcript": "Write an inequality that fits the graph shown below. So here they've graphed a line in red, and the inequality includes this line because it's in bold red. It's not a dashed line. It's going to be all of the area above it. So it's all the area y is going to be greater than or equal to this line. So first we just have to figure out the equation of this line. We can figure out its y-intercept just by looking at it. Its y-intercept is right there. Let me do that in a darker color. Its y-intercept is right there at y is equal to negative 2. That's the point 0, negative 2. So if you think about this line, if you think about its equation as being of the form y is equal to mx plus b in slope-intercept form, we figured out b is equal to negative 2. So that is negative 2 right there. And let's think about its slope. If we move 2 in the x-direction, if delta x is equal to 2, if our change in x is positive 2, what is our Our change in y is equal to negative 1. Slope, or this m, is equal to change in y over change in x, which is equal to, in this case, negative 1 over 2, or negative 1/2. And just to reinforce, you could have done this anywhere. You could have said, hey, what happens if I go back 4 in x? So if I went back 4, if delta x was negative 4, if delta x is equal to negative 4, then delta y is equal to positive 2. And once again, delta y over delta x would be positive 2 over negative 4, which is also negative 1/2. I just want to reinforce that it's not dependent on how far I move along in x or whether I go forward or backward. You're always going to get or you should always get, the same slope. So the equation of that line is y is equal to the slope, negative 1/2x, plus the y-intercept, minus 2. That's the equation of this line right there. Now, this inequality includes that line and everything above it for any x value. Let's say x is equal to 1. This line will tell us-- well, let's take this point so we get to an integer. Let's say that x is equal to 2. Let me get rid of that 1. When x is equal to 2, this value is going to give us negative 1/2 times 2, which is negative 1, minus 2, is going to give us negative 3. But this inequality isn't just y is equal to negative 3. y would be negative 3 or all of the values greater than I know that, because they shaded in this whole area up here. So the equation, or, as I should say, the inequality that fits the graph here below is-- and I'll do it in a bold color-- is y is greater than or equal to"
    },
    {
        "Q": "\nAt 2:50 why is it x- -3 as oppose to just x-3?",
        "A": "Because in the standard formula of a circle, it should be like the following: (x - h)^2 + (y - k)^2 = r^2 And the problem gives you (x + 3)........................................ Mr. Khan just likes to rewrite it in the subtraction like in the standard formula: (x - -3) Let me write this way, it be clearer to you: (x - (-3)). Now, you can see -3 is the x component of the center of the circle in the problem. For the y , you don t need to rewrite, because it is already in the format of subtraction (y - 4).",
        "video_name": "JvDpYlyKkNU",
        "timestamps": [
            170
        ],
        "3min_transcript": "and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. In fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here, that essentially is the change in the vertical axis between these two points, up here, we're at y, here we're k, so this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle, because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here, so these two things are perpendicular. And so from the Pythagorean theorem, be equal to our distance squared, and this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set-- this describes any x and y that satisfies this equation will sit on this circle. Now, with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote, we just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus-- well this is already in the form-- plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. that our h is negative 3-- I want to do that in the red color-- that our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. Make sure to get-- you know you might say, hey, there's a negative 4 here, no. But look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3, but no you're subtracting the h. So you'd say minus negative 3, and similarly, the radius is 7."
    },
    {
        "Q": "@ 09:02 he multiplies by 15 but shouldn't it be 15/64 ?\nI think he made a typo\n",
        "A": "The 1/64 comes from flipping a fair coin (1/2 chance of heads or tails each time) 6 times, because (1/2)^6 = 1/64. At 9:02, he s talking about the unfair coin (4/5 chance of heads & 1/5 chance of tails each time,) so instead of (1/2)^6, he uses (4/5)^4 * (1/5)^2. That number (~1.64%) represents the chance of a particular permutation e.g. HTHHTH, but since we don t care about the order of the flips, he multiplies by 15 because there are 15 ways to arrange the 4 heads within the 6 flips.",
        "video_name": "xw6utjoyMi4",
        "timestamps": [
            542
        ],
        "3min_transcript": "So the probability of getting four out of six heads, given a fair coin, is 15 out of 64. And if you look at it, based on our definition of b and a, this is the probability of b given a. Right? b is four out of six heads, given a fair coin. Fair enough. So let's figure out the probability of-- because there's two ways of getting four out of six heads. One, that we picked a fair coin, and then times 15 out of 64. And then there's the probability that we picked an unfair coin. So what's the probability of the unfair coin? Of getting four out of six heads, given the unfair coin? Well, once again, what's the probability of each of the combinations where we got four out of six? So in this situation, let's do the same one. Heads, tails, heads, tails, heads, heads. But in this situation, it's not a 50% chance of getting heads. It's 80%. So it would be .8 times .2 times .8 times .2 times .8 times .8. Now, essentially, we have-- you know, this multiplication, we can rearrange it, because it doesn't matter what order you multiply things in. So it's .8 to the fourth power times .2 squared. And it doesn't matter. You know, any of the unique combinations will each have Because we can just rearrange the order in which we multiply. And then how many of these combinations are there? If we are, once again, the God of probability, and out of six flips we're picking four-- we're choosing four that are going to end up heads? How many ways can I pick a group of four? Well, once again, that's times 6 choose 4. We figured out what that is. 6 choose 4 is 15. And this is the probability of four out of six heads, given an unfair coin. So what's the total probability of getting four out of six heads? Well, it's going to be the probability of getting the fair coin-- which is 1/3-- times the probability of getting four out of six heads, given the fair coin-- and that's this 15/64. Plus the probability of getting an unfair coin-- 2/3-- times the probability of getting four out of six heads, given the unfair coin-- and that's what we figured out here. Times 15 times .8 to the fourth, times .2 squared."
    },
    {
        "Q": "\nAt 5:22, he switches from \"-2cos2x\" to \"+4sin2x\". Where exactly did he get the \"4\" from?",
        "A": "It came from the derivative of the inner function, 2x. d/dx -2 cos 2x = 2 * 2 sin 2x = 4 sin 2x",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            322
        ],
        "3min_transcript": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8."
    },
    {
        "Q": "\n@4:48 why is the right side of the numerator 4sin2x?\nwhy I use chain rule i get -4sin(2cos)?\n\nf'(x) = g'(x).h'(g(x))\nso\ng(x)= 2cos and h(x)=2x\n\nthen\n-2sin*2(2cos) is the same as -4sin(2cos)\n\nIf some replies, could you workout the steps.",
        "A": "Hello excuse me for butting- I see what you ve done but could you explain this in more mathematical terms? Ideally Sal could explain this. I know how to use the chain rule but it seems the usage is different in this case. Maybe I m missing something about trigonometry? Chain Rule: f(x)=h[g(x)] -> f (x)=g (x) * h [g(x)] I am quite confused.",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            288
        ],
        "3min_transcript": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8."
    },
    {
        "Q": "\nat 5:37 are you allowed to cancel out the sin(x) to make it -2+4(2)/1 which is -2+8 which also equals 6 to find your answer?",
        "A": "No, because they are not all sines of the same value. Two of them are sin(x) the other is sin(2x).",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            337
        ],
        "3min_transcript": "So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. So it looks like we've made some headway or maybe L'Hopital's rule stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0/0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6."
    },
    {
        "Q": "Hi!\n\nAround 12:12, Sal starts transforming 0.5 to 5. Why can't we just leave the 0.5 as is, given that, once multiplied by 10 to the 17th power, we will get the same answer? I mean, isn't 0.5 times 10 to the 17th power just as correct as 5 times 10 to the 16th power?\n\nCheers!\n",
        "A": "Scientific notation requires that the mantissa (the first bit!) is greater than or equal to 1, and less than 10. So if you put 0.5 x 10^17, that is not valid scientific notation because 0.5 is not greater than or equal to 1. You are correct that 0.5 x 10^17 is equal to 5 x 10^16, but one is valid scientific notation and the other is not.",
        "video_name": "0Dd-y_apbRw",
        "timestamps": [
            732
        ],
        "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail."
    },
    {
        "Q": "At 12:00 I thought you had to multiply or divide equaly on both sides of the equation. Here you multiply on one side and divide on the other.\n",
        "A": "He is making two changes to the same side of the equation. The only way that is legal is when the changes cancel each other.",
        "video_name": "0Dd-y_apbRw",
        "timestamps": [
            720
        ],
        "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail."
    },
    {
        "Q": "At 2:27, at the top left says 10^-3. Since the number is getting larger due to the decimal moving to the right, why is the exponent a negative?\n",
        "A": "0.00852 = 8.52* 10^(-3) the exponent -3 is negative because the number 8.52 is 1000 times larger than 0.00852 . The negative exponent -3 means you divide by 1000 and is equal to the multiplication with 1/1000 some examples: 8.52* 10^(-3) = 8.52 * 1/1000 = 8.52 / 1000 = 0.00852 hope this will help you",
        "video_name": "0Dd-y_apbRw",
        "timestamps": [
            147
        ],
        "3min_transcript": "It always helps me to see a lot of examples of something so I figured it wouldn't hurt to do more scientific notation examples. So I'm just going to write a bunch of numbers and then write them in scientific notation. And hopefully this'll cover almost every case you'll ever see and then at the end of this video, we'll actually do some computation with them to just make sure that we can do computation with scientific notation. Let me just write down a bunch of numbers. 0.00852. That's my first number. My second number is 7012000000000. I'm just arbitrarily stopping the zeroes. The next number is 0.0000000 I'll just draw a couple more. If I keep saying 0, you might find that annoying. 500 The next number -- right here, there's a The next number I'm going to do is the number 723. The next number I'll do -- I'm having a lot of 7's here. Let's do 0.6. And then let's just do one more just for, just to make sure we've covered all of our bases. Let's say we do 823 and then let's throw some -- an arbitrary number of 0's there. So this first one, right here, what we do if we want to write in scientific notation, we want to figure out the largest exponent of 10 that fits into it. So we go to its first non-zero term, which is that right there. We count how many positions to the right of the decimal point we have including that term. So we have one, two, three. So it's going to be equal to this. So it's going to be equal to 8 -- that's that guy right there -- 0.52. So everything after that first term is going to be behind the decimal. So 0.52 times 10 to the number of terms we have. 10 to the minus 3. Another way to think of it: this is a little bit more. This is like 8 1/2 thousands, right? Each of these is thousands. We have 8 1/2 of them. Let's do this one. Let's see how many 0's we have. We have 3, 6, 9, 12. So we want to do -- again, we start with our largest term that we have. Our largest non-zero term. In this case, it's going to be the term all the way to the left. That's our 7. So it's going to be 7.012. It's going to be equal to 7.012 times 10 to the what? Well it's going to be times 10 to the 1 with this many 0's. So how many things? We had a 1 here. Then we had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 0's. I want to be very clear. You're not just counting the 0's. You're counting everything after this first"
    },
    {
        "Q": "At at 6:14, why would x^2\u00c2\u00b72 = 4x? Wouldn't it be 2x^2?\n",
        "A": "x^(2) * 2 is equal to 2x^(2). However, in this case we are also taking \u00e2\u0088\u0082/\u00e2\u0088\u0082x(x^(2) * 2) which is equal to 4x. \u00e2\u0088\u0082/\u00e2\u0088\u0082x(x^(2) * 2) = \u00e2\u0088\u0082/\u00e2\u0088\u0082x(2x^(2)) = 4x",
        "video_name": "AXqhWeUEtQU",
        "timestamps": [
            374
        ],
        "3min_transcript": "Again, it depends on the function. And I'll show you how you can compute something like this in just a moment here. But, first there's kind of an annoying thing associated with partial derivatives, where we don't write them with D's in DX/DF. People came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multi-variable function involved. And what you do, is you say, you write a D, but it's got kind of a curl at the top. It's this new symbol and people will often read it as partial. So, you might read like partial F, partial Y. If you're wondering, by the way, why we call these partial derivatives, it's sort of like, this doesn't tell the full story of how F changes 'cause it only cares about the X direction. Neither does this, this only cares about the Y direction. So, each one is only a small part of the story. So, let's actually evaluate something like this. I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, if you're actually evaluating something like this, here, I'll write it up here again up here. Partial derivative of F, with respect to X, and we're doing it at one, two. It only cares about movement in the X direction, so it's treating Y as a constant. It doesn't even care about the fact that Y changes. As far as it's concerned, Y is always equal to two. So, we can just plug that in ahead of time. So, I'm gonna say partial, partial X, this is another way you might write it, put the expression in here. And I'll say X squared, but instead of writing Y, I'm just gonna plug in that constant ahead of time. 'Cause when you're only moving in the X direction, this is kind of how the multi-variable function sees the world. And I'll just keep a little note that we're evaluating this whole thing at X equals one. And here, this is actually just an ordinary derivative. This is an expression that's an X, you're asking how it changes as you shift around X and you know how to do this. This is just taking the derivative is gonna be 4x 'cause X squared goes to 2x. And then the derivative of a constant, sin of two is just a constant, is zero. And of course we're evaluating this at X equals one, so your overall answer is gonna be four. And as for practice, let's also do that with derivative with respect to Y. So, we look over here, I'm gonna write the same thing. You're taking the partial derivative of F with respect to Y. We're evaluating it at the same point one, two. This time it doesn't care about movement in the X direction. So, as far as it's concerned, that X just stays constant at one. So, we'd write one squared times Y, plus sin(Y). Sin(Y). And you're saying, oh, I'm keeping track of this at Y=2. So, it's kind of, you're evaluating at Y=2. When you take the derivative, this is just 1xY."
    },
    {
        "Q": "at 1:41, how do you divide 1/6 by side and dividing both side also with 6 can someone explain.\n",
        "A": "He says multiply both sides by 1/6, or another way you could think about it is we re dividing both sides by 6.",
        "video_name": "u5dPUHjagSI",
        "timestamps": [
            101
        ],
        "3min_transcript": "We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things"
    },
    {
        "Q": "At 1:53, how can we figure out what number to multiply by the equation? Every equation isnt this simple, Khan should do one easy equation and one really hard equation to show that its possible.\n",
        "A": "multiplying everything by 1/6 is the same as dividing it all by 6. The equation was 6x-6y=24. 6 goes into all the numbers equally therefore we are able to divide it by 6. The end product is x-y=4",
        "video_name": "u5dPUHjagSI",
        "timestamps": [
            113
        ],
        "3min_transcript": "We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things"
    },
    {
        "Q": "\nAt 0:08 why was a dot there insted of a X?",
        "A": "Because some people use dots for multiplication instead of an X!",
        "video_name": "0WUXQNjdRvM",
        "timestamps": [
            8
        ],
        "3min_transcript": "Compute 23 times 44. And maybe the hardest part of this problem, or maybe the first hard part, is to recognize that that dot even means multiplication. This could have also been written as 23 times 44, or they could have written it as 23 in parentheses times 44, so you just put the two parentheses That also implies multiplication. So now that we know we're multiplying, let's actually do the problem. So we're going to multiply 23-- I'll write it bigger. We're going to multiply 23 by 44. I'll write the traditional multiplication sign there, just so that we know we're multiplying. When you write it vertically like this, you very seldom put a dot there. So let's do some multiplication. Let's start off multiplying this 4 in the ones place times 23. So you have 3 times 4 is 12. We can write 2 in the ones place, but then we want to carry the 1, or we want to regroup that So it's 12, so you put the 1 over here. And now you have 4 times 2 is 8 plus 1 is 9. So you can think about it as 4, this 4 right here, times 23 is 92. That's what we just solved for. Now, we want to figure out what this 4 times 23 is. Now what we do here is, when you just do it mechanically, when you just learn the process, you stick a 0 here. But the whole reason why you're putting a 0 here is because you're now dealing with a 4 in the tens place. If you had another-- I don't know, a 3 or a 4 or whatever digit, and you're dealing with the hundreds place, you'd put more zeroes here, because we're going to find out 4 times 23 is 92. We just figured that out. If we just multiplied this 4 times 23 again, we would get 92 again. But this 4 is actually a 40, so it actually should be 920, and that's why we're putting that 0. Now you're going to see it in a second. So we have-- so let me put this in a different color. 4 times 3 is 12. Let's put the 2 right here. It should be in the tens place because this is really a 40 times the 3. Just think about it, or you could just think of the process. It's the next space that's free. 4 times 3 is 12. Carry the 1. This blue 1 is from last time. You ignore it now. You don't want to make that mess it up. That's when we multiplied this 4. So now we have 4 times 2 is 8 plus 1 is 9. So what we figured out so far is 4 times 23 is 92, and this green 4 times 23 is 920, and that's because this green 4 actually represents 40. It's in the tens place. So when you multiply 44 times 23, it's going to be 4 times 23, which is 92, plus 40 times 23, which is 920. I just want to make sure we understand what we're doing here. And so we can take their sum now. Let's add them up."
    },
    {
        "Q": "At about 1:44 Sal said when we move to the tens place we put a zero under the quitiont. But can you also put an X?\n",
        "A": "Yes. Both are correct and valid strategies. Some people use a 0 and some use an X, both are meant to help you not to mix up which column to put your current step s number in - otherwise known as a placeholder. I learned the 0 method, but if you re comfortable with X, or that s what your teacher/tutor/coach expects, then that s fine, too.",
        "video_name": "0WUXQNjdRvM",
        "timestamps": [
            104
        ],
        "3min_transcript": "Compute 23 times 44. And maybe the hardest part of this problem, or maybe the first hard part, is to recognize that that dot even means multiplication. This could have also been written as 23 times 44, or they could have written it as 23 in parentheses times 44, so you just put the two parentheses That also implies multiplication. So now that we know we're multiplying, let's actually do the problem. So we're going to multiply 23-- I'll write it bigger. We're going to multiply 23 by 44. I'll write the traditional multiplication sign there, just so that we know we're multiplying. When you write it vertically like this, you very seldom put a dot there. So let's do some multiplication. Let's start off multiplying this 4 in the ones place times 23. So you have 3 times 4 is 12. We can write 2 in the ones place, but then we want to carry the 1, or we want to regroup that So it's 12, so you put the 1 over here. And now you have 4 times 2 is 8 plus 1 is 9. So you can think about it as 4, this 4 right here, times 23 is 92. That's what we just solved for. Now, we want to figure out what this 4 times 23 is. Now what we do here is, when you just do it mechanically, when you just learn the process, you stick a 0 here. But the whole reason why you're putting a 0 here is because you're now dealing with a 4 in the tens place. If you had another-- I don't know, a 3 or a 4 or whatever digit, and you're dealing with the hundreds place, you'd put more zeroes here, because we're going to find out 4 times 23 is 92. We just figured that out. If we just multiplied this 4 times 23 again, we would get 92 again. But this 4 is actually a 40, so it actually should be 920, and that's why we're putting that 0. Now you're going to see it in a second. So we have-- so let me put this in a different color. 4 times 3 is 12. Let's put the 2 right here. It should be in the tens place because this is really a 40 times the 3. Just think about it, or you could just think of the process. It's the next space that's free. 4 times 3 is 12. Carry the 1. This blue 1 is from last time. You ignore it now. You don't want to make that mess it up. That's when we multiplied this 4. So now we have 4 times 2 is 8 plus 1 is 9. So what we figured out so far is 4 times 23 is 92, and this green 4 times 23 is 920, and that's because this green 4 actually represents 40. It's in the tens place. So when you multiply 44 times 23, it's going to be 4 times 23, which is 92, plus 40 times 23, which is 920. I just want to make sure we understand what we're doing here. And so we can take their sum now. Let's add them up."
    },
    {
        "Q": "At 2:13 how can you just split the triangle into two right triangles? Are you allowed to do that for a equilateral triangle too?\n",
        "A": "You can do it for any triangle. To do it, pick a side of the triangle. Obviously, we can draw perpendiculars to that side wherever we like. And surely one of those perpendiculars will go through the opposite vertex. This divides the triangle into two right triangles.",
        "video_name": "pGaDcOMdw48",
        "timestamps": [
            133
        ],
        "3min_transcript": "In the last video, we had a word problem where we had-- we essentially had to figure out the sides of a triangle, but instead of, you know, just being able to do the Pythagorean theorem and because it was a right triangle, it was just kind of a normal triangle. It wasn't a right triangle. And we just kind of chugged through it using SOHCAHTOA and just our very simple trig functions, and we got the right answer. What I want to do now is to introduce you to something called the law of cosines, which we essentially proved in the last video, but I want to kind of prove it in a more-- you know, without the word problem getting in the way, and I want to show you, once you know the law of cosines, so you can then apply it to a problem, like we did in the past, and you'll do it faster. I have a bit of a mixed opinion about it because I'm not a big fan of memorizing things. You know, when you're 40 years old, you probably won't have the law of cosines still memorized, but if you have that ability to start with the trig functions and just move forward, then you'll always be set. And I'd be impressed if you're still doing trig at 40, but who knows? So let's go and let's see what this law of cosines is all about. So let's say that I know this angle theta. No, let's call this side b. I'm being a little arbitrary here. Actually, let me stay in the colors of the sides. Let's call that b and let's call this c, and let's call this side a. So if this is a right triangle, then we could have used the Pythagorean theorem somehow, but now we can't. So what do we do? So we know a-- well, let's assume that we know b, we know c, we know theta, and then we want to solve for a. But, in general, as long as you know three of these, you can solve for the fourth once you know the law of cosines. So how can we do it? Well, we're going to do it the exact same way we did that last problem. We can drop a line here to make-- oh, my I thought I was using the line tool. Edit, undo. So I can drop a line like that. So I have two right angles. And then once I have right triangles, then now I can start to use trig functions and the Pythagorean theorem, et cetera, et cetera. So, let's see, this is a right angle, this is a right angle. So what is this side here? Let me pick another color. I'm probably going to get too involved with all the colors, but it's for your improvement. So what is this side here? What is the length of that side, that purple side? Well, that purple side is just, you know, we use SOHCAHTOA. I was just going to write SOHCAHTOA up here. So this purple side is adjacent to theta, and then this blue or mauve side b is the hypotenuse of this right triangle."
    },
    {
        "Q": "\nAt 4:40, how does Sal know to solve for m (the magenta line) in terms of b and theta? For example, we can also say that tan theta = m/c, no?",
        "A": "maybe because tan theta = m/d actually, and that would introduce another variable that he doesn t want. He wants to relate the existing variables to keep it simple. Also possibly because he knows sin and cos functions are complimentary and so hopes they will cancel or simplify easier down the track in the math..",
        "video_name": "pGaDcOMdw48",
        "timestamps": [
            280
        ],
        "3min_transcript": "because it'll take me forever if I keep switching colors. We know that cosine of theta-- let's call this side, let's call this kind of subside-- I don't know, let's call this d, side d. We know that cosine of theta is equal to d over b, right? And we know b. Or that d is equal to what? It equals b cosine theta. Now, let's call this side e right here. Well, what's e? Well, e is this whole c side-- c side, oh, that's interesting-- this whole c side minus this d side, right? So e is equal to c minus d. minus b cosine of theta. So that's e. We got e out of the way. Well, what's this magenta side going to be? Well, let's call this magenta-- let's call it m from magenta. Well, m is opposite to theta. Now, we know it. We've solved for c as well, but we know b, and b is simple. So what relationship gives us m over b, or involves the opposite and the hypotenuse? Well, that's sine: opposite over hypotenuse. So we know that m over b is equal to sine of theta. We know that-- let me go over here. m over b, right, because this is the hypotenuse, is equal to sine of theta, or that m is equal to b sine So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you. We have two sides of a right triangle. We want to figure out the hypotenuse. We can use the Pythagorean theorem. The Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color just to be arbitrary. a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared. Well, e we figure out is this. So it's plus c minus b cosine theta squared."
    },
    {
        "Q": "\nI was able to follow his process up until 6:24 when he wrote Csquared-2cbcos(theta). Where did he get that part of the equation from? To me it looked like he pulled it out of nowhere.",
        "A": "He simply FOILed the (c-bcos(theta))^2, i.e. multipled out the (c-bcos(theta))^2, which is just (c-bcos(theta))(c-bcos(theta)). That multiples out to c^2-cbcos(theta)-cbcos(theta)+(bcos(theta))^2, and there are two -cbccos(theta), so it simplifies to c^2-2bcos(theta)+(bcos(theta)^2.",
        "video_name": "pGaDcOMdw48",
        "timestamps": [
            384
        ],
        "3min_transcript": "minus b cosine of theta. So that's e. We got e out of the way. Well, what's this magenta side going to be? Well, let's call this magenta-- let's call it m from magenta. Well, m is opposite to theta. Now, we know it. We've solved for c as well, but we know b, and b is simple. So what relationship gives us m over b, or involves the opposite and the hypotenuse? Well, that's sine: opposite over hypotenuse. So we know that m over b is equal to sine of theta. We know that-- let me go over here. m over b, right, because this is the hypotenuse, is equal to sine of theta, or that m is equal to b sine So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you. We have two sides of a right triangle. We want to figure out the hypotenuse. We can use the Pythagorean theorem. The Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color just to be arbitrary. a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared. Well, e we figure out is this. So it's plus c minus b cosine theta squared. So that equals b sine-- b squared sine squared of theta. theta squared, right? Plus, and we just foiled this out, although I don't like using foil. I just multiply it out. c squared minus 2cb cosine theta plus b squared cosine theta, right? I just expanded this out by multiplying it out. And now let's see if we can do anything interesting. Well, if we take this term and this term, we get-- those two terms are b squared sine squared of theta plus b squared cosine-- this should be squared there, right, because we squared it. b squared cosine squared of theta, and then we have plus c"
    },
    {
        "Q": "What happens to the -1 = 15 at 8:37? He glances his cursor towards the direction -1=15 and utters the number 16 but he loses me at that point.\n",
        "A": "He adds 1 to both sides, resulting in 16 for the right hand side, and then subtracts 8 from both sides, which yields 8 for the right hand side. He could have subtracted 1 from 8 for the left hand side and then subtracted 7 from both sides. Same result.",
        "video_name": "A52fEdPn9lg",
        "timestamps": [
            517
        ],
        "3min_transcript": "And now for B, there's no obvious way to make the other two cancel out without making the B cancel out. But we've solved for everything else, so we can really just pick an arbitrary value for x that'll make things easy to compute. So if we just pick x is equal to 0, that's always something that can clear out a lot of the hairiness of an algebra problem. x is equal to 0, not 3. x is equal to 0. Let me do that in a different color. If x is equal to 0, then what do we have? We have A, but A is 2, right? 2 times 0 minus 2 squared. So that's minus 2 squared, so that's 2 times 4. Plus B-- that's what we're trying to solve for; we already solved for everything-- B times minus 1, right? 0 minus 1 is minus 1. Times minus 2, plus C times minus 1 is equal to-- well, Then we just solve for B. We get 8 plus 2B, right? Minus 1 times minus 2 is plus 2. Oh, and we shouldn't have written the C here, we know what C is. C is 1. So this is just a 1. So 1 times minus 1 is minus 1, is equal to 15. And let's see, we have 2B is equal to, lets see, 16 is equal to 8, B is equal to 4, dividing both sides by 2. So we're done. So the partial fraction decomposition of this right here is A, which we've solved for, which is 2. So it equals 2 over x minus 1 plus B, which is 4-- plus 4 And what we did in this with the repeated factor is true if we went to a higher degree term. So if we had blah blah blah, some polynomial there, and it was all over x minus-- I don't know, some number. x minus a to the 10th power. Well, to the-- well, yeah, sure. To the 10th power. If we wanted to decompose this partial fraction, or expand it, it would be A over x minus a-- it's a different a. I'm just showing this. Plus B times x minus a squared, plus, and you go blah blah blah. You'd have 10 terms, plus-- I don't know what the 10th letter of the alphabet is; maybe it's H or I or something. Maybe it's J. J over x minus a to the 10th."
    },
    {
        "Q": "\nWhy is it (x-2)^2? It should be 2(x-2)! If (x-2)^2=(x-2)(x-2). 2(x-2)=(x-2)+(x-2). (at 00:56):D",
        "A": "(x-2)^2 = (x-2)*(x-2) = x^2-4x+4 what you wrote is: 2(x-2) = 2x-4 (x-2)+(x-2) = 2x-4 these are completely different polynomials from the one in the problem.",
        "video_name": "A52fEdPn9lg",
        "timestamps": [
            56
        ],
        "3min_transcript": "There's one more case of partial fraction expansion or decomposition problems that you might see, so I thought I would cover it. And that's the situation where you have a repeated factor in the denominator. So let's see, I've constructed a little problem here. It's 6x squared. Let me make sure my pen is right. 6x squared minus 19x plus 15. All of that over x minus 1 times-- and this is the thing that makes this one especially interesting-- x minus 2 squared. So you might say, gee, this is a little bit different. Because what do I do here? I have this first degree factor, but it shows up twice. It doesn't make sense, for example, it wouldn't make sense to do this. A over x minus 1 plus B over x minus 2 plus C over x minus 2. Because if you did this, the B and the C would just add You could just view them as one variable. You wouldn't have to have two separate variables here. So this wouldn't make sense as a partial fraction expansion of this. You might want to maybe square this and view it as a second degree term, and do it like you did in the previous example. But then you wouldn't have fully decomposed this problem. And so the answer here-- and I'll just show you how to do it, I'll maybe leave it to you a little bit to think about why it works-- is to decompose it almost like this, but instead of having C over x minus 2, you're going to have C over x minus 2 squared. And I'll try to give you a little intuition about why that happens. So the decomposition here is going to be A over x minus 1 plus B over x minus 2 plus C over x minus 2 squared. And the intuition here is that if you were to-- let's ignore this term right here. But if you were to add these two terms right there, you would get a rational function that would end up having And so it would be consistent with what we did in the second partial fraction video, where if you would have a second degree term on the bottom, and the numerator, when you add just these two parts-- let me clarify what I'm saying. And this is just for intuition. If you were to add just these two parts here, you would get something times x plus something, all of that over x minus 2 squared. And that's consistent with what we did in the previous video, where we said if we have a second degree term in the denominator-- which this really is if you were to expand this-- you should have a first degree term in the numerator. So I'll leave that there. That's a little bit of a nuance, and it's good to think about why this works. But with that said, let's just solve this problem. Let me erase some of this stuff that I've written."
    },
    {
        "Q": "\nconfused at 0:55 NEED HELP!",
        "A": "If you re really having difficulty, please try to stay on topic (refrain from gratuitous woodchuck references in the future). Are you having trouble understanding why b<c or why -b is NOT less than -c?",
        "video_name": "kQkeayQke8A",
        "timestamps": [
            55
        ],
        "3min_transcript": "- [Voiceover] So we have a number line here with zero at the center and then on that number line we've marked off some numbers. So, to the left of the number line we have the number a, and then we have the number b here, a little bit closer to zero, and then on the right side of zero we have the number c. And then after that, we have a bunch of statements dealing with inequalities. Now, what I want you to do is pause the video and think about which of these statements are true, which of these statements are false, and maybe which of these statements you don't have enough information to figure out. So, I'm assuming you've had a go at it. You've tried to figure out which of these are true, which of these are false, and which of these you can't figure out. So, let's do them together. So, this first statement says negative b is less than negative c. So, we do know for a fact that b is less than c. We know that b is less than c. How do we know that? Well, b is to the left of c on the number line. It's that straight forward. But what about negative b? Is negative b less than negative c? So, let's think about where negative b is on this number line. So, negative b I will do in yellow. So, negative, negative b means the opposite of b. So if b is one hash mark to the left of zero, negative b is going to be one hash mark to the right of zero. So, that right over there, is going to be negative b. And now, where is negative c? Once again, negative c, this literally means the opposite of c. C is one, two, three, four, five hash marks to the right of zero. And so negative c is going to be one, two, three, four, five hash marks to the left of zero. And actually, let me do this in a different color. So, negative c I will do in purple. This right over here is negative c. So, let's compare, is negative b less than negative c? negative c on the number line. Negative b is greater than negative c. So, this is not true. Negative b is to the right of negative c. Negative b is greater than negative c. And if this is a little confusing, just think about it. Since b is a negative number, negative b is going to be a positive number. And since c is a positive number, negative c is going to be a negative number. So, it makes complete sense that a positive number is going to be greater than a negative number. And you see it here, negative b is to the right of negative c on the number line. So, we can rule this one out. So the next question, is negative b greater than zero? Well, we already plotted negative b, it's going to be one to the right, or one hash mark to the right, we don't know how much each of these hash marks represent, but it's going to be to the right of zero. So, it is greater than zero. This is true. That is true."
    },
    {
        "Q": "\nSince at 8:44 Sal proves that x_1 and x_2 can be expressed as linear combinations of x_3 and x_4, why does he need to (at 19:44) eliminate x_3 and x_4 to make sure that they are linearly dependent on x_1 and x_2?",
        "A": "That s not what he did. x is a vector of 4 variable scalars. He uses rref to solve Ax=0 for the scalars x1 and x2 in terms of x3 and x4. For v1 = column vector 1 , etc., you have, for Ax = 0 , x1v1 + x2v2 + x3v3 + x4v4 = 0 . For every possible pair of values for x3 and x4, we can find x1 and x2. When we make either x3 or x4 0 , we get either v3 or v4 as a linear combination of v1 and v2.",
        "video_name": "_uTAdf_AsfQ",
        "timestamps": [
            524,
            1184
        ],
        "3min_transcript": "Minus 2 times x3. Minus x4 is equal to 0. And then this gives me no information. 0 times all this is equal to 0. So it just turns into 0 equals 0. So let's see if we can solve for our pivot entries, or our pivot variables. What are our pivot entries? This is a pivot entry. That's a pivot entry. That's what reduced row echelon form is all about, getting these entries that are 1 and they're the only non-zero term in their respective columns. And that every pivot entry is to the right of a pivot entry above it. And then the columns that don't have pivot entries? These columns represent the free variables. So this column has no pivot entry. And so when you take the dot product, this column turned into this column in our system of equations. So we know that x3 is a free variable. x3 is free. We can set it equal to anything. Likewise x4 is a free variable. columns in our reduced row echelon form have pivot entries in them. Fair enough. So let's see if we can simplify this into a form we know. And we've seen this before. So if I solve for x1 -- this 0 I can ignore. That 0 I can ignore -- I could say that x1 is equal to minus 3x3 minus 2x4. I just subtracted these two from both sides of the equation and I can say that x2 is equal to 2x3 plus x4. And if we want to write our solution set now, so if I wanted to find the null space of A, which is the same thing as the null space of the reduced row echelon form of A, Maybe I'll do blue -- is equal to all of the vectors x1, x2, x3, x4 that are equal to -- So what are they going to be equal to? X1 has to be equal to minus 3x3 minus 2x4. Just to be clear, these are free variables because I can set these to be anything. And these are pivot variables because I can't just set them to anything. When I determine what my x3's and my x4's are, they determine what my x1's and my x2's have to be. So these are pivoted variables. These are free variables. I can make this guy pi. And I can make this guy minus 2. We can set them to anything. So x1 is equal to -- let's see, let me write it this way -- they're equal to x3 -- let me do it in a different color -- do x3 like this. So it's equal to x3 times some vector plus x4 times some"
    },
    {
        "Q": "so at 2:03 would you just leave it at 2 (2x+9) ?\n",
        "A": "It depends on why you have factored it. If the goal is simply factor this expression, then yes, you re done at that point. Most often, though, you have factored it in order to make it easier to do something else to it.",
        "video_name": "mbb3msmX2xs",
        "timestamps": [
            123
        ],
        "3min_transcript": "What I want to do is start with an expression like 4x plus 18 and see if we can rewrite this as the product of two Essentially, we're going to try to factor this. And the key here is to figure out are there any common factors to both 4x and 18? And we can factor that common factor out. We're essentially going to be reversing the distributive property. So for example, what is the largest number that is-- or I could really say the largest expression-- that is divisible into both 4x and 18? Well, 4x is divisible by 2, because we know that 4 is divisible by 2. And 18 is also divisible by 2, so we can rewrite 4x as being 2 times 2x. If you multiply that side, it's obviously going to be 4x. And then, we can write 18 as the same thing as 2 times 9. you apply the distributive property, you'll usually end up with a step that looks something like this. Now we're just going to undistribute the two right over We're going to factor the two out. Let me actually just draw that. So we're going to factor the two out, and so this is going to be 2 times 2x plus 9. And if you were to-- wanted to multiply this out, it would be 2 times 2x plus 2 times 9. It would be exactly this, which you would simplify as this, right up here. So there we have it. We have written this as the product of two expressions, 2 times 2x plus 9. Let's do this again. So let's say that I have 12 plus-- let me think of something interesting-- 32x. Actually since we-- just to get a little bit of variety here, Well, what's the largest number that's divisible into both 12 and 32? 2 is clearly divisible into both, but so is 4. It doesn't look like anything larger than 4 is divisible into both 12 and 32. The greatest common factor of 12 and 32 is 4, and y is only divisible into the second term, not into this first term right over here. So it looks like 4 is the greatest common factor. So we could rewrite each of these as a product of 4 and something else. So for example, 12, we can rewrite as 4 times 3. And 32, we can rewrite-- since it's going to be plus-- 4 times. Well if you divide 32y by 4, it's going to be 8y. And now once again, we can factor out the 4. So this is going to be 4 times 3 plus 8y."
    },
    {
        "Q": "\nSo at 1:16, Sal said that you factor out the 2. So After you do that 2x2 equals 4 and 2 times x will be 2x right. I don't get how it will go back to 4x. I'm confused",
        "A": "If it had simply been 2\u00c3\u00972 then it would be 4, but it is 2\u00c3\u00972x which would equal 4x.",
        "video_name": "mbb3msmX2xs",
        "timestamps": [
            76
        ],
        "3min_transcript": "What I want to do is start with an expression like 4x plus 18 and see if we can rewrite this as the product of two Essentially, we're going to try to factor this. And the key here is to figure out are there any common factors to both 4x and 18? And we can factor that common factor out. We're essentially going to be reversing the distributive property. So for example, what is the largest number that is-- or I could really say the largest expression-- that is divisible into both 4x and 18? Well, 4x is divisible by 2, because we know that 4 is divisible by 2. And 18 is also divisible by 2, so we can rewrite 4x as being 2 times 2x. If you multiply that side, it's obviously going to be 4x. And then, we can write 18 as the same thing as 2 times 9. you apply the distributive property, you'll usually end up with a step that looks something like this. Now we're just going to undistribute the two right over We're going to factor the two out. Let me actually just draw that. So we're going to factor the two out, and so this is going to be 2 times 2x plus 9. And if you were to-- wanted to multiply this out, it would be 2 times 2x plus 2 times 9. It would be exactly this, which you would simplify as this, right up here. So there we have it. We have written this as the product of two expressions, 2 times 2x plus 9. Let's do this again. So let's say that I have 12 plus-- let me think of something interesting-- 32x. Actually since we-- just to get a little bit of variety here, Well, what's the largest number that's divisible into both 12 and 32? 2 is clearly divisible into both, but so is 4. It doesn't look like anything larger than 4 is divisible into both 12 and 32. The greatest common factor of 12 and 32 is 4, and y is only divisible into the second term, not into this first term right over here. So it looks like 4 is the greatest common factor. So we could rewrite each of these as a product of 4 and something else. So for example, 12, we can rewrite as 4 times 3. And 32, we can rewrite-- since it's going to be plus-- 4 times. Well if you divide 32y by 4, it's going to be 8y. And now once again, we can factor out the 4. So this is going to be 4 times 3 plus 8y."
    },
    {
        "Q": "\nAt 4:22 Sal said\" let's try doing 50 townhouses.\" Do you really have to draw 50 townhomes or is there a different way to do it?",
        "A": "No you don t have to do that! Be creative and make shapes at least more than 1 and to find the pattern watch the video!",
        "video_name": "mFftY8Y_pyY",
        "timestamps": [
            262
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\ni dont get it on 3:06",
        "A": "He essentially completed the square and that allowed him to factor. If you multiply out (x-2)^2 it will get you back to x^2-4x+4. Completing the square is addressed in the video after next, if you need to learn it.",
        "video_name": "dfoXtodyiIA",
        "timestamps": [
            186
        ],
        "3min_transcript": ""
    },
    {
        "Q": "@ 6:00 through 6:10 Sal says \"when x is equal to three it's also going to be equal to 8\" ... he meant \"when x is equal to 4\" because we are calling each line in the graph 2, not one... I know this isn't a question, but it confused me since -2(3)^2+8(3)\u00e2\u0089\u00a00... always check your (and the teacher's) work!\n",
        "A": "Thanks, that was confusing me :)",
        "video_name": "dfoXtodyiIA",
        "timestamps": [
            360,
            370
        ],
        "3min_transcript": ""
    },
    {
        "Q": "why does x need to be zero in 4:35\n",
        "A": "He is actually making (x-2) equal to zero. This would mean that x is 2, and since the -2(x-2)^2 term equal to zero and thus, y=16. This gives the vertex of the parabola at the coordinate (2,16). It also gives the axis of symmetry as the vertical line x=2.",
        "video_name": "dfoXtodyiIA",
        "timestamps": [
            275
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:57, why does Sal write 20 as 4 times 5?\n",
        "A": "Because 20 is within the square root sign, Sal wants to show its perfect square factors (that he ll be able to take the square root of) and its non-perfect square factors (that will have to remain under the square root sign). He does the same process for the s^5 that s inside the square root sign.",
        "video_name": "4F6cFLnAAFc",
        "timestamps": [
            57
        ],
        "3min_transcript": "We're asked to simplify r to the 2/3 s to the third, that whole thing squared. Times the square root of 20r to the fourth s to the fifth. Now this looks kind of daunting, but I think if we take it step by step it shouldn't be too bad. So first we can look at this first expression right here where we're taking this product to the second power. We know that instead we can take each of the terms in the product to the second power and then take the product. So this is going to be the same thing as r to the 2/3 squared times s to the third squared. And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1/2 power. So this is equal to-- so times this part. Let me do this in a different color. This part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as the product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5. Times r to the fourth times s to the fifth. Now let me write s to the fifth also as a product of a perfect square and a non-perfect square. r to the fourth is obviously a perfect square. Its square root is r squared. But let's write s to the fifth in a similar way. So s to the fifth we can rewrite as s to the fourth times s. S to the fourth times s to the first, that is s to the fifth. And of course, all of this has to be raised to the 1/2 power. Now let's simplify this even more. If we're taking something to the 2/3 power and then to the second power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4/3 power. And just as a bit of review, taking something to the 4/3 power, you can view it as either taking-- finding its cube root, taking it to the 1/3 power, and then taking its cube root to the fourth power. Or you can view it as taking it to the fourth power and then finding the cube root of that. the 4/3 power. So you have r to the 4/3 times s to the 3 times 2. Times s to the sixth power. And then we could raise each of these terms right here to the 1/2 power. So times-- let me color code it a little bit. And we actually wouldn't need the parentheses once we do that. Times 4 to the 1/2 times 5 to the 1/2. That term right there. Times r to the fourth to the 1/2 power. Times-- I might run out of colors-- s to the fourth to the 1/2 power. We're raising each of these terms to that 1/2 power. Times s to the 1/2 power. There's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect"
    },
    {
        "Q": "\nWhy does he take the 20 and factor it into 4 and 5 at 1:00, likewise factoring of the s ?",
        "A": "The reason he factored the 20 into 4 and 5 was to simply the terms under the radical sign. Since 20 is not a perfect square, it is composed of a perfect square (the 4) multiplied by another number (the 5). Since we are trying to simplify this expression, our goal is to break down the 20 into simpler components. When we break the 20 down to 4 times 5, we are able to take the square root of the 4, simplifying the term. Hope that makes sense.",
        "video_name": "4F6cFLnAAFc",
        "timestamps": [
            60
        ],
        "3min_transcript": "We're asked to simplify r to the 2/3 s to the third, that whole thing squared. Times the square root of 20r to the fourth s to the fifth. Now this looks kind of daunting, but I think if we take it step by step it shouldn't be too bad. So first we can look at this first expression right here where we're taking this product to the second power. We know that instead we can take each of the terms in the product to the second power and then take the product. So this is going to be the same thing as r to the 2/3 squared times s to the third squared. And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1/2 power. So this is equal to-- so times this part. Let me do this in a different color. This part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as the product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5. Times r to the fourth times s to the fifth. Now let me write s to the fifth also as a product of a perfect square and a non-perfect square. r to the fourth is obviously a perfect square. Its square root is r squared. But let's write s to the fifth in a similar way. So s to the fifth we can rewrite as s to the fourth times s. S to the fourth times s to the first, that is s to the fifth. And of course, all of this has to be raised to the 1/2 power. Now let's simplify this even more. If we're taking something to the 2/3 power and then to the second power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4/3 power. And just as a bit of review, taking something to the 4/3 power, you can view it as either taking-- finding its cube root, taking it to the 1/3 power, and then taking its cube root to the fourth power. Or you can view it as taking it to the fourth power and then finding the cube root of that. the 4/3 power. So you have r to the 4/3 times s to the 3 times 2. Times s to the sixth power. And then we could raise each of these terms right here to the 1/2 power. So times-- let me color code it a little bit. And we actually wouldn't need the parentheses once we do that. Times 4 to the 1/2 times 5 to the 1/2. That term right there. Times r to the fourth to the 1/2 power. Times-- I might run out of colors-- s to the fourth to the 1/2 power. We're raising each of these terms to that 1/2 power. Times s to the 1/2 power. There's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect"
    },
    {
        "Q": "\nAround 3:10 why does those two angles have to both be equal to x? Can you explain?",
        "A": "Because it is an isosceles triangle (notice the 2 legs are the radius of the circle, so they are the same length), and by definition the base angles of an isosceles triangle have equal measures and are congruent. Does that make sense?",
        "video_name": "b0U1NxbRU4w",
        "timestamps": [
            190
        ],
        "3min_transcript": "Well, we have in our tool kit the notion of an inscribed angle, it's relation to a central angle that subtends the same arc. So let's look at that. So let's say that this is an inscribed angle right here. Let's call this theta. Now let's say that that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius -- actually this distance is the same. But we've learned several videos ago that look, this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2theta. Now, this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I flipped it over it would look like that, that, and then the green side would be down like that. And both of these sides are of length r. This top angle is 2theta. So all I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these to base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see, I already used theta, maybe I'll use x for these angles. So this has to be x, and that has to be x. Well, x plus x plus 2theta have to equal 180 degrees. They're all in the same triangle. So let me write that down. We get x plus x plus 2theta, all have to be equal to 180 degrees, or we get 2x plus 2theta is equal to 180 degrees, or we get 2x is equal to 180 minus 2theta. Divide both sides by 2, you get x is equal to 90 minus theta. So x is equal to 90 minus theta. Now let's see what else we could do with this. Well we could look at this triangle right here. This triangle, this side over here also has this distance right here is also a radius of the circle."
    },
    {
        "Q": "\nAt 1:10, what is an inscribed angle and a central angle?",
        "A": "A CENTRAL angle has its vertex at the CENTER of the circle, and its rays cut the circle in 2 places.. The measure of a central angle EQUALS the measure of the arc that is cut off by its rays. An INSCRIBED angle has its vertex on the EDGE of the circle, and its rays cut the circle in 2 places. The measure of an inscribed angle is HALF the measure of the arc that is cut off by its rays.",
        "video_name": "b0U1NxbRU4w",
        "timestamps": [
            70
        ],
        "3min_transcript": "Let's say we have a circle, and then we have a diameter of the circle. Let me draw my best diameter. This right here is the diameter of the circle or it's a diameter of the circle. That's a diameter. Let's say I have a triangle where the diameter is one side of the triangle, and the angle opposite that side, it's vertex, sits some place on the circumference. So, let's say, the angle or the angle opposite of this diameter sits on that circumference. So the triangle looks like this. The triangle looks like that. What I'm going to show you in this video is that this triangle is going to be a right triangle. The 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Well, we have in our tool kit the notion of an inscribed angle, it's relation to a central angle that subtends the same arc. So let's look at that. So let's say that this is an inscribed angle right here. Let's call this theta. Now let's say that that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius -- actually this distance is the same. But we've learned several videos ago that look, this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2theta. Now, this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I flipped it over it would look like that, that, and then the green side would be down like that. And both of these sides are of length r. This top angle is 2theta. So all I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these to base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see, I already used theta, maybe I'll use x for these angles. So this has to be x, and that has to be x."
    },
    {
        "Q": "at 0:01 sal says \" i have drawn an Arbitrary triangle\" what's an Arbitrary Triangle?\n",
        "A": "an Arbitrary triangle is a triangle that has no definite side lengths, no definite angles, and the vertices have no definite position. In other words, it is equally likely to be ANY POSSIBLE TRIANGLE.",
        "video_name": "i0VS3eEGjiQ",
        "timestamps": [
            1
        ],
        "3min_transcript": "I've drawn an arbitrary triangle right over here, and I've also drawn its three medians: median EB, median FC, and median AD. And we know that where the three medians intersect at point G right over here, we call that the centroid. What I want to do in this video is prove to you that the centroid is exactly 2/3 along the way of each median. Or another way to think about it, we can pick any one of these medians, and let's say let's pick EB. What I want to do is I want to prove that EG is equal to 2 times GB. So whatever distance this is it's twice this distance there. Or another way to think about it is EG is 2/3 along the way of EB. And the logic that I'm using to prove this you can use for any of the medians to show that the centroid is exactly 2/3 along the way of any median, or divides it into a segment that's twice as long as the other segment. And to do that, let's focus-- I want to focus on triangle ABE And I'm going to draw this median as essentially the base. So let me draw it that way. I'm going to try to color code it similarly. So we draw it a little bit flatter than that. So it's like that. And then we have the two yellow sides, so it looks something like this. It looks something like that. And then we have the centroid, right over here at G. That is our centroid, and then we have this magenta line going to A. Let me draw it a little bit neater than that. We have that line going to A, and then we have this blue line going to F right over here. And let me label all the points. Go back to the orange color. So this is going to be E, this is going to be B, this is going to be A, this is going to be F right over here. little marking there is that marking, these two markings, these two markings are on this side right over there. And the whole way that I'm going to prove that EG is twice as long as GB is just refer to the result that we did, I think, a couple of videos ago that the medians divide this triangle into six smaller triangles that all have equal area. So another way to think about it is each of these three small triangles have equal area. These are three of the total of six smaller triangles. So these three all have equal area. So let's think about this triangle right over here. Let's think about this triangle, triangle AGB. This is triangle AGB right over there. Those are the same triangles. And let's compare that to triangle EAG right over here. Let's compare it to this triangle, which is this triangle right over here on the original drawing."
    },
    {
        "Q": "1:05 Is when it gets confusing\n",
        "A": "Hey i m doing this to is there any way I could help you",
        "video_name": "_btQus9HV_I",
        "timestamps": [
            65
        ],
        "3min_transcript": "Let's try to evaluate 7 and 6/9 minus 3 and 2/5. So like always, I like to separate out the whole number parts from the fractional parts. This is the same thing as 7 plus 6/9 minus 3 minus 2/5. And the reason why I'm saying minus 3 minus 2/5 is this is the same thing as minus 3 plus 2/5. And so you distribute the negative sign. You're subtracting a 3, and then you're subtracting the 2/5. And so now we can worry about the whole number parts, 7 minus 3. Well, 7 minus 3 is going to give us 4. So that's going to give us 4. And then we're going to have 6/9 minus 2/5. So let me think about what 6/9 minus 2/5 are. 6/9 minus 2/5, well, we're going to have to find a common denominator. So this is going to be the same thing. And I think the least common multiple of 9 and 5 is going to be 45. They have no common factors. So it's going to be over 45. To go from 9 to 45, I have to multiply by 5. So I'm going to have to multiply the numerator by 5. So 6 times 5 is 30. Then I'm going to subtract. To go from 5 to 45, I had to multiply by 9. So I have to multiply the numerator by 9 if I don't want to change the value. So 2 times 9 is 18. And 30/45 minus 18/45 is going to be something over 45. 30 minus 18 is 12. If I subtract these two fractions right over here, I get 12/45. So it's 4 plus 12/45. Or if we wanted to write it as a mixed number, this is equal to 4 and 12/45. But we're not done yet. We can simplify this further. 12 and 45 have common factors. They're both divisible by 3. I think we can divide more after that. If we divide the numerator by 3 and the denominator by 3, we end up with 4. And 12 divided by 3 is 4. And 45 divided by 3 is 15. 4 and 4/15. And actually, we're done. These two can't be simplified anymore. 4 and 4/15."
    },
    {
        "Q": "At 2:12, why does the square root of 3/4 become the square root of 3 over 2?\n",
        "A": "Because 4 is a perfect square and becomes 2 when you take its square root, while 3 is not a perfect square, so it stays root 3 as the numerator while root 4 is simplified to 2.",
        "video_name": "soIt2TwV6Xk",
        "timestamps": [
            132
        ],
        "3min_transcript": "Let's say that we're told that some angle theta, which is going to be expressed in radians, is between negative 3 pi over 2 and negative pi. It's greater than negative 3 pi over 2 and it's less than negative pi. And we're also told that sine of theta is equal to 1/2. Just from this information can we figure out what the tangent of theta is going to be equal to? And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta, plus cosine squared theta is equal to 1. So let's do it. So we know the Pythagorean identity, sine squared theta, plus cosine squared theta is equal to 1. We know what sine squared theta is. Sine theta is 1/2. So this could be rewritten as 1/2 squared, plus cosine squared theta, is equal to 1. theta is equal to 1. Or we could subtract 1/4 from both sides, and we get cosine squared theta is equal to-- let's see. You subtract 1/4 from the left hand side, then this 1/4 goes away. That was the whole point. 1 minus 1/4 is 3/4. So what could cosine of theta be? Well, when I square it, I get positive 3/4. So it could be the positive or negative square root of 3/4. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it, the unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2."
    },
    {
        "Q": "2:42\n\nWhat is the variable s?\nIs it 1/2P?\n",
        "A": "Actually, Amogh was originally correct. S is the semiperimeter of a shape. It is mainly used for Heron s Formula.",
        "video_name": "UmiZK6Hgm6c",
        "timestamps": [
            162
        ],
        "3min_transcript": "trigonometry-- and if the word \"trigonometry\" scares you, you'll just need to know maybe the first two or three videos in the trigonometry playlist to be able to understand what I do here. What I want to do is figure out the area of the region inside the circle and outside of the triangle. So I want to figure out the area of that little space, that space, and this space combined. So the obvious way to do this is to say, well I can figure out the area of the circle pretty easily. Area of the circle. And that's going to be equal to pi r squared. Or pi times 2 squared, which is equal to 4 pi. And I could subtract from 4 pi the area of the triangle. So we need to figure out the area of the triangle. What is the area of the triangle? where if you know the lengths of the sides of a triangle you can figure out the area. But we don't know the lengths of the sides just yet. Once we do maybe we can figure out the area. Let me apply Heron's formula not knowing it. So let me just say that the lengths of this equilateral-- the lengths of the sides-- are a. Applying Heron's formula, we first define our variable s as being equal to a plus a plus a, over 2. Or that's the same thing as 3a over 2. And then the area of this triangle, in terms of a. So the area is going to be equal to the square root of s, which is 3a over 2, times s minus a. So that's 3a over 2 minus a. Right? a is the same thing as 2a over 2. You could cancel those out and get a. And then I'm going to do that three times. So instead of just multiplying that out three times for each of the sides, by Heron's formula I could just say to the third power. So what's this going to be equal to? This is going to be equal to the square root of 3a over 2. And then this right here is going to be equal to 3a minus 2a, is a. So a/2 to the third power. And so this is going to be equal to-- I'll arbitrarily switch colors. We have 3a times a to the third, which is 3a to the fourth, over 2 times 2 to the third. Well that's 2 to the fourth power, or 16."
    },
    {
        "Q": "\nI don't think you even need trig to solve the problem.\nAll you have to do is follow the video until around 6:06 and from there, use the rules of the side of a 30:60:90 right triangle. You can find the side a/2 and the line from the center of the circle to where it reaches side a perpendicularly. With these you can find the eight of the triangle and the base. Then you can just subtract the area of the triangle to the circle.",
        "A": "Yes but you find out about this 30:60:90 rule in trig",
        "video_name": "UmiZK6Hgm6c",
        "timestamps": [
            366,
            1860
        ],
        "3min_transcript": "And then if we take the square root of the numerator and the denominator, this is going to be equal to the square root of a to the fourth is a squared. a squared times, well I'll just write the square root of 3, over the square root of the denominator, which is just 4. So if we know a, using Heron's formula we know what the area of this equilateral triangle is. So how can we figure out a? So what else do we know about equilateral triangles? Well we know that all of these angles are equal. And since they must add up to 180 degrees, they all must be 60 degrees. That's 60 degrees, that's 60 degrees, and that is 60 degrees. Now let's see if we can use the last video, where I talked about the relationship between an inscribed angle So this is an inscribed angle right here. It's vertex is sitting on the circumference. And so it is subtending this arc right here. And the central angle that is subtending that same arc is this one right here. The central angles subtending that same arc is that one right there. So based on what we saw in the last video, the central angle that subtends the same arc is going to be double of the inscribed angle. So this angle right here is going to be 120 degrees. Let me just put an arrow there. 120 degrees. It's double of that one. Now, if I were to exactly bisect this angle right here. So I go halfway through the angle, and I want to just go straight down like that. Well, they're going to be 60 degrees. I'm bisecting that angle. That is 60 degrees, and that is 60 degrees right there. And we know that I'm splitting this side in two. This is an isosceles triangle. This is a radius right here. Radius r is equal to 2. This is a radius right here of r is equal to 2. So this whole triangle is symmetric. If I go straight down the middle, this length right here is going to be that side divided by 2. That side right there is going to be that side divided by 2. Let me draw that over here. If I just take an isosceles triangle, any isosceles triangle, where this side is equivalent to that side. Those are our radiuses in this example. And this angle is going to be equal to that angle. If I were to just go straight down this angle right here, I would split that opposite side in two. So these two lengths are going to be equal. In this case if the whole thing is a, each of these are going to be a/2."
    },
    {
        "Q": "\nWould you be able to do the part at (5:25) with any of the inscribed angles?",
        "A": "Yes. Remember that the triangle is equilateral, meaning that all its sides are equal and all its angles are equal.",
        "video_name": "UmiZK6Hgm6c",
        "timestamps": [
            325
        ],
        "3min_transcript": "Right? a is the same thing as 2a over 2. You could cancel those out and get a. And then I'm going to do that three times. So instead of just multiplying that out three times for each of the sides, by Heron's formula I could just say to the third power. So what's this going to be equal to? This is going to be equal to the square root of 3a over 2. And then this right here is going to be equal to 3a minus 2a, is a. So a/2 to the third power. And so this is going to be equal to-- I'll arbitrarily switch colors. We have 3a times a to the third, which is 3a to the fourth, over 2 times 2 to the third. Well that's 2 to the fourth power, or 16. And then if we take the square root of the numerator and the denominator, this is going to be equal to the square root of a to the fourth is a squared. a squared times, well I'll just write the square root of 3, over the square root of the denominator, which is just 4. So if we know a, using Heron's formula we know what the area of this equilateral triangle is. So how can we figure out a? So what else do we know about equilateral triangles? Well we know that all of these angles are equal. And since they must add up to 180 degrees, they all must be 60 degrees. That's 60 degrees, that's 60 degrees, and that is 60 degrees. Now let's see if we can use the last video, where I talked about the relationship between an inscribed angle So this is an inscribed angle right here. It's vertex is sitting on the circumference. And so it is subtending this arc right here. And the central angle that is subtending that same arc is this one right here. The central angles subtending that same arc is that one right there. So based on what we saw in the last video, the central angle that subtends the same arc is going to be double of the inscribed angle. So this angle right here is going to be 120 degrees. Let me just put an arrow there. 120 degrees. It's double of that one. Now, if I were to exactly bisect this angle right here. So I go halfway through the angle, and I want to just go straight down like that."
    },
    {
        "Q": "1:11-1:36 Sal says tangent=sin x/cos x. But he also says that tangent is the slope of the ray.\nI thought the line=1 because the radius =1. How does radius relate to ray?\n",
        "A": "tanx is confusing for the unit circle in the xy plane, so I ll use tan(t). tan(t) = sint/cost = (y/r)/(x/r) = y/x = slope (m) of the radius, or of a line (ray) through the radius. The length of the radius in the unit circle is 1, but that has nothing to do with how it s oriented (the angle t), or its slope, m, = tan(t).",
        "video_name": "FK6-tZ5D7xM",
        "timestamps": [
            71,
            96
        ],
        "3min_transcript": "- [Voiceover] What I hope to do in this video is get some familiarity with the graph of tangent of theta. And to do that, I'll set up a little unit circle so that we can visualize what the tangent of various thetas are. Let's say that's a y-axis and this is my x-axis. That is my x-axis, and the unit circle would look something like this. We already know, this is all a refresher of the unit circle definition of trig functions, that if I have an angle, an angle theta, where one side is a positive x-axis and then the other side, so this is the other side, so the angle is formed like this, that where this ray intersects the unit circle, the coordinates of that, the x and y coordinates are the sine of theta. Sorry, the x-coordinate is the cosine of theta. It's the cosine of theta, sine of theta. The x-coordinate here is cosine theta. But we're concerned about tangent of theta. We know that tangent of theta is the same thing as the sine of theta over the cosine of theta. Or if you're starting from the origin and you're going and you're taking the value of essentially the y-coordinate, the y-coordinate over the x-coordinate, it's essentially the slope of this line. This is going to be your change in y over change in x. This right over here is going to be the slope, the slope, I guess you could say, of this ray right over here. That's going to help us visualize what the tangents of different thetas are. Let me clean up my unit circle a little bit just so that we can ... All right, there we go. So now let's make a table. So let's make a table. So for various thetas, let's think about what tangent of theta is going to be, So maybe the easiest one, if theta is zero radians, so if it's zero radians, what is the slope of this ray? That ray's, the slope is zero. As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take pi over four radians. This one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes, it's easier thinking degrees. It's a 45-degree angle. This here, your x-coordinate and your y-coordinate is the same. You might remember, it's square root of two over two, square root of two over two. But the important thing is whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it,"
    },
    {
        "Q": "\nAt 4:05, where did Sal get square root of two over two from?",
        "A": "For 45\u00c2\u00ba or pi/4 radians, the x component and the y component are equal. In the unit circle, r = 1: Pythogorean Identity: a^2 + b^2 = c^2 Let a = x, b = y and c = 1 x^2 + y^2 = 1^2 x^2 + y^2 = 1 Recall: x = y x^2 + x^2 = 1 2\u00e2\u0080\u00a2x^2 = 1 x^2 = 1/2 x = sqrt(1/2) x = sqrt(1)/sqrt(2) x = 1/sqrt(2) x = 1/sqrt(2)\u00e2\u0080\u00a2sqrt(2)/sqrt(2) x = sqrt(2)/(sqrt(2)\u00e2\u0080\u00a2sqrt(2)) x = sqrt(2)/(sqrt(2)^2) x = sqrt(2)/2",
        "video_name": "FK6-tZ5D7xM",
        "timestamps": [
            245
        ],
        "3min_transcript": "So maybe the easiest one, if theta is zero radians, so if it's zero radians, what is the slope of this ray? That ray's, the slope is zero. As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take pi over four radians. This one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes, it's easier thinking degrees. It's a 45-degree angle. This here, your x-coordinate and your y-coordinate is the same. You might remember, it's square root of two over two, square root of two over two. But the important thing is whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, Or sine of theta over cosine of theta, they're the same thing, so you're going to get one. Let me just clean that up here just because I'm going to keep reusing the same unit circle. So if I have theta is pi over four, then the tangent of theta is going to be equal to one. Now what if theta is equal to negative pi over four? That is this right over here. Let me just draw a little triangle here. So when x, this x-coordinate over here, is square root of two over two, we know that. We've seen that multiple times, square root of two ... Actually, let me label it a little bit better. So here, our theta is equal to negative pi over four radians. Or if you like to think in degrees, this would be negative 45 degrees. And now, your sine and cosine of this angle The cosine is square root of two over two. The x-coordinate of where this intersects is square root of two over two. The y-coordinate here is negative square root of two, negative square root of two over two. So what's the tangent? Well, it's going to be your sine over your cosine, which is going to be negative one, and you see that. For however much you move in the x direction, you move the opposite of that. You move the negative of that in the y direction. Let me clean this up a little bit because I want to keep reusing my unit circle. So there you go. And so this is going to be negative one. This is going to be negative one. Actually, let's just start plotting a few of these points. If we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y-axis, that's the y-axis, we immediately see tangent of zero is zero."
    },
    {
        "Q": "\nAt 0:33, would the equation y = -2+ 4/3y be equivalent?",
        "A": "yes, except for your typo. It is not y = -2+ 4/3y as you wrote, but y = -2+ 4/3x Typically this would be written as y = 4/3x - 2",
        "video_name": "d3VMo1VWFvc",
        "timestamps": [
            33
        ],
        "3min_transcript": "- [Voiceover] \"Which of the ordered pairs is a \"solution of the following equation?\" 4x minus one is equal to 3y plus five. Now, when we look at an ordered pair we wanna figure out whether it's a solution, we just have to remind ourselves that in these ordered pairs the convention, the standard, is is that the first coordinate is the x coordinate, and the second coordinate is the y coordinate. So they're gonna, if this is a solution, if this ordered pair is a solution, that means that if x is equal to three and y is equal to two, that that would satisfy this equation up here. So let's try that out. So, we have four times x. Well we're saying x needs to be equal to three, minus one, is going to be equal to three times y. Well, if this ordered pair is a solution then y is going to be equal to two, so three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true. Is this really the same thing as three times two which is six, plus five? See, 12 minus one is 11, six plus five is also 11. This is true, 11 equals 11. This pair three, two does satisfy this equation. Now let's see whether this one does, two, three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now gonna see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y, now y we're testing to see if it can be three. Three times three plus five, let's see if this is true. Four times two is eight, minus one, is this equal to three times three? So that's nine plus five. So is seven equal to 14? So these things are not equal to each other. So this is not a solution, when x equals two y cannot be to three and satisfy this equation. So only three, two is a solution."
    },
    {
        "Q": "\nAt 0:45 Sal said that 2y minus -2y would be zero. There was nothing on the other side. Is that possible to do? I thought that there always had to be something on both sides and you couldn't add to an empty side. Could anyone help me understand?",
        "A": "Yes, what Sal did is possible to do. As long as you perform the same operation to both sides, you create an equivalent version of the inequality. FYI... Zero is a valid number. So, the side does have something on that side, a zero.",
        "video_name": "d2cnQ5ahHgE",
        "timestamps": [
            45
        ],
        "3min_transcript": "Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this \"and\" is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side, and we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7. Those cancel out. We just have y is less than 0 minus 7, which is negative 7. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be a 56, so this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4, and we don't have to swap the inequality since we're dividing by a positive number. So it's divide both sides by 4 over here. So we get y is greater than-- what is 56/4, or negative 56/4? So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this \"and\" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7."
    },
    {
        "Q": "\nShouldn't -14 be a closed dot beacause it has a greater than sign? (at 3:56)",
        "A": "No because if it is not a greater than or equal to sign, then the dot should be open.",
        "video_name": "d2cnQ5ahHgE",
        "timestamps": [
            236
        ],
        "3min_transcript": "So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this \"and\" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So not including negative 7, so we'll do an open circle around negative 7, and less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint-- and y has to be greater than negative 14. So you make a circle around negative 14, and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. And we can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here, 8, 9, this is negative 10. So let's try it out. So we'd have 3 times negative 10 plus 7 should So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction, so we're getting less negative. And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works. Now, let's just verify some things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We've got 3 times is 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this condition right over here if we put a 0 over here."
    },
    {
        "Q": "What was that division trick he did around 2:23 ?? How did he divide without a) doing long division and b) a calculator? Could you do that with all numbers?\n",
        "A": "He s solving for y: 4\u00e2\u0080\u00a2y + 8 > -48 4\u00e2\u0080\u00a2y > -48 - 8 4\u00e2\u0080\u00a2y > -56 4\u00e2\u0080\u00a2y/4 > -56/4 y > -14",
        "video_name": "d2cnQ5ahHgE",
        "timestamps": [
            143
        ],
        "3min_transcript": "Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this \"and\" is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side, and we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7. Those cancel out. We just have y is less than 0 minus 7, which is negative 7. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be a 56, so this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4, and we don't have to swap the inequality since we're dividing by a positive number. So it's divide both sides by 4 over here. So we get y is greater than-- what is 56/4, or negative 56/4? So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this \"and\" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7."
    },
    {
        "Q": "\nAt 0:58 Sal talks about \"when x is 0, y is 3 - that's our y intercept\" and then talks about how the slope goes down from there. I've been following everything I can on geometry but I seemed to have missed exactly how these slopes work. Is there another unit I can look at that describes how the whole y = 3 - x thing works?",
        "A": "I would suggest looking up equations of a line and slope-intercept form on the KA search bar.",
        "video_name": "-nufZ41Kg5c",
        "timestamps": [
            58
        ],
        "3min_transcript": "Two of the points that define a certain quadrilateral are 0 comma 9 and 3 comma 4. The quadrilateral is left unchanged by a reflection over the line y is equal to 3 minus x. Draw and classify the quadrilateral. Now, I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point 0 comma 9, that's one of the vertices of the quadrilateral. So 0 comma 9. That's that point right over there. And another one of the vertices is 3 comma 4. That's that right over there. And then they tell us that the quadrilateral is left unchanged by reflection over the line y is equal to 3 minus x. So when x is 0, y is 3-- that's our y-intercept-- and it has a slope of negative 1. You could view this as 3 minus 1x. So the line looks like this. So every time we increase our x by 1, we decrease our y by 1. So the line looks something like this. y is equal to 3 minus x. Try to draw it relatively, pretty carefully. So that's what it looks like. y is equal to 3 minus x. So that's my best attempt at drawing it. y is equal to 3 minus x. So the quadrilateral is left unchanged by reflection over this. So that means if I were to reflect each of these vertices, I would, essentially, end up with one of the other vertices on it, and if those get reflected you're going to end up with one of these so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to 3 minus x. So if we were to try to drop a perpendicular to this line-- notice, we have gone diagonally across one, need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. This is the reflection of this point across that line. Now, let's do the same thing for this blue point. To drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares just like that to get to that point right over there. And now we've defined our quadrilateral. Our quadrilateral looks like this. Both of these lines are perpendicular to that original line, so they're going to have the same slope. So that line is parallel to that line over there. And then we have this line and then we have this line. So what type of quadrilateral is this? Well, I have one pair of parallel sides,"
    },
    {
        "Q": "Isn't the remainder 21 instead of 31 at 7:12?\n",
        "A": "It is 31, 31 + 291 = 322",
        "video_name": "omFelSZvaJc",
        "timestamps": [
            432
        ],
        "3min_transcript": "So it seems that we can get pretty close if we do 291 times 6, so if you do a 1,746 and then add two zeros to it. This is going to be times 6 with two zeros, so this is times 600. Once again, you subtract. And let's say I'm only using the sixes and the threes, because I figured those out ahead of time, so I didn't have to do any extra math. So 2 minus 0 is 2. 5 minus 0 is 5. 9 minus 6 is 3. 0 minus 4-- well, there's a couple of ways you could think about doing this. You could borrow from here. That will become a 6. This becomes a 10. 10 minus 4 is 6. Now this one's lower, so it has to borrow as well. Make this into a 16. 16 minus 7-- and I have multiple videos on how to borrow, if I'm doing that part too fast. But the idea here is to show you a different way of long division. So 16 minus 7 is 9. So now we're at 96,352. is about as close as we can get. So let me put a 873 over here with two zeros. So that would literally be 291 times 3 with two zeros times 300. And so once again, we want to subtract here. 2 minus 0-- you get a 2, a 5, a 0. Make this a 16. Make this an 8. 16 minus 7 is 9. And then we have to get close to 9,052. Once again, that 873, those digits look pretty good, 873. We have to multiply 3 and then 10, so this is going to be times 30 right over here. We subtract again. 2 minus 0 is 2. 5 minus 3 is 2. And then you have 90 minus 87 is 3. I'm doing the subtraction a little fast, just so that we can get the general idea. Then we have to go into 322. And how can we get close to that? Well, actually, 291 is pretty darn close to that. 1 times 291 is 291. 2 minus 1 is 1. 32 minus 29 is 3. So you have a remainder of 31. 291 cannot go into 31 any more, so that's our remainder. But how many times did it actually go into this big, beastly number? This 9,873,952? Well there, we just have to add up all of these right over here. 30 plus 3,000-- we can even do it in our head-- 30 plus 3,000 is 33,000. 33,600, 33,900, 33,931, and we are done, assuming I haven't made some silly mistake. 291 goes into this thing 33,931 times with a remainder of 31."
    },
    {
        "Q": "\nAt 6:55: isn't that shape an octagon instead of a decagon? Isn't Sal supposed to connect the far right purple line with the far right yellow line? (And do the same with the left?)",
        "A": "It s just, if you drew a square with the top first and the bottom next, you wouldn t say you have a 6 sided figure.",
        "video_name": "qG3HnRccrQU",
        "timestamps": [
            415
        ],
        "3min_transcript": "So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles. I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2."
    },
    {
        "Q": "at 4:46 the hexagon looks like cone! : )\n",
        "A": "Hexagons can be in any form as long as it has the right amount of sides and is connected",
        "video_name": "qG3HnRccrQU",
        "timestamps": [
            286
        ],
        "3min_transcript": "of the polygon as a whole. And to see that, clearly, this interior angle is one of the angles of the polygon. This is as well. But when you take the sum of this one and this one, then you're going to get that whole interior angle of the polygon. And when you take the sum of that one and that one, you get that entire one. And then when you take the sum of that one plus that one plus that one, you get that entire interior angle. So if you take the sum of all of the interior angles of all of these triangles, you're actually just finding the sum of all of the interior angles of the polygon. So in this case, you have one, two, three triangles. So three times 180 degrees is equal to what? 300 plus 240 is equal to 540 degrees. Now let's generalize it. And to generalize it, let's realize that just to get our first two triangles, we have to use up four sides. We have to use up all the four sides in this quadrilateral. We had to use up four of the five sides-- One, two, and then three, four. So four sides give you two triangles. And it seems like, maybe, every incremental side you have after that, you can get another triangle out of it. Let's experiment with a hexagon. And I'm just going to try to see how many triangles I get out of it. So one, two, three, four, five, six sides. I get one triangle out of these two sides. One, two sides of the actual hexagon. I can get another triangle out of these two sides of the actual hexagon. And it looks like I can get another triangle out of each of the remaining sides. So one out of that one. And then one out of that one, right over there. So in general, it seems like-- let's say. So let's say that I have s sides. s-sided polygon. And I'll just assume-- we already So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles."
    },
    {
        "Q": "where did the 4 come from in the s-4 @ 8:07?\n",
        "A": "He states that the first 4 sides of the shapes will produce 2 triangles, then after the 4 sides, every side will add an additional triangle, So he adds two in the beginning (my two triangles) then the 5th side would add one (5-4), a 6th side would add two (6-4), etc. The s-4 just shows the first four sides creating the 2 triangles.",
        "video_name": "qG3HnRccrQU",
        "timestamps": [
            487
        ],
        "3min_transcript": "I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2. So the remaining sides I get a triangle each. So the remaining sides are going to be s minus 4. So the number of triangles are going to be 2 plus s minus 4. 2 plus s minus 4 is just s minus 2. So if I have an s-sided polygon, I can get s minus 2 triangles that perfectly cover that polygon and that don't overlap with each other, which tells us that an s-sided polygon, if it has s minus 2 triangles, that the interior angles in it are going to be s minus 2 times 180 degrees. Which is a pretty cool result. So if someone told you that they had a 102-sided polygon-- so s is equal to 102 sides. You can say, OK, the number of interior angles are going to be 102 minus 2. So it's going to be 100 times 180 degrees, which is equal to 180 with two more zeroes behind it."
    },
    {
        "Q": "At about 5:20 Sal says \"Mu of X\", this makes it sound like Mu is some function of X, which may or may not end up being true, he should be saying \"Mu times X\". Sal makes this mistake twice when verbalizing the equation.\n",
        "A": "You are right, it s a slight mistake when verbalizing the equations.",
        "video_name": "jJyRrIZ595c",
        "timestamps": [
            320
        ],
        "3min_transcript": "algebraic stamina, but as long you don't make careless mistakes you'll find it reasonably rewarding, because you'll see where things are coming from. So we get y-- the general solution is y is equal to e to the lambda x, times-- let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x. And let's add the two sine of mu x terms. So plus i-- we could call that c1i-- that's that-- minus c2i times sine of mu x. And we're almost done simplifying. And the last thing we can simplify is-- well you know c1 and c2 are arbitrary constants. I don't know, let's call it-- I'll just call it c3, just to not confuse you by using c1 twice, I'll call this c3. And now this might be a little bit of a stretch for you, but if you think about it, it really makes sense. This is still just a constant, right? Especially if I say, you know what, I'm not restricting the constants to the reals. c could be an imaginary number. So if c is an imaginary number, or some type of complex number, we don't even know whether this is necessarily an imaginary number. So we're not going to make any assumptions about it. Let's just say that this is some other arbitrary constant. Call this c4, and we can worry about it when we're actually given the initial conditions. But what this gives us, if we make that simplification, we actually get a pretty straightforward, general solution to our differential equation, where the characteristic equation has complex roots. And that I'll do it in a new color. That is y is equal to e to the lambda x, times some It could be c a hundred whatever. Some constant times cosine of mu of x, plus some other constant-- and I called it c4, doesn't have to be c4, I just didn't want to confuse it with these-- plus some other constant times the sine of mu of x. So there's really two things I want you to realize. One is, we haven't done anything different. At the end of the day, we still just took the two roots and substituted it back into these equations for r1 and r2. The difference is, we just kept algebraically simplifying it so that we got rid of the i's. There was really nothing new here except for some algebra, and the use of Euler's formula. But when r1 and r2 involved complex numbers, we got to this simplification."
    },
    {
        "Q": "Why \"dv\" at 8:30?\n",
        "A": "because you re summing very small bits of volume. (v for volume)",
        "video_name": "XyiQ2dwJHXE",
        "timestamps": [
            510
        ],
        "3min_transcript": "have positive divergence of our vector field within this region right over here. So we have positive divergence. So you can imagine that it's kind of-- the vector field within the region, it's a source of the vector field, or the vector field is diverging out. That's just the case I drew right over here. And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing, so outward normal vector. So the normal, it's oriented so that the surface-- the normal vector is like that. The other option is that you have an inward-facing normal vector. But we're assuming it's an outward-facing N. Well, then we just extrapolate this to three dimensions. We essentially say the flux across the surface. So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, so multiply that times a little chunk of surface, and then sum it up along the whole surface, so sum it up. So it's going to be a surface integral. So this is flux across the surface. It's going to be equal to-- if we were to sum up the divergence, if we were to sum up across the whole volume, so now if we're summing up things on every little chunk of volume over here in three dimensions, we're going to have to take integrals along each dimension. So it's going to be a triple integral over the region of the divergence of F. So we're going to say, how much is F? What is the divergence at F at each point? And then multiply it times the volume of that little chunk to sense of how much is it totally diverging in that volume. That should be equal to the flux. It's completely analogous to what's here. Here we had a flux across the line. We had essentially a two-dimensional-- or I guess we could say it's a one-dimensional boundary, so flux across the curve. And here we have the flux across a surface. Here we were summing the divergence in the region. Here we're summing it in the volume. But it's the exact same logic. If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux. On the other side, you would have a positive flux, and they would roughly cancel out. And that makes sense, because there would be no diverging going on. If you had a converging vector field, where it's coming in, the flux would be negative, because it's going in the opposite direction of the normal vector. And so the divergence would be negative as well, because essentially the vector field would be converging. So hopefully this gives you an intuition of what the divergence theorem is actually saying something very, very, very, very-- almost common sense or intuitive."
    },
    {
        "Q": "At 6:46 Sal says -45/45 = 1. Is -45/45=-1?\n",
        "A": "Listen more carefully to what he says. He says you divide 45/45 and you get 1 and you are left with -1 ... When you are doing operations, it is easier to think about the signs and numbers separate from each other and then combine them in the answer which is what he did",
        "video_name": "O3jvUZ8wvZs",
        "timestamps": [
            406
        ],
        "3min_transcript": "(typing) We care about how many radians there are per degree. We'll do that same green color. Per degree. How many radians are there per degree? Well, we already know, there's pi radians for every 180 degrees, or there are pi... Let me do that yellow color. There are pi over 180 radians per degree. And so, if we multiply, and this all works out because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? This becomes, let me just rewrite it. 150 times pi. All of that over 180, And so, if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like, 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or 5/6 pi radians, depending how you wanna do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Same exact process. You have negative, and I'll do this one a little quicker. Negative 45 degrees. I'll write down the word. Times, times pi radians, pi radians for every 180 degrees. The degrees cancel out, and you're left So this is equal to negative 45 pi over 180, over 180 radians. How can we simplify this? Well it looks like they're both, at minimum, divisible by nine, nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see... Actually, they're both divisible by 45. What am I doing? If you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times. You're left with negative pi over four radians. This is equal to negative pi over four radians. And we are done."
    },
    {
        "Q": "At 6:20, why does Salman call X and B vectors?\n",
        "A": "ya why does it call it that.... is it just to confuse us all",
        "video_name": "EC2mgUZyzoA",
        "timestamps": [
            380
        ],
        "3min_transcript": "is equal to 7. All I did is I multiplied, I dealt with the first row, first column and said, when I take essentially the dot product of those, and if you don't know what a dot product is, don't worry. We'll explain it at other places. It's essentially what I just did here, the first entry here times the first entry, the second entry here times the second entry, and we add them together, that that must be equal to 7, but when you do that, you essentially construct this first equation. Now when you do it with the second row and this column, you construct the second equation. You get negative 2 times s, negative 2 times s plus 4 times t, 4 times t, is equal to negative 6. Hopefully you appreciate that this contains the same information as that. And there is other ways that I could have done it. For example, you could have, instead of writing it this way, this system is obviously the same thing, and actually, let me just copy and paste it, is the same thing as ... so copy and paste, is the same thing as this system, where I'm really just swapping. So once again, copy and paste. Obviously, I've written the second one first, and I've written the first one second, so this is obviously the same system. If I wanted to construct a matrix equation with this system, I would just swap all of the rows. The first row here would be negative 2, 4. I would swap the rows for the coefficients, but I would still keep the s and ts in the same order, and you could do that. Try to represent this right over here as a matrix equation. You would have the matrix here would be negative 2, 4, 2, negative 5, and this would be negative 6, 7. But now that we have set this up, how do we actually solve something like this? Why do we even do this? To think about this, let's actually think about it in terms of literally a matrix equation. This thing over here is the matrix A. Let's say that this right over here, this is the column vector x. I'll write it as a vector x right over here. You have the column vector x, and then this right over here, you could say that this is equal, and let's call this the column vector b. This is equal to the column vector b. We're essentially saying that A, the matrix A times the column vector x is equal to, is equal to the column vector b. Let me write that again right over here, just to emphasize it. The matrix A times the column vector x is going to be equal to, is equal to the column vector B. This is what they're talking about when they say a matrix equation. Actually, before we even think about computation and computer graphics and all of that, you will see a lot of things like this in physics,"
    },
    {
        "Q": "I don't understand why Khan wrote at 05:34 wrote 3*(7/16) after 55/16, would someone please explain it?\n",
        "A": "Sal just converted the improper fraction 55/16 into the mixed number 3 plus 7/16 . If that process is confusing to you, try looking up the videos and exercise about improper fractions and mixed numbers.",
        "video_name": "pPnxPrhf6Ww",
        "timestamps": [
            334
        ],
        "3min_transcript": "plus 7. And now, how many data points did we have? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 data points. So we're going to divide this by 16. So what is this going to be? This is just 0. Let's see. This is, just right over, that's 0. This is 4. This is 15. This is 12. This is 10. So we have 1 plus 4 is 5 plus 15 is 20 plus 12 is 32 plus 10 is 42. 42 plus 6 is 48, 48. Am I doing ... 42 plus 6 is 48 plus 7, 48 plus 7 is 55. Did I do that right? 1 plus 4 is 5 plus 15 is 20, 42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as 3 and 3 that ... 3 times 16 is 48, so 3 and 7/16. So the mean for the seniors, 3 and 7/16, that's right around ... This is 3, that's 4, so 7/16, it's a little less than a half. It's right around there. So the mean number of fruits is defnitely greater for the freshmen. They have 4 ... Their mean number of fruit eaten per day is 4 versus 3 and 7/16. The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, the mean is fairly sensitive to when you have outliers here. 19 pieces of fruit per day. That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger number, if someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is a middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these really, these points that are really, really high, really, really low. And because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors, or a better measure for the center of distribution for the seniors. Let's fill both of those out. The mean number of fruit is greater for the freshmen, and the mean is a good measure for the center of distribution for the seniors."
    },
    {
        "Q": "\nDid you (Khan) mean to say the median is the middle number at 6:24? It's hard to tell what you meant, because the mode is the same as the median: 3. But it seems odd to call the mode the middle number.",
        "A": "Median is the middle number, and the mode is the most commonly occurring number. (Occurs the most in a data set) The mode can be the same as the median if the middle number is also the most commonly occurring number. Does this clear things up?",
        "video_name": "pPnxPrhf6Ww",
        "timestamps": [
            384
        ],
        "3min_transcript": "plus 7. And now, how many data points did we have? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 data points. So we're going to divide this by 16. So what is this going to be? This is just 0. Let's see. This is, just right over, that's 0. This is 4. This is 15. This is 12. This is 10. So we have 1 plus 4 is 5 plus 15 is 20 plus 12 is 32 plus 10 is 42. 42 plus 6 is 48, 48. Am I doing ... 42 plus 6 is 48 plus 7, 48 plus 7 is 55. Did I do that right? 1 plus 4 is 5 plus 15 is 20, 42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as 3 and 3 that ... 3 times 16 is 48, so 3 and 7/16. So the mean for the seniors, 3 and 7/16, that's right around ... This is 3, that's 4, so 7/16, it's a little less than a half. It's right around there. So the mean number of fruits is defnitely greater for the freshmen. They have 4 ... Their mean number of fruit eaten per day is 4 versus 3 and 7/16. The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, the mean is fairly sensitive to when you have outliers here. 19 pieces of fruit per day. That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger number, if someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is a middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these really, these points that are really, really high, really, really low. And because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors, or a better measure for the center of distribution for the seniors. Let's fill both of those out. The mean number of fruit is greater for the freshmen, and the mean is a good measure for the center of distribution for the seniors."
    },
    {
        "Q": "in 3:20-4:05 in the video, why is 6 and 7 not with a decimal because 2 had a . which is the amount behind it so why not 6 or 7? Is there a reason for it or is it just because the number has only one dot?\n",
        "A": "There isn t any decimal points, those dots are multiplication as a different symbol.",
        "video_name": "pPnxPrhf6Ww",
        "timestamps": [
            200,
            245
        ],
        "3min_transcript": "And then we have two data points at 2, so you write plus 2 times 2. And then, let's see, we have a bunch of data. We have four data points at 3, so we could say we have four 3s. Let me circle that. So we have four 3s, plus 4 times 3. And then we have three 4s, so plus 3 times 4. And then we have a 5, so plus 5, and then we have a 6. Let me do this in a color that you can see. And then we have a 6 right over here, plus 6. How many total points did we have? We had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, oh, actually, be careful. We had 15 points and I didn't put that one in there. So we have 15 points, and I can't forget this one over here, so plus ... my pen is acting a little funny right now, but we'll power through that, plus 19. So what is this going to be? This is just going to be 0. This is going to be 2. This is going to be 4. This is going to be 12. My pen is really acting up. It's almost like it's running out of digital ink or something. This is going to be another 12, and then we have 5, 6, and 19. 2 plus 4 is 6, plus 24 is 30, plus 11 is 41, plus 19 gets us to 60. 60 divided by 15 is 4, so the mean number of fruit per day for the freshmen is 4 pieces of fruit per day. This right over here, that right over there is our mean for the ... Now let's do the same calculation for the seniors. We have one data point where they didn't eat any fruit at all each day, not too healthy. Then you have one 1, so I'll just write that as, we could actually write that as 1 times 1, but I'll just write that as 1. Then we have two 2s, so plus 2 times 2. Then we have one, two , three, four, five 3s, five 3s, so plus 5 times 3. And then we have three 4s, so plus 3 times 4. And then we have two 5s, plus 2 times 5, and then we have a 6. We have a 6, plus 6, and we have a 7, someone eats 7 pieces of fruit each day,"
    },
    {
        "Q": "\nWhy does he have to multiply from 1:04 to 1:36?",
        "A": "at those points, multiple people ate the same number of fruit. so he was calculating each person s contribution to the arithmetic mean.",
        "video_name": "pPnxPrhf6Ww",
        "timestamps": [
            64,
            96
        ],
        "3min_transcript": "Voiceover:Kenny interviewed freshmen and seniors at his high school, asking them how many pieces of fruit they eat each day. The results are shown in the 2 plots below. The first statement that we have to complete is the mean number of fruits is greater for, and actually, let me go down the actual screen, is greater for, we have to pick between freshmen and seniors. Then they said the mean is a good measure for the center of distribution of, and we pick either freshmen or seniors. Let me go back to my scratch pad here, and let's think about this. Let's first think about the first part. Let's just calculate the mean for each of these distributions. I encourage you to pause the video and try to calculate it out on your own. Let's first think about the mean number of fruit for freshmen. Essentially, we're just going to take each of these data points, add them all together, and then divide by the number of data points that we have. We have one data point at 0. We have one data point at 0, so I'll write 0. And then we have two data points at 1, And then we have two data points at 2, so you write plus 2 times 2. And then, let's see, we have a bunch of data. We have four data points at 3, so we could say we have four 3s. Let me circle that. So we have four 3s, plus 4 times 3. And then we have three 4s, so plus 3 times 4. And then we have a 5, so plus 5, and then we have a 6. Let me do this in a color that you can see. And then we have a 6 right over here, plus 6. How many total points did we have? We had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, oh, actually, be careful. We had 15 points and I didn't put that one in there. So we have 15 points, and I can't forget this one over here, so plus ... my pen is acting a little funny right now, but we'll power through that, plus 19. So what is this going to be? This is just going to be 0. This is going to be 2. This is going to be 4. This is going to be 12. My pen is really acting up. It's almost like it's running out of digital ink or something. This is going to be another 12, and then we have 5, 6, and 19. 2 plus 4 is 6, plus 24 is 30, plus 11 is 41, plus 19 gets us to 60. 60 divided by 15 is 4, so the mean number of fruit per day for the freshmen is 4 pieces of fruit per day. This right over here, that right over there is our mean for the ..."
    },
    {
        "Q": "At 1:01 how does he know whether to subtract or add the terms? Would you use the sign in front of or behind the term?\n",
        "A": "Yes you use the sign in front of the term. In the example Sal uses 5x^2 + 2x^2 it results in 7x^2 however later in the question we get a +8x, -7x, +13x using the signs 8x-7x = x, x+13x = 14x hope that helps :)",
        "video_name": "ahdKdxsTj8E",
        "timestamps": [
            61
        ],
        "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done."
    },
    {
        "Q": "At 0:41 he says that he would have to distribute the negative; what is the process/operation for doing that?\n",
        "A": "Multiply the polynomial by -1. This is what he is talking about, I think. :)",
        "video_name": "ahdKdxsTj8E",
        "timestamps": [
            41
        ],
        "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done."
    },
    {
        "Q": "\ndo yo know about the math error starting at about 1:11 in the video?",
        "A": "There is no error at that point in the video. Sal briefly makes an error later on at about 1:28 when he writes 15x, but he quickly corrects it to 14x. Other than that, there are no errors in the video. What do you think is in error? Maybe I can help clarify.",
        "video_name": "ahdKdxsTj8E",
        "timestamps": [
            71
        ],
        "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done."
    },
    {
        "Q": "\nAt 6:42, why did Sal put a negative sign for cosine but not sine, aren't the ratios the same?",
        "A": "In the second and third quadrant, cosine values are always negative. When sine is being taken in the third and fourth quadrant, the sine values will be negative. Does this help?",
        "video_name": "tzQ7arA917E",
        "timestamps": [
            402
        ],
        "3min_transcript": "That's a pretty interesting result. But what about their sines? Well, here, the sine of theta is this distance above the X-axis, and here, the sine of negative theta is the same distance below the X-axis, so they're going to be the negatives of each other. We could say that sine of negative theta, sine of negative theta is equal to, is equal to the negative sine of theta, equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the X-axis, you're going to get the negative value for the sine. We could do the same thing over here. How does this one relate to that? These two are going to have the same sine values. The sine of this, the Y-coordinate, is the same as the sine of that. We see that this must be equal to that. Let's write that down. Now let's think about how do the cosines relate. The same argument, they're going to be the opposites of each other, where the X-coordinates are the same distance but on opposite sides of the origin. We get cosine of theta is equal to the negative of the cosine of ... let me do that in same color. Actually, let me make sure my colors are right. We get cosine of theta is equal to the negative of the cosine of pi minus theta. Now finally, let's think about how this one relates. Here, our cosine value, our X-coordinate is the negative, and our sine value is also the negative. We've flipped over both axes. Let's write that down. which is the same thing as pi plus theta, is equal to the negative of the sine of theta, and we see that this is sine of theta, this is sine of pi plus theta, or sine of theta plus pi, and we get the cosine of theta plus pi. Cosine of theta plus pi is going to be the negative of cosine of theta, is equal to the negative of cosine of theta. Even here, and you could see, You could try to relate this one to that one or that one to that one. You can get all sorts of interesting results. I encourage you to really try to think this through on your own and think about how all of these are related to each other based on essentially symmetries or reflections around the X or Y-axis."
    },
    {
        "Q": "At 2:43, you said, 'this angle right over here is theta'. I don't understand why, shouldn't it be negative theta? Thank you for the video anyway; it was really informative!\n",
        "A": "It s going down, and it s below the x axis, but notice that it s still going counterclockwise, so it is positive. On the unit circle, the angle is not x or y in the xy plane. It s just stuck in there and goes around. It has its own relationships to x and y. It s not graphed in the xy plane like the circle is.",
        "video_name": "tzQ7arA917E",
        "timestamps": [
            163
        ],
        "3min_transcript": "so we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray. Let's flip it over the positive Y-axis. If you flip it over the positive Y-axis, we're going to go from there all the way to right over there then we can draw ourselves a ray. My best attempt at that is right over there. What would be the measure of this angle right over here? What was the measure of that angle in radians? We know if we were to go all the way from the positive X-axis to the negative X-axis, that would be pi radians because that's halfway around the circle. This angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus, Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta."
    },
    {
        "Q": "at 4:35 how can the angle in yellow be equal to cos(pie minus theta),sin( pie minus theta)\nand how can it be equal to sin of theta and cos of theta at 6:09\n",
        "A": "Be careful. Sal isn t saying that the angle equals cos or sine. Sal is labelling the x and y co-ordinates for points where the terminal ray of the angle in question intersects the unit circle. So, for the yellow angle of ( \u00cf\u0080 - theta ), its x co-ordinate is cos ( \u00cf\u0080 - theta) and its x co-ordinate is sin ( \u00cf\u0080 - theta. Likewise for the green angle of theta. Its x co-ordinate is cos (theta) and its y co-ordinate is sin (theta).",
        "video_name": "tzQ7arA917E",
        "timestamps": [
            275,
            369
        ],
        "3min_transcript": "let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta. from the positive X-axis. This is cosine of pi minus theta. And the Y-coordinate is the sine of pi minus theta. Then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say theta plus pi or pi plus theta. Let's write pi plus data and sine of pi plus theta. Now how do these all relate to each other? Notice, over here, out here on the right-hand side, our X-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. That's pretty interesting. Let's write that down. Cosine of theta is equal to ... let me do it in this blue color, That's a pretty interesting result. But what about their sines? Well, here, the sine of theta is this distance above the X-axis, and here, the sine of negative theta is the same distance below the X-axis, so they're going to be the negatives of each other. We could say that sine of negative theta, sine of negative theta is equal to, is equal to the negative sine of theta, equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the X-axis, you're going to get the negative value for the sine. We could do the same thing over here. How does this one relate to that? These two are going to have the same sine values. The sine of this, the Y-coordinate, is the same as the sine of that. We see that this must be equal to that. Let's write that down."
    },
    {
        "Q": "\nAt 0:33 , a constant , '9' ,has been used .Can we use a variable for the value of 'c'\nEx: what will be the value of 'c' when using the quadratic equation to solve the equation\nx^2 - x - 12 = 0",
        "A": "What exactly do you mean by a variable for c? In your example, a=1, b= - 1, and c= - 12. To use the quadratic formula, all of a, b, and c have to be numbers (a and b are coefficients, and c is the constant). c can equal 0 such as x^2 + 2x, but then it is easier to factor x(x+2) so solutions are 0 and -2 than it is to use the quadratic formula. No, none of a, b or c can be a variable.",
        "video_name": "iulx0z1lz8M",
        "timestamps": [
            33
        ],
        "3min_transcript": "Use the quadratic formula to solve the equation, 0 is equal to negative 7q squared plus 2q plus 9. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. 0 is equal to ax squared plus bx plus c. And we generally deal with x's, in this problem we're dealing with q's. But the quadratic formula says, look, if you have a quadratic equation of this form, that the solutions of this equation are going to be x is going to be equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. And this is actually two solutions here, because there's one solution where you take the positive square root and there's another solution where you take the negative root. So it gives you both roots of this. So if we look at the quadratic equation that we need to solve We're dealing with q's, not x's, but this is the same general idea. It could be x's if you like. And if we look at it, negative 7 corresponds to a. That is our a. It's the coefficient on the second degree term. 2 corresponds to b. It is the coefficient on the first degree term. And then 9 corresponds to c. It's the constant. So, let's just apply the quadratic formula. The quadratic formula will tell us that the solutions-- the q's that satisfy this equation-- q will be equal to negative b. b is 2. Plus or minus the square root of b squared, of 2 squared, minus 4 times a times negative 7 times c, which is 9. All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21."
    },
    {
        "Q": "\nAt 4:00, hwy did he put + instead of - ? Was there a certain reason? Also, thanks for the video! :-) Oh, nevermind! XD",
        "A": "with the \u00c2\u00b1 sign, the logical way of doing it is to do the + first since it is on top, then do the - second, but it really does not matter. \u00c2\u00b1 reads as plus or minus and is two separate operations",
        "video_name": "iulx0z1lz8M",
        "timestamps": [
            240
        ],
        "3min_transcript": "All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21. 252. So this becomes 4 plus 252. Remember, you have a negative 7 and you have a minus out front. Those cancel out, that's why we have a positive 252 for that part right there. And then our denominator, 2 times negative 7 is negative 14. Now what does this equal? Well, we have this is equal to negative 2 plus or minus the square root of-- what's 4 plus 252? It's just 256. All of that over negative 14. And what's 256? What's the square root of 256? It's 16. You can try it out for yourself. This is 16 times 16. So the square root of 256 is 16. So we can rewrite this whole thing as being equal to negative 2 plus 16 over negative 14. This is plus 16 over negative 14. Or minus 16 over negative 14. If you think of it as plus or minus, that plus is that plus right there. And if you have that minus, that minus is that minus right there. Now we just have to evaluate these two numbers. Negative 2 plus 16 is 14 divided by negative 14 is negative 1. So q could be equal to negative 1. Or negative 2 minus 16 is negative 18 divided by negative 14 is equal to 18 over 14. The negatives cancel out, which is equal to 9 over 7. So q could be equal to negative 1, or it could be equal to 9 over 7. And you could try these out, substitute these q's back into"
    },
    {
        "Q": "at 2:01 i dont get it\n",
        "A": "Simple division. 42/3 is 14",
        "video_name": "DqeMQHomwAU",
        "timestamps": [
            121
        ],
        "3min_transcript": "Solve for x and check your solution. We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1/3 x is equal to 14. Obviously, x times 1/3 is going to be x/3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent. That's equivalent to multiplying or dividing by 1. These guys cancel out. Remember, if you do it to the left-hand side, you also have to do it to the right-hand side. at the same time, because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12. So it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1/3 is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of-- I guess we call it medium-long division. It's not really long division. 3 into 4. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. 3 goes into 42 14 times. So this right over here simplifies to 14. And it all checks out, so we're done."
    },
    {
        "Q": "From 0:56 to 1:40 you explain the whole process but, wouldn't it be easier if we just multiplied 14 times 3 and done?? Can someone respond to this because doing the whole process seems sorta like unnecessary in a way.\n",
        "A": "The whole point of the video isn t just to intuit the entire process, but to break the process, and, as contrary as it seems, to analyse your intuition. That s why 0:50 to 1:40 exists.",
        "video_name": "DqeMQHomwAU",
        "timestamps": [
            56,
            100
        ],
        "3min_transcript": "Solve for x and check your solution. We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1/3 x is equal to 14. Obviously, x times 1/3 is going to be x/3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent. That's equivalent to multiplying or dividing by 1. These guys cancel out. Remember, if you do it to the left-hand side, you also have to do it to the right-hand side. at the same time, because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12. So it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1/3 is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of-- I guess we call it medium-long division. It's not really long division. 3 into 4. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. 3 goes into 42 14 times. So this right over here simplifies to 14. And it all checks out, so we're done."
    },
    {
        "Q": "\nWhen Sal used the variables A (1:27-1:49) and X (3:39-4:19). Sal says that the two variables are the same. What if (for example) we had 3exp-3 * 2exp-3. Then would we multiply the base and then add the exponents?",
        "A": "The rules for exponents only work if you have a common base. In your expression: 3^(-3) * 2^(-3) you have different bases (the 3 and 2). You can only add exponents if you are multiplying with a common base. Since the exponents match, you can do: (3*2)^(-3) = 6^(-3) This only works because the exponents match. If the exponents were different, you would be required to follow PEMDAS rules and complete the exponents before doing any multiplication. Hope this helps.",
        "video_name": "CZ5ne_mX5_I",
        "timestamps": [
            87,
            109,
            219,
            259
        ],
        "3min_transcript": "- [Narrator] Let's get some practice with our exponent properties, especially when we have integer exponents. So, let's think about what four to the negative three times four to the fifth power is going to be equal to. And I encourage you to pause the video and think about it on your own. Well there's a couple of ways to do this. See look, I'm multiplying two things that have the same base, so this is going to be that base, four. And then I add the exponents. Four to the negative three plus five power which is equal to four to the second power. And that's just a straight forward exponent property, but you can also think about why does that actually make sense. Four to the negative 3 power, that is one over four to the third power, or you could view that as one over four times four times four. And then four to the fifth, that's five fours being multiplied together. So it's times four times four times four times four times four. And so notice, when you multiply this out, and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive,"
    },
    {
        "Q": "\nWhy does the division problem at 2:40 become a multiplication problem when he's finished with it?",
        "A": "12^(-5) can be converted to have a positive exponent if you use its reciprocal. 12^(-5) = 1/12^5 Divide: 12^(-7) / 1/12^5 = 12^(-7) * 12^5/1 Hope this helps.",
        "video_name": "CZ5ne_mX5_I",
        "timestamps": [
            160
        ],
        "3min_transcript": "and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive, And once again, we just have to think about, why does this actually make sense? Well, you could actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of this right over here, you would make exponent positive and then you would get exactly what we were doing in those previous examples with products. And so, let's just do one more with variables for good measure. Let's say I have X to the negative twentieth power divided by X to the fifth power. Well once again, we have the same base and we're taking a quotient. So, this is going to be X to the negative 20 minus five cause we have this one right over here in the denominator. So, this is going to be equal to X to the negative twenty-fifth power."
    },
    {
        "Q": "At 1:18 of the video why did Sal say that he figured out the sum of 35 and another number? He multiplied 7 by 5 to get part of the answer. Isn't the answer to a multiplication problem known as a product?\n",
        "A": "Yes, however........ He multiplied 7 x 5 to get the product. Then 7 x 11 to get that product. Then he added the SUM of those products. 35+77",
        "video_name": "xC-fQ0KEzsM",
        "timestamps": [
            78
        ],
        "3min_transcript": "We're asked to rewrite the expression 7 times open parentheses 5 plus 11 close parentheses as the sum of 35 and another whole number. So really what they're asking us to do is just apply the distributive property. We have 7 times the quantity 5 plus 11. Now this is easy to calculate. You could just say 5 plus 11 is 16 and then 16 times 7 is what? That's 70 plus 42 which would be 112. But that's not what they're asking us to do. They're not saying just calculate this. They're saying express this as a sum of 35 and another whole number. So let's apply the distributive property and let's see if we can get a sum of 35 and another whole number. So 7 times 5 plus 11, that's the same thing as 7 times 5 plus 7 times 11. And you can see with this expression editor right over here, it tells you-- it puts it kind of the nice math formatting for what it looks like for the computer. So if we're distributing the 7 over the 5 and the 11, Well, 7 times 5 is 35. And 7 times 11 is 77. Now, have we done what they're asking us? They said rewrite this expression as the sum of 35 and another whole number. Well, we've done that. We've written it as a sum of 35 and another whole number and we were able to do it using the distributive property. So let's check to make sure that we got the right answer. Yes, we did. Let's do one more of these. Rewrite the expression 12 plus 75 in the following form-- a times 4 plus c where a and c represent whole numbers. Now this might look complex, but they're really asking us to factor out an a-- factoring out an a out of this expression right over here. Seeing how much we can factor out so that one of these two numbers becomes a 4. So let's think about how to do that. If we look at these two numbers, the greatest common divisor Both of them are divisible by 3. So you can write 12. 12 is the same thing as 3 times 4. And 75 is the same thing as 3 times 25. Now what we could do is we could essentially factor out the 3. So this is where you could say we're undistributing the 3. So that's the same thing. 3 times 4 plus 3 times 25, that's the same thing as 3 times 4 plus 25. And it looks like we've actually put it in the form that they want us to put it in, where 3 is a and 4 is right there. And then c is 25. So we've put it in the right form. Let's check our answer. We got it right."
    },
    {
        "Q": "For question 35 can't you use the side ratios for 30:60:90 triangles?\n",
        "A": "yes, you can.",
        "video_name": "BJSk1joCQsM",
        "timestamps": [
            1860
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:55, what does it mean when it says 8 liters per fish?\n",
        "A": "It is saying how many liters there are for one fish. They are trying to find the unit rate.",
        "video_name": "jOZ98FDyl2E",
        "timestamps": [
            55
        ],
        "3min_transcript": "- [Voiceover] Let's get some practice comparing and computing rates. So they tell us the pet store has three fish tanks, each holding a different volume of water and a different number of fish. So Tank A has five fish, and it has 40 liters of water, Tank B, 12 fish, 100 liters of water, and Tank C, 23 fish, and it has 180 liters of water. Order the tanks by volume per fish from least to greatest. So let's think about what volume per fish, and we could think about this as volume divided by fish. Volume per fish. All right, so here for Tank A, it's going to be 40 liters for every five fish. 40 for every five fish, and let's see, 40 over five is eight, so you have eight liters per fish, is the rate at which they have to add water per fish for Tank A. Now, Tank B, you have 100 liters so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up,"
    },
    {
        "Q": "\nDid anyone notice that at 2:54 Sal accidentally put greatest on the middle number?",
        "A": "Yeah, but the video explained for him",
        "video_name": "jOZ98FDyl2E",
        "timestamps": [
            174
        ],
        "3min_transcript": "so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up, C was the least, and B was the most, and we check our answer and we got it right."
    },
    {
        "Q": "\nWhy (at 4:58) is (x-4)\u00c2\u00b2 \u00e2\u0089\u00a5 0?? Then why, when you multiply it by -3 is it \u00e2\u0089\u00a4 0? Am I missing something?",
        "A": "Squares are always positive or equal to 0. Hence, (x-4)\u00c2\u00b2 \u00e2\u0089\u00a5 0. Yes, when you multiply by -3, it becomes \u00e2\u0089\u00a4 0 as +ve * -ve = -ve. But, there is also 0. Anything *0 = 0.",
        "video_name": "IbI-l7mbKO4",
        "timestamps": [
            298
        ],
        "3min_transcript": "And we just have to remind ourselves that if I have x plus a squared, that's going to be x squared plus 2ax plus a squared. So if I want to turn something that looks like this, 2ax, into a perfect square, I just have to take half of this coefficient and square it and add it right over here in order to make it look like that. So I'm going to do that right over here. So if I take half of negative 4, that's negative 2. If I square it, that is going to be positive 4. I have to be very careful here. I can't just willy nilly add a positive 4 here. I have equality here. If they were equal before adding the 4, then they're not going to be equal after adding the 4. So I have to do proper accounting here. I either have to add 4 to both sides or I should be careful. I have to add the same amount to both sides or subtract the same amount again. didn't just add 4 to the right hand side of the equation. Remember, the 4 is getting multiplied by 5. I have added 20 to the right hand side of the equation. So if I want to make this balance out, if I want the equality to still be true, I either have to now add 20 to y or I have to subtract 20 from the right hand side. So I'll do that. I'll subtract 20 from the right hand side. So I added 5 times 4. If you were to distribute this, you'll see that. I could have literally, up here, said hey, I'm adding 20 and I'm subtracting 20. This is the exact same thing that I did over here. If you distribute the 5, it becomes 5x squared minus 20x plus 20 plus 15 minus 20. Exactly what's up here. The whole point of this is that now I can write this in an interesting way. I could write this as y is equal to 5 times x minus 2 squared, and then 15 minus 20 is minus 5. So the whole point of this is now to be able to inspect this. Well, we know that this term right over here is always going to be non-negative. Or we could say it's always going to be greater than or equal to 0. This whole thing is going to hit a minimum value when this term is equal to 0 or when x equals 2. When x equals 2, we're going to hit a minimum value. And when x equals 2, what happens? Well, this whole term is 0 and y is equal to negative 5. The vertex is 2, negative 5."
    },
    {
        "Q": "\nI don't get how he got 16 around 2:10",
        "A": "You have to think about how to achieve a perfect square. In this case, he looked at the coefficient of the second term (-8), divided it by 2 (which equals -4), and then squared it to get 16. You can also double check afterwards to make sure you have correctly calculated a perfect square. Is (x-4)*(x-4) equal to x2 -8x +16? It is, so you re good to move forward with the next steps of simplification.",
        "video_name": "IbI-l7mbKO4",
        "timestamps": [
            130
        ],
        "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here."
    },
    {
        "Q": "\nAt 2:09, Sal completes the square by adding and subtracting 16. Could he have also done this by adding 27 to both sides and finding the perfect square?",
        "A": "Your thinking is correct, but your method is not as good. Since we are looking to get in vertex form, moving the 27 to the other side serves no purpose since we would have to move it back later on. However (this is what you are getting at), leaving the -27 outside of the perfect square is acceptable and how a lot of math thinkers would do it. The answer would end up the same.",
        "video_name": "IbI-l7mbKO4",
        "timestamps": [
            129
        ],
        "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here."
    },
    {
        "Q": "\nAt 1:50 Sal has broken down the expression 2 * 4/3 to 1/3+1/3+1/3+1/3+1/3+1/3+1/3+1/3. I am learning the process of thinking of it in this way for more difficult problems?\n\nI can do the mental math to know that 2 * 4/3 = 8/3 = 2 2/3. What is the importance of the process of breaking it down to 1/3+1/3+1/3+1/3+1/3+1/3+1/3+1/3 that I am not understanding?",
        "A": "yah me too i never thought of it as this way",
        "video_name": "ZlhrXO1-osA",
        "timestamps": [
            110
        ],
        "3min_transcript": "So we have here, it says 2 times 4/3 is equal to 8 times blank. And what I encourage you to do is pause the video right now and try to think about what should go in this blank. So I'm assuming you've given your try. Now, let's think through this. So 2 times 4/3, we can literally view that as the same thing as-- if we rewrite the 4/3, this is the same thing as 2 times-- instead of writing 4/3 like this, I'm literally going to write it as four 1/3's. And I know it sounds like I just said the same thing over again. But I'm literally going to write 1/3 four times-- 1/3 plus 1/3 plus 1/3 plus 1/3. If you call each of these 1/3, you literally have four of them. This is four 1/3's. 2 times 4/3 is the same thing as 2 times, literally, four 1/3's. Now, what would this be? Well, this is going to be equal to-- let me just copy and paste So copy, and then let me paste it. So that's one group of those 1/3's, of those four 1/3's, or one group of one of these four 1/3's. And then, we'll have another one. And then, we'll have another one. And we're going to add them together. That's literally 2 times 4/3. So let's add these together. Now what do we have? Well, we have a bunch of 1/3's. And we need to count them up. We have one, two, three, four, five, six , seven, eight 1/3's. This is literally equal to-- and we could, just to make it clear what I've just done, we could ignore the parentheses and just add up all of these things together. So that might make it a little bit clearer. So let me do that just to make it clear that I literally take-- I've taken eight 1/3's and I'm adding them together, which is the exact same thing as 8/3. So let me clear that, and let me clear that, let me clear that. clearly, equal to 8 times 1/3. I have 8 1/3's there. So going back to the original question, what is this equal to? 2 times 4/3 is the same thing as 8 times 1/3. And we've already seen that 8 times 1/3, well, that's literally 8/3. So we could also write it like this-- 8 over 3. Let me do that 3 in that other color-- 8 over 3."
    },
    {
        "Q": "there is a extra decimal at 0:01 the question has a extra decimal\n",
        "A": "no,actually that is just the period at the end of the sentence . :)",
        "video_name": "Eq4mVCd-yyo",
        "timestamps": [
            1
        ],
        "3min_transcript": "We need to calculate 9.005 minus 3.6, or we could view it as 9 and 5 thousandths minus 3 and 6 tenths. Whenever you do a subtracting decimals problem, the most important thing, and this is true when you're adding decimals as well, is you have to line up the decimals. So this is 9.005 minus 3.6. So we've lined up the decimals, and now we're ready to subtract. Now we can subtract. So we start up here. We have 5 minus nothing. You can imagine this 3.6, or this 3 and 6 tenths, we could add two zeroes right here, and it would be the same thing as 3 and 600 thousandths, which is the same thing as 6 tenths. And when you look at it that way, you'd say, OK, 5 minus 0 is nothing, and you just write a 5 right there. Or you could have said, if there's nothing there, it would have been 5 minus nothing is 5. Then you have 0 minus 0, which is just 0. And you can't subtract 6 from 0. So we need to get something into this space right here, and what we essentially are going to do is regroup. We're going to take one 1 from the 9, so let's do that. So let's take one 1 from the 9, so it becomes an 8. And we need to do something with that one 1. We're going to put it in the tenths place. Now remember, one whole is equal to 10 tenths. This is the tenths place. So then this will become 10. Sometimes it's taught that you're borrowing the 1, but you're really taking it, and you're actually taking 10 from the place to your left. So one whole is 10 tenths, we're in the tenths place. So you have 10 minus 6. Let me switch colors. 10 minus 6 is 4. You have your decimal right there, and then you have 8 minus 3 is 5. So 9.005 minus 3.6 is 5.405."
    },
    {
        "Q": "At 3:15 - How do you get y=0? Did you divide or multiply 5/2 ? Why?\n",
        "A": "He got y=0 by substituting x=4 into our function rule. y=10-(5/2)x x=4. y=10-(5/2)(4) We multiply 4 by 5/2 because that s how our rule works. (5/2)(4)=10. y=10-10 10-10=0. y=0 I hope this explains what Sal did and why!",
        "video_name": "86NwKBcOlow",
        "timestamps": [
            195
        ],
        "3min_transcript": "is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white."
    },
    {
        "Q": "Why did he change the 4 to a 2 3:40\n",
        "A": "I don t entirely understand your question, but at 3:40, he was simplifying his multiplication",
        "video_name": "86NwKBcOlow",
        "timestamps": [
            220
        ],
        "3min_transcript": "So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white. And I could keep going in both directions. So there you have it. That is the graph of our linear equation. Let me make my line a little bit bolder, just in case you found that first line hard to read. So let me make it a little bit bolder. And I think you get the general idea."
    },
    {
        "Q": "\nlike at 3:16, those x in the table could be replace with numbers -2,-1, 0, 1, 2 in this kind of order can we solve this given equation? Our teacher said that we can only use this numbers to have a uniform answers in our problem.",
        "A": "Linear equations have an infinite number of possible values for X. You can use any value of X to calculate Y and create a point on the line. Your teacher likely wants everyone finding the same points so there is consistency in the way the class is approaching the problem.",
        "video_name": "86NwKBcOlow",
        "timestamps": [
            196
        ],
        "3min_transcript": "is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white."
    },
    {
        "Q": "\nAt 1:08 could Sal have simplified y = 10 - 5/2x even more?\n\ny = 5- 5x (cross reducing the numerator 10 and the denominator 2)",
        "A": "Cross cancelling only works when the fractions are being multiplied. In this case, the problem is subtracting not multiplication. So, you can t do it.",
        "video_name": "86NwKBcOlow",
        "timestamps": [
            68
        ],
        "3min_transcript": "Create a graph of the linear equation 5x plus 2y is equal to 20. So the line is essentially the set of all coordinate, all x's and y's, that satisfy this relationship right over here. To make things simpler, what we're going to do is set up a table where we're going to put a bunch of x values in and then figure out the corresponding y value based on this relationship. But to make it a little bit simpler, I'm going to solve for y here. So it becomes easier to solve for y for any given x. So we have 5x plus 2y is equal to 20. If we want to solve for y, let's just get rid of the 5x on the left-hand side. So let's subtract 5x from both sides of this equation. The left-hand side, these guys cancel out, so we get 2y is equal to the right hand side, you have 20 minus 5x. And then you can divide both sides of this equation by 2. So you divide both sides by 2. The left-hand side, we just have a y, and then the right-hand side, we could leave it that way. That actually would be a pretty straightforward way is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points."
    },
    {
        "Q": "\nat 1:17 he takes 8 away from 9 and calls it 10/10 what dos that mean?",
        "A": "10/10 is equal to 1 because 10 10 s is",
        "video_name": "lDXaju6JoQ0",
        "timestamps": [
            77
        ],
        "3min_transcript": "Let's try to calculate 39.1 minus 0.794, and so pause the video and try this on your own. All right, I'm assuming you've given a go at it, so now let's work through it together. So I'm going to rewrite this. It's 39.1 minus-- I'm going to line up the decimals so that I have the right place values below the right place values-- minus-- this 0 is in the ones place, so I'll put it in the ones place-- 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here, and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this one tenth and try to regroup it into the hundredths place So let's think about this. If we make this-- actually that's not actually going to solve our problem. Well we could do it, but then we're going to have zero tenths, and we're still going to have a problem here. So actually let me go to the ones place. So let me get rid of a ones, so that's eight ones, which is going to be 10 tenths. So that's going to now-- we're going to have 11 tenths. The 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths, and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths-- so now we have nine hundredths-- and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10-- let me do this in yellow-- 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8. And then we have 3 minus nothing is 3. So we're done, 38.306."
    },
    {
        "Q": "\nin 0:40... he said nothingness... what does nothingness mean??",
        "A": "Basically nothingness is the equivalent of the word nothing . Hope this helps!",
        "video_name": "lDXaju6JoQ0",
        "timestamps": [
            40
        ],
        "3min_transcript": "Let's try to calculate 39.1 minus 0.794, and so pause the video and try this on your own. All right, I'm assuming you've given a go at it, so now let's work through it together. So I'm going to rewrite this. It's 39.1 minus-- I'm going to line up the decimals so that I have the right place values below the right place values-- minus-- this 0 is in the ones place, so I'll put it in the ones place-- 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here, and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this one tenth and try to regroup it into the hundredths place So let's think about this. If we make this-- actually that's not actually going to solve our problem. Well we could do it, but then we're going to have zero tenths, and we're still going to have a problem here. So actually let me go to the ones place. So let me get rid of a ones, so that's eight ones, which is going to be 10 tenths. So that's going to now-- we're going to have 11 tenths. The 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths, and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths-- so now we have nine hundredths-- and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10-- let me do this in yellow-- 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8. And then we have 3 minus nothing is 3. So we're done, 38.306."
    },
    {
        "Q": "\nAt 1:02 it says meth why would she say that?",
        "A": "She said math, not meth. --Blue Leaf",
        "video_name": "WkmPDOq2WfA",
        "timestamps": [
            62
        ],
        "3min_transcript": "So in my sphere flakes video, I joked about folding and cutting space time, but then I thought, hey, why not? So, how do you do that? Well, when we wanted to fold and cut only space, we chose a medium that takes place in space. That is, static paper cut outs, or sphere sculpture. But to fold and cut time, we need a medium that happens over time. I choose music. Music has two easily recognizable dimensions. One is time, and the other is pitch space. Not quite the same as space-space, but it's one dimensional, which makes things easier. But let's not be confused with the notation. There's a few things to notice about written music. Firstly, that it's not music. You can't listen to this. Or, well, you can, but it'll be like-- [PAPER RUSTLING] It's not music, it's music notation. And you can only interpret it into the beautiful music it represents. Kind of like how a book is squiggles on a page that your brain interprets into a meaningful story. And maybe you don't understand it at all, or understand just a literal surface meaning of the action. Or maybe you can read deep and critically into a story that's simple on the surface, and get more from it than even the author intended. Secondly, written music represents a two-dimensional space of pitch and time, but only represents it. Like, there's the suggestion that this is the time axis, but it's not. This is exactly the same as this. Even though you're changing the values on the x-axis. At least, in standard music notation. Some more modern composers do make use of spatial notation, just like some poets do intentionally stretch out words or play with formatting. But in standard notation, a stretched out word only means your text editor is terrible at justified margins, and has nothing to do with the word itself. Pitch also doesn't entirely depend on the notes placed on the y-axis. So I'm going to use something a little closer to reality. I've got this music box that plays a paper tape. As you put the strip through the box, it plays the punched holes. Hold on, it's too quiet. This is why music boxes usually come attached to wooden boxes. I don't have a wooden box, maybe this nice wooden bowl. Um. Music bowl. Anyway, here distance along the strip does translate directly to time, assuming constant crank speed. [MUSIC PLAYING] The box also has a set C major scale on one staff, so pitch space is represented pretty directly. You don't have to worry about sharps and flats, or the space between staves, or the infinite possibilities on a continuous logarithmic frequency scale. The box magically ignores all notational elements, except for where and when the holes are. Each hole, each note, is a point on this strip of space time. So now, let's fold and cut it. It's easy to fold time so that it goes both forwards and backwards simultaneously. Then we can punch in some notes, and unfold it, into a symmetric melody that goes first forwards, and then backwards. Or first backwards and then forwards. Point is, it's reversible. It sounds the same whether I play it like this-- [MUSIC PLAYING]"
    },
    {
        "Q": "At 1:23 how come 6 times 1/6 cancels?\n",
        "A": "6 multiplied by 1/6 is the same thing as 6 divided by 6, which is equal to 1. You will learn how to do this in later classes. I hope this helps.",
        "video_name": "CJyVct57-9s",
        "timestamps": [
            83
        ],
        "3min_transcript": "Solve for x. And we have x minus 8 is equal to x/3 plus 1/6. Now the first thing I want to do here-- and there's multiple ways to do this problem-- but what I want to do is just to simplify the fraction. I'm going to multiply everything times the least common multiple of all of these guys' denominators. This is essentially x/1. This is 8/1, x/3, 1/6. The least common multiple of 1, 3, and 6 is 6. So if I multiply everything times 6, what that's going to do is going to clear out these fractions. So these weren't fractions to begin with, so we're just multiplying them by 6. So it becomes 6x minus 6 times negative 8, or 6 times 8 is 48. And we're subtracting it right over there. And then we have x/3 times 6. Let me just write it out here. So that's going to be 6 times x/3 plus 6 times 1/6. Or we get 6x minus 48 is equal to 6 times something divided That's the same thing as 6 divided by 3 times that something. That's just going to be equal to 2x plus 6 times 1/6 or 6 divided by 6 is just going to be 1. So that first step cleared out all of the fractions and now this is just a straightforward problem with all integer coefficients or integers on either side of the equation. And what we want to do is we want to isolate all of the x's on one side or the other. And we might as well isolate them all on the left hand side. So let's subtract 2x from both sides. We want to get rid of this 2x here. That's why I'm subtracting the 2x. So let's subtract 2x from both sides. And on the right hand side, I have 2x plus 1 minus 2x. Those cancel out. That was the whole point. So I'm left with just this 1 over here. On the left hand side I have 6x minus 2x. Well, that's just going to be 4x. If I have 6 of something minus 2 of that something, I have 4 of that something. Minus 48. And now I can-- let's see, I want to get rid of this 48 So let me add 48 to both sides of the equation. I'll do this in a new color. So let me add 48 to both sides of this equation. And on the left hand side 4x minus 48 plus 48, I'm left with just a 4x. And on the right hand side, 1 plus 48 is going to be 49. And now I've isolated the x but it's still multiplied by a 4. So to make that a 1 coefficient, let's multiply both sides by 1/4. Or you could also say, let's divide both sides by 4. Anything you do to one side you have to do to the other. And so you have-- what do we have over here? 4x/4 is just x. x is equal to 49 over 4. And that's about as far as we can simplify it because these don't have any common factors, 49 and 4. Let's check to see whether 49/4 is indeed the answer. So let's put it into the original equation. Remember, the original equation is"
    },
    {
        "Q": "At 2:51, he says that there's no way to simplify that. Why doesn't he use mixed numbers like 12 1/4?\n",
        "A": "In algebra, it is easier to use improper fractions.",
        "video_name": "CJyVct57-9s",
        "timestamps": [
            171
        ],
        "3min_transcript": "That's the same thing as 6 divided by 3 times that something. That's just going to be equal to 2x plus 6 times 1/6 or 6 divided by 6 is just going to be 1. So that first step cleared out all of the fractions and now this is just a straightforward problem with all integer coefficients or integers on either side of the equation. And what we want to do is we want to isolate all of the x's on one side or the other. And we might as well isolate them all on the left hand side. So let's subtract 2x from both sides. We want to get rid of this 2x here. That's why I'm subtracting the 2x. So let's subtract 2x from both sides. And on the right hand side, I have 2x plus 1 minus 2x. Those cancel out. That was the whole point. So I'm left with just this 1 over here. On the left hand side I have 6x minus 2x. Well, that's just going to be 4x. If I have 6 of something minus 2 of that something, I have 4 of that something. Minus 48. And now I can-- let's see, I want to get rid of this 48 So let me add 48 to both sides of the equation. I'll do this in a new color. So let me add 48 to both sides of this equation. And on the left hand side 4x minus 48 plus 48, I'm left with just a 4x. And on the right hand side, 1 plus 48 is going to be 49. And now I've isolated the x but it's still multiplied by a 4. So to make that a 1 coefficient, let's multiply both sides by 1/4. Or you could also say, let's divide both sides by 4. Anything you do to one side you have to do to the other. And so you have-- what do we have over here? 4x/4 is just x. x is equal to 49 over 4. And that's about as far as we can simplify it because these don't have any common factors, 49 and 4. Let's check to see whether 49/4 is indeed the answer. So let's put it into the original equation. Remember, the original equation is But in theory, we should be able to put it into any of these steps and the x should satisfy. But let's do it in our original equation. So we have x minus 8. So we have 49/4 minus 8 should be equal to 49/4 over 3 plus 1/6. So let's see what we can do here. So we can multiply. Well, like we did before. We can multiply both sides of this equation by 6. That'll help simplify a lot of the fractions here. So if we multiply both sides of this equation by 6-- so we're going to multiply everything by 6-- what do we get on the left hand side? 6/4 is the same thing as 3/2, right? So this is going to be 3 times 49 over 2. 3 times 49 over 2 minus 48 will be equal to 6 divided by 3"
    },
    {
        "Q": "\nHes making no sense when he wrote down the exponent why did he bring down the numbers? can u just right down (2x2x2)x(2x2x2x2x2)? then solve? but do you HAVE to put parenthese?at 0:37",
        "A": "You can do it that way but that was just a simplified example. It may be as complex as 2^67 x 2^89. Exponents make it easier. The parenthesis is needed if the number is negatice or if you need to find the opposite of the answer ex: -(x)^2 or if multiplying with another exponent",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            37
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:58, why is it 2^8 and not 2^15?\n",
        "A": "Because you add 3+5 and not multiply 3x5.",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            118
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 7:16 why does this work",
        "A": "Because it is the rule ofcancellation. x^200/x^50 if you wrote all these x s out and then cancelled them out, you d be able the cancel off 50 from each from before being left with no more x s on the bottom. So you d be left with x^150 as 50x s from the top would leave 150 x s.",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            436
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I don't understand at 8:08 how the 7^-5 can change into 7^5...?\nThanks!\n",
        "A": "when you have a negative exponent in the denominator of the fraction you have to move the exponent and its base to the numerator and the exponent is now positive. If that doesn t help then try to watch a video on negative exponents. its complicated",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            488
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:11 he says 2^9/2^10 is the same as 2^9*1/2^10 . Could someone explain how this works?\n",
        "A": "Anything times one is going to be that number. 9*1 = 9. 54*1 = 54. That means that 2^9*1 = 2^9, so you might as well not have the *1 there.",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            371
        ],
        "3min_transcript": ""
    },
    {
        "Q": "How in the world does he do that at 6:21 ?!!!?\n",
        "A": "One of the earlier videos explains it, but i cant remember which one... 1/2^10 is the same as 2 to the 10th root, which can be written as 2^-10",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            381
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 1:22 he says \"the easiest way to split this whole into tenths is to take each of those fifths and turn them into 2 tenths.\" What's the other way to do it? (Even if it's not the easiest.)",
        "A": "Not really, because the only way to turn fifths to tenths is to multiply 5 by 2. Hope this helps!",
        "video_name": "XHLgY7Z3cb8",
        "timestamps": [
            82
        ],
        "3min_transcript": "Let's see if we can write 3/5 as a decimal. And I encourage you to pause this video and think about if you can do it on your own. And I'll give you a hint here. Can we rewrite this fraction so, instead of it being in terms of fifths, it can be in terms of tenths? So I'm assuming you've given a go at it. Let's try to rewrite this as a fraction with 10 as the denominator. But let's just first visualize this. So we have fifths. So let's say that's 1/5. Actually, let me just copy and paste this. That is 2/5. That is 3/5, and that is 4/5. And that is 5/5, or this would be a whole now. So that is our whole. And we want to color in 3 of those 5, so we want to think about what 3/5 are. So let me get my magenta out. I can actually make this bigger even-- 2/5 and 3/5. There you go. Color that in. That is 3/5. Now, how could I write this in terms of tenths-- instead of 3/5, a certain number of tenths? Well, let's split this whole into tenths. And the easiest way to split this whole into tenths is to take each of those fifths and turn them into 2/10. So let's do that. So If we were to do this right over here, we now have twice as many sections. So another way of thinking about it, we are multiplying the number of sections by 2. We now have 10 sections. Each of these is a tenth. And the 3 of those sections are now going to be twice as many. What we have in magenta, we now have twice as many sections in magenta. Notice we just multiplied the numerator and the denominator by 2. But hopefully it makes conceptual sense. Every piece, when we're talking about fifths, we've now doubled so that instead of every 1/5 is now 2/10. You have a 1/10 now and a 1/10 now. And we could just keep writing 1/10 if we like. Each of these things right over here are a tenth. And then each of the 3 are now twice as many tenths. So the 3/5 is now 6/10. So let's write that down. So this is going to be equal to 6/10. Now why is this interesting? You can literally view this as 6/10-- let me write it this way-- 6 times 1/10."
    },
    {
        "Q": "At the end of the video: 5:25 Sal says \"converges to zero\"\n\nWhat makes this epsilon proof valid?\n\nWould not we be forced to count all the possibilities to prove this to be right??\nAnd as there are infinitely many cases for epsilon it seems to be impossible.\n\nWhat is the axiom behind this limit proof or\nin what is this epsilon limit proof based on??\nCommon sense??\n",
        "A": "Yeah, but as we know for example that infinity +1 = infinity, one could ask then that if epsilon is infinitesimally small, would not then |an-L|= roughly epsilon and not strictly smaller than (<epsilon) as the formula goes. So in what mathematical logic/axiom/principle is this statement based on?: For ANY epsilon>0, there IS a positive M such that if n>M then |a (sub)n-L|<epsilon",
        "video_name": "-XRQovTI04Q",
        "timestamps": [
            325
        ],
        "3min_transcript": "because this negative 1 to the n plus 1, this numerator just swaps us between a negative and a positive version of 1 over n. But if you take the absolute value of it, this is always just going to be positive. So this is the same thing as 1 over n, as the absolute value of 1 over n has to be less than epsilon. Now, n is always going to be positive. n starts at 1 and goes to infinity. So this value is always going to be positive. So this is saying the same thing that 1 over n has to be less than epsilon in order for this stuff to be true. And now we can take the reciprocal of both sides. And if you take the reciprocal of both sides of an inequality, you would have that n-- if you take the reciprocal of both sides of an inequality, you swap the inequality. So for this to be true, n has to be greater than 1 over epsilon. And we essentially have proven it now. So now we've said, look, for this particular sequence, to set M to be 1 over epsilon. Because if n is greater than M, which is 1 over epsilon, then we know that this right over here is going to be true. That is going to be true. So the limit does definitely exist. And so over here, for this particular epsilon, it looks like we've picked 0.5 or 1/2 as our epsilon. So as long as n is greater than 1 over 1/2, which is 2, so in this case we could say, look, you gave me 1/2. My M is going to be a function of epsilon. It's going to be defined for any epsilon you give me greater than 0. So here, 1 over 1/2 is right over here. I'm going to make my M right over here. And you see it is indeed the case that my sequence is within the range So for n is equal to 3 it's in the range. For n is equal to 4 it's in the range. For n equals 5-- and it keeps going and going. And we're not just taking our word for it. We've proven it right over here. So we've made the proof. You give me any other any epsilon, I said M is equal to 1 over that thing. And so for n greater than that, this is going to be true. So this is definitely the case. This sequence converges to 0."
    },
    {
        "Q": "\nThe ratio of the area to circumference of a circle is 30:1. How do you find the circumference of the circle?",
        "A": "You would do PI times the diameter of the circle.",
        "video_name": "Rcb7ZUTOQ1I",
        "timestamps": [
            1801
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:05, he said altitude. What is that?\n",
        "A": "An altitude is a line that runs from the vertex of a triangle to the opposite side intersecting at a right angle.",
        "video_name": "UKQ65tiIQ6o",
        "timestamps": [
            65
        ],
        "3min_transcript": "Let's say that this triangle right over here is equilateral, which means all of its sides have the same length. And let's say that that length is s. What I want to do in this video is come up with a way of figuring out the area of this equilateral triangle, as a function of s. And to do that, I'm just going to split this equilateral in two. I'm just going to drop an altitude from this top vertex right over here. This is going to be perpendicular to the base. And it's also going to bisect this top angle. So this angle is going to be equal to that angle. And we showed all of this in the video where we proved the relationships between the sides of a 30-60-90 triangle. Well, in a regular equilateral triangle, all of the angles are 60 degrees. So this one right over here is going to be 60 degrees, let me do that in a different color. This one down here is going to be 60 degrees. This one down here is going to be 60 degrees. And then this one up here is 60 degrees, but we just split it in two. So this angle is going to be 30 degrees. And then the other thing that we know is that this altitude right over here also will bisect this side down here. So that this length is equal to that length. And we showed all of this a little bit more rigorously on that 30-60-90 triangle video. But what this tells us is well, if this entire length was s, because all three sides are going to be s, it's an equilateral triangle, then each of these, so this part right over here, is going to be s/2. And if this length is s/2, we can use what we know about 30-60-90 triangles to figure out this side right over here. So to figure out what the actual altitude is. And the reason why I care about the altitude is because the area of a triangle is 1/2 times the base times the height, or times the altitude. So this is s/2, the shortest side. The side opposite the 30 degree angle is s/2. Then the side opposite the 60 degree angle is going to be square root of 3 times that. And we know that because the ratio of the sides of a 30-60-90 triangle, if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is going to be square root of 3 times that. And the side opposite the 90 degree side, or the hypotenuse, is going to be 2 times that. So it's 1 to square root of 3 to 2. So this is the shortest side right over here. That's the side opposite the 30 degree side. The side opposite the 60 degree side is going to be square root of 3 times this. So square root of 3 s over 2. So now we just need to figure out what the area of this triangle is, using area of our triangle is equal to 1/2 times the base, times the height of the triangle. Well, what is the base of the triangle? Well, the entire base of the triangle right over here is s. So that is going to be s. And what is the height of the triangle?"
    },
    {
        "Q": "\nAt 1:15 Sal mentions that this material was previously covered in greater depth in the 30,60,90 video yet there is no video listed in Geometry by that name any clues to what the actual name might be?",
        "A": "There are a number of videos concerning 30-60-90 triangles. try searching for geometry>right triangles>special right triangles.",
        "video_name": "UKQ65tiIQ6o",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's say that this triangle right over here is equilateral, which means all of its sides have the same length. And let's say that that length is s. What I want to do in this video is come up with a way of figuring out the area of this equilateral triangle, as a function of s. And to do that, I'm just going to split this equilateral in two. I'm just going to drop an altitude from this top vertex right over here. This is going to be perpendicular to the base. And it's also going to bisect this top angle. So this angle is going to be equal to that angle. And we showed all of this in the video where we proved the relationships between the sides of a 30-60-90 triangle. Well, in a regular equilateral triangle, all of the angles are 60 degrees. So this one right over here is going to be 60 degrees, let me do that in a different color. This one down here is going to be 60 degrees. This one down here is going to be 60 degrees. And then this one up here is 60 degrees, but we just split it in two. So this angle is going to be 30 degrees. And then the other thing that we know is that this altitude right over here also will bisect this side down here. So that this length is equal to that length. And we showed all of this a little bit more rigorously on that 30-60-90 triangle video. But what this tells us is well, if this entire length was s, because all three sides are going to be s, it's an equilateral triangle, then each of these, so this part right over here, is going to be s/2. And if this length is s/2, we can use what we know about 30-60-90 triangles to figure out this side right over here. So to figure out what the actual altitude is. And the reason why I care about the altitude is because the area of a triangle is 1/2 times the base times the height, or times the altitude. So this is s/2, the shortest side. The side opposite the 30 degree angle is s/2. Then the side opposite the 60 degree angle is going to be square root of 3 times that. And we know that because the ratio of the sides of a 30-60-90 triangle, if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is going to be square root of 3 times that. And the side opposite the 90 degree side, or the hypotenuse, is going to be 2 times that. So it's 1 to square root of 3 to 2. So this is the shortest side right over here. That's the side opposite the 30 degree side. The side opposite the 60 degree side is going to be square root of 3 times this. So square root of 3 s over 2. So now we just need to figure out what the area of this triangle is, using area of our triangle is equal to 1/2 times the base, times the height of the triangle. Well, what is the base of the triangle? Well, the entire base of the triangle right over here is s. So that is going to be s. And what is the height of the triangle?"
    },
    {
        "Q": "At 5:48 , how does Sal know that the intersecting point is exactly at pi/4? You cannot just assume that since the intersecting point is in between two points, that it is located in the middle. Can someone please help clarify this for me. Thank you.\n",
        "A": "You can get an exact result: sin(x) = cos(x) sin^2(x) + cos^2(x) = 1 sin^2(x) = 1 - cos^2(x) sin(x) = sqrt(1 - cos^2(x)) sqrt(1 - cos^2(x)) = cos(x) sqrt(1 - cos^2(x))^2 = cos^2(x) 1 - cos^2(x) = cos^2(x) 1 = 2\u00e2\u0080\u00a2cos^2(x) 1/2 = cos^2(x) sqrt(1/2) = cos(x) sqrt(1)/sqrt(2) = cos(x) sqrt(2)/2 = cos(x) arccos(sqrt(2)/2) = x pi/4 = x",
        "video_name": "fp9DZYmiSC4",
        "timestamps": [
            348
        ],
        "3min_transcript": "The look of these curves should look somewhat familiar at this point. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to 0, sine theta is 0. When theta is pi over 2, sine of theta is 1. When theta is equal to pi, sine of theta is 0. When theta is equal to 3 pi over 2, sine of theta is negative 1. When theta is equal to 2 pi, sine of theta is equal to 0. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it. So just visually, we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this 2 pi range for theta, you get two points of intersection. Now let's think about what they are, because they look to be pretty close between 0 and pi over 2. And right between pi and 3 pi over 2. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over 4. So let's verify that. So let's think about what these values are at pi over 4. So pi over 4 is that angle, or that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same thing as a 45 degree angle. So we have to figure out what this point is what. What the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here, to make it a little clear. This is a typical type of right triangle. So it's good to get some familiarity with it. So let me draw my best attempt. Alright. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse? Well this is a unit circle. It has radius 1. So the length of the hypotenuse here is 1. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees, because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean Theorem"
    },
    {
        "Q": "At 5:17 Sal found that the graphs of y=sin(theta) and y=cos(theta) intersect at two points in the given interval. So couldn\u00e2\u0080\u0099t the answer to the question asked (which is 2) be given then and there without doing anything else?\n",
        "A": "yes, the answer could have been found easily but he is trying to explain why it is",
        "video_name": "fp9DZYmiSC4",
        "timestamps": [
            317
        ],
        "3min_transcript": "and think about where they might intersect. So first let's do cosine of theta. When theta is 0-- and let me mark this off. So this is going to be when y is equal to 1. And this is when y is equal to negative 1. So y equals cosine of theta. theta equals 0. Cosine of theta equals 1. So cosine of theta is equal to 1. When theta is equal to pi 2, cosine of theta is 0. When theta is equal to pi, cosine of theta is negative 1. When theta is equal to 3 pi over 2, cosine of theta is equal to 0. That's this right over here. And then finally when theta is 2 pi, cosine of theta is 1 again. And the curve will look something like this. My best attempt to draw it. Make it a nice smooth curve. The look of these curves should look somewhat familiar at this point. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to 0, sine theta is 0. When theta is pi over 2, sine of theta is 1. When theta is equal to pi, sine of theta is 0. When theta is equal to 3 pi over 2, sine of theta is negative 1. When theta is equal to 2 pi, sine of theta is equal to 0. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it. So just visually, we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this 2 pi range for theta, you get two points of intersection. Now let's think about what they are, because they look to be pretty close between 0 and pi over 2. And right between pi and 3 pi over 2. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over 4. So let's verify that. So let's think about what these values are at pi over 4. So pi over 4 is that angle, or that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same thing as a 45 degree angle."
    },
    {
        "Q": "\nSo basically in the whole row I can make any changes which would be equivalent in standard equation? (At 1:25 is mentioned the addition)",
        "A": "Yes, it s like solving a system of linear equations using the elimination method. Same operations.",
        "video_name": "obts_JDS6_Q",
        "timestamps": [
            85
        ],
        "3min_transcript": "I will now show you my preferred way of finding an inverse of a 3 by 3 matrix. And I actually think it's a lot more fun. And you're less likely to make careless mistakes. But if I remember correctly from Algebra 2, they didn't teach it this way in Algebra 2. And that's why I taught the other way initially. But let's go through this. And in a future video, I will teach you why it works. Because that's always important. But in linear algebra, this is one of the few subjects where I think it's very important learn how to do the operations first. And then later, we'll learn the why. Because the how is very mechanical. And it really just involves some basic arithmetic for the most part. But the why tends to be quite deep. So I'll leave that to later videos. And you can often think about the depth of things when you have confidence that you at least understand the hows. So anyway, let's go back to our original matrix. And what was that original matrix that I did in the last video? It was 1, 0, 1, 0, 2, 1, 1, 1, 1. So this is what we're going to do. It's called Gauss-Jordan elimination, to find the inverse of the matrix. And the way you do it-- and it might seem a little bit like magic, it might seem a little bit like voodoo, but I think you'll see in future videos that it makes a lot of sense. What we do is we augment this matrix. What does augment mean? It means we just add something to it. So I draw a dividing line. Some people don't. So if I put a dividing line here. And what do I put on the other side of the dividing line? I put the identity matrix of the same size. This is 3 by 3, so I put a 3 by 3 identity matrix. So that's 1, 0, 0, 0, 1, 0, 0, 0, 1. All right, so what are we going to do? What I'm going to do is perform a series of elementary row operations. And I'm about to tell you what are valid elementary row But whatever I do to any of these rows here, I have to do to the corresponding rows here. And my goal is essentially to perform a bunch of operations on the left hand side. And of course, the same operations will be applied to the right hand side, so that I eventually end up with the identity matrix on the left hand side. And then when I have the identity matrix on the left hand side, what I have left on the right hand side will be the inverse of this original matrix. And when this becomes an identity matrix, that's actually called reduced row echelon form. And I'll talk more about that. There's a lot of names and labels in linear algebra. But they're really just fairly simple concepts. But anyway, let's get started and this should become a little clear. At least the process will become clear. Maybe not why it works. So first of all, I said I'm going to perform a bunch of operations here. What are legitimate operations? They're called elementary row operations. So there's a couple things I can do. I can replace any row with that row"
    },
    {
        "Q": "\nAt 0:08, why does Sal convert to an improper fraction, instead of just adding the 2/3 to the 8 1/3 to get 9. Wouldn't that be much easier? I just think it's a waste of time.",
        "A": "becauseit s easier to multiply and divide with it, and he can convert at the end.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            8
        ],
        "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12."
    },
    {
        "Q": "At 2:10\nCould you add +12y to both sides and subtract 2 you would get\n12y > -27\n= y > -27/12\n= y > -9/4\n",
        "A": "That is correct. You could certainly do it that way.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            130
        ],
        "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12."
    },
    {
        "Q": "\nAt 0:56, do you have to multiply by 3, or can you add 25/3 to both sides? Thanks!!",
        "A": "If Sal had added 25/3 on both sides, he would still be left with a fraction (getting him no where, just moving the number) multiplying by 3 gets rid of the denominator of 3.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            56
        ],
        "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12."
    },
    {
        "Q": "\nat 1:38 Sal says '3 times negative 25 over 3 is just negative 25'. I don't remember it ever being -25. Did Sal just say it wrong, or I'm I missing something?",
        "A": "The equation starts out with -25/3. The minus in front of the fraction makes that fraction negative. -25/3 *3/1 = -25 Hope this helps.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            98
        ],
        "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12."
    },
    {
        "Q": "can you explain it briefly? I couldn't understand from 2:31 to 4:04. pl. explain it again in a theoretical way.\n",
        "A": "the third point is the right answer because the limit of f(x) when x approaches k from the negative direction = to the limit of f(x) when x approaches k from the positive direction.",
        "video_name": "_bBAiZhfH_4",
        "timestamps": [
            151,
            244
        ],
        "3min_transcript": "But then we jump right at x equals 8. And then we continue from 1 again. So this is our other candidate. So these are the three candidates where the function is not continuous. Now let's think about which of these points, which of these x values, does f of k exist. So if one of these is k, does f of k exist? Well f of negative 2 exists. f of 3 exists, right over here. That's f of 3. This is f of negative 2. And f of 8, all exist. So all of these potential k's meet this constraint-- f of k exists, and f is not continuous at k. So that's true for x equals 8, 3, or negative 2. Now let's look at this first constraint. The limit of f of x as x approaches k needs to exist. Well, if we tried to look at x equals 2, the limit of f as x approaches negative 2 here-- the limit from the left, the limit from values lower than negative 2, it looks like our function is approaching something a little higher. It looks like it's a little higher than 3. And the limit from the right, it looks like our function is approaching negative 3. So this one, the limit does not exist. You get a different limit from the left and from the right. Same thing for x equals positive 3. The limit from the left seems like it's approaching 4 and 1/2, while the limit from the right looks like it's approaching negative 4. So this is also not a candidate. So we only have one left. So for this one, the limit should exist. And we see the limit as f of x as x approaches 8 from the negative direction, it looks like f of x is approaching 1. And it looks like, as we approach 8 from the positive direction, the limit of f It's also equal to 1. So your left- and your right-sided limits approach the same value. So the limit of f of x as x approaches 8 is equal to 1. This limit exists. Now, the reason why the function isn't continuous there is that the limit of f of x as x approaches 8, which is equal to 1, it does not equal the value of f of 8. f of 8, we're seeing, is equal to 7. So that's why it meets the last constraint. The function is not continuous there. The function exists. It's defined, f of 8 is equal to 7. And the limit exists. But the limit of f of x as x approaches k is not the same thing, or is not the same as the value of the function evaluated at that point. And so x equals 8 meets all of our constraints. So we could say k is equal to 8."
    },
    {
        "Q": "\nTo make sure x=4 is a minimum value (at 4:22) can you also use the first derivative test and test around the point x=4?\nIf the derivative goes from decreasing to increasing at x=4 it would indicate a minimum at that point.",
        "A": "Correct. I prefer your technique too.",
        "video_name": "1TK8V_qmqrk",
        "timestamps": [
            262
        ],
        "3min_transcript": "is where the derivative is either 0 or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative s prime-- let me do this in a different color-- s prime of x. I'll do it right over here, actually. The derivative s prime of x with respect to x is going to be equal to 2x times negative 2 times 2x plus 256 times negative 2. So that's minus 512x to the negative 3 power. Now, this is going to be undefined when x is equal to 0. But if x is equal to 0, then y is undefined. So this whole thing breaks down. So that isn't a useful critical point, x equals 0. Well, it's defined everywhere else. So let's think about where the derivative is equal to 0. So when does this thing equal 0? So when does 2x minus 512x to the negative 3 equal 0? Well, we can add 512x to the negative 3 to both sides. So you get 2x is equal to 512x to the negative third power. We can multiply both sides times x to the third power so all the x's go away on the right-hand side. So you get 2x to the fourth is equal to 512. We can divide both sides by 2, and you get x to the fourth power is equal to 256. And so what is the fourth root of 256? Well, we could take the square root of both sides just to help us here. So let's see. So it's going to be x squared is going to be equal to 256 So this is 16. This is going to be x squared is equal to 16 or x is equal to 4. Now that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here. But let's make sure it's a minimum value. And to do that, we can just do our second derivative test. So let's figure out. Let's take the second derivative s prime prime of x and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2. And then we're going to have negative 3 times negative 512. So I'll just write that as plus 3 times 512. That's going to be 1,536. Yeah, 3 times 500 is 1,500, 3 times 12 is 36, x to the negative 4 power. And this thing right over here is actually going to be positive for any x."
    },
    {
        "Q": "Sal in 7:34 86-30=56,and you said it is 50.That does not make any sense\n",
        "A": "Yes, he did, and you are correct, but if you continue to watch, you will realize that he catches his mistake.",
        "video_name": "ko-cYG3d6ec",
        "timestamps": [
            454
        ],
        "3min_transcript": "And maybe we can do it, we can do it in a good color. Maybe we can do it in a couple of different ways. So one thing we could do is we could figure out what this angle is, so we could just subtract this green angle from 86 and we would get our answer. Well, this angle's easy, right, because we know two angles of this triangle, so we could figure that out. Let's just call this, I don't know, let's call this y. So y plus 122 plus 28 degrees is going to equal 180. So y plus 150 is equal to 180. So y is equal to 30 degrees, right? So this is equal to 30 degrees. So this is 30 degrees, and this big angle here is 86. So our goal, let's call that x, so x is going to just be equal out, minus 30. So x is going to be equal to 50 degrees. That was a pretty straightforward problem. Let's see if we could figure that out any other way. Well, we could say instead of doing it that way -- let's forget we just solved it that way. We could say this angle here is supplementary to this 122 degree angle, right, so it has to add up to 180. So this plus 122 is 180, so what does that make this? It makes this 58 degrees, right? This plus this is going to be 180. So we figured out this. If we could figure out this, then we could use this triangle. How do we figure out this angle? Well, we could look at this big triangle here, and we know Let's call this z. So we know that z plus this angle, plus 28, plus this big angle, plus 86 is equal to 180. So z plus, what is this, 106, 114 is equal to 180. So z is equal to, what is this, 66 degrees. I don't know if I'm doing any of my math correctly, but let's just hope. z equals 66. So z is 66, this angle is 58, and now we can use this triangle here to figure out what this angle is, our x. So x plus 66 plus 58 is equal to 180. I already think I might have made a mistake some"
    },
    {
        "Q": "\nAt 7:17, Sal said X=50' but probably meant X=56'.",
        "A": "he fixed it around 7:20 there is a little note that pops up",
        "video_name": "ko-cYG3d6ec",
        "timestamps": [
            437
        ],
        "3min_transcript": "Let me draw some more. So this is going to be a pretty straightforward drawing. It's pretty much just two triangles next to each other. Like that and then let me draw another line that goes like that, and then we draw a line that goes like that, and I think I have done my drawing. There you go. I'm have done my drawing. So let's see. What do we know about this triangle and what do we need to figure out? I'm going to tell you that this angle here, this big angle here, is 86 degrees. We also know that this angle here is 28 degrees. And we also know that this angle here is 122 degrees. And our goal, our mission in this round is to figure And maybe we can do it, we can do it in a good color. Maybe we can do it in a couple of different ways. So one thing we could do is we could figure out what this angle is, so we could just subtract this green angle from 86 and we would get our answer. Well, this angle's easy, right, because we know two angles of this triangle, so we could figure that out. Let's just call this, I don't know, let's call this y. So y plus 122 plus 28 degrees is going to equal 180. So y plus 150 is equal to 180. So y is equal to 30 degrees, right? So this is equal to 30 degrees. So this is 30 degrees, and this big angle here is 86. So our goal, let's call that x, so x is going to just be equal out, minus 30. So x is going to be equal to 50 degrees. That was a pretty straightforward problem. Let's see if we could figure that out any other way. Well, we could say instead of doing it that way -- let's forget we just solved it that way. We could say this angle here is supplementary to this 122 degree angle, right, so it has to add up to 180. So this plus 122 is 180, so what does that make this? It makes this 58 degrees, right? This plus this is going to be 180. So we figured out this. If we could figure out this, then we could use this triangle. How do we figure out this angle? Well, we could look at this big triangle here, and we know"
    },
    {
        "Q": "\nAt 2:25 Sal mentions that factorial shows up in other topics of math. What might these other topics be?",
        "A": "Some of the things where factorial comes up is really advanced math, but it does appear in what are known as Taylor Polynomials.",
        "video_name": "HGoZfzz6dU0",
        "timestamps": [
            145
        ],
        "3min_transcript": "If you've been paying particularly close attention to our use of the factorial operator and videos on permutations and combinations. You may or may not have noticed something that might be interesting. So let's just review factorial a little bit. So if I were to say n factorial, that of course is going to be n times n minus one times n minus two. And I would just keep going down, until I go to times one. So I'd keep decrementing n until I get to one and then I would multiply all of those things together. So for example: if I were to say three factorial: that's going to be three times two times one. If I were to say two factorial: that's going to be two times one. One factorial: well I just keep decrementing until I get to one but I don't even have to decrement here I'm already at one. So I just multiply one. So one logical thing is to say: \"Oh hey you know maybe zero factorial is zero.\" I'm just starting with itself, it's already below one, and maybe it is zero. Now what we will see is that this is not the case, that mathematics and mathematicians have decided. Remember this is what's interesting that the factorial operation is something that humans have invented. That they think it's just an interesting thing, it's a useful notation. So they can define what it does, and mathematicians have found it far more useful to define zero factorial as something else. To define zero factorial as -- and there's a little drum roll here... They believe zero factorial should be one, and I know, based on the reasoning, the conceptional reasoning of this, this doesn't make any sense. why this is a useful concept, especially in the world of permutations and combinations, which is frankly where factorial shows up the most. Most of the cases that I've ever seen factorial in anything has been in situations of combinations & permutations, and in a few other things but mainly permutations and combinations. So let's review a little bit, we've said if we have n things and we want to figure out the number of ways to permute them into k spaces it's going to be n factorial over n minus k factorial. Now we've also said that if we had n things that we wanna permute into n places, well this really should just be n factorial, the first place has -- Let's do this, so this is the first place, this is the second place,"
    },
    {
        "Q": "Can, at 1:30, this be done if the coefficient in front is a negative one? If not, would we just use long division?\n",
        "A": "It can be done just negative will become positive and same procedure applies.",
        "video_name": "1byR9UEQJN0",
        "timestamps": [
            90
        ],
        "3min_transcript": "In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12."
    },
    {
        "Q": "\nAround 2:24 in the video u mentioned turning the 4 in the denominator in -4. So what if the number in the denominator is already negative. would you keep it or change it positive.",
        "A": "If you had x-4 in the denominator, you use +4. The X and the minus are ignored. Hope this helps.",
        "video_name": "1byR9UEQJN0",
        "timestamps": [
            144
        ],
        "3min_transcript": "In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12."
    },
    {
        "Q": "\nAt 1:48, why did Sal add 9 to each number and then cross out the 9's?",
        "A": "This is because 9-9 equals 0. Crossing it out is another way to show that it is 0.",
        "video_name": "qzsR7cChujg",
        "timestamps": [
            108
        ],
        "3min_transcript": "We have the proportion x minus 9 over 12 is equal to 2/3. And we want to solve for the x that satisfies this proportion. Now, there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they want to cross multiply. They want to say, hey, 3 times x minus 9 is going to be equal to 2 times 12. And that's completely legitimate. You would get-- let me write that down. So 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides, and you would get-- let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to-- let's see, 51. And then x would be equal to 17. 17 minus 9 is 8. 8/12 is the same thing as 2/3. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication , you could say look, I want to get rid of this 12 in the denominator right over here. Let's multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2/3 times 12. Well, 2/3 of 12 is just 8. And you could do the actual multiplication. 2/3 times 12/1. 12 and 3, so 12 divided by 3 is 4. 3 divided by 3 is 1. So it becomes 2 times 4/1, which is just 8. And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. and both sides by 3, and then that would be functionally equivalent to cross multiplying. Let's do one more. So here another proportion. And this time the x in the denominator. But just like before, if we want we can cross multiply. And just to see where cross multiplying comes from, it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true."
    },
    {
        "Q": "at 5:24 why does vi wear the hilbert curve\n",
        "A": "1. Because she SO rocks it. 2. I m not sure.",
        "video_name": "ik2CZqsAw28",
        "timestamps": [
            324
        ],
        "3min_transcript": "Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other, You can keep going until you run out of room, or you can make each new version the same size by making each line half the length. Or you can make it out of snakes. Or if you have friends, you can each make an iteration of the same size, and put them together. Or invent your own fractal curve so that you could be cool like Hilbert. Who was like, mathematics? I'm going to invent meta-mathematics like a boss."
    },
    {
        "Q": "\nIs there any design other than the one at 4:19 (The one that that person made?) Also, is Vi her real name, or is it Violet?",
        "A": "There are uncountably infinitely many space-filling curves. Her name is Victoria. She prefers to be called Vi.",
        "video_name": "ik2CZqsAw28",
        "timestamps": [
            259
        ],
        "3min_transcript": "down a squiggle, up, wop, all the way over here. OK, but say you're me and you're in math class. This mean that you have graph paper. Opportunity for precision. You could draw that first curve like this. Squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a. The second iteration to fit squiggles going up and down will have a line three boxes across on top and bottom, if you want the squiggles as close on the grid as possible without touching. You might remind yourself by saying, three a-squig, a-squig, a-squiggle, three, a-squig, a-squig, a-squiggle. The next iteration has a woop, and you have to figure out how long that's going to be. Meanwhile, other lengths change to keep everything close. And, two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle, two, nine. We could write the pattern down like this. So what would the next pattern be? Five. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other,"
    },
    {
        "Q": "At 1:26, when he divides -5 times the cubed root of y by -5, why can the fives cancel and you are left with the cubed root of y when -5 times the cubed root of y divided by five is equal to -5/-5 times the cubed root of y/-5?\n",
        "A": "[ 1/(-5) ] * [ (-5)*cbrt(y) ] = [ [ 1/(-5) ] * (-5) ] * cbrt(y) (associative law) = 1*cbrt(y).",
        "video_name": "b6WtwQddAcY",
        "timestamps": [
            86
        ],
        "3min_transcript": "We're asked to solve for y. So we're told that the negative of the cube root of y is equal to 4 times the cube root of y plus 5. So in all of these it's helpful to just be able to isolate the cube root, isolate the radical in the equation, and then solve from there. So let's see if we can isolate the radical. So the simplest thing to do, if we want all of the radical onto the left-hand side equation, we can subtract 4 times the cube root of y from both sides of this equation. So let's subtract 4. We want to subtract 4 times the cube root of y from both sides of this equation. And so your left-hand side, you already have negative 1 times the cube root of y, and you're going to subtract 4 more of the cube root of y. So you're going to have negative 5 times the cube root of y. That's your left-hand side. Now the right-hand side-- these two guys-- cancel out. So that cancels out and you're just left with a 5 there. You're just left with this 5 right over there. Now, we've almost isolated this cube root of y. We just have to divide both sides of the equation by negative 5. So you just divide both sides of this equation by negative 5. And these cancel out. That was the whole point. And we are left with the cube root of y is equal to-- 5 divided by negative 5 is negative 1. Now, the cube root of y is equal to negative 1. Well the easiest way to solve this is, let's take both sides of this equation to the third power. This statement right here is the exact same statement as saying y to the 1/3 is equal to negative 1. These are just two different ways of writing the same thing. This is equivalent to taking the 1/3 power. So if we take both sides of this equation to the third third power. And you can see here, y to the 1/3 to the third-- y to the 1/3 and then to the third, that's like saying y to the 1/3 times 3 power. Or y to the first power. That's the whole point of it. If you take the cube root of y to the third power, that's just going to be y. So the left-hand side becomes y. And then the right-hand side, what's negative 1 Negative 1 times negative 1 is 1. Times negative 1 again is negative 1. So we get y is equal to negative 1 as our solution. Now let's make sure that it actually works. Let's go back to our original equation. And I'll put negative 1 in for our y's. We had the negative of the cube root of-- this time, negative 1-- has to be equal to 4 times the cube root of negative 1 plus 5."
    },
    {
        "Q": "At 2:43, the third and fourth probability terms are identical. Sal says the fourth term correctly but writes it incorrectly. Am I correct?\n",
        "A": "just saw that myself. it has to be a typo/writo(so). it should read ttth, not ttht.",
        "video_name": "8TIben0bJpU",
        "timestamps": [
            163
        ],
        "3min_transcript": "So we're going to have one flip, then another flip, then another flip, then another flip. And this first flip has two possibilities. It could be heads or tails. The second flip has two possibilities. It could be heads or tails. The third flip has two possibilities. It could be heads or tails. And the fourth flip has two possibilities. It could be heads or tails. So you have 2 times 2 times 2 times 2, which is equal to 16 possibilities. 16 possible outcomes when you flip a coin four times. And any one of the possible outcomes would be 1 of 16. So if I wanted to say, so if I were to just say the probability, and I'm just going to not talk about this one heads, if I just take a, just maybe this thing that has three heads right here. This exact sequence of events. This is the first flip, second flip, third flip, fourth flip. Getting exactly this, this is exactly one out of a possible of 16 events. Now with that out of the way, let's think possibilities, involve getting exactly one heads? Well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first flip, plus the probability of getting the heads in the second flip, plus the probability of getting the heads in the third flip. Remember, exactly one heads. We're not saying at least one, exactly one heads. So the probability in the third flip, and then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events, and each of these are one of those 16 possible events. So this is going to be 1 over 16, 1 over 16, 1 over 16, And so we're really saying the probability of getting exactly one heads is the same thing as the probability of getting heads in the first flip, or the probability of getting heads-- or I should say the probability of getting heads in the first flip, or heads in the second flip, or heads in the third flip, or heads in the fourth flip. And we can add the probabilities of these different things, because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. 1/16 plus 1/16 plus 1/16 plus 1/16. Did I say that four times? Well, assume that I did. And so you would get 4/16, which is equal to 1/4. Fair enough. Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly two heads. And there's a couple of ways we can think about it."
    },
    {
        "Q": "I'm confused at 1:15, because I thought that (4x^2)/5 was equivalent to 4/5 and (x^2)/5. However, Sal sets it up as 4/5x^2 where x^2 is not being divided by 5. Any reasons why this is so?\n",
        "A": "It s the difference between addition and multiplication. (\u00f0\u009d\u0091\u008e + \u00f0\u009d\u0091\u008f) \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 = \u00f0\u009d\u0091\u008e \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 + \u00f0\u009d\u0091\u008f \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 (\u00f0\u009d\u0091\u008e \u00e2\u0088\u0099 \u00f0\u009d\u0091\u008f) \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 = (\u00f0\u009d\u0091\u008e \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090)\u00f0\u009d\u0091\u008f",
        "video_name": "6nALFmvvgds",
        "timestamps": [
            75
        ],
        "3min_transcript": "- [Voiceover] So up here, we are multiplying two rational expressions. And here, we're dividing one rational expression by another one. Now what I encourage you do is pause this video and think about what these become when you multiply them. I don't know, maybe you simplify it a little bit, and I also want you to think about what constraints do you have to put on the x values in order for your resulting expression to be algebraically equivalent to your original expression. So let's work it out together just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two, and our denominator, we're going to have five times three x. And we can see both the numerator and the denominator are divisible by x, so let's divide the denominator by x. We get one there. Let's divide x to the third by x. We get x squared. And we can also see the both the numerator and denominator are divisible by three, so divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, over five times one times one over five. And we can also write that as 4/5 x squared. Now someone just presented you on the street with the expression 4/5 x squared and say, for what x is this defined? I could put any x here, x could be zero because zero squared is zero times 4/5 is just going to be zero, so it does seems to be defined for zero, and that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well then, you'd have to say, well, this first expression is not defined for all x. For example, if x were equal to zero, then you would be dividing by zero right over here, which would make this undefined. So you could explicitly call it out, x can not be equal to zero. And so if you want this to be algebraically equivalent, you would have to make that same condition, x cannot be equal to zero. defined this way, if you said, if you said f of x is equal to six x to the third over five times two over, times two over three x; and if someone said, well what is f of zero, you would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're going to get two divided by zero and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4/5 times x squared. But if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make them, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero."
    },
    {
        "Q": "\nAt 1:56, Sal writes that the limit of g(x) as x approaches 6 from the negative direction = \"not exist\". Is there an actual notation for it not existing? Maybe something like the zero with a slash through it would do it? A preemptive thanks to anyone who may have an answer to my question.",
        "A": "it s kind of common to write DNE",
        "video_name": "5f1-Rg3MmKs",
        "timestamps": [
            116
        ],
        "3min_transcript": "- [Voiceover] Over here we have the graph of y is equal to g of x. What I wanna do is I wanna figure out the limit of g of x as x approaches positive six from values that are less than positive six or you could say from the left, from the, you could say the negative direction. So what is this going to be equal to? And if you have a sense of it, pause the video and give a go at it. Well, to think about this, let's just take different x-values that approach six from the left and look at what the values of the function are. So g of two, looks like it's a little bit more than one. G of three, it's a little bit more than that. G of four, looks like it's a little under two. G of five, it looks like it's around three. G of 5.5, looks like it's around five. G of, let's say 5.75, looks like it's like nine. And so, as x gets closer and closer to six becomes unbounded, it's just getting infinitely large. And so in some context, you might see someone write that, maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number. If we're talking technically about limits the way that we've looked at it, what is, you'll sometimes see this in some classes. But in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded. Now this is interesting because the left-handed limit here doesn't exist, but the right-handed limit does. If I were to say the limit of g of x as x approaches six from the right-hand side, well, let's see. We have g of eight is there, go of seven is there, g of 6.5, looks like it's a little less than negative three. G of 6.0000001 is very close to negative three. So it looks like this limit right over here, at least looking at it graphically, it looks like when we approach six from the right, looks like the function is approaching negative three. But from the left, it's just unbounded, so we'll say it doesn't exist."
    },
    {
        "Q": "\nAt 0:58, what is a p/2-gon?",
        "A": "She said that wrong. She meant 2p-gon....I think. I think this because if you have two of the same Polygons (as in p) and overlap them. you have twice as many sides as the first polygons",
        "video_name": "CfJzrmS9UfY",
        "timestamps": [
            58
        ],
        "3min_transcript": "Let's say you're me, and you're in math class, and you're supposed to be learning about factoring. Trouble is, your teacher is too busy trying to convince you that factoring is a useful skill for the average person to know, with real-world applications ranging from passing your state exams all the way to getting a higher SAT score. And unfortunately, does not have the time to show you why factoring is actually interesting. It's perfectly reasonable for you to get bored in this situation. So like any reasonable person, you start doodling. Maybe it's because your teacher's soporific voice reminds you of a lullaby, but you're drawing stars. And because you're me, you quickly get bored of the usual five-pointed star and get to wondering, why five? So you start exploring. It seems obvious that a five-pointed star is the simplest one, the one that takes the least number of strokes to draw. Sure, you can make a start with four points, but that's not really a star the way you're defining stars. Then there's a six-pointed star, which is also pretty familiar, but totally different from the five-pointed star because it takes two separate lines to make. And then you're thinking about how, much like you can put two triangles together to make a six-pointed star, you can put two squares together to make an eight-pointed star. And any even-numbered star with p points can be made out of two p/2-gons. It is at this point that you realize maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways."
    },
    {
        "Q": "\nAt 4:24 how is -9 times 1 = -8. Doesn't he mean -9 + 1?",
        "A": "He multiplied (10^1)*(10^-9). When you multiply numbers that are the same base to an exponent, you can add the exponents. Because of that, while he s multiplying 10^1 with 10^-9 to get 10^-8, he s adding 1 with -9 to get -8. I hope this helps!",
        "video_name": "67jn5Zv-myg",
        "timestamps": [
            264
        ],
        "3min_transcript": "and less likely to make a careless mistake-- times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing. And so if we simplify this exponent, you get 40.1534 times 10 to the 8 minus 12 is negative 4, minus 5 is negative 9. 10 to the negative 9 power. Now you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10. But this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to 1 and less than 10. Essentially, for it to be in scientific notation, you want a non-zero digit right over here. And then you want your decimal and then the rest of everything else. So here-- and you want a non-zero single digit over here. Here we obviously have two digits. This is larger than 10-- or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to 1. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to 4, we have to move this decimal over to the left. Moving a decimal over to the left to go from 40 to 4 you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it, is 4.0 and all this stuff times 10 is going to be 40.1534. to the first power, that's the same thing as 10-- times this thing-- times 10 to the negative ninth power. And then once again, powers of 10, so it's 10 to the first times 10 to the negative 9 is going to be 10 to the negative eighth power. And we still have this 4.01534 times 10 to the negative 8. And now we have written it in scientific notation. Now, they wanted us to express it in both decimal and scientific notation. And when they're asking us to write it in decimal notation, they essentially want us to multiply this out, expand this out. And so the way to think about it-- write these digits out. So I have 4, 0, 1, 5, 3, 4. And if I'm just looking at this number, I start with the decimal right over here."
    },
    {
        "Q": "\nHi at 3:16 in the video he had the value (40.1534x10^-9) I was wondering why you cant simply move the decimal one to the left and add a -1 to the exponent. Is it because moving the decimal to the left and adding the negative exponent to the ten would change the true value of the numer by a tenth?",
        "A": "He does move the decimal over and add a -1 to the exponent. First, he multiplied the numbers 1.45, 9.2 and 3.01. That s where he got the 40.1534. Then he multiplied the 10s. So then he had (40.1534x10^-9). Finally, he puts it into scientific notation in the way you described.",
        "video_name": "67jn5Zv-myg",
        "timestamps": [
            196
        ],
        "3min_transcript": "in that purple color-- times 10 to the eighth times 10 to the negative 12th power times 10 to the negative fifth power. And this is useful because now I have all of my powers of 10 right over here. I could put parentheses around that. And I have all my non-powers of 10 right over there. And so I can simplify it. If I have the same base 10 right over here, so I can add the exponents. This is going to be 10 to the 8 minus 12 minus 5 power. And then all of this on the left-hand side-- let me get a calculator out-- I have 1.45. and less likely to make a careless mistake-- times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing. And so if we simplify this exponent, you get 40.1534 times 10 to the 8 minus 12 is negative 4, minus 5 is negative 9. 10 to the negative 9 power. Now you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10. But this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to 1 and less than 10. Essentially, for it to be in scientific notation, you want a non-zero digit right over here. And then you want your decimal and then the rest of everything else. So here-- and you want a non-zero single digit over here. Here we obviously have two digits. This is larger than 10-- or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to 1. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to 4, we have to move this decimal over to the left. Moving a decimal over to the left to go from 40 to 4 you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it, is 4.0 and all this stuff times 10 is going to be 40.1534."
    },
    {
        "Q": "\nAt 3:13, what does Sal mean when he says this:\n\"When arithmetic is a noun, we call it arithmetic. When arithmetic is an adjective like this, we call it arithmetic, arithmetic mean.\"\n?",
        "A": "He is telling you how to pronounce the word, arithmetic, as in an arithmetic mean. Listen carefully to how he says the two words differently. The word has two different pronounciations, depending on their meaning. He puts the emphasis on a different syllable so that we know which way he is using the word.",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            193
        ],
        "3min_transcript": "in our garden. And let's say we have six plants. And the heights are 4 inches, 3 inches, 1 inch, 6 inches, and another one's 1 inch, and another one is 7 inches. And let's say someone just said-- in another room, not looking at your plants, just said, well, you know, how tall are your plants? And they only want to hear one number. They want to somehow have one number that represents all of these different heights of plants. How would you do that? Well, you'd say, well, how can I find something that-- maybe I want a typical number. Maybe I want some number that somehow represents the middle. Maybe I want the most frequent number. Maybe I want the number that somehow represents the center of all of these numbers. And if you said any of those things, you would actually have done the same things that the people who first came up with descriptive statistics said. They said, well, how can we do it? And we'll start by thinking of the idea of average. has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean. I'll write in yellow, arithmetic mean. When arithmetic is a noun, we call it arithmetic. When it's an adjective like this, we call it arithmetic, arithmetic mean. And this is really just the sum of all the numbers divided by-- this is a human-constructed definition that we've found useful-- the sum of all these numbers divided by the number of numbers we have. So given that, what is the arithmetic mean of this data set? Well, let's just compute it. It's going to be 4 plus 3 plus 1 plus 6 plus 1 plus 7 over the number of data points we have. So we have six data points. So we're going to divide by 6. And we get 4 plus 3 is 7, plus 1 is 8, plus 6 is 14,"
    },
    {
        "Q": "At 5:40, Sal says you have to repeat the one since it is repeated. Why? Why can't you just put one 1?\n",
        "A": "In the example that Khan is doing, the numbers are 431617. There are two ones in this sequence so the first 1 is repeated. If I was trying to find the average of 58972, I wouldn t have to repeat any number because every number in that sequence was different. Hope this helps!",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            340
        ],
        "3min_transcript": "15 plus 7 is 22. Let me do that one more time. You have 7, 8, 14, 15, 22, all of that over 6. And we could write this as a mixed number. 6 goes into 22 three times with a remainder of 4. So it's 3 and 4/6, which is the same thing as 3 and 2/3. We could write this as a decimal with 3.6 repeating. So this is also 3.6 repeating. We could write it any one of those ways. But this is kind of a representative number. This is trying to get at a central tendency. Once again, these are human-constructed. No one ever-- it's not like someone just found some religious document that said, this is the way that the arithmetic mean must be defined. It's not as pure of a computation as, say, finding the circumference of the circle, which there really is-- that was kind of-- we studied the universe. And that just fell out of our study of the universe. It's a human-constructed definition Now there are other ways to measure the average or find a typical or middle value. The other very typical way is the median. And I will write median. I'm running out of colors. I will write median in pink. So there is the median. And the median is literally looking for the middle number. So if you were to order all the numbers in your set and find the middle one, then that is your median. So given that, what's the median of this set of numbers going to be? Let's try to figure it out. Let's try to order it. So we have 1. Then we have another 1. Then we have a 3. Then we have a 4, a 6, and a 7. So all I did is I reordered this. And so what's the middle number? Well, you look here. Since we have an even number of numbers, we have six numbers, there's not one middle number. You actually have two middle numbers here. You have two middle numbers right over here. And in this case, when you have two middle numbers, you actually go halfway between these two numbers. You're essentially taking the arithmetic mean of these two numbers to find the median. So the median is going to be halfway in-between 3 and 4, which is going to be 3.5. So the median in this case is 3.5. So if you have an even number of numbers, the median or the middle two, the-- essentially the arithmetic mean of the middle two, or halfway between the middle two. If you have an odd number of numbers, it's a little bit easier to compute. And just so that we see that, let me give you another data set. Let's say our data set-- and I'll order it for us-- let's say our data set was 0, 7, 50, I don't know, 10,000, and 1 million. Let's say that is our data set. Kind of a crazy data set. But in this situation, what is our median?"
    },
    {
        "Q": "\nat 7:55pm est...what is the shape, center, and mean in statistics?",
        "A": "It is the number that is the answer when 2-3 or more are put together in an equation?",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            475
        ],
        "3min_transcript": "And in this case, when you have two middle numbers, you actually go halfway between these two numbers. You're essentially taking the arithmetic mean of these two numbers to find the median. So the median is going to be halfway in-between 3 and 4, which is going to be 3.5. So the median in this case is 3.5. So if you have an even number of numbers, the median or the middle two, the-- essentially the arithmetic mean of the middle two, or halfway between the middle two. If you have an odd number of numbers, it's a little bit easier to compute. And just so that we see that, let me give you another data set. Let's say our data set-- and I'll order it for us-- let's say our data set was 0, 7, 50, I don't know, 10,000, and 1 million. Let's say that is our data set. Kind of a crazy data set. But in this situation, what is our median? We have an odd number of numbers. So it's easier to pick out a middle. The middle is the number that is greater than two of the numbers and is less than two of the numbers. It's exactly in the middle. So in this case, our median is 50. Now, the third measure of central tendency, and this is the one that's probably used least often in life, is the mode. And people often forget about it. It sounds like something very complex. But what we'll see is it's actually a very straightforward idea. And in some ways, it is the most basic idea. So the mode is actually the most common number in a data set, if there is a most common number. If all of the numbers are represented equally, if there's no one single most common number, then you have no mode. But given that definition of the mode, what is the single most common number in our original data set, in this data set right over here? Well, we only have one 4. We only have one 3. We have one 6 and one 7. So the number that shows up the most number of times here is our 1. So the mode, the most typical number, the most common number here is a 1. So, you see, these are all different ways of trying to get at a typical, or middle, or central tendency. But they do it in very, very different ways. And as we study more and more statistics, we'll see that they're good for different things. This is used very frequently. The median is really good if you have some kind of crazy number out here that could have otherwise skewed the arithmetic mean. The mode could also be useful in situations like that, especially if you do have one number that's showing up a lot more frequently. Anyway, I'll leave you there. And we'll-- the next few videos, we will explore statistics even deeper."
    },
    {
        "Q": "At 2:11 I don't get how you move the x+1 over to the numerator. What happens to the denominator? Does the numerator become x+2 times x+1? Thanks.\n",
        "A": "how do you know which one the denominator is?",
        "video_name": "9IUEk9fn2Vs",
        "timestamps": [
            131
        ],
        "3min_transcript": "Welcome to the presentation on level four linear equations. So, let's start doing some problems. Let's say I had the situation-- let me give me a couple of problems-- if I said 3 over x is equal to, let's just say 5. So, what we want to do -- this problem's a little unusual from everything we've ever seen. Because here, instead of having x in the numerator, we actually have x in the denominator. So, I personally don't like having x's in my denominators, so we want to get it outside of the denominator into a numerator or at least not in the denominator as So, one way to get a number out of the denominator is, if we were to multiply both sides of this equation by x, you see that on the left-hand side of the equation these two x's will cancel out. And in the right side, you'll just get 5 times x. So this equals -- the two x's cancel out. And you get 3 is equal to 5x. Now, we could also write that as 5x is equal to 3. We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on"
    },
    {
        "Q": "At 4:58 why is it Kt and not K^2/2?\n",
        "A": "It s because K is a constant and we re integrating *with respect to t - the dt bit in \u00e2\u0088\u00ab K dt If we were integrating with respect to (a variable) K, then we d have \u00e2\u0088\u00ab K dK and you d be right, the integral would be K\u00c2\u00b2/2",
        "video_name": "IYFkXWlgC_w",
        "timestamps": [
            298
        ],
        "3min_transcript": "or differential equations in general, you can treat, you can treat these, this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So, we want to put all the Ps and dPs on one side and all the, all the things that involve t or that I guess don't involve P on the other side. So, we could divide both sides by P. We could divide both sides by P, and so we'll have one over P, and you have one over P here and then those will cancel, and then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity which isn't, it really isn't a quantity. You really have to view this as a limit of change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. we can do this, and when we do that we would be left with one over P dP is equal to, is equal to k dt, is equal to k dt. Now we can integrate, integrate both sides. Because this was a separable differential equation, we were able to completely separate the Ps and dPs from the things involving ts or, I guess, the things that aren't involving Ps, and then if we integrate this side, we would get the natural log, the natural log of the absolute value of our population, and we could say plus some constant if we want but we're going to get a constant on this side as well so we could just say that's going to be equal to, that's going to be equal to k, it's going to be equal to k times t, k times t plus some constant. I'll just call that C one, and once again I could've put a plus C two here, but I could've then subtracted the constant from both sides and I would just get the constant on the right hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means that's the same thing. That means that the absolute value of P, that means that the absolute value of P is equal to e to all this business, e to the, e to the, let me do in the same colors, kt, kt plus, plus C one, plus C one. Now this right over here is the same thing. Just using our exponent properties, this is the same thing as e to the kt, e to the k times t times e, times e to the C one,"
    },
    {
        "Q": "0:08 you said they skipped 7 and 8 why did they skip them are their no tricks for 7 or 8\n",
        "A": "sorry I did t read below",
        "video_name": "2G_Jr_XpnY4",
        "timestamps": [
            8
        ],
        "3min_transcript": "Determine whether 380 is divisible by 2, 3, 4, 5, 6, 9 or 10. They skipped 7 and 8 so we don't have to worry about those. So let's think about 2. So are we divisible by 2? Let me write the 2 here. Well, in order for something to be divisible by 2, it has to be an even number, and to be an even number, your ones digit-- so let me rewrite 380. To be even, your ones digit has to be even, so this has to be even. And for this to be even, it has to be 0, 2, 4, 6 or 8, and this is 0, so 380 is even, which means it is divisible by 2, so it works with 2. So 2 works out. Let's think about the situation for 3. Now, a quick way to think about 3-- so let me write just And if the sum that you get is divisible by 3, then you are divisible by 3. So let's try to do that. So 380, let's add the digits. 3 plus 8 plus 0 is equal to-- 3 plus 8 is 11 plus 0, so it's just 11. And if you have trouble figuring out whether this is divisible by 3, you could then just add these two numbers again, so you can actually add the 1 plus 1 again, and you would get a 2. Regardless of whether you look at the 1 or the 2, neither of these are divisible by 3. So not divisible by 3, and maybe in a future video, I'll explain why this works, and maybe you want to think about why this works. So these aren't divisible by 3, so 380 is not divisible. 380, not divisible by 3, so 3 does not work. Now, I'll think about the situation for 4, so we're thinking about 4 divisibility. So let me write it in orange. So we are wondering about 4. Now, something you may or may not already realize is that 100 is divisible by 4. It goes evenly. So this is 380. So the 300 is divisible by 4, so we just have to figure out whether the leftover, whether the 80, is divisible by 4. Another way to think about it is are the last two digits divisible by 4? And this comes from the fact that 100 is divisible by 4, so everything, the hundreds place or above, it's going to be divisible by 4. You just have to worry about the last part. So in this situation, is 80 divisible by 4?"
    },
    {
        "Q": "\nAt 3:19, why can we simply assume that cos(theta) is positive?",
        "A": "Sal has explained this in one of the past videos, we have assumed x=asin(theta) and we have no problem with that, now based on this assumption sin(theta)=x/a now apply the arcsine function, theta= arcsine(x/a), the range of arcsine function is (-pi/2, pi/2) therefore based on our first assumption theta is between -pi/2 and pi/2, in this domain cos function has a positive value, thus cos(theta) is always positive, hope this helped",
        "video_name": "nMrJ6nbOQhQ",
        "timestamps": [
            199
        ],
        "3min_transcript": "So let's make the substitution that x is going to be equal to a sine theta, or three times sine of theta. And then we're going to also have to figure out what dx is equal to. So if you take the derivative, we will get dx. We could have dx d theta is equal to 3 cosine theta. Or if we wanted to write it in differential form, we could write that dx is equal to 3 times cosine theta d theta. This is just the derivative of this with respect to theta. And we're ready to substitute back. Our original expression now becomes-- I'll write it in that original green-- 3 sine theta to the third power, which is the same thing as 27 sine-- actually, let me color code it just so you know what parts I'm doing. So this part right over here, x to the third, is now going to become 27 sine to the third power of theta, And then all of this business, this is going to be the square root of 9 minus x squared, so minus 9 sine squared theta. And then dx-- let me do this in a new color-- dx right over here is going to be equal to-- that's not a new color. dx is going to be equal to all of this business, so times 3 cosine theta d theta. And now let's see if we can simplify this business a little bit. Let me do this over to the side. This thing right over here can be written as 9 times 1 minus sine squared theta, which is equal to the square root of 9 times cosine squared theta. as we did in the last video. And so this is going to be equal to 3 cosine theta in orange. So this right over here is 3 cosine theta. And so what does this simplify to? We have a 27 times 3 is 81 times 3 is going to be 243. So this is going to be 243-- I'll put it out front-- times the integral of-- let's see, we're going to have sine cubed theta. And then you're going to have cosine theta times cosine theta, or we could say cosine squared theta. That's this term right over here and this term right over there, and of course, d theta."
    },
    {
        "Q": "at 2:42 in the video, he is able to rule out cos and pick sin because he says one of them evaluates to 0. What does that mean, its very confusing.\n",
        "A": "sin(kx) = 0 if x is 0. cos(kx) = 1 if x is 0. So where the function crosses the y-axis, the function will be at the midline for sine and the maximum for cosine. Since it s at the midline, we can eliminate cosine.",
        "video_name": "yHo0CcDVHsk",
        "timestamps": [
            162
        ],
        "3min_transcript": "Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be? at the period of this function. Let's see. If we went from this point-- where we intersect the midline-- and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let us just remind ourselves what the period of sine of x is. So the period of sine of x-- so I'll write \"period\" right over here-- is 2pi. You increase your angle by 2 pi radians or decrease it. you're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now, your x, your input is increasing k times faster."
    },
    {
        "Q": "At 0:17 what does Sal mean by when he says the midline and the amplitude are not just the plain vanilla function?\n",
        "A": "In the plain versions of the sine and cosine functions, the midline would be at y = 0, and the amplitude would be 1. Since this is not the case in the given example, these considerations will come into play.",
        "video_name": "yHo0CcDVHsk",
        "timestamps": [
            17
        ],
        "3min_transcript": "Write the equation of the function f of x graphed below. And so we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or a cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum points, it's a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4. Divided by 2 is negative 2. So this right over here is the midline. So this is y is equal to negative 2. So it's clearly shifted down. Actually, I'll talk in a second about what type of an expression it might be. But now, also, let's think about its amplitude. Its amplitude-- that's how far it might get away from the midline-- we see here. Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be?"
    },
    {
        "Q": "Sal begins explaining which function to choose @ about 2:00 .\nI'm still not sure why Cos(0) = 1 and Sin(0) = 0\nIs it because in the unit circle when an angle is 0 Cos = the x value, would still be one, but Sin = the y value which would be 0?\n",
        "A": "That s exactly why, yes. For any angle you can refer to the unit circle definition of sine and cosine.",
        "video_name": "yHo0CcDVHsk",
        "timestamps": [
            120
        ],
        "3min_transcript": "Write the equation of the function f of x graphed below. And so we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or a cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum points, it's a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4. Divided by 2 is negative 2. So this right over here is the midline. So this is y is equal to negative 2. So it's clearly shifted down. Actually, I'll talk in a second about what type of an expression it might be. But now, also, let's think about its amplitude. Its amplitude-- that's how far it might get away from the midline-- we see here. Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be?"
    },
    {
        "Q": "\nat 3:01 how is it 625 i am getting confused",
        "A": "It s 625 because it s 5^4 (he says 5 x 5 x 5 x 5) 5 x 5 (5 squared) =25 25 x 5 (5 cubed) =125 125 x 5 (5^4) = 625",
        "video_name": "XZRQhkii0h0",
        "timestamps": [
            181
        ],
        "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)"
    },
    {
        "Q": "\n2:03 , Is there a simple way to do exponents?",
        "A": "Well, think of the exponent like this: the raised number is how many times you multiply the the other number. For example, 2 to the power of 3 is just 2x2x2.",
        "video_name": "XZRQhkii0h0",
        "timestamps": [
            123
        ],
        "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)"
    },
    {
        "Q": "At 2:46 we multipy 25 times 5 then 5 again?\n",
        "A": "YES!! It is 5 x 5 = 25 25 x 5 = 125 125 x 5 = 625",
        "video_name": "XZRQhkii0h0",
        "timestamps": [
            166
        ],
        "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)"
    },
    {
        "Q": "at 7:12. when sal was copy pasting the diagram in the middle, what about the two right triangles sticking out. aren't those a part of the diagram too?\n",
        "A": "Sal only used those triangles to show that the heights of the parallelograms are equal to \u00f0\u009d\u0091\u008e and \u00f0\u009d\u0091\u008f respectively, thereby showing that their respective areas are definitely equal to \u00f0\u009d\u0091\u008e\u00c2\u00b2 and \u00f0\u009d\u0091\u008f\u00c2\u00b2, which is of course imperative to the proof. After that the triangles can be discarded because they are no longer needed for the rest of the proof.",
        "video_name": "rcBaqkGp7CA",
        "timestamps": [
            432
        ],
        "3min_transcript": "That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I wanted to do that b in blue. So let me do the b in blue. And then this right angle once we've rotated is just sitting right over here. Now, let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now, what is the area of this parallelogram right over there? Well, once again to help us visualize it, we can draw it sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. That's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives us right there. We know that this is 90 degrees. We did a 90-degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now, things are starting to get interesting. And what I'm going to do is I'm going to copy and paste the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I'm going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a few parts of it. I want to do that in black so that it cleans it up. So let me clean this thing up, so we really get the part that we want to focus on. So cleaning that up and cleaning this up, cleaning this up right over there. And actually, let me delete this right down here as well,"
    },
    {
        "Q": "\nIn around 0:02, Sal mentions \"hypotenuse\". What douse that mean?",
        "A": "also the equation being A squared + B squared = C squared",
        "video_name": "rcBaqkGp7CA",
        "timestamps": [
            2
        ],
        "3min_transcript": "In case you haven't noticed, I've gotten somewhat obsessed with doing as many proofs of the Pythagorean theorem as I can do. So let's do one more. And like how all of these proofs start, let's construct ourselves a right triangle. So I'm going to construct it so that its hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible, so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We can have two sides that are equal. But I'll just draw it so that it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle. So maybe it goes right over there. That's side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90-degree angle. So the first thing that I'm going to do it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale as much as I can eyeball it. This side of length a will now look something like this. It'll actually be parallel to this over here. So let me see how well I can draw it. So this is the side of length a. And if we cared, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. That's going to be 90 degrees. That's going to be 90 degrees. Now, let me draw side b. So it's going to look something like that And this and the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now, what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially-- and let me label. So this is height c right over here. Let me do that white color. This is height c. Now, what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line."
    },
    {
        "Q": "Why x-c at an approximate 6:30???\n",
        "A": "Do you remember how a parabola f(x) = x^2 can be translated? f(x) = x^2 is a parabola opening up, centered at 0 . A new function g(x) = (x - 2)^2 is just like f(x) = x^2, just moved over 2 units to the right on the x-axis. The same idea can be applied to this video. Instead of approximating the function at 0 , we approximate the function at a new point x = c. Hope that helps.",
        "video_name": "1LxhXqD3_CE",
        "timestamps": [
            390
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nSal Uses The F.O.I.L Method at 1:11 right? Just Making Sure!",
        "A": "Yes! He is using the F.O.I.L Method!",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            71
        ],
        "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9."
    },
    {
        "Q": "\ndoes it matter which variable is which number? for example, in 2:57, he says that a could equal 1, and b could equal 9. So, my question is, could he have said a=9 and b=1 and have the equation come out to be the same? (other than switching the values of a and b of course)",
        "A": "Yes, he could have said it that way. The order of factors does not matter.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            177
        ],
        "3min_transcript": "Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. times x plus 9. And if you multiply these two out, using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x, plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form, so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. I think the more examples we do the more sense this'll make."
    },
    {
        "Q": "3:11- can the the number be 5 and 5 instead of 1 and 9\n",
        "A": "5 and 5 would not work because the numbers have to fit two things. They have to add to be 10 and they have to multiply to be 9. 5+5 = 10 is okay but 5*5 = 25 which is not 9. Notice that 1+9 = 10 okay; and 1*9 = 9 okay as well. Good luck with your factoring.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            191
        ],
        "3min_transcript": "Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. times x plus 9. And if you multiply these two out, using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x, plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form, so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. I think the more examples we do the more sense this'll make."
    },
    {
        "Q": "At 15:10, why can't a * b be taken directly as (-1)(-72) ? Is that a wrong method?\n",
        "A": "a and b also have to sum to -(-18) = 18. They don t, so they re wrong.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            910
        ],
        "3min_transcript": "we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17."
    },
    {
        "Q": "is 2x the same thing as x squared? for example, around 1:30 when we use the foil method to do (x+a) (x+b) and we do x times x, does it become 2x or x squared?\n",
        "A": "2x = x + x x^2 = x * x. So, you want x^2",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            90
        ],
        "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9."
    },
    {
        "Q": "\nat 15:30 can't we do this without factoring out negative 1?",
        "A": "Most likely, but factoring out the -1 is easy, and the tools we have for factoring are known to work on positive x-squared s. In the real world all behavior that is not specifically forbidden is allowed... in mathematics all that is not specifically allowed is forbidden. So I would first have to verify that the factoring procedure I was using was certain to work with negative leading coefficients.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            930
        ],
        "3min_transcript": "we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17."
    },
    {
        "Q": "\nI have a question at 6:11 min in the video. I would like to know if the answers for x1 and x2 are negative or positive? I ask because I see that clearly they are -3 and -8 in the binomial answer (x-3)(x-8) but all the online Quadratic Equation Solvers show that the answers are positive numbers i.e. x1=3 and x2=8. thx",
        "A": "They would be positive, because you want x-3 or x-8 to be equal to zero, so the right side is equal to zero. 3-3=0 and 8-8=0, so your answers would be positive.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            371
        ],
        "3min_transcript": "already did 10x, let's do a different number-- x squared plus 15x, plus 50. And we want to factor this. Well, same drill. We have an x squared term. We have a first degree term. This right here should be the sum of two numbers. And then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that, when I multiply them I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50. 2 times 25. Let's see, 4 doesn't go into 50. It could be 5 times 10. Let's try out these numbers, and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5, times x plus 10. And multiply it out. I encourage you to multiply this out, and see that this is indeed x squared plus 15x, plus 10. In fact, let's do it. x times x, x squared. x times 10, plus 10x. 5 times x, plus 5x. 5 times 10, plus 50. Notice, the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x, plus 50. signs in here. Let's say I had x squared minus 11x, plus 24. Now, it's the exact same principle. I need to think of two numbers, that when I add them, need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. When I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive, or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them, I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number, so the fact that their sum is"
    },
    {
        "Q": "\nAt 0:45, where did you get the (x+a)(x+b) from?",
        "A": "He was just showing you the general method, it wasn t directly part of the main problem.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            45
        ],
        "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9."
    },
    {
        "Q": "\nat 0:01 til 4:14 way do we need to emphasise this point in short way do i need this video !",
        "A": "One real life situation that this is used for is exchanging money or making change. Because many people have not studied this idea, they get very nervous when exchanging money. They don t feel confident that Ten $20 bills is the same as Two $100 bills. Similarly they don t feel confident that 35 dimes is the same as 3 $1 bills and 10 nickles.",
        "video_name": "3szFVS5p_7A",
        "timestamps": [
            1,
            254
        ],
        "3min_transcript": "I want to think about all of the different ways we can represent value in the number 21.3. So one is to just look straight up at the place values. This 2 is in the tens place, so it literally represents 2 tens. So this is equal to 20, 2 times 10. This 1 is literally equal to 1. It's 1 one. And then this 3 is 3/10, so plus 3/10. But now I want to rearrange or regroup the value in these places. So, for example, I could take 1 from the ones place and give it to the tenths place. So let's see how that would work. So we're going to take 1 away from the ones place, and so it's going to become a 0. And we're going to give it to the tenths place. And what we're going to see is that that's going to make the tenths place into 13/10. Now, does that actually make sense that I took 1 from here and it essentially added 10 to the tenths place? So we still have 2 tens. So this is still going to be 2 tens. Now we have plus 0 ones. And we essentially wanted to write that 1 that we took away from the ones place in terms of tenths. So if we were to write this in terms of tenths, it would be 10/10 plus the 3/10 that were already there. And so this is going to be equal to 13/10. Let me write that down. So this is equal to 20. That's the color you can't see. This is equal to 20 plus 0 ones, so 2 tens plus 0 ones plus 13 tens. Let's do another example with this exact same number. So once again, 21.3. And I'll write it out again. This is equal to 20 plus 1. Plus 1 plus 3/10, plus 3 over 10. Now, I could take 1 from the tens place so that this becomes just 1. Now what do I do with that 10? Well, let's say with that 10 I give 9 of it to the ones place. So I give 9 of it to the ones place so that this becomes 10. And I still have 1 left over, and I give it to the tenths place, so that's going to become 13/10. So what did I just do? Well, I could rewrite this. Let me be clear what I did. This is the same thing as 1 plus 9. Actually, let me write it this way. 1 plus 9 plus 1. That's obviously the same thing-- 10"
    },
    {
        "Q": "At 8:05, Can anybody say please what means \"slope=6\"? What means that value?\n",
        "A": "Slope is: (change in y)/(change in x) In linear notation: y = m\u00e2\u0080\u00a2x + b m is the slope derived from: (y(2) - y(1))/(x(2) - x(1)) b is the y intercept: b = y - m\u00e2\u0080\u00a2x Slope = 6 means that for every \u00c2\u00b11 in x, results in a \u00c2\u00b16 in y.",
        "video_name": "IePCHjMeFkE",
        "timestamps": [
            485
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nHow did he get 6 delta x as part of the y coordinate at 4:13?",
        "A": "He foiled the (3 + delta X)^2. 3 * 3 + 3 * delta X + delta X * 3 + delta X * delta X = 9 + 3deltaX + 3deltaX + deltaX^2 = 9 + 6deltaX + deltaX^2",
        "video_name": "IePCHjMeFkE",
        "timestamps": [
            253
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 6:33 how does it equal 6+deltax and not 6+deltax^2?",
        "A": "Immediately before that point we ve determined that the slope is equal to (6\u00ce\u0094x + (\u00ce\u0094x)\u00c2\u00b2)/\u00ce\u0094x. Divide 6\u00ce\u0094x by \u00ce\u0094x to get 6 and divide (\u00ce\u0094x)\u00c2\u00b2 by \u00ce\u0094x to get \u00ce\u0094x.",
        "video_name": "IePCHjMeFkE",
        "timestamps": [
            393
        ],
        "3min_transcript": ""
    },
    {
        "Q": "1:58 You can solve the quadratic equation without using the quadratic formula like this:\n4r^2 - 8t + 3 = 0 |Divide by four:\nr^2 - (8/4)r + 3/4 = 0 |Multiply both numerator and denominator of 3/4 by four:\nr^2 - (8/4)r + 12/16 = 0 |Figure out two numbers a,b that satisfy a + b = -8 and ab = 12:\na = -2, b = -6 |Substitute a,b into (r + a/4) (r + b/4) = 0:\n(r - 2/4) (r - 6/4) = 0\n(r - 1/2) (r - 3/2) = 0\nr = 1/2 or r = 3/2\n",
        "A": "It seems easier to just -> (2r - 1)(2r - 3)",
        "video_name": "3uO_uPb9H8w",
        "timestamps": [
            118
        ],
        "3min_transcript": "Let's solve another 2nd order linear homogeneous differential equation. And this one-- well, I won't give you the details before I So the differential equation is 4 times the 2nd derivative of y with respect to x, minus 8 times the 1st derivative, plus 3 times the function times y, is equal to 0. And we have our initial conditions y of 0 is equal to 2. And we have y prime of 0 is equal to 1/2. Now I could go into the whole thing y is equal to e to the rx is a solution, substitute it in, then factor out e to the rx, and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video, just to see where that characteristic equation comes from. But in this video, I'm just going to show you, literally, how quickly you can do these type of problems mechanically. characteristic equation is going to be-- and I'll do this in a different color-- 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick, you just substitute the 2nd derivatives with an r squared, the first derivatives with an r, and then the function with-- oh sorry, no. This is supposed to be a constant-- And then the coefficient on the original function is just a constant, right? I think you see what I did. 2nd derivative r squared. 1st derivative r. No derivative-- you could say that's r to the 0, or just 1. But this is our characteristic equation. And now we can just figure out its roots. This is not a trivial one for me to factor so, if it's not So we could say the solution of this is r is equal to negative b-- b is negative 8, so it's positive 8-- 8 plus or minus the square root of b squared. So that's 64, minus 4 times a which is 4, times c which is 3. All of that over 2a. 2 times 4 is 8. That equals 8 plus or minus square root of 64 minus-- what's 16 times 3-- minus 48. All of that over 8. What's 64 minus 48? Let's see, it's 16, right? 10 plus 48 is 58, then another-- so it's 16. So we have r is equal to 8 plus or minus the square root"
    },
    {
        "Q": "\nAt 6:04 is multiplying only top equation right? It seems intuitive, but doesn't that affect the solution?",
        "A": "No, that one of the most common ways to save a system of equations. Since you are multiplying the whole equation (both sides of the equal sign), you are not modifying the equation at all.",
        "video_name": "3uO_uPb9H8w",
        "timestamps": [
            364
        ],
        "3min_transcript": "This differential equations problem was literally just a problem in using the quadratic equation. And once you figure out the r's you have your general solution. And now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3/2 times c1 e to the 3/2 x, plus-- derivative of the inside-- 1/2 times c2 e to the 1/2 x. And now let's use our actual initial conditions. I don't want to lose them-- let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1/2. So let's use that information. So y of 0-- what happens when you substitute x is equal to 0 here.? You get c1 times e to the 0, essentially, so that's just 1, plus c2-- well that's just e to the 0 again, because x is 0-- is equal to-- so this is, when x is equal to 0, what is y? y is equal to 2. Y of 0 is equal to 2. And then let's use the second equation. So when we substitute x is equal to 0 in the derivative-- so when x is 0 we get 3/2 c1-- this goes to 1 again-- plus 1/2 c2-- this is 1 again, e to the 1/2 half times 0 is e to the 0, which is 1-- is equal to-- so when x is 0 for the derivative, y is equal to 1/2, or the derivative is 1/2 at And now we have two equations and two unknowns, and we could solve it a ton of ways. I think you know how to solve them. Let's multiply the top equation-- I don't know-- let's multiply it by 3/2, and what do we get? We get-- I'll do it in a different color-- we get 3/2 c1 plus 3/2 c2 is equal to-- what's 3/2 times 2? It's equal to 3. And now, let's subtract-- well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1/2 minus 3/2? 1/2 minus 1 and 1/2. Well, that's just minus 1, right? So minus c2 is equal to-- what's 1/2 minus 3?"
    },
    {
        "Q": "\nWhat is the \"a\" he is talking about in 5:00?",
        "A": "Sal is just using a as an arbitrary critical point. It can mean any critical point, and it s analogous to using letters such as x or \u00ce\u00b2 to represent any given quantity.",
        "video_name": "lFQ4kMcODzU",
        "timestamps": [
            300
        ],
        "3min_transcript": "The slope was undefined right at the point. But it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point. So, so far our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make-- and I think we called this x0, this was x1, this was x2. So this is x3. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative. And then as we cross it, our slope is still negative, we're still decreasing. So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with the criteria for a minimum point. And I think you could see where this is likely to go. We can see that. It's a local minimum just by looking at it. And what's the slope of doing as we approach it? So the function is decreasing, the slope is negative as we approach it. f prime of x is less than 0 as we approach that point. And then right after we cross it-- this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than 0. So this seems like pretty good criteria for a minimum point. f prime of x switches signs from negative to positive as we cross a. If we have some critical point a, the function takes on a minimum value from negative to positive as we cross a, from negative to positive. Now once again, this point right over here, this critical point x sub 3 does not meet that criteria. We go from negative to 0 right at that point then go to negative again. So this is not a minimum or a maximum point."
    },
    {
        "Q": "\nIn 2:12 how did you change 5/5 into a whole?",
        "A": "5/5 is equal to 1. (A whole) Imagine a pie cut into 5 pieces. If all five pieces are there, then the pie is whole.",
        "video_name": "U44my48zgFE",
        "timestamps": [
            132
        ],
        "3min_transcript": "I have a square here divided into one, two, three, four, five, six, seven, eight, nine equal sections. And we've already seen that if we were to shade in one of these sections, if we were to select one of these sections, let's say the middle one right over here, this is one out of the nine equal sections. So if someone said, what fraction of the whole does this purple square represent? Well, you would say, well, that represents 1/9 of the whole. This thing right over here represents 1/9. Now what would happen if we shaded in more than that? So let's say we shaded in this one and this one, let me shade it in a little bit better. And this one and this one right over here. Now what fraction of the whole have we shaded in? Well, each of these, we've already seen, each of these represent 1/9. So that's 1/9, that's 1/9. When I say 1/9, I could also say a ninth. So this is 1/9 or a ninth, so each of these But how many of these ninths do we have shaded in? Well we have one, two, three, four shaded in. So now we have a total of 4/9 shaded it. 4 of the 9 equal sections are shaded in. So 4/9 of the whole is shaded in. Now let's make things a little bit more interesting. Let's shade in. So here I have five equal sections. Let me write this down. I have five equal sections. And let me shade in five of them. So one, two, three, four, five. We already know that each of these sections, each of these situated in sections represent 1/5. So 1/5, another way of saying that is a fifth, is 1/5. But now how much do I have shaded in? Well I have five out of the five equal sections shaded, or I have 5/5 shaded in. And you might be saying, wait, wait, if I have 5/5 shaded in, I've got the whole thing shaded in. And you would be absolutely right. 5/5 is equal to the whole. Now what I want you to do is pause this video and write down on a piece of paper or at least think in your head, what fraction of each of these wholes is shaded in? So let's go to this first one. We have one, two, three, four, five, six equal sections. And we see that one, two, three, four are shaded in. So 4/6 of this figure is shaded in. Let's go over here. We have one, two, three, four, five equal sections. And one, two, three, four are shaded in. So here, 4/5 of this circle is shaded in. Now in this figure, I have two equal sections and both of them"
    },
    {
        "Q": "at (0:30), where did the 5t come from? I watched the last video and was it because the car is going 5 m/s?\n",
        "A": "Yes. He is simply putting the base information from the previous video so that we can continue with the same problem.",
        "video_name": "wToSIQJ2o_8",
        "timestamps": [
            30
        ],
        "3min_transcript": "In the last video, we used a set of parametric equations to describe the position of a car as it fell off of a cliff, and the equations were x as a function of t was-- and I will write that-- x as a function of t was equal to 10, I believe. I did this a couple of hours ago, so I think 10 plus 5t. And y as a function of t was equal to 50 minus 5t squared. And the graph looks something like this. Let me redraw it. It never hurts. That was the y-axis. That's the x-axis. And we saw that at t equals 0, and we could try t equals 0 here, and we'll get the point 10 comma 50. So the point 10 comma 50 was right there. That was at t equal to 0, and then we plotted a few points in the last video. I think t equals 1 was there. t equals 2 was, like, there, and t equal 3 looked something it hurtled to the ground. But this parametric equation actually doesn't just describe this part of the curve. It describes a curve that goes in both directions forever. So it describes a curve that does something like this. If you actually plot t equals minus 1, what do you get here? You get minus 5. So 10 minus 5 is 5. If you put a minus 1 here, this becomes a plus 1. So minus 1, so you get 5, 45. So you get that point right there. And if you did minus 2, you're going to get a point that looks something like that. And minus 3, you're going to get a point something like that. So the whole curve described by this parametric equation will look something like this. Let me do it in a different color. It looks something like this. And the direction as x increases-- sorry, as This is t is equal to minus 3, minus 2, minus 1, 0, 1, 2, and so forth and so on. So, in the last example, our path was actually just a subset of the path described by this parametric equation. And I'm saying all of this because sometimes it's useful to just bound your parametric equation and say this is a path only for certain values of t. And in our example, it was when we leave this point, which is at t equals 0, all the way to when we hit the ground. So if we want to know those boundaries, well, we know the first boundary is t equals 0, so this in the last example, for when we actually talked about a car careening off of a cliff, it was between 0, so t is greater than or And then we have to figure out what t-value makes us hit the ground? Or when does y equal 0? Because when y equals 0-- y is our altitude, so when y equals 0, we've hit the ground."
    },
    {
        "Q": "\nI assume that at 4:34 he meant to say \"1.5\" rather than 1.25?",
        "A": "Yeah, but I think he did it correctly, so...",
        "video_name": "dEAk0BHBYCM",
        "timestamps": [
            274
        ],
        "3min_transcript": "I'm giving up that area. I'm giving up that area there. But this is just an approximation, and maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles So the area of this first rectangle is going to be the height, which is f of 1, times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said was f of 1.5, times the base, times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2-- so plus f of 2 times the base, times delta x. And then, finally, the area of the third rectangle, the height is the function evaluated at 2.5, so plus-- that's a different color than what I wanted to use. I wanted to use that orange color-- so This is going to be equal to our approximate area-- let me make it clear-- approximate area under the curve, just the sum of these rectangles. So let's evaluate this. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2, so it's going to be 2 times 1/2. Plus the function evaluated at 1.25. 1.25 squared is 2.25. And then you add 1 to it, it becomes 3.25. So plus 3.25 times 1/2. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5, so it's 5 times 1/2. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 plus 1. And just to make the math simpler, we can factor out the 1/2. So this is going to be equal to-- write 1/2 in a neutral color-- 1/2 times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1/2 times-- let's see if I can do this in my head. 2 plus 5 is easy. That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25, so it's going to be 10.5 plus 7 is 17.5. So 1/2 times 17.5, which is equal to 8.75,"
    },
    {
        "Q": "\nAt 4:58, shouldn't Sal put a comma instead of a addition sign?",
        "A": "Good question! The complex plane isn t the same as the usual x, y plane because you plot complex numbers on it like 3 + 4i (3 in the x direction and 4 in the imaginary direction) while on the x, y plane you plot an ordered pair like (4,5) (4 in the x direction and 5 in the y). So no there shouldn t be a comma because it is one complex number rather than two real numbers. A complex number has a real part and an imaginary part so you plot them separately on the complex plane. I hope that answers your question :)",
        "video_name": "Efoeqb6tC88",
        "timestamps": [
            298
        ],
        "3min_transcript": "Well along the imaginary axis we're going from negative one to three so the distance there is four. So now we can apply the Pythagorean theorem. This is a right triangle, so the distance is going to be equal to the distance. Let's just say that this is x right over here. x squared is going to be equal to seven squared, this is just the Pythagorean theorem, plus four squared. Plus four squared or we can say that x is equal to the square root of 49 plus 16. I'll just write it out so I don't skip any steps. 49 plus 16, now what is that going to be equal to? That is 65 so x, that's right, 59 plus another 6 is 65. x is equal to the square root of 65. There's no factors that are perfect squares here, this is just 13 times five so we can just leave it like that. x is equal to the square root of 65 so the distance in the complex plane between these two complex numbers, square root of 65 which is I guess a little bit over eight. Now what about the complex number that is exactly halfway between these two? Well to figure that out, we just have to figure out what number has a real part that is halfway between these two real parts and what number has an imaginary part that's halfway between these two imaginary parts. So if we had some, let's say that some complex number, let's just call it a, is the midpoint, it's real part is going to be the mean of these two numbers. So it's going to be two plus negative five. Two plus negative five over two, over two, and it's imaginary part is going to be the mean of these two numbers so plus, plus three minus one. and this is equal to, let's see, two plus negative five is negative three so this is negative 3/2 plus this is three minus 1 is negative, is negative two over two is let's see three, make sure I'm doing this right. Three, something in the mean, three minus one is two divided by two is one, so three plus three. Negative 3/2 plus i is the midpoint between those two and if we plot it we can verify that actually makes sense. So real part negative 3/2, so that's negative one, negative one and a half so it'll be right over there and then plus i so it's going to be right over there."
    },
    {
        "Q": "\nAt 3:07 Sal says that A, B, and D are non-collinear. I get that part. But how are they not coplanar? Or am I seeing this wrong? If A and B exist on all plains, shouldn't they be exist on the same plane as D?",
        "A": "They are coplanar because any three non-collinear points create a plane, but they are not coplanar on plane S in the diagram.",
        "video_name": "J2Qz-7ZWDAE",
        "timestamps": [
            187
        ],
        "3min_transcript": "So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line. But A, B, and D does not sit on-- They are non-colinear. So for example, right over here in this diagram, we have a plane. This plane is labeled, S. But another way that we can specify plane S is we could say, plane-- And we just have to find three non-collinear points on that plane. So we could call this plane AJB. We could call it plane JBW. We could call it plane-- and I could keep going-- plane WJA. But I could not specify this plane, uniquely, by saying plane ABW. And the reason why I can't do this is because ABW are all on the same line."
    },
    {
        "Q": "\n1:47 What does sufficient mean?",
        "A": "Sufficient = enough to fulfill one s needs. It s not sufficient (/it s not enough) to say that three points define a plane, because we also need the points to be non-collinear.",
        "video_name": "J2Qz-7ZWDAE",
        "timestamps": [
            107
        ],
        "3min_transcript": "We've already been exposed to points and lines. Now let's think about planes. And you can view planes as really a flat surface that exists in three dimensions, that goes off in every direction. So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. Now the question is, how do you specify a plane? Well, you might say, well, let's see. Let's think about it a little bit. Could I specify a plane with a one point, right over here? Let's call that point, A. Would that, alone, be able to specify a plane? Well, there's an infinite number of planes that could go through that point. I could have a plane that goes like this, where that point, A, sits on that plane. I could have a plane like that. Or, I could have a plane like this. I could have a plane like this where point A sits on it, as well. So I could have a plane like that. So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line."
    },
    {
        "Q": "\nAt 0:18 sal said that the plane is something that is not curved goes off in every direction infinitely. He said that it is two dimensional and exists in three dimensions. How can it not have round edges if it goes off infinitely in every direction? Wouldn't it eventually become round, once it reaches infinity?",
        "A": "Infinity does not have curves. The negative space the curves create are not everything and thus infinity is a square. Think of it like this: If you had a sandbox and wanted to cover all of it, and the sandbox went on forever, there would be no edge for curves to exist on.",
        "video_name": "J2Qz-7ZWDAE",
        "timestamps": [
            18
        ],
        "3min_transcript": "We've already been exposed to points and lines. Now let's think about planes. And you can view planes as really a flat surface that exists in three dimensions, that goes off in every direction. So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. Now the question is, how do you specify a plane? Well, you might say, well, let's see. Let's think about it a little bit. Could I specify a plane with a one point, right over here? Let's call that point, A. Would that, alone, be able to specify a plane? Well, there's an infinite number of planes that could go through that point. I could have a plane that goes like this, where that point, A, sits on that plane. I could have a plane like that. Or, I could have a plane like this. I could have a plane like this where point A sits on it, as well. So I could have a plane like that. So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line."
    },
    {
        "Q": "\nAt about 1:05 Sal said lumber line\nWhen he meant to say number line",
        "A": "Yes, Sal ( the man speaking in the video ) was meant to say number line.",
        "video_name": "LpLnmuAyNWg",
        "timestamps": [
            65
        ],
        "3min_transcript": "Use a number line to compare 11.5 and 11.7. So let's draw a number line here. And I'm going to focus between 11 and 12, because that's where our two numbers are sitting. They're 11, and then something else, some number of 10ths. So this right here is 11. And this right here would be 12. And then let me draw the 10ths. So this would be smack dab in between. So that would be 11 and 5/10, or that would be 11.5. Well, I've already done the first part. I've figured out where 11.5 is. It's smack dab in between 11 and 12. It's 11 and 5/10. But let me find everything else. Let me mark everything else on this number line. So that's 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10, and then 10/10, right on the 12. It's not completely drawn to scale. I'm hand-drawing it as good as I can. So where is 11.7 going to be? Well, this is 11.5, this is 11.6, this is 11.7. 11 and 7/10. This is 11.7. And the way we've drawn our number line, we are increasing as we go to the right. 11.7 is to the right of 11.5. It's clearly greater than 11.5. 11.7 is greater than 11.5. And really, seriously, you didn't have to draw a number line to figure that out. They're both 11 and something else. This is 11 and 5/10. This is 11 and 7/10. So clearly, this one is going to be greater. You both have 11, but this has 7/10, as opposed to 5/10."
    },
    {
        "Q": "At 0:17, the answer is 11! just kidding.... but isn't this basically a common sense video?\n",
        "A": "Well, it s introducing you to the idea of adding variables to simple equations.",
        "video_name": "P6_sK8hRWCA",
        "timestamps": [
            17
        ],
        "3min_transcript": "Let's say that you started off with 3 apples. And then I were to give you another 7, another 7 apples. So my question to you-- and this might be very obvious-- is how many apples do you now have? And I'll give you a second to think about that. Well, this is fairly basic. You had 3 apples. Now, I'm going to give you 7 more. You now have 3 plus 7. You now have 10 apples. But let's say I want to do the same type of thinking, but I'm too lazy to write the word \"apples.\" Let's say instead of writing the word \"apples,\" I just use the letter a. And let's say this is, say, a different scenario. You start off with 4 apples. And to that, I add another 2 apples. How many apples do you now have? Instead of writing apples, I'm just going to write a's here. So how many of these a's do you now have? And once again, I'll give you a few seconds This also might be a little bit of common sense for you. If you had 4 of these apples or whatever these a's represented, if you had 4 of them and then you add 2 more of them, you're now going to have 6 of these apples. But once again, we started off assuming that a's represent apples. But they could have represented anything. If you have 4 of whatever a represents, and then you have another 2 of whatever a represents, you'll now have 6 of whatever a represents. Or if you just think of it if I have 4 a's, and then I add another 2 a's, I'm going to have 6 a's. You can literally think of 4 a's as a plus a plus a plus a. And if to that, I add another 2 a's-- so plus a plus a, that's 2 a's right over there-- how many a's do I now have? Well, that's 1, 2, 3, 4, 5, 6. I now have 6 a's. So thinking of it that way, let's get a little bit more abstract. Let's say that I have 5 x's, whatever x represents. So I have 5 of whatever that number is. And from that, I subtract 2 of whatever that number is. What would this evaluate to? How many of these x's would I now have? So it's essentially 5x minus 2x is going to be what times x? Once again, I'll give you a few seconds to think about it. Well, if I have 5 of something and I subtract 2 of those away, I'm going to have 3 of that something left. So this is going to be equal to 3x. 5x minus 2x is equal to 3x. And if you really think about what that means, five x's are just x plus x plus x plus x plus x. And then we're going to take away two of those x's. So take away one x, take away two x's. You are going to be left with three x's."
    },
    {
        "Q": "At 4:15, how did he make the expression -8 - 5a into the expression 8 + 5a, I don't understand how it is possible to do that, I thought that you can only make that kind of equation into -8 + (-5a)?\n",
        "A": "He stated that he multiplies both sides of the equation by -1. so he changed the signs of all the terms on both sides of the equation which is valid",
        "video_name": "adPgapI-h3g",
        "timestamps": [
            255
        ],
        "3min_transcript": "We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right, So I'm kind of doing two steps at once, here, but hopefully it makes sense. I'm trying to get rid of the b-x here, and I'm trying to get rid of the five-a here. So, I subtract five-a there, and I'll subtract five-a there, and then let's see what this give us. So, the five-a's cancel out. And, on the left-hand side, I have negative a-x, negative a-x, minus b-x, minus, you know, in that same green color, minus b-x. And on the right-hand side, This is going to be equal to, the b-x's cancel out, and I have negative eight minus five-a. Negative eight minus, in that same magenta color, minus five-a. And let's see, I have all my x's on one side, all my non-x's on the other side. And here I can factor out an x, And, actually, one thing that might be nice. Let me just multiply both sides by negative one. If I multiple both sides by negative one, I get a-x plus b-x, plus b-x is equal to eight plus five-a. That just gets rid of all of those negative signs. And now I can factor out an x here. So let me factor out an x, and I get x times a plus b. A plus b is going to be equal to eight plus five-a. Eight plus five-a. And we're in the home stretch now. We can just divide both sides by a plus b. So we could divide both sides by a plus b. A plus b. And we're going to be left with, x is equal to"
    },
    {
        "Q": "at 2:30, Khan said something called Algebraic Manipulations. What does that mean?\n",
        "A": "Algebraic Manipulation. The key to solving simple algebraic equations containing a single unknown (e.g. x + 6 = 10) is to realize that the equation is an equality.",
        "video_name": "2REbsY4-S70",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] Anna wants to celebrate her birthday by eating pizza with her friends. For $42.50 total, they can buy p boxes of pizza. Each box of pizza costs $8.50. Select the equation that matches this situation. So before I even look at these, let's see if I can make sense of the sentence here. So for $42.50 total, and I'll just write 42.5, especially because in all these choices they didn't write 42.50, they just wrote 42.5 which is equivalent. So 42.50 that's the total amount they spent on pizza and if I wanted to figure out how many boxes of pizza they could buy, I could divide the total amount they spend, I could divide that by the price per box. That would give me the number of boxes. So this is the total, total dollars. This right over here is the dollar per box and then this would give me the number of boxes. Now other ways that I could think about it. I could say, well what's the total that they spend? So 42.50, but what's another way of thinking about the total they spend? Well you could have the amount they spend per box, times the number of boxes. So this is the total they spend and this another way of thinking about the total they spend, so these two things must be equal. So let's see, if I can see anything here that looks like this, well actually this first choice, this, is exactly, is exactly what I wrote over here. Let's see this choice right over here. P is equal to 8.5 x 42.5. Well we've already been able to write an equation that has explicitly, that has just a p on one side and so when you solve for just a p on one side, you get this thing over here, not this thing, so we could rule that out. Over here it looks kind of like this, except the p is on the wrong side. This has 8.5p is equal to 42.5, If we try to get the p on the other side here, you could divide both sides by p, but then you would get p divided by p is one. You would get 42.5 is equal to 8.5/p which is not true. We have 8.5 times p is equal to 42.5, so this is, this is not going to be the case. One thing to realize, no matter what you come up with, if you came up with this first, or if you came up with this first, you can go between these two with some algebraic manipulations. So for example, to go from this blue one to what I wrote in red up here, you just divide both sides by 8.5. So you divide by 8.5 on the left, you divide 8.5 on the right. Obviously to keep the equal sign you have to do the same thing to the left and right, but now you would have 42.5/8.5 is equal to, is equal to p. Which is exactly what we have over there. Let's do one more of these. Good practice."
    },
    {
        "Q": "\nSo the angle with a smaller area can have more degrees as long as they equal 180? (as noted at 1:54 in parallel lines 2)",
        "A": "no, I think he just had it backwards, he switched the 60 and 120, I think he meant to do it the other way around",
        "video_name": "0eDwckZOffc",
        "timestamps": [
            114
        ],
        "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one"
    },
    {
        "Q": "at1:54 can you explain what in the world he's talking?\n",
        "A": "He s saying that both of those angles are equal to 180 degrees. We already know one of them is 60, so to figure out what the other one is you just subtract 60 from 180. And that will leave you with 120. So that mystery angle is 120.",
        "video_name": "0eDwckZOffc",
        "timestamps": [
            114
        ],
        "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one"
    },
    {
        "Q": "\nHow did you get those angles for 1:26 ? I'm confused",
        "A": "He made up those angles.",
        "video_name": "0eDwckZOffc",
        "timestamps": [
            86
        ],
        "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one"
    },
    {
        "Q": "at 2:30,sal said 3,6,12,24's pattern is*2,but itcould be+3,+6,+9,+12...\n",
        "A": "I did changer the pattern",
        "video_name": "l-6uEtTBH7g",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] What I want to in this video is get some practice figuring out patterns and numbers. In particular, patterns that take us from one number to a next number in a sequence. So over here, in this magenta color, I go from 4 to 25 to 46 to 67. So what's the pattern here? How did I get from 4 to 25 and can I get the same way from 25 to 46 and 46 to 67, and I could just keep going on and on and on? Well there's a couple of ways to think about it. When I see 4 and 25, let's see, 25 isn't an obvious multiple of 4. Another way to go from 4 to 25, I could add 21. Let's see, if I add 21, 4 plus 21 is 25. If I were to go from 25 to 46, well I could just add 21 again. It looks like to go from one number to the next I'm just adding. I wrote 12 by accident, 21. I'm just adding 21 over and over again. And if I were to keep going, if I add 21 I'm going to get to 89. If I add 21 to that I'm going to get 110, and I could keep going and going and going. I could just keep adding 21 over and over again. The pattern here is I'm adding 21. Now what about over here, in green? When I look at it at first, it's tempting to say, 3 plus 3 is 6. But then I'm not adding 3 anymore to get from 6 to 12, I'm adding 6. And then to get from 12 to 24, I'm not adding 6 anymore, I added 12. So every time I'm adding twice as much. But maybe an easier pattern might be, another way to go from 3 to 6, isn't to add 3, but to multiply it by 2. and if I multiply by 2 again, I go from 6 to 12. 6 times 2 is 12. If I multiply by 2 again, I'll go to 24. 2 times 12 is 24 and I could keep going on and on and on. 2 times 24 is 48, 96, I could go on and on and on. The pattern here, it's not adding a fixed amount, it's multiplying each number by a certain amount, by 2 in this case, to get the next number. So 3 times 2 is 6, 6 times 2 is 12, 12 times 2 is 24. Alright, now let's look at this last one. The first two terms here are the same, 3 and 6. The first two numbers here. I could say, maybe this is times 2, but then to go from 6 to 9, I'm not multiplying by 2. But maybe I am just adding 3 here. So 3 to 6, I just added 3. Then 6 to 9, I add 3 again, and then 9 to 12, I add 3 again."
    },
    {
        "Q": "3:16, what if you put x instead of nothing for the -8x+8x? Will it still be right?\n",
        "A": "No, that would be incorrect as -8 + 8 does not = 1; it equals zero. If you make -8x + 8x = x, the x has a coefficient of 1. Hope this helps.",
        "video_name": "vN0aL-_vIKM",
        "timestamps": [
            196
        ],
        "3min_transcript": "Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared and then we add 8 Xs to it or you can actually swap these around you could view it as you are subtracting 8 x from positive 8x we'll those are just going to cancel out so it's just going to be zero - I could just write plus zero here, but that'd just be redundant it wouldn't change the value and then finally I have a plus 7 minus 3 well that is just clearly 7 minus 3 is 4 so I have plus 4 And we're done! We've simplified it! 2x to the third plus 2 x squared plus 4"
    },
    {
        "Q": "\nAt around 2:30 Sal says that he used the commutative property of 'addition and subtraction' to change the order, but is there even a 'commutative property of subtraction'? (I understand that one use the commutative property of addition to add both positive and negative numbers in any order).",
        "A": "That is what he meant. Treat subtraction as the addition of a negative and it becomes commutative. a - b = a + (-b) = -b + a",
        "video_name": "vN0aL-_vIKM",
        "timestamps": [
            150
        ],
        "3min_transcript": "simply 3x squared minus 8x plus 7 plus 2x to the third minus x squared plus eight x minus 3 so when we simplify this we're essentially going to add up like terms and just as a reminder we can only add or subtract like terms or simplify like terms and just a reminder and what I mean by that if I had an x squared to an x squared these are like terms they're both x terms raised to the same power the same degree so if I have one x squared and another x squared well then I have 2x squared - this is 2x squared If I have an x to the third - let's say I have 3x to thirds plus another 4x to the thirds Well that means I have 7x to the thirds - 7x to the thirds I can't take an x squared and add it to an x to the third I cannot simplify this in any way, so this you cannot simplify cannot simply - these are not like terms just because they both have Xs The Xs are not to - they are not to the same degree Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared"
    },
    {
        "Q": "At 2:52, when Sal was subtracting the terms, does it mean the term without the numerical coefficient is equal to one?\n",
        "A": "The term of -x^2 has a coefficient of -1. Whenever the coefficient is not shown, it s understood to be 1 or -1, depending on its sign.",
        "video_name": "vN0aL-_vIKM",
        "timestamps": [
            172
        ],
        "3min_transcript": "Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared and then we add 8 Xs to it or you can actually swap these around you could view it as you are subtracting 8 x from positive 8x we'll those are just going to cancel out so it's just going to be zero - I could just write plus zero here, but that'd just be redundant it wouldn't change the value and then finally I have a plus 7 minus 3 well that is just clearly 7 minus 3 is 4 so I have plus 4 And we're done! We've simplified it! 2x to the third plus 2 x squared plus 4"
    },
    {
        "Q": "may someone explain to me what he is starting to do at 0:30 in the video. do I have to do that simplifying in all my problems.\nthanks in advanced.\n",
        "A": "thanks again that helped",
        "video_name": "vN0aL-_vIKM",
        "timestamps": [
            30
        ],
        "3min_transcript": "simply 3x squared minus 8x plus 7 plus 2x to the third minus x squared plus eight x minus 3 so when we simplify this we're essentially going to add up like terms and just as a reminder we can only add or subtract like terms or simplify like terms and just a reminder and what I mean by that if I had an x squared to an x squared these are like terms they're both x terms raised to the same power the same degree so if I have one x squared and another x squared well then I have 2x squared - this is 2x squared If I have an x to the third - let's say I have 3x to thirds plus another 4x to the thirds Well that means I have 7x to the thirds - 7x to the thirds I can't take an x squared and add it to an x to the third I cannot simplify this in any way, so this you cannot simplify cannot simply - these are not like terms just because they both have Xs The Xs are not to - they are not to the same degree Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared"
    },
    {
        "Q": "\nAt 1:47 how did you know the slope equaled 1?",
        "A": "When Rise=Run (delta x = delta y), The slope is one",
        "video_name": "EQoNfxToez0",
        "timestamps": [
            107
        ],
        "3min_transcript": "- [Voiceover] Slope is defined as your change in the vertical direction, and I could use the Greek letter delta, this little triangle here is the Greek letter delta, it means change in. Change in the vertical direction divided by change in the horizontal direction. That is the standard definition of slope and it's a reasonable way for measuring how steep something is. So for example, if we're looking at the xy plane here, our change in the vertical direction is gonna be a change in the y variable divided by change in horizontal direction, is gonna be a change in the x variable. So let's see why that is a good definition for slope. Well I could draw something with a slope of one. A slope of one might look something like... so a slope of one, as x increases by one, y increases by one, so a slope of one... Notice, however much my change in x is, so for example here, my change in x is positive two, I'm gonna have the same change in y. My change in y is going to be plus two. So my change in y divided by change in x is two divided by two is one. So for this line I have slope is equal to one. But what would a slope of two look like? Well, a slope of two should be steeper and we can draw that. Let me start at a different point, so if I start over here a slope of two would look like... for every one that I increase in the x direction I'm gonna increase two in the y direction, so it's going to look like... that. If my change in x is equal to one, my change in y is two. So change in y over change in x is gonna be two over one, the slope here is two. And now, hopefully, you're appreciating why this definition of slope is a good one. The higher the slope, the steeper it is, the faster it increases, the faster we increase in the vertical direction as we increase in the horizontal direction. Now what would a negative slope be? So let's just think about what a line with a negative slope would mean. A negative slope would mean, well we could take an example. If we have our change in y over change in x was equal to a negative one. That means that if we have a change in x of one, then in order to get negative one here, that means that our change in y would have to be equal to negative one."
    },
    {
        "Q": "How did you get Negative 4 at 3:22? You are not starting from 0 you are starting from 1 to go down. wouldn't it be -5?\n",
        "A": "Yes, the distance is -5, as Sal goes on to say. The confusion arises because in finding that distance, Sal subtracted the y-values, the first y-value being at -4, the second point lying at +1, So that subtraction gave him ( -4 ) - ( +1 ) = -5.",
        "video_name": "iX5UgArMyiI",
        "timestamps": [
            202
        ],
        "3min_transcript": "Something strange about my pen tool. It's making that very thin. Let me do it one more time. Okay, that's better. (laughs) The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of the Pythagorean Theorem. This is a right triangle. is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me... So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value in the change of x, and once you square it So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one and we are going to y is equal to negative four. So it would be negative four minus one which is equal to negative five. And so our change in y is negative five. You can view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter. The negative sign goes away. And so, this is going to simplify to seven squared, change in x-squared, is 49. Change in y-squared, negative five squared, is 25. So we get r-squared, we get r-squared is equal to 49 plus 25. So what's 49 plus 25?"
    },
    {
        "Q": "at 7:42 I do not understand why it is our lower bound sqrt of y, from were we concluded that that's the funtion of our curved? thanks\n",
        "A": "The curved shape that we are integrating at that moment of the video is bounded by x = 1 on the upper end and by the curve formed by y = x\u00c2\u00b2 at the lower end. As Sal says, we don t know what point we would chose, so we need a value of x . We need to put the bounds in terms of x so the next step is to find the inverse of y = x\u00c2\u00b2 . We can do that by taking the square root of both side of y = x\u00c2\u00b2 which gives \u00e2\u0088\u009ay = x",
        "video_name": "hrIPO8mQqtw",
        "timestamps": [
            462
        ],
        "3min_transcript": "And then you could imagine that we're looking at this thing from above. So the surface is up here some place and we're looking straight down on it, and so this is just this area. So let's say we wanted to take the integral with respect to x first. So we want to sum up, so if we want the volume above this column, first of all, is this area times dx, dy, right? So let's write the volume above that column. It's going to be the value of the function, the height at that point, which is xy squared times dx, dy. This expression gives us the volume above this area, or this column right here. And let's say we want the sum in the x direction first. So we want to sum that dx, sum one here, sum here, So we're going to sum in the x-direction. So my question to you is, what is our lower bound of integration? Well, we're kind of holding our y constant, right? And so if we go to the left, if we go lower and lower x's we kind of bump into the curve here. So the lower bound of integration is actually the curve. And what is this curve if we were to write x is a function of y? This curve is y is equal to x squared, or x is equal to the square root of y. So if we're integrating with respect to x for a fixed y right here-- we're integrating in the horizontal direction first --our lower bound is x is equal to the square root of y. That's interesting. I think it's the first time you've probably seen a variable bound integral. But it makes sense because for this row that we're adding up right here, the upper bound is easy. The upper bound is x is equal to 1. x is equal to the square root of y. Because you go back like, oh, I bump into the curve. Well the curve is x is equal to the square root of y because we don't know which y we picked. Fair enough. So once we've figured out the volume-- so that'll give us the volume above this rectangle right here --and then we want to add up the dy's. And remember, there's a whole volume above what I'm drawing right here. I'm just drawing this part in the xy plane. So what we've done just now, this expression, as it's written right now, figures out the volume above that rectangle. Now if we want to figure out the entire volume of the solid, we integrate along the y-axis. Or we add up all the dy's. This was a dy right here, not a dx. My dx's and dy's look too similar. So now what is the lower bound on the y-axis if I'm summing"
    },
    {
        "Q": "At 1:57, doesn't 5/2 = 2.5 instead of 2.4?\n",
        "A": "Yes, well caught.",
        "video_name": "LoKEPEPaNm4",
        "timestamps": [
            117
        ],
        "3min_transcript": "We're asked to graph y is equal to 2.5 times x. So we really just have to think about two points that satisfy this equation here, and the most obvious one is what happens when x equals 0. When x equals 0, 2.5 times 0 is going to be 0. So when x is 0, y is going to be equal to 0. And then let's just pick another x that will give us a y that is a whole number. So if x increases by 1, y is going to increase by 2.5. It's going to go right over there, and I could graph it just like that. And we see just by what I just said that the unit rate of change of y with respect to x is 2.5. A unit increase in x, an increase of 1 and x, results in a 2.5 increase in y. You see that right over here. x goes from 0 to 1, and y goes from 0 to 2.5. But let's increase x by another 1, and then y is going to increase by 2.5 again to get to 5. Or you could say, hey, look, if x is equal to 2, 2.5 times 2 is equal to 5. but then they also tell us to select the statements that are true. So the first one is the equation does not represent a proportional relationship. Well, this is a proportional relationship. A proportional relationship is one where, first of all, if you have zero x's, you're going to have zero y's, where y is equal to some constant times x. And here, y is equal to 2.5 times x. So this is definitely a proportional relationship, so I'm not going to check that. The unit rate of the relationship is 2/5. So I'm assuming-- this is a little ambiguous the way they stated it. I'm assuming they're saying the unit rate of change of y with respect to x. And the unit rate of change of y with respect to x is, when x increases 1, y changed 2.5. So here they're saying when x changes by 1, y changes by 0.4, 2/5 is the same thing as 0.4. This should be 5/2. 5/2 would be 2.4. So this isn't right as well. The slope of the line is 2.5. Slope is change in y over change in x. When x changes 1, y changes 2.5. So change in y, 2.5, over change in x, 1. 2.5 over 1 is 2.5. And you could also see it looking at the form of this equation. y is equal to-- this is the slope times x. So that's right. A change of 5 units in x results in a change of 2 units in y. Well, let's test that idea. We know when x is 0, y is 0. So if x goes from 0 to 5, what's going to happen to y? Well, y is going to be 2.5 times 5. 2.5 times 5 is 12.5. So y would not just change 2. It actually would change 12.5. So this isn't right. A change of 2 units in x results in a change of 5 units in y. Well, we see that. A change in 2 units of x results in a change of 5 units in y. That's exactly what we graphed right over here."
    },
    {
        "Q": "\nFor the second triangle at 2:30, when he writes that the two base angles are 3x+5 and x+6, does that mean that 3x+5 is equal to x+6, therefore x= 0.5?",
        "A": "When he calculates the second triangle at 3:30, he writes that the base angles are 3x+5 and x+16. This means the value of x = 11/2, or 5.5: 3x+5 = x+16 2x+5 = 16 2x = 11 x = 11/2",
        "video_name": "ceDV0QBpcMA",
        "timestamps": [
            150
        ],
        "3min_transcript": "The measure of two angles of an isosceles triangle are 3x plus 5 degrees, we'll say, and x plus 16 degrees. Find all possible values of x. So let's think about this. Let's draw ourselves an isosceles triangle or two. So it's an isosceles triangle, like that and like that. And actually, let me draw a couple of them just because we want to think about all of the different possibilities here. So we know, from what we know about isosceles triangles, that the base angles are going to be congruent. So that angle is going to be equal to that angle. That angle is going to be equal to that angle. And so what could the 3x plus 5 degrees and the x plus 16, what could they be measures of? Well, maybe this one right over here has a measure of 3x plus 5 degrees. And the vertex is the other one. So maybe this one up here is the x plus 16 degrees. The other possibility is that this is describing both base angles, in which case, So maybe this one is 3x plus 5, and maybe this one over here is x plus 16. And then the final possibility-- actually we haven't exhausted all of them-- is if we swap these two-- if this one is x plus 16, and that one is 3x plus 5. So let me draw ourselves another triangle. And obviously swapping these two aren't going to make a difference because they are equal to each other. And then we could make that one equal to 3x plus 5. But that's not going to change anything either because they're equal to each other. So the last situation is where this angle down here is x plus 16, and this angle up here is 3x plus 5. This is 3x plus 5. So let's just work through each of these. So in this situation, if this base angle is 3x plus 5, so is this base angle. And then we know that all three of these are going to have to add up to 180 degrees. So we get 3x plus 5 plus 3x plus 5 We have 3x. Let's just add up. You have 3x plus 3x, which gives you 6x, plus another x gives you 7x. And then you have 5 plus 5, which is 10, plus 16 is equal to 26. And that is going to be equal to 180. And then we have, let's see, 180 minus 26. If we subtract 26 from both sides, we get 180 minus 20 is 160, minus another 6 is 154. You have 7x is equal to 154. And let's see how many times-- if we divide both sides by 7,"
    },
    {
        "Q": "\nAt 1:51, I still don't understand how you're able to express the integral in terms of \"n\" or \"m\"... could someone explain/refer me to a link? Otherwise, if I were to do practice problems where I convert the improper integral into a definite integral, I'll only have a slight idea of how to do so.",
        "A": "A definite integral from a to b measures the area under a curve from x = a to x = b. If we have an indefinite integral, say from 0 to infinity, we are saying a = 0 and trying to figure out what the area is when b gets VERY LARGE. So in the context of the video, in the red integral he is setting the upper bound of the integral as a variable m, so we are measuring the area under the curve from 0 to m, and taking the limit as m goes to infinity to see what happens when the upper bound becomes infinitely large.",
        "video_name": "9JX2s90_RNQ",
        "timestamps": [
            111
        ],
        "3min_transcript": "Right here we have the graph of y is equal to 250 over 25 plus x squared. And what I'm curious about in this video is the total area under this curve and above the x-axis. So I'm talking about everything that I'm shading in white here, including what we can't see, as we keep moving to the right and we keep moving to the left. So I'm talking about from x at negative infinity all the way to x at infinity. So first, how would we actually denote this? Well, it would be an improper integral. We would denote this area as the indefinite integral from x is equal to negative infinity to x is equal to infinity of our function, 250 over 25 plus x squared, dx. Now, we've already seen improper integrals where one of our boundaries was infinity. But how do you do it when you have one boundary at positive infinity and one boundary at negative infinity? You can't take a limit to two different things. And so the way that we're going to tackle this into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5."
    },
    {
        "Q": "At 03:28: What is the archtangent?\n",
        "A": "It s the inverse of the tangent function. For example, tangent(Pi/4 + kPi) = 1. So arctangent(1)=Pi/4 + kPi. Where k is a whole number.",
        "video_name": "9JX2s90_RNQ",
        "timestamps": [
            208
        ],
        "3min_transcript": "into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5. is equal to a tangent theta, 5 tangent theta. And since we're going to have to reverse substitute later on, we can also put in the constraint-- well, we'd say x/5 is equal to tangent of theta, which is completely consistent with this first statement. And so if we wanted to have theta expressed as a function of x, we can put the constraint that theta is equal to arctangent of x/5. So once again, this is completely consistent with this over here. x can be 5 tangent of theta, and theta can be equal to arctangent of x/5. So now let's do the substitution. Actually, before that, we also have to figure out what dx is equal to. So dx is equal to-- I'll do it right over here-- well, the derivative of this with respect to theta is 5 secant squared theta d theta. So now we're ready to substitute back in. So all of this business is going to be equal to 250 times dx."
    },
    {
        "Q": "Near 2:52 7/9 does not equal 2/9 + 3/9 x 2/9\n",
        "A": "The second sign is an addition sign. You were mistaken. 2/9 + 3/9 + 2/9 = 7/9.",
        "video_name": "_E9fG8BYcBo",
        "timestamps": [
            172
        ],
        "3min_transcript": "2 over 9 plus 3 over 9, that's going to get us to 5 over 9 so we need 2 more, so that's going to be plus another 2 over 9. Plus another 2/9, so what would this look like? So let's just draw another grid here, so this is going to look like (so I'm going to put it right below it so we can look at what it looks like). So we have 2 ninths, this here is 2 ninths well we have nine equal sections. So 2 ninths is going to be, 1 and 2. And then 2 more ninths, that's 1 and 2. So notice, when I added 2 ninths to 3 ninths to 2 ninths, this equals 7 ninths. And we know that when we add a bunch of fractions like this that have the same denominator, we can just add the numerators. And this is why, this is 2/9 and 3/9 and 2/9 and they give me 7/9. Actually, let's do this again, this is a lot of fun. So let me draw my grid again. And then, let's see what we can do. So let me get my pen tool out and make sure my ink isn't too thick. And then we could add, that won't get us, let's see 1/9. I'm going to try to add 4 fractions here. So first I'm going to try to add 1/9 and see where that get's us. 1/9 is going to get us right over here. So that's 1/9, so let's say we add 2/9 to that. So that's 1 and 2, so that still doesn't get us there. So that's 3/9, 1 plus 2 is 3. So let's add 4 ninths."
    },
    {
        "Q": "at 8:03, why substitute t=b\u0002a?\n",
        "A": "So that by 11:06 we would have a form similar to that we were looking for in the first place. Remember that the inequality was valid for any t, so Sal used one that he knew would lead to a simplification.",
        "video_name": "r2PogGDl8_U",
        "timestamps": [
            483
        ],
        "3min_transcript": "So these two terms result in that term right there. And then if you just rearrange these you have a minus 1 times a minus 1. They cancel out, so those will become plus and you're just left with plus x dot x. And I should do that in a different color as well. I'll do that in an orange color. So those terms end up with that term. Then of course, that term results in that term. And remember, all I did is I rewrote this thing and said, look. This has got to be greater than or equal to 0. So I could rewrite that here. This thing is still just the same thing. I've just rewritten it. So this is all going to be greater than or equal to 0. Now let's make a little bit of a substitution just to clean up our expression a little bit. And we'll later back substitute into this. Let's define this as a. So the whole thing minus 2x dot y. I'll leave the t there. And let's define this or let me just define this X dot x as c. So then, what does our expression become? It becomes a times t squared minus-- I want to be careful with the colors-- b times t plus c. And of course, we know that it's going to be greater than It's the same thing as this up here, greater than or equal to 0. I could write p of t here. Now this is greater than or equal to 0 for any t that I put in here. For any real t that I put in there. Let me evaluate our function at b over 2a. I just have to make sure I'm not dividing by 0 any place. So a was this vector dotted with itself. And we said this was a nonzero vector. So this is the square of its length. It's a nonzero vector, so some of these terms up here would end up becoming positively when you take its length. So this thing right here is nonzero. This is a nonzero vector. Then 2 times the dot product with itself is also going to be nonzero. So we can do this. We don't worry about dividing by 0, whatever else. But what will this be equal to? This'll be equal to-- and I'll just stick to the green. It takes too long to keep switching between colors. This is equal to a times this expression squared. So it's b squared over 4a squared. I just squared 2a to get the 4a squared. Minus b times this. So b times-- this is just regular multiplication. b times b over 2a."
    },
    {
        "Q": "\nFrom 0:40 onwards, I notice that whenever Sal has to find a derivative of a log with an arbitrary base, he always chooses to represent the log (via change of base formula) as a base-e log and not a base-10 log. Is that how you should always interpret a log with an arbitrary base when finding its derivative?",
        "A": "It is not about interpreting (what ever you meant by that). The only reason, we transform to base-e is simply that we know its derivative. There is (as far as I know) no other way to do it.",
        "video_name": "ssz6TElXEOM",
        "timestamps": [
            40
        ],
        "3min_transcript": "We already know that the derivative with respect to x of the natural log of x is equal to 1/x. But what about the derivative, not of the natural log of x, but some logarithm with a different base? So maybe you could write log base b of x where b is an arbitrary base. How do we evaluate this right over here? And the trick is to write this using the change of base formula. So we could write it in terms of logarithms. We know that log-- I'm just going to restate the change of base formula. And I'm going to change from log base b to log base e, which is essentially the natural log. So the change of base formula, we prove it elsewhere on the site. Feel free to search for it on the Khan Academy. The change of base formula tells us that log base b of x is equal to the natural log, if we want to go to log base e. The natural log of x over the natural-- so it makes it clear what I'm doing. Log base e of x over log base e of b, which is the exact same thing as the natural log of x over the natural log of b. So all we have to do is rewrite this thing. This is equal to the derivative with respect to x of the natural log of x over the natural log of b. Or we could even write it as 1 over the natural log of b times the natural log of x. And now this becomes pretty straightforward. Because what we have right here, 1 over the natural log of b, this is just a constant that's multiplying the natural log of x. So we could take it out of the derivative. So this is the same thing as 1 over the natural log of b times the derivative with respect to x of the natural log of x. This thing right over here is just going to be equal to 1/x. So we end up with 1 over the natural log of b times 1/x. So we end up with 1 over the natural log of b times 1/x, or 1 over the natural log of b, which is just a number times x. So if someone asks you what is the derivative with respect to x of log base 5 of x, well, now you know. It's going to be 1 over the natural log of 5 times x, just like that."
    },
    {
        "Q": "At 2:55 , what is meant by derivative of something \" with respect to\" something else?\n",
        "A": "It means that the variable on top of the derivative symbol is changing due to changes in variable on the bottom (the independent variable). For instance, a lot of calculus in physics uses dt as the bottom, meaning with respect to time. So dV/dt describes how volume changes in response to passing time.",
        "video_name": "ssz6TElXEOM",
        "timestamps": [
            175
        ],
        "3min_transcript": "We already know that the derivative with respect to x of the natural log of x is equal to 1/x. But what about the derivative, not of the natural log of x, but some logarithm with a different base? So maybe you could write log base b of x where b is an arbitrary base. How do we evaluate this right over here? And the trick is to write this using the change of base formula. So we could write it in terms of logarithms. We know that log-- I'm just going to restate the change of base formula. And I'm going to change from log base b to log base e, which is essentially the natural log. So the change of base formula, we prove it elsewhere on the site. Feel free to search for it on the Khan Academy. The change of base formula tells us that log base b of x is equal to the natural log, if we want to go to log base e. The natural log of x over the natural-- so it makes it clear what I'm doing. Log base e of x over log base e of b, which is the exact same thing as the natural log of x over the natural log of b. So all we have to do is rewrite this thing. This is equal to the derivative with respect to x of the natural log of x over the natural log of b. Or we could even write it as 1 over the natural log of b times the natural log of x. And now this becomes pretty straightforward. Because what we have right here, 1 over the natural log of b, this is just a constant that's multiplying the natural log of x. So we could take it out of the derivative. So this is the same thing as 1 over the natural log of b times the derivative with respect to x of the natural log of x. This thing right over here is just going to be equal to 1/x. So we end up with 1 over the natural log of b times 1/x. So we end up with 1 over the natural log of b times 1/x, or 1 over the natural log of b, which is just a number times x. So if someone asks you what is the derivative with respect to x of log base 5 of x, well, now you know. It's going to be 1 over the natural log of 5 times x, just like that."
    },
    {
        "Q": "\nAt 3:56, sal says 25 goes into zero 0 times. How does that work? That number was 60....",
        "A": "You could think it this way... You have no candy, and 25 people want some from you. How are you going to give them candy? So 25 goes into 0, ZERO times!",
        "video_name": "TvSKeTFsaj4",
        "timestamps": [
            236
        ],
        "3min_transcript": "We can divide both sides of this equation by 0.25, or if you recognize that four quarters make a dollar, you could say, let's multiply both sides of this equation by 4. You could do either one. I'll do the first, because that's how we normally do algebra problems like this. So let's just multiply both by 0.25. That will just be an x. And then the right-hand side will be 150 divided by 0.25. And the reason why I wanted to is really it's just good practice dividing by a decimal. So let's do that. So we want to figure out what 150 divided by 0.25 is. And we've done this before. When you divide by a decimal, what you can do is you can make the number that you're dividing into the other number, you can turn this into a whole number by essentially shifting the decimal two to the right. But if you do that for the number in the denominator, you also have to do that to the numerator. So right now you can view this as 150.00. decimal two to the right. Then you'd also have to do that with 150, so then it becomes 15,000. Shift it two to the right. So our decimal place becomes like this. So 150 divided by 0.25 is the same thing as 15,000 divided by 25. And let's just work it out really fast. So 25 doesn't go into 1, doesn't go into 15, it goes into 150, what is that? Six times, right? If it goes into 100 four times, then it goes into 150 six times. 6 times 0.25 is-- or actually, this is now a 25. We've shifted the decimal. This decimal is sitting right over there. So 6 times 25 is 150. You subtract. You get no remainder. Bring down this 0 right here. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. So 150 divided by 0.25 is equal to 600. And you might have been able to do that in your head, because when we were at this point in our equation, 0.25x is equal to 150, you could have just multiplied both sides of this equation times 4. 4 times 0.25 is the same thing as 4 times 1/4, which is a whole. And 4 times 150 is 600. So you would have gotten it either way. And this makes total sense. If 150 is 25% of some number, that means 150 should be 1/4 of that number. It should be a lot smaller than that number, and it is. 150 is 1/4 of 600. Now let's answer their actual question. Identify the percent. Well, that looks like 25%, that's the percent. The amount and the base in this problem."
    },
    {
        "Q": "\n2:10 I still don't understand why it would be x--5, how did he get that from x+5?",
        "A": "x - (-5) is the same as x + 5. The general formula for a circle with center (a, b) and radius r is (x-a)^2 + (y-b)^2 = r^2 for a center at (-5, 7) and a radius of 3, the equation is (x- -5)^2 + (y - 7)^2 = 3^2 or (x+5)^2 + (y - 7)^2 = 9",
        "video_name": "thDrJvWNI8M",
        "timestamps": [
            130
        ],
        "3min_transcript": "- [Voiceover] Whereas to graph the circle x plus five squared plus y minus 5 squared equals four. I know what you're thinking. What's all of this silliness on the right hand side? This is actually just the view we use when we're trying to debug things on Khan Academy. But we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is well what's the center of this equation? Well the standard form of a circle is x minus the x coordinate of the center squared, plus y minus the y coordinate of the center squared is equal to the radius squared. So x minus the x coordinate of the center. So the x coordinate of the center must be negative five. Cause the way we can get a positive five here's by subtracting a negative five. So the x coordinate must be negative five and the y coordinate must be positive five. Cause y minus the y coordinate of the center. So y coordinate is positive five and then the radius squared is going to be equal to four. is equal to two. And the way it's drawn right now, we could drag this out like this, but this the way it's drawn, the radius is indeed equal to two. And so we're done. And I really want to hit the point home of what I just did. So let me get my little scratch pad out. Sorry for knocking the microphone just now. That equation was x plus five squared plus y minus five squared is equal to four squared. So I want to rewrite this as, this is x minus negative five, x minus negative five squared, plus y minus positive five, positive five squared is equal to, instead of writing it as four I'll write it as two squared. So this right over here tells us x equals negative five y equals five and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean Theorem, straight out of the distance formula, which comes out of the Pythagorean Theorem. Remember, if you have some center, in this case is the point negative five comma five, so negative five comma five, and you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them, x comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with radius two around that center. Plus think about how we got that actual formula. Well the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here,"
    },
    {
        "Q": "At 0:26, why tan of a number is equal to sin of the number divided by the cosine of the number?\n",
        "A": "(It s tangent of an a n g l e). Sketch an angle on the unit circle in the 1st quadrant, and include x, y, and r. (y/r)/(x/r) = sin/cos = y/x = tan. Do it for the other 3 quadrants.",
        "video_name": "k_wJsio68D4",
        "timestamps": [
            26
        ],
        "3min_transcript": "Voiceover:The previous video we explored how the cosine and sines of angles relate. We essentially take the terminal ray of the angle and we reflect it about the X or the Y axis, or both axes. What I want to do in this video is think a little bit about the tangent of these different angles. So just as a little bit of a reminder, we know that the tangent of theta is equal to the sine of an angle over the cosine of an angle, and by the Unit Circle Definition, it's essentially saying, \"What is the slope \"of the terminal ray right over here?\" We remind ourselves slope is rise over run. It is our change in the vertical axis over our change in the horizontal axis. If we're starting at the origin, what is our change in the vertical axis if we go from zero to sine theta? Well, our change in the vertical axis is sine theta. What is our change in the horizontal axis? It's cosine of theta. So this is change in Y over change in X for the terminal ray. over cosine of theta, or you could view it as the slope of this ray right over here. Lets think about what other angles are going to have the exact same tangent of theta? This ray is collinear with this ray right over here. In fact if you put them together you get a line. So the tangent of this angle right over here, this pink angle going all the way around, the tangent of pi plus theta, or the tangent of theta plus pi, obviously you could write theta plus pi instead of pi plus theta. This should be, just based on this slope argument, this should be equal to the tangent of theta. Lets see if this actually is the case. So these two things should be equal equal to the slope of the terminal ray. Of course the other side of the angle is going to be the positive X axis based on the conventions that we've set up. Lets think about what it is when the tangent of theta plus pi is in terms of sine and cosine. Let me write this down in the pink color. The tangent. That's not pink. The tangent of pi plus theta, that's going to be equal to, put the parentheses to avoid ambiguity, that's equal to the sine of pi plus theta, or theta plus pi, over the cosine of theta plus pi. And in the previous video we established that the sine of theta plus pi, that's the same thing as negative sine theta. So this is equal to negative sine theta."
    },
    {
        "Q": "Are there any other videos covering the area of equilateral triangles, like why the perpendicular bisector at 2:40 is equal to (\u00e2\u0088\u009a3*s)/2 ?\n",
        "A": "Well you find the area of the circle than find the area of the square than subtract both of them together, in this case if the square is in the circle than the square would be smaller than the circle so you would subtract the area of the square from the area of the circle.",
        "video_name": "QVxqgxVtKbs",
        "timestamps": [
            160
        ],
        "3min_transcript": "the area of each of these equilateral triangles are. And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. But how do we figure out the height of an equilateral triangle? So for example, if I have an equilateral triangle like this-- let me draw it big so I can dissect it little bit-- so I have an equilateral triangle like this. The length of each of the sides are s. And I always have to re-prove it for myself. Just because I always forget the formula. We remember that the angles are 60 degrees, 60 degrees, and 60 degrees. They're all equal. And what I like to do to find out the area of this, in order to figure out the height, is I drop an altitude. So I drop an altitude just like here, and it would split the side in two. I know it doesn't look like it perfectly because I didn't draw it to scale. But it would split it in two. It would form these right angles. split my equilateral triangle into two 30-60-90 triangles. And that's useful because I know the ratio of the sides of a 30-60-90 triangle. If this is s and I've just split this in two, this orange section right over here is going to be s/2. This is also going to be s/2 right over here. They obviously add up to s. And then we know from 30-60-90 triangles, that the side opposite the 60-degree side is square root of 3 times the shortest side. So this altitude right over here, is going to be square root of 3s/2. And now we can figure out a generalized formula for the area of an equilateral triangle. It's going to be equal to 1/2 times the base. Well the base is going to be s. So the base is s. And the height is square root of 3s over 2. see we have in the numerator we have the square root of 3s squared over four. And now we can apply this to figure out the areas of each of these triangles. So this is going to be equal to the area of the larger triangle, is going to be square root of 3/4 times 14 squared. And the area of the smaller triangle is going to be square root of 3/4 times 4 squared. And let's see, we could factor out a square root of 3/4. So this is going to be equal to square root of 3/4 times 14 squared minus 4 squared. Which of course we know is to be, is 16. But now let's actually evaluate this, to actually get a number here. And I could try to simplify it by hand. But instead let me actually just get my-- Actually,"
    },
    {
        "Q": "\nAt 03:15, why is the denominator 4 instead of 2? I am still confused here.",
        "A": "He is calculating the area of the entire equilateral triangle at this point. The formula for the area of an equilateral triangle is 1/2(base x height) He has calculated the height as being the square root of 3 divided by 2. Therefore, Area = 1/2(S x sqr(3)/2), so multiply out the denominators to get (S x sqr(3)) / 4",
        "video_name": "QVxqgxVtKbs",
        "timestamps": [
            195
        ],
        "3min_transcript": "the area of each of these equilateral triangles are. And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. But how do we figure out the height of an equilateral triangle? So for example, if I have an equilateral triangle like this-- let me draw it big so I can dissect it little bit-- so I have an equilateral triangle like this. The length of each of the sides are s. And I always have to re-prove it for myself. Just because I always forget the formula. We remember that the angles are 60 degrees, 60 degrees, and 60 degrees. They're all equal. And what I like to do to find out the area of this, in order to figure out the height, is I drop an altitude. So I drop an altitude just like here, and it would split the side in two. I know it doesn't look like it perfectly because I didn't draw it to scale. But it would split it in two. It would form these right angles. split my equilateral triangle into two 30-60-90 triangles. And that's useful because I know the ratio of the sides of a 30-60-90 triangle. If this is s and I've just split this in two, this orange section right over here is going to be s/2. This is also going to be s/2 right over here. They obviously add up to s. And then we know from 30-60-90 triangles, that the side opposite the 60-degree side is square root of 3 times the shortest side. So this altitude right over here, is going to be square root of 3s/2. And now we can figure out a generalized formula for the area of an equilateral triangle. It's going to be equal to 1/2 times the base. Well the base is going to be s. So the base is s. And the height is square root of 3s over 2. see we have in the numerator we have the square root of 3s squared over four. And now we can apply this to figure out the areas of each of these triangles. So this is going to be equal to the area of the larger triangle, is going to be square root of 3/4 times 14 squared. And the area of the smaller triangle is going to be square root of 3/4 times 4 squared. And let's see, we could factor out a square root of 3/4. So this is going to be equal to square root of 3/4 times 14 squared minus 4 squared. Which of course we know is to be, is 16. But now let's actually evaluate this, to actually get a number here. And I could try to simplify it by hand. But instead let me actually just get my-- Actually,"
    },
    {
        "Q": "\nAt 2:12, Sal mentions that the 1st quartile begins at 14 instead of assuming it would begin at 14 as originally mentioned in the video. Does that mean, that every quartile ends at the end of each one?",
        "A": "In this problem box plot is given, Sal isn\u00e2\u0080\u0099t assuming 14. This data shows age of 100 trees. About 25 trees have the age between 8-14 years. Another 25 have the age between 14-21. You should not see gap between two quartiles(For e.g. Q1 and Q2) in box plots. Hope this helps.",
        "video_name": "b2C9I8HuCe4",
        "timestamps": [
            132
        ],
        "3min_transcript": "An ecologist surveys the age of about 100 trees in a local forest. He uses a box-and-whisker plot to map his data shown below. What is the range of tree ages that he surveyed? What is the median age of a tree in the forest? So first of all, let's make sure we understand what this box-and-whisker plot is even about. This is really a way of seeing the spread of all of the different data points, which are the age of the trees, and to also give other information like, what is the median? And where do most of the ages of the trees sit? So this whisker part, so you could see this black part is a whisker, this is the box, and then this is another whisker right over here. The whiskers tell us essentially the spread of all of the data. So it says the lowest to data point in this sample is an eight-year-old tree. I'm assuming that this axis down here is in the years. And it says at the highest-- the oldest tree right over here is 50 years. So if we want the range-- and when we think of range in a statistics point of view we're thinking of the highest data So it's going to be 50 minus 8. So we have a range of 42. So that's what the whiskers tell us. It tells us that everything falls between 8 and 50 years, including 8 years and 50 years. Now what the box does, the box starts at-- well, let me explain it to you this way. This line right over here, this is the median. And so half of the ages are going to be less than this median. We see right over here the median is 21. So this box-and-whiskers plot tells us that half of the ages of the trees are less than 21 and half are older than 21. And then these endpoints right over here, these are the medians for each of those sections. So this is the median for all the trees that are less than the real median or less than the main median. So this is in the middle of all of the ages of trees that are less than 21. This is the middle age for all the trees that are greater than 21 or older than 21. into four groups. This we would call the first quartile. So I'll call it Q1 for our first quartile. Maybe I'll do 1Q. This is the first quartile. Roughly a fourth of the tree, because the way you calculate it, sometimes a tree ends up in one point or another, about a fourth of the trees A fourth of the trees are between 14 and 21. A fourth are between 21 and it looks like 33. And then a fourth are in this quartile. So we call this the first quartile, the second quartile, the third quartile, and the fourth quartile. So to answer the question, we already did the range. There's a 42-year spread between the oldest and the youngest And then the median age of a tree in the forest is at 21. So even though you might have trees that are as old as 50, the median of the forest is actually closer to the lower end of our entire spectrum of all"
    },
    {
        "Q": "\nWhen he starts grouping @ 2:59 and he switches @ 3:33 and starts to factor out 3f does that mean we can do it both ways and still come out with the correct answer?",
        "A": "Yes. You can do math in quite a few ways, but still end up with the same answers. :) Hope this helps! :)",
        "video_name": "d-2Lcp0QKfI",
        "timestamps": [
            179,
            213
        ],
        "3min_transcript": "times negative 11. So two numbers, so a times b, needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think will work. Right. If we take 22 times negative 3, that is negative 66, and 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, you know, they're going to be of different signs, so the positive versions of them have to be about 19 apart, and that worked out. 22 and negative 3. So now we can rewrite this 19f right here as the sum of That's the same thing as 19f. I just kind of broke it apart. And, of course, we have the 6f squared and we have the minus 11 here. Now, you're probably saying, hey Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6 because they have the common factor of the 3. I like to put the 22 with the negative 11, they have the same common factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And, of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there, but that'll just kind of hang out for awhile. But let's do some grouping. So let's group these first two. And then we're going to group this-- let me get a nice color here-- and then we're going to group this second two. Let me do it in this purple color. And then we can group that second two right there. So these first two, we could factor out a negative 3f, so it's negative 3f times-- 6f squared divided by negative 3f is negative 2f. And then negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f, so we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, can factor out an 11. And if we factor that out, 22f divided by 11 is 2f, and negative 11 divided by 11 is negative 1."
    },
    {
        "Q": "\nAT 3:00 he said that 2-2= 1 but that is not true right ?",
        "A": "It was a mistake",
        "video_name": "uCBm8iDyg1s",
        "timestamps": [
            180
        ],
        "3min_transcript": "So 8 minus 5 is 3 and 9 minus 6 is 3. And now we can bring down the next digit, this 1 here. And now this is where the art is going to come into play because we need to figure out how many times does 65 go into 331 without going over it. And you might just try to look at these numbers, try to approximate them a little bit. You might say, well, maybe 65, let me round this thing up. Maybe this is close to 70. And let's see, this is close to 300. So maybe we say, well, 70 would go into 300. So maybe we think about how many times does 70 go into 300? And we say without going over it, it doesn't go exactly into 300. Well you could say, well how many times does 7 go into 30? Well we know 7 goes into 30 four times. So maybe try a 4 right over here because then this will be 280, 4 times 70 is 280. You're still going to have a little bit left over, but what you have left over is going to be less than 70. It's going to be 20. So you say, well, if this is roughly 70 and if this is roughly 300, then maybe it's going to be the same thing. So let's try that out. Let's see if it goes four times. So 4 times 5 is 20, carry the 2. 4 times 6 is 24 plus 2 is 26. And now let's see how much we had left over. So when we subtract, we are left with-- I'll do this in a new color-- 1 minus 0 is 1. We have a 3 here and a 6 here so we're going to have to do a little regrouping. Let's take 100 from the hundreds place. It becomes 200. Give those 10 tens, that 100, to the tens place. So now we have 13 tens. 13 minus 6 is 7 and then 2 minus 2 is 1. Well no, our remainder, after we said it went in four times, we actually had 71 left over. 71, this right over here, is larger than 65. You don't want a situation where what you have left over is larger than what you're trying to divide into the number. You could have gone into it one more time because you had so much left over. So this 4 was actually too low. We should have probably approximated this as 60, and 60 goes into 300, if we were to estimate, we'd say, well that might be closer to five times. So this is where the art of this comes into play. So it was very reasonable to do what I just did, but it just turned out to not be the right way to think about it. I could just say, well the 4 wasn't enough. I had too much left over. Let me try 5 now. 5 times 5 is 25, carry the 2. 5 times 6 is 30, plus 2 is 32. We got much closer to 331 without going over. Now we can subtract."
    },
    {
        "Q": "At 7:30, in this case we are talking about a Normal Distribution, but what if the coin is an unfair coin? Would we get something completely different? or just something displaced to one of the two sides?\n\nIn other words, what kind of distribution we get when the possible outcomes are not equally likely?\n",
        "A": "Even if the possible outcomes are not equally likely, it will still be a binomial distribution because there are only two possible outcomes from each flip. It doesn t matter what the actual probabilities are of each outcome as long as they sum up to 1.",
        "video_name": "NF0lrkqXIkQ",
        "timestamps": [
            450
        ],
        "3min_transcript": "You could have something that takes on discrete values, but in theory, it could take on an infinite number of discrete values. You could just keep counting higher and higher and higher. But this is discrete, in that it's these whole, these particular values. It can't take on any value in between, and it's also finite. It can only take on x equals zero, x equals one, x equals two, x equals three, x equals four, or x equals five, and you see when you plot its probability distribution, this discrete probability distribution, it starts at 1/32, it goes up, and then it comes back down, and it has this symmetry, and a distribution like this, this right over here, a discrete distribution like this, we call this a binomial distribution, and we'll talk in the future about why it's called a binomial distribution, but a big clue... Actually, I'll tell you why it's called a binomial distribution, is that these probabilities, you can get them using binomial coefficients, using combinatorics. In another video, we'll talk about, why we even call those things binomial coefficients. It's really based on taking powers of binomials in algebra, but this is a very, very, very, very important distribution. It's very important in statistics, because for a lot of discrete processes, one might assume that the underlying distribution is a binomial distribution, and when we get further into statistics, we'll talk why people do that. Now, if you were to have much more than five cases here, if, instead of saying that the number of heads from flipping a coin five times, you said, x is equal to the number of heads of flipping a coin five million times, then, you can imagine, you'd have much, much... The bars would get narrower and narrower relative to the whole hump, and what it would start to do, it would start to approach something that looks really, something that looks really like a bell curve. that you can see better, that I haven't used yet. So if you had more and more of these, if you had more and more of these possibilities, it's going to start approaching what looks like a bell curve, and you've probably heard the notion of a bell curve, and the bell curve is a normal distribution. So one way to think about it, is the normal distribution is a probability density function. It's a continuous case. So, the yellow one, that we're approaching a normal distribution, and a normal distribution, in kind of the classical sense, is going to keep going on and on, normal distribution, and it's related to the binomial. You know, a lot of times in statistics, people will assume a normal distribution, because you can say, okay, it's the product of kind of an almost an infinite number of random processes happening. Here, we're taking a coin, and we're flipping it five times, but if you imagine molecules interacting, or humans interacting, you're saying, oh, there's almost an infinite number of interactions,"
    },
    {
        "Q": "\nAround 8:30, is a normal distribution by definition continuous? How does a very large discrete (binomial?) distribution (flip a coin 5 trillion times) differ from a normal distribution?",
        "A": "It doesn t differ significantly beyond the fact that you have discrete results (which can be easily accounted for using a continuity correction). This is the premise of the Central Limit Theorem which states that the mean of many random variables independently drawn from the same distribution is distributed approximately normally, irrespective of the form of the original distribution (wikipedia.org)",
        "video_name": "NF0lrkqXIkQ",
        "timestamps": [
            510
        ],
        "3min_transcript": "why we even call those things binomial coefficients. It's really based on taking powers of binomials in algebra, but this is a very, very, very, very important distribution. It's very important in statistics, because for a lot of discrete processes, one might assume that the underlying distribution is a binomial distribution, and when we get further into statistics, we'll talk why people do that. Now, if you were to have much more than five cases here, if, instead of saying that the number of heads from flipping a coin five times, you said, x is equal to the number of heads of flipping a coin five million times, then, you can imagine, you'd have much, much... The bars would get narrower and narrower relative to the whole hump, and what it would start to do, it would start to approach something that looks really, something that looks really like a bell curve. that you can see better, that I haven't used yet. So if you had more and more of these, if you had more and more of these possibilities, it's going to start approaching what looks like a bell curve, and you've probably heard the notion of a bell curve, and the bell curve is a normal distribution. So one way to think about it, is the normal distribution is a probability density function. It's a continuous case. So, the yellow one, that we're approaching a normal distribution, and a normal distribution, in kind of the classical sense, is going to keep going on and on, normal distribution, and it's related to the binomial. You know, a lot of times in statistics, people will assume a normal distribution, because you can say, okay, it's the product of kind of an almost an infinite number of random processes happening. Here, we're taking a coin, and we're flipping it five times, but if you imagine molecules interacting, or humans interacting, you're saying, oh, there's almost an infinite number of interactions, which is very, very important in science and statistics. Binomial distribution is the discrete version of that, and one little point of notion, these are where the distributions are, this is where they come from, If you kind of think about as you get more and more trials, the binomial distribution is going to really approach the normal distribution, but it's really important to think about where these things come from, and we'll talk about it much more in a statistics, because it is reasonable to assume an underlying binomial distribution, or normal distribution, for a lot of different types of processes, but sometimes it's not, and even in things like economics, sometimes people assume a normal distribution when it's actually much more likely that the things on the ends are going to happen, which might lead to things like economic crises, or whatever else. But anyway, I don't want to get off-topic. The whole point here is just to appreciate, hey, we started with this random variable, the number of heads from flipping a coin five times, and we plotted it, and we were able to see, we were able to visualize this binomial distribution, and I'm kind of telling you,"
    },
    {
        "Q": "At 0:37, what is the meaning of arbitrary point?\n",
        "A": "An arbitrary point is a point of your choosing. You could pick any point on that line and Sal s proof would still work.",
        "video_name": "KXZ6w91DioU",
        "timestamps": [
            37
        ],
        "3min_transcript": "Let's start off with segment AB. So that's point A. This is point B right over here. And let's set up a perpendicular bisector of this segment. So it will be both perpendicular and it will split the segment in two. So thus we could call that line l. That's going to be a perpendicular bisector, so it's going to intersect at a 90-degree angle, and it bisects it. This length and this length are equal, and let's call this point right over here M, maybe M for midpoint. What I want to prove first in this video is that if we pick an arbitrary point on this line that is a perpendicular bisector of AB, then that arbitrary point will be an equal distant from A, or that distance from that point to A will be the same as that distance from that point to B. So let me pick an arbitrary point on this perpendicular bisector. So let's call that arbitrary point C. And so you can imagine we like to draw a triangle, and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it."
    },
    {
        "Q": "\nAt 2:48 equidistant means equal to right?",
        "A": "Equidistant means equal distance",
        "video_name": "KXZ6w91DioU",
        "timestamps": [
            168
        ],
        "3min_transcript": "and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it. Now, let's go the other way around. Let's say that we find some point that is equidistant from A and B. Let's prove that it has to sit on the perpendicular bisector. So let's do this again. So I'll draw it like this. So this is my A. This is my B, and let's throw out some point. We'll call it C again. So let's say that C right over here, and maybe I'll draw a C right down here. So this is C, and we're going to start with the assumption that C is equidistant from A and B. So CA is going to be equal to CB. This is what we're going to start off with. This is going to be our assumption, and what we want to prove is that C sits on the perpendicular bisector of AB. So we've drawn a triangle here, and we've done this before."
    },
    {
        "Q": "\nAt 3:10, Sal says that that proof covers all the points C, but what if C is on segment AB?",
        "A": "If C is on segment AB, then triangle CAM and triangle CBM would be degenerate triangles (triangles with the measure 0,0, 180) and will technically be congruent.",
        "video_name": "KXZ6w91DioU",
        "timestamps": [
            190
        ],
        "3min_transcript": "and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it. Now, let's go the other way around. Let's say that we find some point that is equidistant from A and B. Let's prove that it has to sit on the perpendicular bisector. So let's do this again. So I'll draw it like this. So this is my A. This is my B, and let's throw out some point. We'll call it C again. So let's say that C right over here, and maybe I'll draw a C right down here. So this is C, and we're going to start with the assumption that C is equidistant from A and B. So CA is going to be equal to CB. This is what we're going to start off with. This is going to be our assumption, and what we want to prove is that C sits on the perpendicular bisector of AB. So we've drawn a triangle here, and we've done this before."
    },
    {
        "Q": "\nAt 0:52 why did he use the word (Concerned?)",
        "A": "I guess because he was concerned about something",
        "video_name": "ory05j2jgBM",
        "timestamps": [
            52
        ],
        "3min_transcript": "- [Voiceover] What I wanna do in this video is compare the fractions 3/4 and 4/5, and I wanna do this visually. So what I'm gonna do is I'm gonna have two copies of the same whole, so let me just draw that, but I'm gonna divide the first one, so this is one whole right over here, this rectangle, when we draw the whole thing. So this is a whole, and right below that, we have the same whole. We have a rectangle of exactly the same size. Now you might notice that I've divided them into a different number of equal sections. In the top one, I've divided it into four equal sections because I am concerned with fourths so I've divided this top whole into fourths and I've divided this bottom whole, or this bottom bar or five equal sections. So let's think about what 3/4 represent. So that's gonna be one of the fourths, right over here, two of the fourths, and then three of the fourths. And what is 4/5 going to be? Well, 4/5 is going to be one fifth, two fifths, three fifths, and four fifths. So when you look at them visually, remember, we're taking fractions of the same whole. This is 3/4 of that rectangle, this is 4/5 of a same-sized rectangle. It wouldn't make any sense if you're doing it for different shapes or different sized rectangles. We just divided them into different sections and you see that if you have four of the fifths, than three of the fourths, and so 4/5 is greater than 3/4 or you could say 3/4 is less than 4/5, or any way you wanna think about it. The symbol you wanna use always opens to the larger number. 4/5 is larger than 3/4, so the large end of our symbol is facing the 4/5, so we would say 3/4 is less than 4/5."
    },
    {
        "Q": "at 0:17,what is a tape graph?!\n",
        "A": "its to help you compare fractions.",
        "video_name": "ory05j2jgBM",
        "timestamps": [
            17
        ],
        "3min_transcript": "- [Voiceover] What I wanna do in this video is compare the fractions 3/4 and 4/5, and I wanna do this visually. So what I'm gonna do is I'm gonna have two copies of the same whole, so let me just draw that, but I'm gonna divide the first one, so this is one whole right over here, this rectangle, when we draw the whole thing. So this is a whole, and right below that, we have the same whole. We have a rectangle of exactly the same size. Now you might notice that I've divided them into a different number of equal sections. In the top one, I've divided it into four equal sections because I am concerned with fourths so I've divided this top whole into fourths and I've divided this bottom whole, or this bottom bar or five equal sections. So let's think about what 3/4 represent. So that's gonna be one of the fourths, right over here, two of the fourths, and then three of the fourths. And what is 4/5 going to be? Well, 4/5 is going to be one fifth, two fifths, three fifths, and four fifths. So when you look at them visually, remember, we're taking fractions of the same whole. This is 3/4 of that rectangle, this is 4/5 of a same-sized rectangle. It wouldn't make any sense if you're doing it for different shapes or different sized rectangles. We just divided them into different sections and you see that if you have four of the fifths, than three of the fourths, and so 4/5 is greater than 3/4 or you could say 3/4 is less than 4/5, or any way you wanna think about it. The symbol you wanna use always opens to the larger number. 4/5 is larger than 3/4, so the large end of our symbol is facing the 4/5, so we would say 3/4 is less than 4/5."
    },
    {
        "Q": "Re: 04:06. Is Sal saying that all (emphasis) 3 points cannot be on the same line to have a triangle, though it is true that there will always be 2 of the 3 points which are on the same line.\n",
        "A": "Any two points, no matter how far apart or weirdly placed (even in 3D, if the two points were at any depth), will always form a line. But if you get three points on the same line, all it would form is a line, and not a triangle, which is a polygon and not a line. To better explain, if you had three points that fell on a same line, you couldn t make a triangle out of them, consequently preventing you from taking a circumcenter (because you don t have a triangle), and finally not letting you make a circle.",
        "video_name": "4_xhiP6g2ow",
        "timestamps": [
            246
        ],
        "3min_transcript": "and the length of OB, so OA is equal to OC is equal to OB, which is the c circumradius. And we've learned when we first talked about circles, if you give me a point, and if we find the locus of all points that are equidistant from that point, then that is a circle. And when I say a locus, all I mean is, the set of all points. If you give me any point right over here, so that's an arbitrary point, and you also specify a radius, and say what is the set of all the points on this two dimensional plane that are equidistant, that are that radius away from the center? It uniquely defines a circle. That's how we defined a circle right over here. And similarly, if you say, look, if you start with the center at O, and you say all of the points that are the circumradius away from O, And that circle will contain the points A, B, and C because those are the circumradius away from O. So they are included in that set. So the circle would look something like-- let me draw it. It would look something like this-- trying my best to draw it, just like that. Everything we've talked about, just now within the last few minutes, is all review. We know all of this. But I went over it just to kind of reinstate a pretty interesting idea, that if you give me three points that defines a unique triangle, and if you have a unique triangle-- And let me make it clear. This is three non-collinear points, so three points not on the same line. If you have three points that are not on the same line, that defines a unique triangle. and circumradius. I'll rewrite it, I don't want to get lazy and confuse you-- circumradius. And if you give me any point in space, any unique point, and a radius, the set of all points that are exactly that radius away from it, that defines a unique circle. So we went through all of this business of talking about the unique triangle, and the unique circumcenter, and the unique radius, to really just show you that if you give me any three points that eventually, really, just defines a unique circle. So just as you need three points to define a triangle, you also need three points to define a circle, two points won't do it. And one way to think about it is, if you give me two points,"
    },
    {
        "Q": "At 1:30, why does using the fraction 9/13 work? I think I understand, but sometimes having other people's thoughts make my own clearer.\n",
        "A": "The fraction 9/13 works because there are 13 integers between 36 and 49 (the closest square numbers). 45 is 9 integers greater than 36, so that gives you 9/13. In other words, you want to find out the relationship of the number you have with the close square numbers you can find nearby.",
        "video_name": "EFVrAk61xjE",
        "timestamps": [
            90
        ],
        "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69."
    },
    {
        "Q": "At 0:50 what does he mean when he says \"the principle root of 49 is 7?\" What is a principle root? Why not just say root?\n",
        "A": "A principal (square) root is a unique, positive square root. 49 has two square roots: 7 & -7, but only has one principal root: 7.",
        "video_name": "EFVrAk61xjE",
        "timestamps": [
            50
        ],
        "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69."
    },
    {
        "Q": "\nWhat if you are doing a problem with an octagon, could it be used the same way? Question starting from 3:24",
        "A": "Yes",
        "video_name": "ZqzAOZ9pP9Q",
        "timestamps": [
            204
        ],
        "3min_transcript": "So this right over here, this distance right over here, h, is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle and then multiply it by how much you go deep, so multiply it by this length. So the volume is going to be the area of this triangle-- let me do it in pink-- the area of this triangle. We know that the area of a triangle is 1/2 times the base times the height. So this area right over here is going to be 1/2 times the base times the height. And then we're going to multiply it by our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4, And we get-- let's see, 1/2 times 4 is 2. So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42. And it would be in some type of cubic units. So if these were in-- I don't know-- centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side is of equal length x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with the cube, where you really just want to find the area of this surface right over here. Now, this is pretty straightforward. This is just a square, or it would Or essentially the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep, so times 3. And so we get 3 times 3 times 3, which is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it. Because, literally, to find the volume of a cube, you take the length of one side, and you multiply that number by itself three times, one for each dimension-- one for the length, the width, and-- or I guess the height, the length, and the depth, depending on how you want to define them. So it's literally just 3 times 3 times 3."
    },
    {
        "Q": "At 13:26, shouldn't it have been sqrt(b2-a2)? He added them, but I thought that was for finding the foci of an ellipse.\n",
        "A": "For finding the foci of hyperbolas, the equation is like the Pythagorean theorum, sqrt(a2+b2) For an ellipse, its sqrt(a2-b2)",
        "video_name": "S0Fd2Tg2v7M",
        "timestamps": [
            806
        ],
        "3min_transcript": "the focal length is the same on either side of the center of the hyperbola depending on how you may view it, but I think that's not too much of a stretch of a statement for you to for you to accept. So if this distance is the same as this distance, then the magenta distance minus this blue distance is going to be equal to this green distance. And this green distance is what? That's 2a. We saw that at the beginning of this video. So this, once again, is also equal to 2a. Anyway, I'll leave you there right now. Actually, let's actually just do one problem, just because I like to make one concrete. Because I told you at the beginning that if you wanted to find the-- so if you have an ellipse-- so if you have-- this is an ellipse, x squared over a squared plus y squared over b squared is equal to 1, we learned that the-- that's over b squared-- this is an ellipse. square root of a squared minus b squared. Now for a hyperbola, you kind of see that there's a very close relation between the ellipse and the hyperbola, but it is kind of a fun thing to ponder about. And a hyperbola's equation looks like this. x squared over a squared minus y squared over b squared, or it could be y squared over b squared minus x squared over a square is equal to 1. It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. So if I were to give you-- so notice the difference. It's just a difference in sign. You're taking the difference of those two denominators, and now you're taking the sum of the two denominators. So if I were to give you the following hyperbola. x squared over 9 plus y squared over 16 is equal to 1. we could just figure out the focal length just by plugging into the formula. The focal length is equal to the square root of a squared plus b squared. This is squared, right? a is three. b is 4. So 9 plus 16 is 25, which is equal to 5. And so if we were to graph this-- that's my y-axis, that's my x-axis-- and the focal length is the distance to, in this case, to the left and the right of the origin. If it was kind of an up and down opening hyperbola, it would be above and below the origin, so this is a-- oh sorry, this should be a plus. We're doing with a hyperbola, that should be a minus. Don't want to confuse you. What I had written before, with a plus, that would have been an ellipse. A minus is the hyperbola. So the two asymptotes-- this is centered at the origin, it hasn't been shifted-- are going to be 16 over 9, so it's going"
    },
    {
        "Q": "\nI understand the difference between a leading coefficient and coefficient. In the expression at 4:40 wouldn't the leading coefficient be 1 since the x^2 is being multipled by 1.",
        "A": "It appears you didn t quite understand it. Leading coefficient means the coefficient of the term with the highest exponent. It is usually the first term, but not necessarily always the case. As you see in the video, 7x\u00e2\u0081\u00b5 was not in the first order. If you were to rewrite them in order as Khan did, you d see it be in the first order. Just remember that the leading is the highest power, not the necessarily the one standing in first order.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            280
        ],
        "3min_transcript": "properly is to understand the coefficients of a polynomial. So let me write a fifth degree polynomial here. And I'm going to write it in maybe a non-conventional form right here. I'm going to not do it in order. So let's just say it's x squared minus 5x plus 7x to the fifth minus 5. So, once again, this is a fifth degree polynomial. Why is that? Because the highest exponent on a variable here is the 5 So this tells us this is a fifth degree polynomial. And you might say, well why do we even care about that? And at least, in my mind, the reason why I care about the degree of a polynomial is because when the numbers get large, the highest degree term is what really dominates all of the other terms. It will grow the fastest, or go negative the fastest, depending on whether there's a But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared. idea of the standard form of a polynomial. Now none of this is going to help you solve a polynomial just yet, but when we talk about solving polynomials, I might use some of this terminology, or your teacher might use some of this terminology. So it's good to know what we're talking about. The standard form of a polynomial, essentially just list the terms in order of degree. So this is in a non-standard form. If I were to list this polynomial in standard form, I would put this term first. So I would write 7x to the fifth, then what's the next smallest degree? Well, they have this x squared term. I don't have an x to the fourth or an x to the third here. So that'll be plus 1-- well I don't have to write 1-- plus x squared. And then I have this term, minus 5x. And then I have this last term right here, minus 5."
    },
    {
        "Q": "at 1:42 Sal said x have to be raised to a non negative power or positive so that means 1/2 power or positive or principal square root should be acceptable as polynomial?\n",
        "A": "The definition of a polynomial is a rational expression in which no variable occurs as a denominator. Perhaps Sal should have been more explicit stating that the powers a variable is raised to is limited to whole numbers which excludes negative numbers and fractional numbers. True that 4/2 is a rational and whole number but it is not allowed when writing a polynomial.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            102
        ],
        "3min_transcript": "In this video I want to introduce you to the idea of a polynomial. It might sound like a really fancy word, but really all it is is an expression that has a bunch of variable or constant terms in them that are raised to non-zero exponents. So that also probably sounds complicated. So let me show you an example. If I were to give you x squared plus 1, this is a polynomial. This is, in fact, a binomial because it has two terms. The term polynomial is more general. It's essentially saying you have many terms. Poly tends to mean many. This is a binomial. If I were to say 4x to the third minus 2 squared plus 7. This is a trinomial. I have three terms here. Let me give you just a more concrete sense of what is and is not a polynomial. For example, if I were to have x to the negative 1/2 plus 1, That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term."
    },
    {
        "Q": "\nAt 6:16 on the video, what would you do if you were in a situation where 2 of the numbers had no exponents?",
        "A": "Most likely that will not happen and if so just add the two numbers in each term",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            376
        ],
        "3min_transcript": "But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared. idea of the standard form of a polynomial. Now none of this is going to help you solve a polynomial just yet, but when we talk about solving polynomials, I might use some of this terminology, or your teacher might use some of this terminology. So it's good to know what we're talking about. The standard form of a polynomial, essentially just list the terms in order of degree. So this is in a non-standard form. If I were to list this polynomial in standard form, I would put this term first. So I would write 7x to the fifth, then what's the next smallest degree? Well, they have this x squared term. I don't have an x to the fourth or an x to the third here. So that'll be plus 1-- well I don't have to write 1-- plus x squared. And then I have this term, minus 5x. And then I have this last term right here, minus 5. it in descending order of degree. Now let's do a couple of operations with polynomials. And this is going to be a super useful toolkit later on in your algebraic, or really in your mathematical careers. So let's just simplify a bunch of polynomials. And we've kind of touched on this in previous videos. But I think this will give you a better sense, especially when we have these higher degree terms over here. So let's say I wanted to add negative 2x squared plus 4x minus 12. And I'm going to add that to 7x plus x squared. Now the important thing to remember when you simplify these polynomials is that you're going to add the terms of the same variable of like degree. I'll do another example in a second where I have multiple variables getting involved in the situation."
    },
    {
        "Q": "At 1:19 why did Sal say \"negative and FRACTIONAL exponent\", isn't it supposed to be RATIONAL exponent? Just need a clarification.\n\nThanks,\nAnkit Thumma\n",
        "A": "No. A rational exponent is anything that does NOT have a square root in it. A fractional exponent is any fraction. When exponents are fractions, they have a square root portion. For example, x^(1/2)= \u00e2\u0088\u009ax You will most likely learn more about this later.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            79
        ],
        "3min_transcript": "In this video I want to introduce you to the idea of a polynomial. It might sound like a really fancy word, but really all it is is an expression that has a bunch of variable or constant terms in them that are raised to non-zero exponents. So that also probably sounds complicated. So let me show you an example. If I were to give you x squared plus 1, this is a polynomial. This is, in fact, a binomial because it has two terms. The term polynomial is more general. It's essentially saying you have many terms. Poly tends to mean many. This is a binomial. If I were to say 4x to the third minus 2 squared plus 7. This is a trinomial. I have three terms here. Let me give you just a more concrete sense of what is and is not a polynomial. For example, if I were to have x to the negative 1/2 plus 1, That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term."
    },
    {
        "Q": "\nat 14:00 it doesnt seem very mathematically sound after working out this whole proof to just assume that (f^2-a^2) = b^2 is there no other way to connect the two without saying that looks like that other thing so lets assume this part of the equation equals that other part?",
        "A": "How do we know f = a + b?",
        "video_name": "HPRFmu7JsKU",
        "timestamps": [
            840
        ],
        "3min_transcript": "Right, I divided both sides by this, so I just get a 1 on this right hand side. Let's see if I can simplify this. If I multiply the numerator and the denominator by a squared, right? As long as I multiply the numerator and the denominator by the same number I'm just multiplying my 1, so I'm not changing anything. So if I do that, the numerator becomes f-- if I multiply it, it becomes f squared minus a squared. I'm just multiplying that times a squared. And the denominator becomes a squared times f squared minus a squared. And all that times x squared. Minus y squared minus a squared mine is a squared is equal to 1. This cancels with this. And we get something to starting to look like the My energy is coming back! It seems like I see the light at the end of the tunnel. We get x squared over a squared minus y squared over f minus a squared is equal to 1. Now this looks a lot like our original equation of the hyperbola, which was x squared over a squared minus y squared over b squared is equal to 1. In fact this is equation of the hyperbola but instead set of writing b squared, since we wrote it, we essentially said, what is the locus of all points where the difference of the distances to those two foci is equal to 2a? And we just played with the algebra for while. It was pretty tiring, and I'm impressed if you've gotten this far into the video, and we got this equation, which should be the equation of the hyperbola, and it is the equation of the hyperbole. It is this equation. So this is the same thing is that. So f squared minus a square. Or the focal length squared minus a squared is You add a squared to both sides, and you get f squared is equal to b squared plus a squared or a squared plus b squared. Which tells us that the focal length is equal to the square root of this. Of a squared plus b squared. And that's what we set out to figure out in the beginning. So hopefully you're now satisfied that the focal length of a hyperbola is the sum of these two denominators. And it's also truth if it is an upward or vertical hyperbola. And if we're dealing with an ellipse, it's the difference of these two-- the square root of the difference of these two numbers. Anyway, I'll leave you there. That was an exhausting problem. I have to go get a glass of water now."
    },
    {
        "Q": "I don't understand how this proof can apply for an ellipse. I've tried the same proof, while naturally adding both d's instead of subtracting one from the other.\nUp until 9:07 the proof works fine, and in the ellipse's case fx and a^2 switch places. But once we raise the equation to the power of two, (fx-a^2)^2 is the exact same thing as (a^2-fx)^2 - and from that point on we can prove that the ellipse is - a hyperbola?!\n",
        "A": "after you factor out a -1 from the denominator of the y term and simplify you get + y^2/(a^2-f^2). With an ellipse f^2 = a^2-b^2. or b^2=a^2-f^2. So y^2/b^2 = y^2/(a^2-f^2). It works. I did the entire proof only to end up with the same equation too. I was also confused about how the same equation could be both an ellipse and a hyperbola but it s the value of f that ultimately determines which one. If f > a, then it s a hyperbola.",
        "video_name": "HPRFmu7JsKU",
        "timestamps": [
            547
        ],
        "3min_transcript": "We have x squared on both sides of this, we subtract x squared from both sides of the equation. We have an f squared on both sides of the equation so let's cancel that out. And let's see, what can we do to simplify it. So we have a minus 2xf and a plus 2xf. Let's add 2xf to both sides this equation or bring this term over here. So if you add 2xf to both sides of this equation-- let's see, my phone is ringing, let me just turn it off --if you add 2xf to both sides of this equation, what do you get? You get 4xf-- remember I just brought this term over this left hand side --is equal to 4a squared plus 4a times a square root of x minus f squared plus y squared. It's easy to get lost in the algebra. Remember all we're doing, just to kind of remind you of what of the distances between these two points, and then see how it relates to the equation of the hyperbola itself. The a's and the b's. Let's take this 4a put it on this side, so you get 4xf minus 4a squared is equal to 4a times the square root of-- well let's just multiply this out 'cause we'll probably have to eventually --x squared minus 2xf plus f squared plus y squared. That's this just multiplied out. That's the y squared right there. We could divide both sides of this by 4. All I'm trying to do is just simplify this as much as possible, so then this becomes xf minus a squared is equal to a times the square root of this whole thing. x squared minus 2xf plus f squared plus why squared. equation right here. And then if you square both sides, this side becomes x squared f squared minus 2a squared xf plus a to the fourth. That's this side squared. And that's equal to, if you square the right hand side, a squared times the square of a square root is just that expression, x squared minus 2xf plus f squared plus y squared. This really is quite hairy. And let's see what we can do now. Let's divide both sides of this equation by a squared, and then you get x squared-- I'm really just trying to simplify this as much as possible --over a squared minus-- so the a"
    },
    {
        "Q": "\nAt 3:02 he devides both sides of the equation by nine, but doesn`t he only have to devide once on the left side of the equation in stead of deviding all numbers?",
        "A": "Since the equation has an equals sign (=), the equation must be kept equal at all times. Multiply the left by 11, you have to multiply the right by 11. If not, then the equation will be false since both sides are no longer equal. However, if you set 1 side equal to 0, you can divide or multiply both sides and still have the 0 side set to 0 since 0 times/divided by anything is 0.",
        "video_name": "vl9o9XEfXtw",
        "timestamps": [
            182
        ],
        "3min_transcript": "That's, literally, the volume of the box. Now they also tell us that the volume of the box is 405 cubic units, is equal to 405. So now we just solve for x. So what do we get here? If we distribute this x into this x plus 4. Actually, if we distribute a 9x. Let me just rewrite it. This is the same thing as 9x times x plus 4 is equal to 405. 9x times x is equal to 9x squared. 9x times 4 is equal to 36x, is equal to 405. Now we want our quadratic expression to be equal to 0. So let's subtract 405 from both sides of this equation. So when you do that, your right-hand side equals 0, and Now, is there any common factor to these numbers right here? Well 405, 4 plus 0 plus 5 is 9, so that is divisible by 9. So all of these are divisible by 9. Let's just figure out what 405 divided by 9 is. So 9 goes into 405-- 9 goes into 40 4 times. 4 times 9 is 36. Subtract you get 45. 9 goes into 45 5 times. 5 times 9 is 45. Subtract, you get 0. So it goes 45 times. So if we factor out a 9 here, we get 9 times x squared-- actually even better, you don't even have to factor out of 9. If you think about it, you can divide both sides of this equation by 9. So if you can divide all of the terms by 9, it won't You're doing the same thing to both sides of equations, which we've learned long ago is a very valid thing to do. So here you get x squared-- if you just had this expression, here, and someone told you to factor it, then you'd have to factor out the 9. But because this is an equation, it equals 0, let's just divide everything by 9. It'll simplify things. So you get x squared plus 4x minus 45 is equal to 0. And now we can try to factor this right here. And this fits the pattern, where we don't have a leading 1 out here. So we don't even have to do it by grouping. You just have to think, what 2 numbers, when I take their product I get negative 45, and when I take their sum, I get positive 4. They are 4 apart. 1 has to be positive, 1 has to be negative. Their positive versions have to be 4 apart. Because when you take the sum, you are really taking their difference because 1 of them is negative. So let's think about it. When you have positive 9 and negative 5, I think that'll work."
    },
    {
        "Q": "at 9:27 what would the answer be if he had done 1000 instead of 100 what would it be?\n",
        "A": "0.00865556 is the answer if sal had done it right. I used a calculator so that may be an abreviated version but that s what I got.",
        "video_name": "ScvuRb6vsz4",
        "timestamps": [
            567
        ],
        "3min_transcript": "This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out. So I'll just keep-- let me just write it like this. So there's going to be 11/1,000 times 0.9 milliliters of our solution per kilogram. So we've gotten this far. So this is per kilogram of patient body weight. And then finally, they tell us how many kilograms our patient weighs. So let's do that last multiplication, and then we can actually get our calculator out and do it all at once. So let's multiply this times-- we want to know how many milliliters per patient. We want the kilograms to cancel out. So we want kilograms per patient. Now we're talking about this particular patient. Not every patient is going to be the same number of kilograms. But if we do this, kilograms will cancel out. We'll have milliliters per patient-- milliliters of solution per patient-- which is exactly what we want. We want milliliters of solution per dose per patient. But everything we've assumed so far has been per dose. Well, there's 72.7 kilograms per patient. That's how much the patient weighs. So we just do this final multiplication and we'll be done. So our answer-- and as these two things are going to cancel out-- so our final answer is going to be 11 times 72.7 divided by 100 times-- actually, 100 times 0.9 is pretty easy to figure out. That's 900. Divided by 900 milliliters per patient. Or you can just say milliliters per dose. However you want to say it. Per dose per patient. Let's get our calculator out and do this. So we have 11 times 72.7 is equal to 799 divided by 900. Is equal to 0.88-- well, we could round up. 0.889. Hopefully the doctor won't mind. So that is equal to-- I'll write it in a nice, vibrant"
    },
    {
        "Q": "\nNot sure what to do with this. \"The 1st unit of PRBCs 250mL was started by the nurse at 7:30PM to be infused at 120 mL/hr in the first 15 minutes; then increased to 225mL/hr after the first 15 minutes. How long will the entire infusion take? (Provide answer in hours and minutes) \"",
        "A": "you need to know the volume of the unit of blood. Typically it s written on the bag. In the equation they should have provided the volume in the unit (usually for an adult) between 280 and a bit over 300ml/unit of blood",
        "video_name": "ScvuRb6vsz4",
        "timestamps": [
            450
        ],
        "3min_transcript": "So let's do that. Pounds in the numerator, pounds in the denominator cancel out. And you multiply 5 times 2.2. This is equal to-- let's see. 5 times 2 is 10. 5 times 0.2 is 1. So this is equal to 11. And then in our numerator, we have milligrams. 11 milligrams per kilogram. So we just converted our dosage information to a pure metric system. It was actually a mix between the metric and the English system before. Now let's see what we can do. Well, let's see if we can get it in terms of how many milliliters we have to deliver per pound. So once again, we want this-- well, actually, let's go to grams first. Because we have milligrams here. We have grams up here. So let's see if we can convert this thing to grams. So just like we did before, we want a milligrams I'll do it in orange. We want a milligrams in the denominator and we want a gram in the numerator. Why did I say that? Because I want this and this to cancel. And I want a grams in the numerator. So how many grams are there per milligram? You can just think it through. There's 1 gram per 1,000 milligrams. Or 1,000 milligrams per gram. And you just multiply it out. So the milligrams cancels with the milligrams, and then we get-- this is equal to 11/1,000 grams per kilogram. So now we have everything in terms of grams, but we want it in terms of milliliters. The question is, how many milliliters of solution per dose? So let me go down here on this line right here. So we had this result. We have 11/1,000-- I won't do the division just yet-- grams This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out."
    },
    {
        "Q": "Do you mean 1000 times .9 @9:28?\n",
        "A": "yes A do bleave that is true!",
        "video_name": "ScvuRb6vsz4",
        "timestamps": [
            568
        ],
        "3min_transcript": "This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out. So I'll just keep-- let me just write it like this. So there's going to be 11/1,000 times 0.9 milliliters of our solution per kilogram. So we've gotten this far. So this is per kilogram of patient body weight. And then finally, they tell us how many kilograms our patient weighs. So let's do that last multiplication, and then we can actually get our calculator out and do it all at once. So let's multiply this times-- we want to know how many milliliters per patient. We want the kilograms to cancel out. So we want kilograms per patient. Now we're talking about this particular patient. Not every patient is going to be the same number of kilograms. But if we do this, kilograms will cancel out. We'll have milliliters per patient-- milliliters of solution per patient-- which is exactly what we want. We want milliliters of solution per dose per patient. But everything we've assumed so far has been per dose. Well, there's 72.7 kilograms per patient. That's how much the patient weighs. So we just do this final multiplication and we'll be done. So our answer-- and as these two things are going to cancel out-- so our final answer is going to be 11 times 72.7 divided by 100 times-- actually, 100 times 0.9 is pretty easy to figure out. That's 900. Divided by 900 milliliters per patient. Or you can just say milliliters per dose. However you want to say it. Per dose per patient. Let's get our calculator out and do this. So we have 11 times 72.7 is equal to 799 divided by 900. Is equal to 0.88-- well, we could round up. 0.889. Hopefully the doctor won't mind. So that is equal to-- I'll write it in a nice, vibrant"
    },
    {
        "Q": "at the point 0:38, how are we expected to know which is concave and which is not?\n",
        "A": "All curves are concave. The question is if they are upward or downward.",
        "video_name": "UK2shgCXALo",
        "timestamps": [
            38
        ],
        "3min_transcript": "If you were paying close attention in the last video, an interesting question might have popped up in your brain. We have talked about the intervals over which the function is concave downwards. And then we talked about the interval over which the function is concave upwards. But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing and then the slope started increasing. So that's one way to look at it. Right here in our function we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name you're probably And you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has a extrema point, like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u. Or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive. But we could have also switched from being positive to negative. So inflection point, your second derivative f prime prime of x switches signs. Goes from being positive to negative or negative to positive. Switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards, the slope was increasing. So the second derivative would to be positive. And then the slope is decreasing, so your second derivative would be negative. So here your second derivative is going from positive to negative. Here your second derivative is going from negative to positive. In either case, you are talking about an inflection point."
    },
    {
        "Q": "At 3:09 you described each of a function and an equation what would an example be of something that falls into both a function and an equation category?\n",
        "A": "Here are some examples: y=f(x)=x\u00c2\u00b2 y=f(x)=x+1 y=f(x)=2 y=f(x)=cos(x) y=f(x)=sin(x) and so on. These equations are functions because they assign each real number x to ONE real number f(x).",
        "video_name": "l3iXON1xEC4",
        "timestamps": [
            189
        ],
        "3min_transcript": "So let me think of it that way. So I'm going to draw-- if this is the world of equations right over here, so this is equations. And then over here is the world of functions. That's the world of functions. I do think there is some overlap. We'll think it through where the overlap is, the world of functions. So an equation that is not a function that's sitting out here, a simple one would be something like x plus 3 is equal to 10. I'm not explicitly talking about inputs and outputs or relationship between variables. I'm just stating an equivalence. The expression x plus 3 is equal to 10. So this, I think, traditionally would just be an equation, would not be a function. Functions essentially talk about relationships between variables. You get one or more input variables, and we'll give you only one output variable. And you can define a function. And I'll do that in a second. You could define a function as an equation, but you can define a function a whole bunch of ways. You can visually define a function, maybe as a graph-- so something like this. And maybe I actually mark off the values. So that's 1, 2, 3. Those are the potential x values. And then on the vertical axis, I show what the value of my function is going to be, literally my function of x. And maybe that is 1, 2, 3. And maybe this function is defined for all non-negative values. So this is 0 of x. And so let me just draw-- so this right over here, at least for what I've drawn so far, defines that function. I didn't even have to use an equal sign. If x is 2, at least the way I drew it, y is equal to 3. You give me that input. I gave you the value of only one output. So that would be a legitimate function definition. Another function definition would be very similar to what you do in a computer program, And if day is equal to Monday, maybe you output cereal. So that's what we're going to eat that day. And otherwise, you output meatloaf. So this would also be a function. We only have one output. For any one day of the week, we can only tell you cereal or meatloaf. There's no days where you are eating both cereal and meatloaf, which sounds repulsive. And then if I were to think about something that could be an equation or a function, I guess the way I think about it is an equation is something that could be used to define a function. So for example, we could say that y is equal to 4x minus 10."
    },
    {
        "Q": "Is that suppose to be a '=' instead of a minus sign at 6:00 ?\n",
        "A": "Yes, it s intended to be a =, not a -.",
        "video_name": "4PCktDZJH8E",
        "timestamps": [
            360
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 4:59 he says \"you can always construct a faster-expanding function.\" Is that really true? What if you had a function that was only defined for the domain consisting of two x-values: 0, 1. And f(0) was defined to be negative infinity, and f(1) was defined to be infinity. How could you construct a faster-expanding function than that?",
        "A": "Sal is specifically talking about real-valued functions. Since infinity is not defined in the context of real numbers, it would not make sense to define a function to have an infinite value. Infinites only make sense as a limiting value, not a functional value.",
        "video_name": "6WMZ7J0wwMI",
        "timestamps": [
            299
        ],
        "3min_transcript": "That's 10. That's 30. That'll be good for approximation. And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Let's say this is negative 1, negative 2, negative 3, negative 4. Then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is right around there. When x is equal to 1, y is equal to 3, which is right around there. When x is equal to 2, y is equal to 9, which is right around there. When x equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. If I did 5, we'd go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1/9. Eventually, you're not even going to see this. It's going to get closer and closer to zero. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative thousand, 3 to the negative million, we're getting numbers closer and closer to zero without actually ever approaching zero. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to zero, we're slowly getting our way ourselves away from zero, but then bam! Once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. Well, you can always construct a faster expanding function. faster expanding, but out of the ones that we deal with in everyday life, this is one of the fastest. So given that, let's do some word problems that just give us an appreciation for exponential functions. So let's say that someone sends out a chain letter in week 1. In week 1, someone sent a chain letter to 10 people. And the chain letter says you have to now send this chain letter to 10 more new people, and if you don't, you're going to have bad luck, and your hair is going to fall out, and you'll marry a frog, or whatever else. So all of these people agree and they go and each send it to 10 people the next week. So in week 2, they go out and each send it to 10 more people. So each of those original 10 people are each sending out 10 more of the letters. So now 100 people have the letters, right?"
    },
    {
        "Q": "\nAt 4:28 Sal says \"were getting getting number closer and closer to 0, without actually ever approaching 0.\", yet the function does approach 0 when counting towards the negative direction of x.",
        "A": "Poor choice of words on Sal s part. But, a box does pop up at 4:33 in the video and tells you that Sal meant without ever reaching 0 .",
        "video_name": "6WMZ7J0wwMI",
        "timestamps": [
            268
        ],
        "3min_transcript": "quickly we're exploding. Let me draw my axes here. So that's my x-axis and that is my y-axis. And let me just do it in increments of 5, because I really want to get the general shape of the graph here. So let me just draw as straight a line as I can. Let's say this is 5, 10, 15. Actually, I won't get to 81 that way. I want to get to 81. Well, that's good enough. Let me draw it a little bit differently than I've drawn it. So let me draw it down here because all of these values, you might notice, are positive values because I have a positive base. So let me draw it like this. Good enough. And then let's say I have 10, 20, 30, 40, 50, 60, 70, 80. That's 10. That's 30. That'll be good for approximation. And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Let's say this is negative 1, negative 2, negative 3, negative 4. Then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is right around there. When x is equal to 1, y is equal to 3, which is right around there. When x is equal to 2, y is equal to 9, which is right around there. When x equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. If I did 5, we'd go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1/9. Eventually, you're not even going to see this. It's going to get closer and closer to zero. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative thousand, 3 to the negative million, we're getting numbers closer and closer to zero without actually ever approaching zero. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to zero, we're slowly getting our way ourselves away from zero, but then bam! Once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. Well, you can always construct a faster expanding function."
    },
    {
        "Q": "at 5:51: Why can f'(a) also be undefined?\n",
        "A": "He is saying that it can either be 0 or undefined. Not 0 and also undefined. At any no endpoint max or min, the derivative has to be either 0 or undefined. This is part of the definition of a non endpoint max/min.",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            351
        ],
        "3min_transcript": "We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting-- and once again, I'm not rigorously proving it to you, I just want you to get the intuition here. We see that if we have some type of an extrema-- and we're not talking about when x is at an endpoint of an interval, just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is where you have an interval from there. So let's say a function starts right over there, and then This would be a maximum point, but it would be an end point. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that, or points like this. We're talking about the points in between. it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3."
    },
    {
        "Q": "\nAt 0:46, what does he mean by x0? Does he mean x to the 0th power?",
        "A": "The subscript appended x_0 signifies a specific value of x. If we are trying to talk about specific values of x, we will just use the subscripts to differentiate them (x_0, x_1, x_2...). This is much easier and makes much more sense than assigning new variables/letters to these values of x.",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            46
        ],
        "3min_transcript": "I've drawn a crazy looking function here in yellow. And what I want to think about is when this function takes on the maximum values and minimum values. And for the sake of this video, we can assume that the graph of this function just keeps getting lower and lower and lower as x becomes more and more negative, and lower and lower and lower as x goes beyond the interval that I've depicted right over here. So what is the maximum value that this function takes on? Well we can eyeball that. It looks like it's at that point right over there. So we would call this a global maximum. Function never takes on a value larger than this. So we could say that we have a global maximum at the point x0. Because f of of x0 is greater than, or equal to, f of x, for any other x in the domain. And that's pretty obvious, when you look at it like this. Now do we have a global minimum point, Well, no. This function can take an arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have-- let me write this down-- we have no global minimum. Now let me ask you a question. Do we have local minima or local maxima? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here, or local minimum here? Well, a local minimum, you could imagine means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving a very rigorous definition here. But one way to think about it is, we can say that we have a local minimum point at x1, is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it, right over there. Now do we have any other local minima? Well it doesn't look like we do. Now what about local maxima? Well this one right over here-- let me do it in purple, I don't want to get people confused, actually let me do it in this color-- this point right over here looks like a local maximum. Not lox, that would have to deal with salmon. Local maximum, right over there. So we could say at the point x1, or sorry, at the point x2, we have a local maximum point at x2. Because f of x2 is larger than f of x for any"
    },
    {
        "Q": "On 5:17, Sal said that there aren't any other solutions. Can't the values for a and b switch around? So what I am basically asking is that instead of 3 solutions, aren't there supposed to be 6 solutions because a and b will switch their values and be a repeat?\n",
        "A": "He s assuming that such things are accounted for (i.e. order is irrelevant, because multiplication and addition are commutative).",
        "video_name": "JPQ8cfOsYxo",
        "timestamps": [
            317
        ],
        "3min_transcript": "So this is pretty interesting here. We can divide both sides by a plus b. We know that a plus b cannot be equal to 0 since all of these numbers have to be positive numbers. And the reason why I say that is if it was 0, dividing by 0 would give you an undefined answer. So if we divide both sides by a plus b, we get a times b is equal to 12. So all the constraints that they gave us boiled down to this right over here. The product of a and b is equal to 12. And there's only so many numbers, so many positive integers where you if you take the product, Let's try them out. So let me write some columns here. Let's say a, b, c. And then we care about their product, so I'll write that over here, so abc. c is the sum of those two, so c is going to be 13, 1 times 12 times 13. 12 times 12 is 144 plus another 12 is going to be 156. And just for fun, you can verify that this is going to be equal to 6 times their sum. Their sum is 26, 26 times 6 is 156. So this one definitely worked. It definitely worked for the constraints. And it should because we boiled down those constraints to a times b need to be equal to 12. So let's try another one. 2 times 6, their sum is 8. And then if I were to take the product of all of these, you get 2 times 6 is 12 times 8 is 96. And then we could try 3 and 4. 3 plus 4 is 7. Actually, I should have known the a times b is always 12, so you just have to multiply 12 times this last column. 12 times 7 is 84. And there aren't any others. You definitely can't go above 12 because then you would have to deal with non-integers. You would have to deal with fractions. You can't do the negative versions of these because they all have to be positive integers. So that's it. Those are all of the possible positive integers where if you take their products you get 12. We've essentially just factored 12. So they want us to find the sum of all possible values of N. Well, these are all the possible values of N. N was the product of those integers. So let's just take the sum. 6 plus 6 is 12 plus 4 is 16, 1 plus 5 is 6 plus 9 is 15 plus 8 is 23, 2 plus 1 is 3."
    },
    {
        "Q": "\nAt around 0:06, if Consistent solution #1 is independent, Consistent solution #2 is dependent, then what is an INconistent solution?",
        "A": "He draws an inconsistent system at 1:20 - it s 2 parallel lines.",
        "video_name": "WSpF5uvApLA",
        "timestamps": [
            6
        ],
        "3min_transcript": "Is the system of linear equations below dependent or independent? And they give us two equations right here. And before I tackle this specific problem, let's just do a little bit a review of what dependent or independent means. And actually, I'll compare that to consistent and inconsistent. So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes. So that's my first x-axis and y-axis. Let me draw another one. That is x and that is y. Let me draw one more, because there's only three possibilities in two dimensions. x and y if we're dealing with linear equations. So you can have the situation where the lines just intersect in one point. Let me do this. So you could have one line like that and they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation-- actually let me draw it over here-- where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept, so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line, so that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking would be considered consistent. But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place. And here they're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent and this one over here dependent. So independent-- both lines are doing their own thing. They're not dependent on each other. They're not the same line. They will intersect at one place. Dependent-- they're the exact same line. Any point that satisfies one line will satisfy the other. Any points that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're going to intersect in one place or going to intersect in an infinite number of places."
    },
    {
        "Q": "\nat 3:06, when 7 doesn't go into 5, why doesn't he write 0, and instead he immediately uses \"51\"?",
        "A": "This is a stylistic preference. Placing a zero to the left of a number will not change its value, so you are free to place a zero there if you like. However, when reporting your final answer, I would not include superfluous digits, so you would have to change it again.",
        "video_name": "cfr-yZxTH8Y",
        "timestamps": [
            186
        ],
        "3min_transcript": "So if we move the decimal over to the right with the 0.7 to turn into a 7, we also need to move the decimal over to the right for 518. Now, you're probably saying, well, I don't see a decimal in 518. Well, there is one. You just didn't have to write it, because it's 518.00-- and we can add as many zeroes as we want. So if we move the decimal to the right, it becomes 5,180. So really what we're saying is 518 divided by 0.7 is the same thing as 5,180 divided by 7. Notice all we did by moving the decimal one place to the right, is we multiplied both of these numbers by 10, which is not going to change the actual value of the decimal. One other way of thinking about this, if you wanted to write this as a fraction, You multiply both the numerator and denominator by 10, you will get 5,180 over 7. So let's clean this up a little bit, just so we remember what we did. So we moved the decimal over to the right, one. So now this is just a 7. The decimal is there. In fact, we really don't have to write the decimal anymore. It's just a 7.0-- you could imagine 7.0, so we can just write this as a 7. And then the 518, the decimal is now out here. So this is 5,180. And let's increase the sign right over here. Now, this is just a straight-up long division problem. How many times does 7 go into 5? Well, it goes 0 times. 0 times 7-- actually, let's just cut to the chase. It does go into 51. 7 times 7 is 49. So it goes 7 times. 7 times 7 is 49. Subtract 51 minus 49 is 2. And now we can bring down this 8. 7 goes into 28 four times. 4 times 7 is 28. Subtract, you get a 0. Now we can bring down another 0. We want at least get to the decimal place. So we bring down another 0 right over here. When I say get to the decimal place, we could put the decimal place up here too, just to make sure we're keeping track of the right place values or that we can have the decimal in the right place. So notice, I'm very particular. When I'm doing 7 goes into 51, I put the 7 right above the 1 in the 51's place. When I'm saying 7 goes into 28, I'm putting the 4 right above the 8 in this one's place when we're doing the division."
    },
    {
        "Q": "At 2:57,why was it 1/2 of the whole triangle?\n",
        "A": "The 1/2 is not based on the triangle, it is based on a triangle being exactly 1/2 of the rectangle that it is enclosed within. In the case that he is using, he is actually showing 1/2 of the parallelogram, not the triangle.",
        "video_name": "rRTXKQpblEc",
        "timestamps": [
            177
        ],
        "3min_transcript": "that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So let's look at some triangles here. So that is a triangle, and we're given the base and the height, and we're gonna try to think about what's the area of this triangle going to be, and you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So let me copy, and then let me paste it, and what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram. that has twice the area of our original triangle, 'cause I have two of our original triangles right over here, you saw me do it, I copied and pasted it, and then I flipped it over and I constructed the parallelogram. Now why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base times this height. You also have height written with the \"h\" upside down over here. It's going to be base times height. That's going to be for the parallelogram, for the entire-- let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So our original triangle is just going to have half the area. So this area right over here is going to be one half the area of the parallelogram. One half base times height. One half base times height. And you might say, \"OK, maybe it worked for this triangle, \"but I wanna see it work for more triangles.\" And so, to help you there, I've added another triangle right over here, you could do this as an obtuse triangle, this angle right over here is greater than 90 degrees, but I'm gonna do the same trick. We have the base, and then we have the height. if you start at this point right over here, and if you drop a ball, the length that the ball goes, if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height, it's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, \"Well, this is the height.\" How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine, it's going to be one half base times height. How do we feel good about that? Well, let's do the same magic here."
    },
    {
        "Q": "What was that thing that popped up at 2:08?\n\n(It seems to be something called ArtRage. . .)\n",
        "A": "I agree because it said ART, and the person was drawing",
        "video_name": "rRTXKQpblEc",
        "timestamps": [
            128
        ],
        "3min_transcript": "- [Voiceover] We know that we can find the area of a rectangle by multiplying the base times the height. The area of a rectangle is equal to base times height. In another video, we saw that, if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, then the area is still going to be base times height. Now, it's not as obvious when you look at the parallelogram, but in that video, we did a little manipulation of the area. We said, \"Hey, let's take this \"little section right over here.\" So we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that, as long as the base and the height is the same, as this rectangle here, I'm able to construct the same rectangle by moving that area over, and that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So let's look at some triangles here. So that is a triangle, and we're given the base and the height, and we're gonna try to think about what's the area of this triangle going to be, and you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So let me copy, and then let me paste it, and what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram. that has twice the area of our original triangle, 'cause I have two of our original triangles right over here, you saw me do it, I copied and pasted it, and then I flipped it over and I constructed the parallelogram. Now why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base times this height. You also have height written with the \"h\" upside down over here. It's going to be base times height. That's going to be for the parallelogram, for the entire-- let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So our original triangle is just going to have half the area. So this area right over here is going to be one half the area of the parallelogram."
    },
    {
        "Q": "\nAt 2:18, why is there -B on the right side?",
        "A": "It is the same thing as saying: A=50-B It also equals: A=50+-B",
        "video_name": "3mimxluSVBo",
        "timestamps": [
            138
        ],
        "3min_transcript": "Use graphing to solve the following problem. Abby and Ben did household chores last weekend. Together they earned $50, and Abby earned $10 more than Ben. How much did they each earn? So let's define some variables here. Let's let A equal Abby's earnings. And let's let B equal Ben's earnings. Then they tell us how these earnings relate. They first tell us that together they earned $50. So that statement can be converted mathematically into-- well, together, that means the sum of the two earnings. So A plus B needs to be equal to $50. Abby's plus Ben's earnings is $50. And then they tell us Abby earned $10 more than Ben. is equal to Ben's earnings plus 10. Abby earned $10 more than Ben. So we have a system of two equations and actually with two unknowns. And then they say, how much did each earn? So to do that, and they want us to solve this graphically. There's multiple ways to solve it, but we'll do what they ask us to do. Let me draw some axes over here. And I'll be in the first quadrant since we're dealing with earnings, so neither of their earnings can be negative. And let me just define the vertical axis as Abby's axis or the Abby's earnings axis. And let me define the horizontal axis as Ben's axis or Ben's earnings axis. And let me just graph each of these equations. And to do that, I'm going to take this first equation, and I'm going to put it in the equivalent of slope-intercept form. It might look a little unfamiliar to you, but it really is slope-intercept form. Let me rewrite it first. So we have A plus B is equal to 50. So let's subtract B from both sides. And then we get A is equal to negative B plus 50. So if you think about it this way, when B is equal to 0, A is going to be 50. So we know our A intercept, we could call it. We normally would call that a y-intercept, but now this is the A axis. So this right here, let me call this 10, 20, 30, 40, and 50. So if Ben made $0, then Abby would have to make $50 based on that first constraint. So we know that that's a point on the line right over there. And we also know that the slope is negative 1, that B is the independent variable, the way I've written it over here, and this coefficient is negative 1. Or another way to think about it is if A is 0, then B is going to be 50. If Abby made no money, then Ben would have to make $50."
    },
    {
        "Q": "\nAt 3:47 Sal said factor but didn't he mean simplify?",
        "A": "Hi Knight of Stone / Ninjormon (Mormon ninja), Sal did say factor, but meant simplify. They have added the correction to the video in the box which shows up at 3:45. Hope that helps! - JK",
        "video_name": "f-wz_ZzSDdg",
        "timestamps": [
            227
        ],
        "3min_transcript": "In the numerator, we took care of our a plus 2's. That's the only one that's common, so in the numerator, we also have an a minus 2. Actually, we have an a plus 1-- let's write that there, too. We have an a plus 1 in the numerator. We also have an a plus 1 in denominator. In the numerator, we have an a minus 2, and in the denominator, we have an a minus 1. So all I did is I rearranged the numerator and the denominator, so if there was something that was of a similar-- if the same expression was in both, I just wrote them on top of each other, essentially. Now, before we simplify, this is a good time to think about the domain or think about the a values that aren't in the domain, the a values that would invalidate or make this expression undefined. Like we've seen before, the a values that would do that are the ones that would make the denominator equal 0. So the a values that would make that equal to 0 is a is equal to negative 2. You could say a plus 2 is equal to 0, or a is equal to negative 2. a plus 1 is equal to 0. Subtract 1 from both sides. a is equal to negative 1. Or a minus 1 is equal to 0. Add one to both sides, and you get a is equal to 1. For this expression right here, you have to add the constraint that a cannot equal negative 2, negative 1, or 1, that a can be any real number except for these. We're essentially stating our domain. We're stating the domain is all possible a's except for these things right here, so we'd have to add that little caveat right there. Now that we've done that, we can factor it. We have an a plus 2 over an a plus 2. We know that a is not going to be equal to negative 2, so that's always going to be defined. When you divide something by itself, that is going to just be 1. The same thing with the a plus 1 over the a plus 1. That's going to be 1. over a minus 1. So the simplified rational is a minus 2 over a minus 1 with the constraint that a cannot equal negative 2, negative 1, or 1. You're probably saying, Sal, what's wrong with it equaling, for example, negative 1 here? Negative 1 minus 1, it's only going to be a negative 2 here. It's going to be defined. But in order for this expression to really be the same as this expression up here, it has to have the same It has to have the same domain. It cannot be defined at negative 1 if this guy also is not defined at negative 1. And so these constraints essentially ensure that we're dealing with the same expression, not one that's just close."
    },
    {
        "Q": "at 2:10, to think of it as 4/8 is that the easiest way to think about 1/2?\n",
        "A": "Yes, 4/8 is equal to 1/2, because 4 is half of 8 and 1 is half of 2. 1/2 is the lowest simplified form of 4/8.",
        "video_name": "erZe85NrsK0",
        "timestamps": [
            130
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhy did Sal take anti derivative at 8:25?",
        "A": "Because derivative of arctan(2x) = f(x) at 2:46 So, antiderivative of f(x) = arctan(2x) at 8:25",
        "video_name": "NgYrsqoKXpM",
        "timestamps": [
            505
        ],
        "3min_transcript": "If f of x is equal to that then f of x is power series representation its just going to be taking this power series or at least the first four terms of it and replacing the x's with four x squares and then multiplying the whole thing times two, so let's do that. We can write that f of x, so we could write the f of x is going to be approximately equal to two times this thing, a value when x is equal to four x squared. It is one minus instead of an x, I'm going to write a four x squared plus x squared but instead of an x, I have a four x squared squared. This is plus four x squared squared. Well that's going to be 16 x to the fourth, so let me just write that, that's going to be plus 16 x to the fourth but now an x is four x squared so it's minus four x squared to the third power. Well that's going to be 64 x to the sixth. Let me write that, minus 64 x to the sixth power and then we could say that this is going to be so f of x is approximately equal to, distribute the two, two minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Just like that with a little bit of a substitution, I was able to reasonably a non hairy fashion find the first four nonzero terms of the power series of two over one plus four x squared which is the derivative, which is going to be the derivative of the power series of arctangent of two x. Let's write this down, so I'm going to write, I'm going to write down, arctangent of two x which is equal to an antiderivative of f of x, dx which is going to be equal to an antiderivative of this whole, all of this business. It's the antiderivative of two minus eight, minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Actually let me make this approximate because now of course we're doing approximation with the power series. Dx and what is this going to be equal to? We get approximately equal to well I'm going to have a constant there. Let me write the constant first because when we write our power series or Maclaurin Series, the first term is the constant term. It's our function evaluated to add zero. We're going to have a constant,"
    },
    {
        "Q": "\nDear Sal, I think you made a mistake at 1:20 - 1:21 when you said that the third term is ar to the third power. Didn't you write ar^2/ar squared? I am slightly confused there. It might've been a careless mistake.",
        "A": "yeah it was just a careless mistake",
        "video_name": "dIGLhLMsy2U",
        "timestamps": [
            80,
            81
        ],
        "3min_transcript": "Let's talk about geometric sequences, which is a class of sequences where we start at some number, then each successive number is the previous number multiplied by the same thing. So what am I talking about? Well let's multiply a times r. And then I'm going to get ar. Let's multiply it times, but to get the third term, let's multiply the second term times r. And then what am I going to have? I'm going to have-- it's a different shade of yellow-- I'm going to have ar squared. Multiply by r again, you're going to get ar to the third power, and you just keep on going like that. And this is, the way I've denoted this, this is an infinite geometric sequence. We just keep going on and on and on and on. And the different ways we can denote it, we can denote it explicitly. We could say that our sequence is a sub n starting with the first term going all the way to infinity, with a sub for any term-- is going to be a times r. And just to be clear, this right over here, a is the same thing as a times r to the zeroth power, r to the 0 is just 1. This second term is ar to the first power. The third term is ar to the third power. It looks like the nth term is going to be ar to the n minus 1 power. So ar to the n minus 1. And you could verify it. If you want the second term, you say a times r to the 2 minus 1, a times r to the first power. It works out. This is defining it explicitly. We could also define it recursively. We could say a sub n from n equals 1 to infinity, with a sub 1 being equal to a. That's the base case. a sub 1 is equal to a, ar to the 0 is just a. Or we could say for n equals 1, and then we could say a-- that because we're making it very clear that a sub 1 is equal to a-- and then we could say a sub n is equal to the previous term, a sub n minus 1, times r, for n is greater than or equal to 2. So this is saying, look, our first term is going to be a, that right over there is a, ar to the 0 is just a, and then each successive term is going to be the previous term times r, which is exactly what we did over there. So let's look at some geometric sequences. So I could have a geometric sequence like this. I could have a sub n, n is equal to 1 to infinity with, let's say, a sub n is equal to, let's say our first term is, I don't know, let's say it is equal to 20. And then r, the number that we're multiplying to get each successive term, let's say it's equal to 1/2. 1/2 to the n minus 1."
    },
    {
        "Q": "\nAt 3:47, what was that other sequence that wasn't geometric, and how do you solve it/put it into explicit form? For instance, my sequence is 1, 3, 6, 10, 15, and 21. How could I solve to find the 100th term?\nSide note just found out this is a triangular sequence. no idea wat that means...",
        "A": "In general, there s no easy way to do this. But for a lot of cases, taking the sequence of differences or the sequence of quotients is a good way to approach this. In your case, the sequence of differences is 2,3,4,5,6,... From there we can say that a\u00e2\u0082\u0099 = a\u00e2\u0082\u0099\u00e2\u0082\u008b\u00e2\u0082\u0081 + n+1 and a\u00e2\u0082\u0081 = 1 so a\u00e2\u0082\u0099 = 1+2+3+4+...+n+(n+1) = n(n+1)/2.",
        "video_name": "dIGLhLMsy2U",
        "timestamps": [
            227
        ],
        "3min_transcript": "that because we're making it very clear that a sub 1 is equal to a-- and then we could say a sub n is equal to the previous term, a sub n minus 1, times r, for n is greater than or equal to 2. So this is saying, look, our first term is going to be a, that right over there is a, ar to the 0 is just a, and then each successive term is going to be the previous term times r, which is exactly what we did over there. So let's look at some geometric sequences. So I could have a geometric sequence like this. I could have a sub n, n is equal to 1 to infinity with, let's say, a sub n is equal to, let's say our first term is, I don't know, let's say it is equal to 20. And then r, the number that we're multiplying to get each successive term, let's say it's equal to 1/2. 1/2 to the n minus 1. Well let's think about it. The first term is 20. If you say, if n is 1, this is going to be 1/2 to the 0-th power. So it's going to be 1 times 20. So the first term is 20, and then each time we're multiplying by what? Well here each time we're multiplying by 1/2. So this could be 20 times 1/2 is 10, 10 times 1/2 is 5, 5 times 1/2 is 2.5-- actually let me just write that as a fraction, is 5/2, 5/2 times 1/2 is 5/4, and you can just keep going on and on and on. This is a geometric sequence. Now let me give you another sequence, and tell me if it is geometric. So let's say we start at 1, so then I'm going to go to 2, and then I'm going to go to 6, and then I'm going to go to-- let me see what I want to do-- I want to go to 24. And then I could go to 120, and I go on and on and on. Well let's think about what's going on. To go from 1 to 2, I multiplied by 2. To go from 2 to 6, I multiplied by 3. To go from 6 to 24, I multiplied by 4. So I'm always multiplying not by the same amount. You have to multiply by the same amount in order for it to be a geometric sequence. Here I'm multiplying it by a different amount. So this sequence that I just constructed has the form, I have my first term, and then my second term is going to be 2 times my first term, and then my third one is going to be 3 times my second term, so 3 times 2 times a. My fourth one is 4 times the third term, so 4 times 3 times 2 times a."
    },
    {
        "Q": "\ni dont get what you sal said at 5:36",
        "A": "because he is smart",
        "video_name": "t8m0NalQtEk",
        "timestamps": [
            336
        ],
        "3min_transcript": "So you put the 10 there. There's nothing left. You put the 0 there. There's nothing left to put the 1, so you put the 10 there. So you essentially have solved the problem that 36-- let me do this is another color. That 36 times 3 is equal to 108. That's what we've solved so far, but we have this 20 sitting out here. We have this 20. We have to figure out what 20 times 360 is. Or sorry, what 20 times 36 is. This 2 is really a 20. And to make it all work out like that, what we do is we throw a 0 down here. We throw a 0 right there. In a second I'm going to explain why exactly we did that. So let's just do the same process as we did before with the 3. Now we do it with a 2, but we start filling up here and move to the left. So 2 times 6. That's 12. We put the 1 up here and we have to be very careful because we had this 1 from our previous problem, which doesn't apply anymore. So we could erase it or that 1 we could get rid of. If you have an eraser get rid of it, or you can just keep track in your head that the 1 you're about to write is a different 1. So what were we doing? We wrote 2 times 6 is 12. Put the 2 here. Put the 1 up here. And I got rid of the previous 1 because that would've just messed me up. Now I have 2 times 3. 2 times 3 is equal to 6. But then I have this plus 1 up here, so I have to add plus 1. So I get 7. So that is equal to 7. 2 times 3 plus 1 is equal to 7. So this 720 we just solved, that's literally-- let me write that down. That is 36 times 20. 36 times 20 is equal to 720. throw this 0 here. If we didn't throw that 0 here we would have just a 2-- we would just have a 72 here, instead of 720. And 72 is 36 times 2. But this isn't a 2. This is a 2 in the 10's place. This is a 20. So we have to multiply 36 times 20, and that's why we got 720 there. So 36 times 23. Let's write it this way. Let me get some space up here. So we could write 30-- well, actually, let me just finish the problem and then I'll explain to you why it worked. So now to finish it up we just add 108 to 720. So 8 plus 0 is 8. 0 plus 2 is 2. 1 plus 7 is 8. So 36 times 23 is 828. Now you're saying Sal, why did that work? why were we able to figure out separately 36 times 3 is equal to 108, and then 36 times 20 is equal to 720, and then"
    },
    {
        "Q": "At 1:39 Sal says that -10 and -4 will work for both 40 and -14. Would it matter if you used 8 and 5 too?\n",
        "A": "It matters... The 2 numbers you pick must add to -14 AND multiply to 40. While 8 and 5 multiply to 40, they do not add to -14. 8+5 = +13. So, the 8 and 5 will not work. Hope this helps.",
        "video_name": "1kfq0aR3ASs",
        "timestamps": [
            99
        ],
        "3min_transcript": "To better understand how we can factor second degree expressions like this, I'm going to go through some examples. We'll factor this expression and we'll factor this expression. And hopefully it'll give you a background on how you could generally factor expressions like this. And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx plus a times b. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about well, what's our coefficient on our x term, and can I figure out two numbers that when I take their sum, are equal to that coefficient, and what's my constant term, and can I think of two numbers, those same two numbers, that when I take the product equal that constant term? If we look at our coefficient on x, can we think of an a plus ab that is equal to that number negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40? So what's an a and a b that would work over here? Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. What happens if we make them both negative? If we have negative 4 plus negative 10, well that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive, this number right over here is positive, tells you that these are going to be the same sign. then we would have different signs. And so if you have 2 numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12?"
    },
    {
        "Q": "\nI Don't Get what he said at 2:03.",
        "A": "He is writing the remainder part as a single fraction instead of two separate fractions.",
        "video_name": "WqNc6My1aNU",
        "timestamps": [
            123
        ],
        "3min_transcript": "The quotient of two polynomials-- a of x and b of x-- can be written in the form a of x over b of x is equal to q of x plus r of x over b of x-- where q of x and r of x are polynomials and the degree of r of x is less than the degree of b of x. Write the quotient 7x to the sixth plus x to the third plus 2x plus 1 over x squared in this form. Well, this one is pretty straightforward because we're dividing by x squared. So you could literally view this as 7x to the sixth divided by x squared plus x to the third divided by x squared plus 2x divided by x squared plus 1 divided by x squared. So we could just do this term by term. What's 7x to the sixth divided by x squared? Well, x to the sixth divided by x squared is x to the fourth. So it's going to be 7x to the fourth power. And then, same thing right over here. Plus x to the third divided by x squared. So plus x. And then, we're going to have 2x divided by x squared. But remember, we want to write it in a form of r of x over b of x-- where r of x has a lower degree than b of x. Well, 2x has a lower degree than x squared. Here this is degree 1. This is degree 2. So you could write it as plus 2x over x squared. Like that. And then, you could write plus 1 over x squared. So you could do this-- plus 1 over x squared. So you could write it like that. But that's not exactly the form that they want. They want us to write it q of x-- and you could view that as 7x to the fourth plus x. And then, they want plus r of x over b of x So plus some polynomial over x squared in this case. So instead of writing it as 2x over x squared plus 1 over x squared, we could just write it as 2x plus 1 over x squared. So let me just put some parentheses here so that it interprets my typing correctly. So notice, this part of the polynomial, these terms have an equal or higher degree than x squared. So I just divided those. 7x to the sixth divided by x squared is 7x to the fourth. x to the third divided by x squared is x. And then, once I got two terms that had a lower degree than x squared, I just left on there. I just said plus whatever 2x plus 1 divided by x squared is. And that's the form that they wanted us to write it in. We'll check our answer. And we got it right."
    },
    {
        "Q": "\nAt 2:11 You factor the 15x and not the 8 but at 4:45 you factor the 16 and 10x. Why was the 8 not a factor in the first part?",
        "A": "In the first part we need two numbers such that their product is 16(8*2)(coefficient of x^2 and constant term) and their sum is 15(coefficient of x). The nos. are 16 and -1 Similarly, in the 2nd part. But, we factor out only 16 as coefficient of x^2 over there is just 1.",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            131,
            285
        ],
        "3min_transcript": "f of x is equal to 2x squared plus 15x minus 8. g of x is equal to x squared plus 10x plus 16. Find f/g of x. Or you could interpret this is as f divided by g of x. And so based on the way I just said it, you have a sense of what this means. f/g, or f divided by g, of x, by definition, this is just another way to write f of x divided by g of x. You could view this as a function, a function of x that's defined by dividing f of x by g of x, by creating a rational expression where f of x is in the numerator and g of x is in the denominator. And so this is going to be equal to f of x-- we have right up here-- is 2x squared 15x minus 8. And g of x-- I will do in blue-- is right over here, g of x. So this is all going to be over g And you could leave it this way, or you could actually try to simplify this a little bit. And the easiest way to simplify this would see if we could factor the numerator and the denominator expressions into maybe simpler expressions. And maybe some of them might be on-- maybe both the numerator and denominator is divisible by the same expression. So let's try to factor each of them. So first, let's try the numerator. And I'll actually do it up here. So let's do it. Actually, I'll do it down here. So if I'm looking at 2x squared plus 15x minus 8, we have a quadratic expression where the coefficient is not 1. And so one technique to factor this is to factor by grouping. You could also use the quadratic formula. And when you factor by grouping, you're going to split up this term, this 15x. And you're going to split up into two terms where the coefficients are, if I were to take the product of those coefficients, of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared"
    },
    {
        "Q": "\nAt 3:35, how did Sal get 2x(x+8) from 2x^2 + 16x?",
        "A": "Sal is distributing the 2x and multiplying with the x and the 8. To get a better visualization, learn about the distributive property.",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            215
        ],
        "3min_transcript": "of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared All I did here is I took this middle term and, using this technique right over here, I split it into 16x minus x, which is clearly still just 15x. Now what's useful about this-- and this is why we call it factoring by grouping-- is we can see, are there any common factors in these first two terms right over here? Well, both 2x squared and 16x, they are both divisible by 2x. So you could factor out a 2x of these first two terms. This is the same thing as 2x times x plus x plus 8. 16 divided by 2 is 8, x divided by x is 1. So this is 2x times x plus 8. And then the second two terms right over here-- this is the whole basis of factoring by grouping-- we can factor out a negative 1. So this is equal to negative 1 times x plus 8. Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8."
    },
    {
        "Q": "\nwhat is Sal doing, @ 2:28?",
        "A": "He is trying to factor out f(x).",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            148
        ],
        "3min_transcript": "f of x is equal to 2x squared plus 15x minus 8. g of x is equal to x squared plus 10x plus 16. Find f/g of x. Or you could interpret this is as f divided by g of x. And so based on the way I just said it, you have a sense of what this means. f/g, or f divided by g, of x, by definition, this is just another way to write f of x divided by g of x. You could view this as a function, a function of x that's defined by dividing f of x by g of x, by creating a rational expression where f of x is in the numerator and g of x is in the denominator. And so this is going to be equal to f of x-- we have right up here-- is 2x squared 15x minus 8. And g of x-- I will do in blue-- is right over here, g of x. So this is all going to be over g And you could leave it this way, or you could actually try to simplify this a little bit. And the easiest way to simplify this would see if we could factor the numerator and the denominator expressions into maybe simpler expressions. And maybe some of them might be on-- maybe both the numerator and denominator is divisible by the same expression. So let's try to factor each of them. So first, let's try the numerator. And I'll actually do it up here. So let's do it. Actually, I'll do it down here. So if I'm looking at 2x squared plus 15x minus 8, we have a quadratic expression where the coefficient is not 1. And so one technique to factor this is to factor by grouping. You could also use the quadratic formula. And when you factor by grouping, you're going to split up this term, this 15x. And you're going to split up into two terms where the coefficients are, if I were to take the product of those coefficients, of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared"
    },
    {
        "Q": "Hey at 4:24, does the perpendicular sign have to be turned upside-down?\n",
        "A": "no it is written correctly",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            264
        ],
        "3min_transcript": "And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up. that's kind of related to right angles, when a right angle is formed, the two rays that form the right angle or the two lines that form that right angle or the two line segments that form that right angle are called perpendicular. So because we know that measure of angle DBC is 90 degrees or that angle DBC is a right angle, this tells us, we know that the line segment DB is perpendicular to line segment BC. Or we could even say ray BD is-- instead of using the word perpendicular, there's sometimes this symbol right here, which really just shows two perpendicular lines-- perpendicular to BC. And these come out of the fact that the angle formed between DB and BC, that is a 90-degree angle. Now, we have other words when our two angles add up to other things. So let's say, for example, I have one angle over here. Let me put some letters here so we can specify it. So let's say this is X, Y, and Z. And let's say that the measure of angle XYZ is equal to 60 degrees. And let's say that you have another angle that looks like this. And I'll call this, let's say, maybe MNO."
    },
    {
        "Q": "\nat 1:19 ,do adjacent angles have to add up to 90 degrees?",
        "A": "No. They can, but they don t have to. Adjacent angles can sum up to any valid angle.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            79
        ],
        "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up."
    },
    {
        "Q": "\nWhy at 0:21 did Sal change the angle name from DAB to DBA? Does it matter because the subtitles don't change from DBA when he changes it in the first place?",
        "A": "There is a difference: angle DAB is the angle made by lines DA and AB. The angle itself is at point (or vertex , see my answer to your other question) A. Angle DBA on the other hand is the angle made by lines DB and BA, and the vertex is at point B. Sal changed his mind and drew line DB (not line DA as he first intended), so DBA is the correct name for that angle. One side note: angle DBA is the same as angle ABD; the order doesn t matter as long as the point where the angle is remains in the middle.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            21
        ],
        "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up."
    },
    {
        "Q": "\nstarting at 0:34, there is an lowercase 'm' in front of all of the angles. What is it for, and why is it there?",
        "A": "That means measure of . For example: m\u00e2\u0088\u00a0ABC = 50\u00cb\u009a That means the measure of angle ABC is 50 degrees.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            34
        ],
        "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up."
    },
    {
        "Q": "during 2:31 are complementary angles only made up of 2 different types angles?\n",
        "A": "Yes. They only consist of two types of angles.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            151
        ],
        "3min_transcript": "or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up. that's kind of related to right angles, when a right angle is formed, the two rays that form the right angle or the two lines that form that right angle or the two line segments that form that right angle are called perpendicular. So because we know that measure of angle DBC is 90 degrees or that angle DBC is a right angle, this tells us, we know that the line segment DB is perpendicular to line segment BC. Or we could even say ray BD is-- instead of using the word perpendicular, there's sometimes this symbol right here, which really just shows two perpendicular lines-- perpendicular to BC."
    },
    {
        "Q": "\nAt 1:45, Sal begins to talk about Unit Vectors using the variables 'i' and 'j' with what he describes as 'little hats' on top. Could someone please explain the importance or relevance of these symbols? Can one only use letters i and j? Or does it matter?",
        "A": "That notation means that those vectors are unit vectors , (the meaning is that does vectors have direction but their lengh is alwayes equal to 1 ). Any letter can be used to represent a vecor .",
        "video_name": "9ylUcCOTH8Y",
        "timestamps": [
            105
        ],
        "3min_transcript": "We've already seen that you can visually represent a vector as an arrow, where the length of the arrow is the magnitude of the vector and the direction of the arrow is the direction of the vector. And if we want to represent this mathematically, we could just think about, well, starting from the tail of the vector, how far away is the head of the vector in the horizontal direction? And how far away is it in the vertical direction? So for example, in the horizontal direction, you would have to go this distance. And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector-- and let's call this vector v. We could represent vector v as an ordered list or a 2-tuple of-- so we could say we and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2 comma 3, like that. And what I now want to introduce you to-- and we could come up with other ways of representing this 2-tuple-- is another notation. And this really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that. So let's define a unit vector i. And the way that we denote that is the unit vector we put this hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes 1 unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like this. That is the unit vector i. And then we can define another unit vector. And let's call that unit vector-- or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And not in the horizontal direction, and it goes 1 unit in the vertical direction. So this went 1 unit in the horizontal. And now j is going to go 1 unit in the vertical. So j-- just like that. Now any vector, any two dimensional vector, we can now represent as a sum of scaled up versions of i and j."
    },
    {
        "Q": "At 3:31, could x be zero? Or would the square root of zero be a non-real number?\n",
        "A": "Yes... you can do sqrt(0), it just = 0 because 0^2 = 0.",
        "video_name": "qFFhdLlX220",
        "timestamps": [
            211
        ],
        "3min_transcript": "And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it, And then at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this-- if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open it up to complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x, because it's not going to be a negative number. And so if we're assuming that the domain of x is-- or if this expression is going to be evaluatable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers-- and if you don't know what a complex number is, or an imaginary number, But if you were dealing with those, then you would have to keep the absolute value of x there. Because then this would be defined for numbers that are less than 0."
    },
    {
        "Q": "\nAround 2:18 of the video Sal mentions the absolute value of | x | which he got from the x^2 under the radical. if he were to use the absolute value for the x^2 why would he not use the absolute value of the lone | x | under the radical?",
        "A": "Because the lone x was not a perfect square, he could not simplify the radical. Only if you are taking the principle root and you are SIMPLIFYING the radical can you put in the absolute value. If he would have put the absolute value sign for the x under the radical it would ve become: sqrt(|x|) = +/- (x^1/2) Which still gives you the negative root which is extraneous for principle roots. Therefore putting the absolute value under the radical is never done when taking the principle root as it yields unwanted answers.",
        "video_name": "qFFhdLlX220",
        "timestamps": [
            138
        ],
        "3min_transcript": "What I want to do in this video is resimplify this expression, 3 times the principal root of 500 times x to the third, and take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you could simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500-- we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5. Or even better, we could rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it,"
    },
    {
        "Q": "is there a mathematical pattern for 1:46 or is that just random?\n",
        "A": "There is no pattern because it s the digits of tau, which like pi is transcendental (which is just a fancy term meaning that not only is it irrational, but you can t even have it as a solution for a polynomial).",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            106
        ],
        "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211."
    },
    {
        "Q": "At 1:20, why does she start talking about the forth dimension?\n",
        "A": "It s to show that the third dimension isn t the highest we can go.",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            80
        ],
        "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211."
    },
    {
        "Q": "At around 1:30, you can see how (ab)^4 is the same as (a^4) * (b^4). Why is it that when you add different terms and square them, you need to use FOIL, but when multiplying you can simply do that? I mean, sure, it works on paper but a textbook explanation would be nice to think about. They just look so similar.\n",
        "A": "Expanding (ab)^2, for example, yields (ab)(ab), which is (a^2)(b^2). Expanding (a+b)^2, however gives you (a+b)(a+b), which is aa+ab+ba+bb, or a^2+2ab+b^2",
        "video_name": "SwqOrUWzDY8",
        "timestamps": [
            90
        ],
        "3min_transcript": "And now I want to go over some of the other core exponent properties. But they really just fall out of what we already know about exponents. Let's say I have two numbers, a and b. And I'm going to raise it to-- I could do it in the abstract. I could raise it to the c power. But I'll do it a little bit more concrete. Let's raise it to the fourth power. What is that going to be equal to? Well that's going to be equal to-- I could write it like this. Copy and paste this, copy and paste. That's going to be equal to ab times ab times ab times ab times ab. But what is that equal to? Well when you just multiply a bunch of numbers like this it doesn't matter what order you're going to multiply it in. This right over here is going to be equivalent to a times a times a times a times-- We have four b's as well that we're multiplying together. And what is that equal to? Well this right over here is a to the fourth power. And this right over here is b to the fourth power. And so you see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent. And then taking their product. And here I just used the example with 4, but you could do this really with any arbitrary-- actually any exponent. This property holds. And you could satisfy yourself by trying different values, and using the same logic right over here. But this is a general property. That-- let me write it this way-- that if I have a to the b, to the c power, And we'll use this to throughout actually mathematics, when we try to simplify things or rewrite an expression in a different way. Now let me introduce you to another core idea here. And this is the idea of raising something to some power. And I'll just use example of 3. And then raising that to some power. What could this be simplified as? Well let's think about it. This is the same thing as a to the third-- let me copy and paste that-- as a to the third times a to the third. And what is a to the third times-- So this is equal to a to the third times a to the third. And that's going to be equal to a to the 3 plus 3 power."
    },
    {
        "Q": "At 10:59, why is Sal dividing 7:7 instead of multiplying (35*1)/(4*7) = (35/28) and then dividing the whole fraction by 7?\n",
        "A": "its kinda late but hat 11:00 when he said you divide 7 by 7. 35/4 * 1/7 = 35/28. simplify it by dividing 7. sames as 5/4 so yeah.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            659,
            427
        ],
        "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4."
    },
    {
        "Q": "\nHow do you know which number to multiply it by at 2:20",
        "A": "Because we can t see a clear variable to eliminate right away, we want to find the lowest common multiple of 5 and 7, in order to eliminate x. lcm(5, 7) = 5 x 7 = 35 Therefore, in order to get to 35x and -35x (one can be negative to cancel out when added together), we need to multiply the top equation by 7 and the bottom equation by -5. 7(5x+7) = 7(15) -5(7x-3y) = -5(5)",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that If we added these two left-hand sides, you would get 8x minus 12y. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand"
    },
    {
        "Q": "\nat about 3:20, he wrote 15, but it is supposed to be five. Just so you all know.",
        "A": "Please review the video again. That did not happen. The video starts with the following system. . . . 5x - 10y = 15 3x - 2y = 3 Sal chooses to isolate the x term by multiplying the -2y in equation two by -5, so that the 10y produced, cancels with the -10y in equation one. Naturally, you need to keep the system balanced by multiplying both sides of equation two by -5 -5(3x - 2y) = (-5)3 and that gives -15x + 10y = -15 Everything is correct as he did it.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            200
        ],
        "3min_transcript": "And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15."
    },
    {
        "Q": "Sal makes a mistake at 3:21\n",
        "A": "At 3:27: 5x + (-15x) = -10x - this is correct -10y + 10y = 0 - this is correct 15 + (-15) = 0 - this is also correct. No errors there.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            201
        ],
        "3min_transcript": "And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15."
    },
    {
        "Q": "\nAt 10:31 you said 7x = 20/4 + 15/4 but I thought we got rid of the 15/4 when we added it to both sides of the equation 7x - 15/4 = 5. It would look like this 7x-15/4+15/4 = 5+15/4. The 15/4's on the left side cancel out. 5+15/4 on the right side becomes 20/4. Why do you then add 15/4 to the 20/4?",
        "A": "To add a whole number and a fraction, you have to get a common denominator, so 5 \u00e2\u0080\u00a2 4/4 = 20/4. What you are trying to do is (5+15)/4 which is incorrect order of operations. If switched to decimals, it would be 5 + 3.75 which is 8.75 or 35/4.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            631
        ],
        "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4."
    },
    {
        "Q": "\nWhat does Sal mean at 0:08 when he says \"Massage the Equation\"?",
        "A": "Sal meant that to solve the systems of equations, you first must manipulate or change the equations. Once this is done, it will be much easier to solve the system.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            8
        ],
        "3min_transcript": "Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that If we added these two left-hand sides, you would get 8x minus 12y. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand"
    },
    {
        "Q": "at 2:50 sal talks about a \"zero matrix\" what does he mean by it?\n",
        "A": "A zero matrix is a matrix all of whose entries are zero.",
        "video_name": "OjF765iVuF8",
        "timestamps": [
            170
        ],
        "3min_transcript": "Here you have Bernard, who says A times, C times B. We already know that that's going to be the equivalent to A, C, B which once again they've swapped the order between the B and the C, matrix multiplication is not commutative. You can't just swap order and expect to get the same product for any square matrices A, B, and C so we could rule that one out. A times, B, C, so we've already seen matrix multiplication is associative, so this is the same thing as A times B, times C which of course is the same thing as A, B, C. What Caren has right over here, that is right, that is equivalent for any square matrices A, B, and C that is equivalent to A, B, C. Now Ducheval, let's see, now this looks like a bit of a crazy expression but let's think it through a little bit. First of all matrix multiplication, the distributive property does hold. This first part right over here is equivalent to ... Let me write this down, this one's interesting. We have A times B, C plus A, minus A squared. You can actually distribute this A and I encourage you to prove it for yourself maybe using some two by two matrices for simplicity. This is going to be equal to, this part over here is going to be A, B, C plus A A, A times A which we could write as A squared and then we're going to subtract A squared. These two things are going to cancel out, they're going to end up being the zero matrix and if you take the zero, so these are going to be the zero matrix right over here. If you take the zero matrix and add it to A, B, C you're just going to end up with A, B, C. this one actually is equivalent. This one is right and this one is right. Here this is A times B, plus C so this is kind of not-y right over here. They're not even multiplying B and C, so this one's definitely not going to be true for all square matrices A, B, and C."
    },
    {
        "Q": "\nAt 1:00 why is it +4 and not *4? I don't understand above it was y^2/4. So wouldn't you multiply both sides by 4? And instead of getting (4/9)x^2 + 4 wouldn't it be (4/36)4x^2?\nI'm not doing to well in this chapter, so someone please explain this!",
        "A": "Sal isn t showing the multiplication on each side. He is taking each term times 4. y^2/4 * 4 = y^2; x^2/9 * 4 = 4/9 x^2; 1 * 4 = 4. The +4 comes from taking the +1 times 4.",
        "video_name": "hl58vTCqVIY",
        "timestamps": [
            60
        ],
        "3min_transcript": "In the last hyperbola video I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the hyperbola y squared over 4 minus x squared over, I don't know, let me think of a good number. Let's say, x squared over 9 is equal to 1. So the first thing to figure out about this hyperbola is, what are its asymptotes? And, once again, I always forget the formulas. And I just try to solve for y and see what happens when x approaches positive or negative infinity. So if you solve for y, you can add x squared over 9 to both sides. And you get y squared over 4 is equal to x squared over 9 plus 1. Now, I can multiply 4 times both sides. And you get y squared is equal to 4 over 9 times I distribute the 4, take the positive and negative square root both sides. y is equal to the plus or minus square root of 4 over 9x squared plus 4. And you can't really simplify this anymore. But we can think about, what does this approach as x approaches positive or negative infinity. So, as x approaches plus or minus infinity, what does this roughly equal? What does this approximate? What does the graph get a lot closer to? Well, then, y is approximately equal to just the square Because this becomes super huge and relative to this term, this starts to matter less and less and less. And that's why we get closer and closer to the asymptotes. Because when this number is, like, a trillion, or a google, then this number is almost insignificant. square root of this, and you'll just be a little bit above the graph. Because you have this extra plus-4 there. So as you approach positive or negative infinity, this equation is approximately equal to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would be approximately equal We can take the square root of this. Plus or minus the square root of 4/9 is 2/3, right? Square root of 4 over square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just to make things interesting. So let me draw the first one."
    },
    {
        "Q": "1:55 The +4 disappears just because it \"becomes irrelevant\" compared to the other term?\n",
        "A": "The +4 is reverent to the equation, Its what makes the asymptotes that are the basis of the hyperbolas shape but when finding the asymptotes we want to find the line it get increasingly closer to. Since the +4 is what separates the function from the asymptotes if you remove the +4 you get the asymptotes",
        "video_name": "hl58vTCqVIY",
        "timestamps": [
            115
        ],
        "3min_transcript": "In the last hyperbola video I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the hyperbola y squared over 4 minus x squared over, I don't know, let me think of a good number. Let's say, x squared over 9 is equal to 1. So the first thing to figure out about this hyperbola is, what are its asymptotes? And, once again, I always forget the formulas. And I just try to solve for y and see what happens when x approaches positive or negative infinity. So if you solve for y, you can add x squared over 9 to both sides. And you get y squared over 4 is equal to x squared over 9 plus 1. Now, I can multiply 4 times both sides. And you get y squared is equal to 4 over 9 times I distribute the 4, take the positive and negative square root both sides. y is equal to the plus or minus square root of 4 over 9x squared plus 4. And you can't really simplify this anymore. But we can think about, what does this approach as x approaches positive or negative infinity. So, as x approaches plus or minus infinity, what does this roughly equal? What does this approximate? What does the graph get a lot closer to? Well, then, y is approximately equal to just the square Because this becomes super huge and relative to this term, this starts to matter less and less and less. And that's why we get closer and closer to the asymptotes. Because when this number is, like, a trillion, or a google, then this number is almost insignificant. square root of this, and you'll just be a little bit above the graph. Because you have this extra plus-4 there. So as you approach positive or negative infinity, this equation is approximately equal to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would be approximately equal We can take the square root of this. Plus or minus the square root of 4/9 is 2/3, right? Square root of 4 over square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just to make things interesting. So let me draw the first one."
    },
    {
        "Q": "\n3:43 Inverse sin? Is there a video explaining raising trig functions to exponents?",
        "A": "Inverse sine and sine to the power of -1 are different things. Sine Inverse: sin(x) = y arcsin(y) = x (the inverse function is commonly notated with arcsin) Sine to an exponent: sin^n(x) = [sin(x)]^n",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            223
        ],
        "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40,"
    },
    {
        "Q": "\ni don't understand why did he use inverse sine to solve for it\nat around 4:15",
        "A": "Inverse sin basically negates sin. Think of it like this: sin(90)=1 so arcsin(1)=90. It takes the output and gives the input. So in order to solve for sin(x)=(4/3)(sin(40), you have to take the inverse sin of both sides to get rid of the sin(x) and solve for x. Remember, (4/3)sin(40) is simply just a number.",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            255
        ],
        "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40,"
    },
    {
        "Q": "\ncan someone please elaborate on why Sal took the inverse sine instead of just dividing both side of the equation by sine to isolate theta at 3:50? Thank you!",
        "A": "sine is a function, not a variable",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            230
        ],
        "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40,"
    },
    {
        "Q": "\nin 0:44 - 0:46 Sal says \"law of cosines \" twice (Though the second time the captions say \"sines\") does anyone know why?? I'm not sure if he just made a mistake or if I am hearing incorrectly.",
        "A": "In this case, all Sal was trying to do was say that you could use EITHER the law of cosines or the law of sines for the problem shown above, but for the purpose of the video, he decided to use the law of sines. But, the second time he said the law of cosines was a mistake and it was corrected.",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            44,
            46
        ],
        "3min_transcript": "Voiceover:Say you're out flying kites with a friend and right at this moment you're 40 meters away from your friend and you know that the length of the kite's string is 30 meters, and you measure the angle between the kite and the ground where you're standing and you see that it's a 40 degree angle. What you're curious about is whether you can use your powers of trigonometry to figure out the angle between the string and the ground. I encourage you to pause the video now and figure out if you can do that using just the information that you have. Whenever I see, I guess, a non right triangle where I'm trying to figure out some lengths of sides or some lengths of angles, I immediately think maybe the Law of Cosine might be useful or the Law of Sines might be useful. So, let's think about which one could be useful in this case. Law of Cosines, and I'll just rewrite them here. The Law of Cosine is c squared is equal to a squared plus b squared minus 2ab cosine of theta. 3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta."
    },
    {
        "Q": "\nWhat is the troll he's talking about in 0:02?",
        "A": "Watch the 2 previous videos Sal has made in that catagory. Hope that helps!",
        "video_name": "VhH2nEDCd68",
        "timestamps": [
            2
        ],
        "3min_transcript": "Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked."
    },
    {
        "Q": "\n1:05 you were counting, why , I dont understand where you were counting?",
        "A": "The slope of the equation was 7/5. Therefore, for every increase of 7 in y there s an increase of 5 in x. Sal counted the 7/5 slope and adjusted the line to follow it.",
        "video_name": "VhH2nEDCd68",
        "timestamps": [
            65
        ],
        "3min_transcript": "Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked."
    },
    {
        "Q": "\nDoesn't \"constant\" begin with a c, so why do we use K? (0:17) I really do not get it.",
        "A": "It does not matter what number you use. Unless you are dealing with a problem that has specific words like baseball, or cookies then you would probably want to use what ever letter it starts with. At least thats what I have always been told and I have an A in math.",
        "video_name": "lkP-E2LUnjA",
        "timestamps": [
            17
        ],
        "3min_transcript": "We're told in this question that on a string instrument, the length of a string-- so let's call that l-- the length of a string l varies inversely as the frequency, so varies inversely as the frequency. So l is going to be equal to some constant times the inverse of the frequency. And I'll use f for frequency-- the frequency of its vibrations. And then they tell us the vibrations are what give the string instruments their sound. That's nice. An 11-inch-- and it's actually the vibrations of the string affecting the air, and then the air compressions eventually get to our eardrum. And that's actually what gives us the perception of the sound. But we don't want to delve too much into the physics of it. An 11-inch inches string has a frequency. So 11-inch, so this is its length. So the 11-inch string has a frequency of 400 cycles per second. So this right here is the frequency. Find the constant of proportionality, and then find the frequency of a 10-inch string. So they say an 11-inch string. So 11-inch string is equal to some constant of proportionality times 1/400 cycles per second. So one over 400 cycles per second. I'll write second as sec. So to solve for the constant of proportionality, we need to multiply both sides by 400 cycles per second. Multiply the left hand side by 400 cycles per second, and the left hand side becomes 400 times 11. Well, 4 times 11 is 44. So 400 times 11 is 4,400. And then we have in our units, just for out of interest, So it's cycles times inches in the numerator of our units divided by seconds. And that is equal to our constant of proportionality. So we can say that the length is equal to 4,400 times-- or 4,400 cycles times inches per second-- I want to get the units right-- per second, times 1 over the frequency. So we solved for our constant of proportionality. And then we can use this to find the frequency of a 10-inch string. So now we're talking about a situation where our length is 10 inches, so 10 inches. So we get 10 inches are equal to 4,400--"
    },
    {
        "Q": "At 5:01, Sal uses three numbers for his example, do you NEED to have three numbers?\n",
        "A": "Yes, in the scope of standard multivariable calculus, you can only use the cross product if you have vectors with three components.",
        "video_name": "pJzmiywagfY",
        "timestamps": [
            301
        ],
        "3min_transcript": "So it's a2 times b3 minus a3 times b2. That was hopefully pretty straightforward. Now, not to make your life any more complicated, when you do the second, when you do the middle row, when you do this one right here, so you cross that out. And you might want to do a1 times b3 minus a3 times b1. And that would be natural because that's what we did up there. But the middle row you do the opposite. You do a3 times b1 minus a1 times b3. Or you can kind of view it as the negative of what you would have done naturally. So you would have done a1 b3 minus a3 b1. Now we're going to do a3 b1 minus a1 b3. And then, for the bottom row, we cross that out again or ignore it. And we do a1 times b2, just like we do with the first row. Times a2 b1. Or minus a2 b1. This seems all hard to-- and it is hard to remember. That's why I kind of have to get that system in place like I just talked to you about. But this might seem pretty bizarre and hairy. So let me do a couple of examples with you, just so you get the hang of our definition of the dot product in R3. So let's say that I have the vector-- let's say I'm crossing the vector. I have the vector 1, minus 7, and 1. And I'm going to cross that with the vector 5, 2, 4. So this is going to be equal to a third vector. So for the first element in this vector, the first component, we just ignore the first components of these vectors and we say minus 7 times 4 minus 1 times 2. And these are just regular multiplication. I'm not taking the dot product. These are just regular numbers. Then for the middle term, we ignore the middle terms here and then we do the opposite. We do 1 times 5 minus 1 times 4. Remember, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the But the middle term is the opposite. And then finally, the third term you ignore the third terms here and then you do it just like the first term. You start in the top left. 1 times 2 minus 7."
    },
    {
        "Q": "\nAt 0:33 how does she do that? Can anyone help me I really want to try it?!",
        "A": "Yeah, I don t know but it is so cool. I am afraid you will have to explore it yourself.... I wish I knew.",
        "video_name": "VIVIegSt81k",
        "timestamps": [
            33
        ],
        "3min_transcript": "So say you just moved from England to the US and you've got your old school supplies from England and your new school supplies from the US and it's your first day of school and you get to class and find that your new American paper doesn't fit in your old English binder. The paper is too wide, and hangs out. So you cut off the extra and end up with all these strips of paper. And to keep yourself amused during your math class you start playing with them. And by you, I mean Arthur H. Stone in 1939. Anyway, there's lots of cool things you do with a strip of paper. You can fold it into Shapes and more shapes. Maybe spiral it around snugly like this. Maybe make it into a square. Maybe wrap it into a hexagon with a nice symmetric sort of cycle to the flappy parts. In fact, there's enough space here to keep wrapping the strip, and the your hexagon is pretty stable. and you're like. \"I don't know, hexagons aren't too exciting, but I guess it has symmetry or something.\" Maybe you could kinda fold it so the flappy parts are down and the unflappy parts are up. That's symmetric, and it collapses down into these three triangles, which collapse down into one triangle, and collapsible hexagons are, you suppose, cool enough to at least amuse you a little but during your class. you decide to try this three-way fold the other way, with flappy parts up, and are collapsing it down when suddenly the inside of your hexagon decides to open right up What, you close it back up and undo it. Everything seems the same as before, the center is not open-uppable. But when you fold it that way again, it, like, flips inside-out. Weird. This time, instead of going backwards, you try doing it again and again and again and again. And you want to make one that's a little less messy, so you try with another strip and tape it nicely into a twisty-foldy loop. You decide that it would be cool to colour the sides, so you get out a highlighter and make one yellow. Now you can flip from yellow side to white side. Yellow side, white side, yellow side, white side Hmm. White side? What? Where did the yellow side go? So you go back and this time you colour the white side green, and find that your piece of paper has three sides. Yellow, white and green. Now this thing is definitely cool. Therefore, you need to name it. And since it's shaped like a hexagon and you flex it and flex rhymes with hex, hexaflexagon it is. That night, you can't sleep because you keep thinking And the next day, as soon as you get to your math class you pull out your paper strips. You had made this sort of spirally folded paper that folds into again, the shape of a piece of paper, and you decide to take that and use it like a strip of paper to make a hexaflexagon. Which would totally work, but it feels sturdier with the extra paper. And you color the three sides and are like, orange, yellow, pink. And you're sort of trying to pay attention to class. Math, yeah. Orange, yellow, pink. Orange, yellow, white? Wait a second. Okay, so you colour that one green. And now it;s orange, yellow, green, Orange, yellow, green. Who knows where the pink side went? Oh, there it is. Now it's back to orange, yellow, pink. Orange, yellow, pink. Hmm. Blue. Yellow, pink, blue. Yellow, pink, blue. Yellow, pink, huh. With the old flexagon, you could only flex it one way, flappy way up. But now there's more flaps. So maybe you can fold it both ways. Yes, one goes from pink to blue, but the other, from pink to orange. And now, one way goes from orange to yellow, but the other way goes from orange to neon yellow."
    },
    {
        "Q": "\nAt 1:17, why didn't he take the (-) sign? Is there a reason?",
        "A": "It s the definition: If (x - a) is a factor, f(a) = 0. Otherwise if (x + a) is a factor, you have to check wether f(-a) = 0. You have to revert the sign.",
        "video_name": "JAdNNJynWM4",
        "timestamps": [
            77
        ],
        "3min_transcript": "- [Voiceover] So we're asked, Is the expression x minus three, is this a factor of this fourth degree polynomial? And you could solve this by doing algebraic long division by taking all of this business and dividing it by x minus three and figuring out if you have a remainder. If you do end up with a remainder then this is not a factor of this. But if you don't have a remainder then that means that this divides fully into this right over here without a remainder which means it is a factor. So if the remainder is equal to zero, the remainder is equal to zero, if and only if, it's a factor. It is a factor. And we know a very fast way of calculating the remainder of when you take some polynomial and you divide it by a first degree expression like this. I guess you could say when you divide it by a first degree polynomial like this. The polynomial remainder theorem, the polynomial remainder theorem tells us that if we take some polynomial, p of x and we were to divide it by some x minus a our polynomial evaluated at our polynomial evaluated at a. So let's just see what's a in this case. Well in this case our a is positive three. So let's just evaluate our polynomial at x equals 3, if what we get is equal to zero that means our remainder is zero and that means that x minus three is a factor. If we get some other remainder that means well we have a non-zero remainder and this isn't a factor, so let's try it out. So, we're gonna have, so I'm just gonna do it all in magenta. It might be a little computationally intensive. So it's going to be two times three to the fourth power, three to the fourth, three to (mumbles), that's 81. 81. Minus 11 Yeah, this is gonna get a little computationally intensive but let's see if we can power through it. 11 times 27, I probably should have picked a simpler example, but let's just keep going. Plus four times three is 12. Minus 12 So lucky for us, at least those last two terms cancel out. And so this is going to be the rest from here is arithmetic. Two times 81 is 162. Now let's think about what 27 times 11 is. So let's see, 27 times 10 is going to be 270. 270 plus another 27 is minus 297. 297, did I do that, yeah, 270 So 27 times 10 is 270 plus 27, 297 Yep, that's right. And then we have, I'm prone to make careless errors here, see 90 plus 45 is 135. So plus 135. And let's see, if I were to take if I were to take 162 and 135, that's going to give me 297"
    },
    {
        "Q": "at 0:37, I am a bit confused on how that worked. How is X 2 and Y 5? How do we assign solution numbers to the variables? How do we know which ones to assign? Couldn't it be X 5 and Y 2? Checking the inequalities is really difficult for me.\n",
        "A": "The question is asking to check for an x,y coordinate. I think the video assumes the watcher to understand the typical structure of a coordinate point: (x, y). So x = 2 and y = 5. Not the other way around. And after that, simply substitute the values into the inequalities.",
        "video_name": "XzYNh2wpO0A",
        "timestamps": [
            37
        ],
        "3min_transcript": "Is two comma five a solution of this system? And we have a system of inequalities right over here. We have Y is greater than or equal to 2x plus 1 and X is greater than 1. In order for two comma five to be a solution of this system, it just has to satisfy both inequalities. So, lets just try it out. So when X is equal to two and Y is equal to five, it has to satify both of these. So lets try it with the first one. So if we assume X is two and Y is five, we would get an inequality that says that five is greater than or equal to two times two plus one. X is two; Y is five. This gives us five is greater than or equal to two times two is four plus one is five. Y is greater than or equal to five. That's true! Five is equal to five. So that equal part of the greater than or equal saves us. That definitely satisfies the first inequality. Lets see the second one. X needs to be greater than one. So in two comma five, X is two. So it actually satisfies both of these inequalities. So two comma five is a solution for this system."
    },
    {
        "Q": "At 5:13, why doesn't he take the reciprocal of (2^-70) and make the exponent positive?\n\nHope that makes sense.\n",
        "A": "I would agree that his final answer should have a positive exponent as that is the preferred form. He did leave his final answer as 2^(-98). This should have been changed to 1 / 2^98. However, while he is simplifying the expression, he can work with negative exponents.",
        "video_name": "dC1ojsMi1yU",
        "timestamps": [
            313
        ],
        "3min_transcript": "times this thing to the second power. Eight to the seventh to the second power, and then here, negative two times two is negative four, so that's A to the negative four times, eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh. Well, you would then add the two exponents, and you would get to eight to the 14th, so however many times you have eight to the seventh, you would just keep adding the exponents, or you would multiply by seven that many times. Hopefully that didn't sound too confusing, but the general idea is if you raise something to exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second,"
    },
    {
        "Q": "\nAt 6:43 Sal says that (3^-8*7^3) is equivalent to (7^3/3^8). Shouldn't it be (7^3/3^-8)?",
        "A": "No. The negative exponent tells you to use the reciprocal to change it to a positive exponent. So: 3^(-8) = 1/3^8. Multiply that with 7^3, and you get: 7^3/3^8 Hope this helps.",
        "video_name": "dC1ojsMi1yU",
        "timestamps": [
            403
        ],
        "3min_transcript": "divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, well, that's two to the 28th power, two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, if I'm taking a quotient with the same base, I can subtract the exponent. So it's gonna be negative 70. It's going to be negative 70 minus 28th power, minus 28, and so this is going to simply two to the negative 98th power, and that's another way of viewing the same expression."
    },
    {
        "Q": "At 10:11, why does 1/2i become the square root of 1/4?\n",
        "A": "because as we square 1/2, the numerator 1^2= 1 and denominator 2^2=4 and take the square root of 1/4, this is base on Pythagorean Theorem.",
        "video_name": "FwuPXchH2rA",
        "timestamps": [
            611
        ],
        "3min_transcript": "Now just to make this tangible, let's actually do this with an actual example. So let's say that I had, I don't know, let's say that I had to z1 is equal to square root of, let's say it's square root of 3/2 plus i. And so we want to figure out its magnitude, and we want to figure out its argument. So let's do that. So the magnitude of z1 is going to be equal to the square root of this squared. So this is going to be equal to 3/4 plus 1 squared-- or I should say plus 4/4. So this is going to be equal to square root of 7/4, which And now let's figure out its argument. So if I were to draw this on an Argand diagram, it would look like this. It's going to be in the first quadrant, so that's all I have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it's going to be square root of 3, actually, let me change this up a little bit, just so the numbers get a little bit cleaner. Sorry about this. Let me make it a little bit, slightly cleaner. So just so that we have a slightly cleaner result, because we want to make our first example a simple one. So let's make this square root of 3/2 plus 1/2i.. So let's figure out the magnitude, the magnitude here is z1 is equal to the square root equal to 3/4 plus 1/2 squared is equal to 1/4,. This makes things a lot nicer. This is equal to the square root of 1, which is 1. And now let's think about it, let's draw it on an Argand diagram to visualize the argument. So this is my imaginary axis. This is my real axis. And so this complex number is square root of 3/2. The square root of 3 is like 1.7. So if we have like 1, it'll be like right over here, someplace right over here. This is square root of 3/2, the real part. The imaginary part is 1/2. So if this is 1, this is 1/2, the imaginary part is right over here, 1/2. And we actually also know its length, its length, or its magnitude is 1."
    },
    {
        "Q": "\nat 8:55 why is i squared +1 or 4/4? shouldn't it be minus 1?",
        "A": "Sal is looking for a and b, which are the amplitude of the real and imaginary parts. b is the amplitude of the imaginary part by definition, so you remove the i before you square the coefficient. I m sure Sal could explain it better, but I hope that makes some sense to you.",
        "video_name": "FwuPXchH2rA",
        "timestamps": [
            535
        ],
        "3min_transcript": "And it's really one of the most profound results and all of mathematics, it still gives me chills. This is Euler's formula. Or this, by Euler's formula, is the same thing. And we show it by looking at the Taylor series representations of e to the x. And the Taylor series representations of cosine of x and sine of x. But this is, if we're dealing with radians, e to the i phi. So z is going to be equal to r times e to the i phi. So there's two ways to write a complex number. You could write it like this, where you have the real and imaginary part, that's maybe what we're used to. Or we can write it in exponential form, where you have the modulus, or the magnitude, being multiplied by a complex exponential. And we're going to see that this going to be super useful, Now just to make this tangible, let's actually do this with an actual example. So let's say that I had, I don't know, let's say that I had to z1 is equal to square root of, let's say it's square root of 3/2 plus i. And so we want to figure out its magnitude, and we want to figure out its argument. So let's do that. So the magnitude of z1 is going to be equal to the square root of this squared. So this is going to be equal to 3/4 plus 1 squared-- or I should say plus 4/4. So this is going to be equal to square root of 7/4, which And now let's figure out its argument. So if I were to draw this on an Argand diagram, it would look like this. It's going to be in the first quadrant, so that's all I have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it's going to be square root of 3, actually, let me change this up a little bit, just so the numbers get a little bit cleaner. Sorry about this. Let me make it a little bit, slightly cleaner. So just so that we have a slightly cleaner result, because we want to make our first example a simple one. So let's make this square root of 3/2 plus 1/2i.. So let's figure out the magnitude, the magnitude here is z1 is equal to the square root"
    },
    {
        "Q": "at 0:56 you mentioned the principal root what does that mean\n",
        "A": "This means ignoring the negative when you take a square root. That is, the number 4 has two square roots; 2 and -2. Principal means you only take the positive one. Cheers",
        "video_name": "P1DJxuG7U9A",
        "timestamps": [
            56
        ],
        "3min_transcript": "Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5. If x does not equal 5, and it's equal to c if x equals 5. Then say, if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now, if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9. The square root of that is positive 3, the principal root is positive 3. 3 minus 3 is 0. So you get a 0 in the numerator. And then you get 5 minus 5 in the denominator, So you get this indeterminate form of 0/0. And in the future, we will see that we do have a tool that allows us, or gives us an option to attempt to find the limits when we get this indeterminate form. It's called L'Hopital's rule. But we can actually tackle this with a little bit of fancy algebra. And to do that, I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical, what you can do is multiply by the radical-- or, if you have a radical minus 3, you multiply by the radical plus 3. So in this situation, you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator so that we actually don't change the value of the expression. If this right over here had a plus 3, then we would do a minus 3 here. This is a technique that we learn in algebra, or sometimes in pre-calculus class, to rationalize usually denominators, but to rationalize numerators or denominators. It's also a very similar technique that we use often times to get rid of complex numbers, usually in denominators. But if you multiply this out-- and I encourage you to do it-- you notice this has the pattern that you learned in algebra class. It's a difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. And the second term is going to be the second something, or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. And in the denominator, you're of course going to have x minus 5 times the square root of x"
    },
    {
        "Q": "At 1:11, shouldn't the unknown angle be 180-90-theta? Just for the sake of clarity.?\n",
        "A": "that just makes it more confusing, we see the 90 degree angle. so the other two add to 90. so one is theta and one is 90-theta.",
        "video_name": "QuZMXVJNLCo",
        "timestamps": [
            71
        ],
        "3min_transcript": "We've got two right triangles here. And let's say we also know that they both have an angle whose measure is equal to theta. So angle A is congruent to angle D. What do we now know about these two triangles? Well for any triangle, if you know two of the angles, you're going to know the third angle, because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer. So if this angle is theta, this is 90. They all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here, so angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well we could use the same logic over here. We already use of 90 degrees over here. So we have a remaining 90 degrees So this angle is going to be 90 degrees minus theta. You have three corresponding angles being congruent. You are dealing with similar triangles. Now why is that interesting? Well we know from geometry that the ratio of corresponding sides of similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out-- when you're dealing with the right triangles-- the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And then we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now this side right over here, side BC, what side does that correspond to? Well if you look at this triangle, you can view it as the side that is opposite this angle theta. So it's opposite. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. So it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is the hypotenuse. We could call this, maybe, the adjacent side to it. And so D corresponds to A, and so this would be the side that corresponds. Now the whole reason I did that is to leverage that, corresponding sides, the ratio between corresponding sides of similar triangles, is always going to be the same. So for example, the ratio between BC and the hypotenuse, BA-- so let me write that down-- BC/BA is going to be equal to EF/ED, the length of segment EF"
    },
    {
        "Q": "At 6:55, is he saying that sine, cosine, and tangent are all equal to each other?\n",
        "A": "Actually, Sal is saying that the ratio of the lengths of two sides of one triangle is equal to the ratio of the corresponding sides of a similar triangle. He is certainly not implying that the sine, cosine and tangent are equivalent.",
        "video_name": "QuZMXVJNLCo",
        "timestamps": [
            415
        ],
        "3min_transcript": "And I keep stating from theta's point of view because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well theta's right over here. Clearly AB and DE are still the hypotenuses-- hypoteni. I don't know how to say that in plural again. And what is AC, and what are DF? Well, these are adjacent to it. They're one of the two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio, in either of these triangles, between the adjacent side-- so this is relative. Once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here-- relative to angle A, Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. And I really want to stress the importance-- and we're going to do many, many more examples of this to make this very concrete-- but for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles. We've just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same, for any of these triangles, as long as it has that angle theta in it. theta, between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same, so the opposite over hypotenuse, they call this the sine of the angle theta. Let me do this in a new color-- by definition-- and we're going to extend this definition in the future-- this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent of theta. And a mnemonic that will help you remember this-- and these really are just definitions. People realized, wow, by similar triangles, for any angle theta, this ratio is always going to be the same."
    },
    {
        "Q": "\nAt 5:43,why is angle 1 a supplement of angle 2?",
        "A": "The angle 1 and the angle that Sal put a magenta arc on are corresponding angles therefore they are equal. Since they are equal then you know that \u00e2\u0088\u00a01 + \u00e2\u0088\u00a02 = 180\u00c2\u00b0 since the magenta arc and \u00e2\u0088\u00a02 form a straight line.",
        "video_name": "h0FFEBHBufo",
        "timestamps": [
            343
        ],
        "3min_transcript": "Copy and paste this. OK. I try to avoid talking when I'm copying and pasting because I think it slows down my computer. In the accompanying diagram, parallel lines L and M are cut by transversal T. So it's a classic parallel line with a transversal problem. And they're parallel. That's why I did those arrows. Which statements about angle 1 and 2 must be true? I don't know if you've seen the Khan Academy videos of the angle game, but that's what we're going to play here, the angle game. So angle 1, if you want to look at its corresponding angle, its corresponding angle on the other parallel line, or with the transversal and the other parallel line, is right there. And they're going to be congruent. Those two angles are going to be congruent. So you could say this is equal to the measure of angle 1. I've picked up their terminology well, I think. And this is obviously angle 2. You see immediately that they have to be supplementary. Because when you add them together you get 180 degrees. Together they go all the way around and they kind of form a line. So you know that if this angle and this angle are supplementary. And this angle is congruent to angle 1, then angle 1 and angle 2 must be supplementary. So what do they say? Angle 1 is definitely not necessarily congruent to angle 2. It's congruent to this angle here. Angle 1 is a complement of angle 2. Complement means you add up to 90. No, we're talking about supplement. So it's not that. Angle 1 is a supplement of angle 2. There you go. And there's nothing that says that they're right angles, that's silly. All right, next problem. Let me copy and paste it. And paste it. Ready to go. What values-- let me pick a good color-- what values of A and B make the quadrilateral MNOP a parallelogram. For this to be a parallelogram, the opposite sides have to be equal. And I challenge you to experiment to draw a parallelogram where opposite sides are parallel where the opposite sides are also not equal. If you make two of the sides not equal, then the other two lines aren't going to be parallel anymore. And you can play with that if you like. But if opposite sides are going to be equal that means 4A plus B is equal to 21. Because they're opposite sides, so they should be equal to each other. Similarly, 3A minus 2B should be equal to 13 because they're opposite sides. So 3A minus 2B is equal to 13. And now we have two linear equations with two unknowns. So this is really an Algebra 1 problem in disguise. So let's see, they want us to solve for both."
    },
    {
        "Q": "\nAt 0:37, doesn't 5*3=3*3*3*3*3? I know it's the same thing, but I was wondering which one was the right way.",
        "A": "5*3 is either 5+5+5 or 3+3+3+3+3. 5*3 = 3*5 because of the commutative property of multiplication. The commutative property of multiplication states that rearranging the order of the factors will not affect the product. Because of that, 5+5+5 = 3+3+3+3+3. They are both correct, but what you are asking is different. In your question, you multiplied 3 by itself 5 times. That would make it 3^5 and not 5*3. 5^3 = 5*5*5 = 125 is very different from 3^5 = 3*3*3*3*3 = 243.",
        "video_name": "5qfOViJda_g",
        "timestamps": [
            37
        ],
        "3min_transcript": "Find the value of 5 to the third power. Let me rewrite that. We have 5 to the third power. Now, it's important to remember, this does not mean 5 times 3. This means 5 times itself three times, so this is equal to 5 times 5 times 5. 5 times 3, just as a bit of a refresher so you realize the difference, 5 times 3-- let me write it over here. 5 times 3 is equal to 5 plus 5 plus 5. So when you multiply by 3, you're adding the number to itself three times. When you take it to the third power, you're multiplying the number by itself three times. So 5 times 3, you've seen that before, that's 15. But 5 to the third power, 5 times itself three times, is equal to-- well, 5 times 5 is 25, and then 25 times 5 is And we're done!"
    },
    {
        "Q": "\nIs the cochlea able to tune out any sound besides the frequency that is coming into the ear? 10:22 Confused me and made me wonder if the cochlea could do such a thing.",
        "A": "Compare this to when you are watching a video, but an air conditioner is running in the background. You will subconsciously tune it out in order to focus on the video. The cochlea translates all frequencies into sounds, but the brain is where they are chosen to be ignored or not.",
        "video_name": "i_0DXxNeaQ0",
        "timestamps": [
            622
        ],
        "3min_transcript": "[PLAYS BACK PITCH] If we play the two at once, do you think we'll hear the two separate pitches? Or will our brain say, Hey, two pure frequencies an octave apart? The higher one must be an overtone of the lower one. So we're really hearing one note. [PLAYS BACK PITCH] Let's add the next overtime. 3 times 220 gives us 660. Here they are all at once. It sounds like a different instrument for the fundamental sine wave but the same pitch. Let's add 880 and now 1000. That sounds wrong. 880 plus 220 is 1100. There, that's better. We can keep going and now we have all these happy overtones. Zooming in to see the individual sine waves, I can highlight one little bump here and see how the first overtone perfectly fits two bumps. And the next has three, then four, and so on. By the way, knowing that the speed of sound is about 340 meters per second, and seeing that this wave takes about 0.0009 seconds to play, I can multiply those out to find that the distance between here and here is about 0.3 meters, or one foot. So C-sharp, 1100 is about a foot long. And each octave down is 1/2 the frequency or twice the length. That means the lowest C on a piano, which is five octaves lower than this C, has a sound wave 1 foot times 2 to the 5, or 32 feet long. OK, now I can play with the timbre of the sound by changing how loud the overtones are relative to each other. What your ears are doing right now is pretty complicated. All these sound waves get added up together into a single wave. And if I export this file, we can see what it looks like. Or I suppose you could graph it. Anyway, your speakers or headphones have this little diaphragm in them that pushes the air to make sound waves. To make this shape, it pushes forward fast here, then does this wiggly thing, and then another big push forwards. The speak, remember, is not pushing air from itself to your ears. It bumps against the air, which bumps against more air, and so on, until some air bumps into your ear drum, which moves in the same way that the diaphragm in the speaker did. And that pushes the little bones that push the cochlea, which pushes the fluid, which, depending is either going to push the basilar membrane in such a way that makes it vibrate a lot and push the little hairs, or it pushes with the wrong timing, just like someone bad at playgrounds. This sound wave will push in a way that makes the A220 part of your ear send off a signal, which is pretty easy to see. Some frequencies get pushed the wrong direction sometimes, but the pushes in the right direction more than make up for it. So now all these different frequencies that we added together and played are now separated out again. And in the meantime, many other signals are being sent out from other noise, like the sound of my voice and the sound of rain and traffic and noisy neighbors and air conditioner and so on. But then our brain is like, Yo, look at these! I found a pattern! And all these frequencies fit together into a series starting at this pitch. So I will think of them as one thing. And it is a different thing than these frequencies, which fit the patterns of Vi's voice. And oh boy, that's a car horn. Somehow this all works. And we're still pretty far from developing technology that can listen to lots of sound and separate it out into things anywhere near"
    },
    {
        "Q": "\nI think that Sal should't make up a word called \"furgle\" at 2:14~3:43\nbecause,then many people would find it very confusing weather or\nnot \"furgle is a real word.",
        "A": "When you think about, someone, long ago, made up the words mile , inch , and pound . Sal is trying to explain an idea. He used a made up word so students wouldn t think the idea only worked with inches or centimeters or whatever real word he could have used instead.",
        "video_name": "O1R4H3Ca82E",
        "timestamps": [
            134,
            223
        ],
        "3min_transcript": "I have two identical rectangles here, and I want to measure how much space each of them take up on the plane of my screen, the screen that you are looking at right now. And I want to do it using two different units. It's clear, since they're two identical rectangles, that they take up the exact same amount of space. They will have the same area. But what we could see it that we can measure that area using different units. So first over here, let's say that this figure is 1 foot in width. And it is also 1 foot in height. So this right over here is equal to 1 square foot. It's clearly a square. It has the same width and the same height. And each of these dimensions are 1 foot, so we could call it 1 square foot. So let's see how many square feet we can get onto one of these rectangles, and essentially we'll be measuring its area And so we want to cover the entire space without overlapping and without going over the boundary. So that's 1, 2, 3, 4, 5, and 6. 6 in that first row, and then I have 7, 8, 9 10, 11, and 12. So that looks like this area, which is the same as this area If I were to measure it in square feet, the area is equal to, let me write this down, the area is equal to, we have-- let me write it down. We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 square feet. Now, I'm going to try to measure that same area in a different unit, and I'm going to just make up this unit. I'm going to call it a furgle. And a furgle in one dimension, so a furgle in one dimension is twice a foot. So that distance right over here, I'm going to call a furgle. That's one furgle. This is something that I made up just for the purpose of this video. Most people will not recognize what a furgle is. So its height is one furgle. Its width is a furgle. And so we could say that this is 1 square furgle. So let's see how many square furgles is this area, that same area that is 12 square feet. So let me copy and paste this. So let me copy, and then paste."
    },
    {
        "Q": "\nAt 2:51, when he was talking about 30 x 40, how did the 4 change to 40?",
        "A": "It didn t. 30*4 is 120. If it was 40, his answer would ve been 1200.",
        "video_name": "8bK-xfh8-rY",
        "timestamps": [
            171
        ],
        "3min_transcript": "Number of times we get an elephant, out of the 210 times. So why don't you to have a go at it? All right, so you've, I'm assuming, like always, pause the video and then had a try. So one way to think about it, is, well, for one spin, what is the probability of getting an elephant? So let's do this, one spin. So for one spin, what is the probability of getting an elephant? Well let's see, we have already talked about, this is a fair spinner. There are seven equally likely possibilities. And then how many involve getting an elephant? Well, so we have one, two, three, four. Four out of the seven equally likely possibilities involve us getting an elephant. So one reasonable thing to do, and this is actually what I would do, is go, look, a 4/7 probability means I should expect that 4/7 of the time, and over, and over again, it's a reasonable expectation that, hey, 4/7 of the time, I will get an elephant. I've just calculated the theoretical probability here, based on this being a fair spinner. And that should inform, that if I were to do a bunch of experiments, that 4/7 of the time that I should see me getting the elephant. So it would be a reasonable prediction to say well, look, I'm going to spin this thing 210 times, and I would expect that 4/7 of those 210 times, I would get an elephant. And so, let's think about what this is. 210 times 4/7, 210 divided by seven is 30. 30 times 40 is 120. So 120 times. My prediction, or maybe your prediction was this as well, I think it's a reasonable prediction, is that if I spin it 210 times, that I'm going to get an elephant 120 times. what this is saying, and this is not saying. Is it possible that I get an elephant 121 times, or maybe 119 times? Sure, sure, it's completely reasonable that you might get something different than this. In fact, there's some probability that you get no elephant. If you consider getting an elephant lucky, that you just happen to keep landing on the monkey or one of the mice, and that's a very low probability that that would happen if you spun it 210 times, but it is possible. So it's important to realize that this is just a prediction. There's actually a possibility that you might get an elephant on all 210 spins. Once again, that's a low probability, but it is possible. So this isn't saying that you're definitely going to get the elephant 120 times. In fact, it's very reasonable, that you might get the elephant 123 times, or 128 times, or 110 times, or even 90 times."
    },
    {
        "Q": "\nCan the expression at 3:40 also be expressed as 5n+1 ?",
        "A": "It was expressed as 5n + 1 when Sal simplified 6 + 5(n - 1) then he observed that 5n + 1 was equal to 1 + 5n. I prefer 5n + 1 because it can be used to visualize the pattern without resorting to a 0 term. 5n toothpicks can be used to almost construct n houses. Almost, because 1 more toothpick is needed for the right wall on the rightmost house.",
        "video_name": "GvbrtnEYRpY",
        "timestamps": [
            220
        ],
        "3min_transcript": "So let's say you are in the nth term of the sequence. If you are in the nth term, however more you are than 1, so if you're the nth term, you're going to be n minus 1 more than 1. If that is confusing, we'll do it with real numbers, so it gets a little bit more tangible. So the nth term, you are n minus 1 more or greater than-- I'll just say more-- than 1. For example, if n is 2, you are 1 more than 1. If n is 3, you are 3 minus 1, which is 2 more than 1. So however much more you are than 1, you multiply that by 5. We're 2 more than 1, so we add 10 to the number of toothpicks we have here. We're 3 more than 1 here, so we add 15 to the number of toothpicks we have there. toothpicks is equal to the number you're more than 1, < n minus 1 times 5. That's how much you're going to add above and beyond the amount of toothpicks in just the first sequence, so times 5. n minus 1 times 5 plus the number of toothpicks that you would just have in the first sequence or just this one house. And we already counted that: plus 6. So that's one way to think about it. And if this looks complicated, you just say, well, look, if I put a 4 here, I'm 3 more than 1, so 4 minus 1 is 3, so that'll be 3 times 5, which is 15. And then you add the number of toothpicks in 1 and then you get 6. Now another way, and many of you all might find this easier to think about, is even in 1, you could imagine a term here-- let me do it this way. You can imagine a 0th term. Let me just draw it here. Imagine a 0th term, and the 0th term would just be kind of a left wall of the house, or in this case, the left toothpick of a house. And then the first one, you're adding 5 toothpicks to that. In the second one, you're adding 5 toothpicks to that. And when you think about it this way, it actually becomes a little simpler to think in terms of n. Here, you could say, well, the nth term-- let me do this in a different color. You could say that the nth term is going to have-- or maybe we should say number of toothpicks in nth figure is going to be equal to 1. So in the 0th figure, which I just made up, you have at least 1 toothpick, and then whatever term in the sequence"
    },
    {
        "Q": "There is no longer a clarification for Sal's mistake at 2:20.\nHow do we fix this? This could really confuse a kid.\n",
        "A": "Request a change, that might work",
        "video_name": "muZmOiiukQE",
        "timestamps": [
            140
        ],
        "3min_transcript": "- [Voiceover] Let's get some practice interpreting graphs of proportional relationships. This says the proportion relationship between the distance driven and the amount of time driving shown in the following graph, so we have the distance driven on the vertical axis, it's measured in kilometers, then we have the time driving and it's measured in hours along the horizontal axis. We can tell just visually that this, indeed, is a proportional relationship, how do we know that? Well, the point (0, 0) is on this graph, the graph goes through the origin. If we have zero time, then we have zero distance, and we can also see that it's a line, then it's a linear relationship. If you have a linear relationship that goes through the origin, you're dealing with a proportional relationship. You could also see that by taking out some points here. Let's see, I'm just eyeballing, so I'm gonna look at where does the graph kind of hit a very well-defined point. Actually, point A right over here, we see that when our time is five hours, our distance travelled or driven Then if we look at time, this point right over here, when our time is 2.5, we see that our distance driven is 200 km. And notice, the ratio between these variables at any one of these points is the same. 400 divided by five is 80, and 200 divided by 2.5 is also going to be 80. Or if you wanna go the other way around, to go from time to distance, we're always multiplying by 80. In fact, we can say that distance divided by time, our proportionality constant is going to be 80. Or if we wanted to include the units there, it might be a little more obvious than dealing with the rate, distance is in kilometers, time is in hours, 80 km per hour, We are going at this speed, 80 km per hour. This is also the proportionality constant. Anyway, with all of that out of the way, let's actually answer the questions. Which statements about the graph are true? Select all that apply. The vertical coordinate of point A represents the distance driven in four hours. The vertical coordinate. So point A is at the coordinate (5, 400). The vertical coordinate tells us how high to go up, how far to move in the vertical direction. That's gonna be the second coordinate right over here, so this is the vertical coordinate. This right over here tells us the distance we've driven in four hours, so yes, the vertical coordinate of point A represents the distance driven in four hours. We've driven 400 km, I like that one, I'll check that one. The distance driven in one hour is 80 km."
    },
    {
        "Q": "at 0:43, why did you use x instead of t for tiles\n",
        "A": "variables represent values that you are solving for, often times the letter used to represent the unknown is arbitrary. he certainly could have used t, but most likely elected to avoid t because it is often used to represent time",
        "video_name": "FZ2APP6-grU",
        "timestamps": [
            43
        ],
        "3min_transcript": "A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to spend less than $1,000. And it's less than $1,000, not less than or equal to $1,000. The size of each tile is one square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than $1,000, we say it right there. sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. And we're done."
    },
    {
        "Q": "@1:00, Sal writes 3x<1000. Can we also write 3x-1000? Is that communicating the same thing.?- Thank You\n",
        "A": "Adriene, If you have 3x<1000 you could subtract 1000 from both sides and then you would have 3x-1000<0",
        "video_name": "FZ2APP6-grU",
        "timestamps": [
            60
        ],
        "3min_transcript": "A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to spend less than $1,000. And it's less than $1,000, not less than or equal to $1,000. The size of each tile is one square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than $1,000, we say it right there. sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. And we're done."
    },
    {
        "Q": "\nat 1:44, you said ' if we subtract these 2 guy's', then said '90 plus 90 is 180'. it is, but weren't you going to subtract?",
        "A": "I believe he said If we subtract... so these two guys.. not If we subtract these two guys . He was talking about subtracting theta from 180, not subtracting 90.",
        "video_name": "iqeGTtyzQ1I",
        "timestamps": [
            104
        ],
        "3min_transcript": "So in this diagram over here, I have this big triangle. And then I have all these other little triangles inside of this big triangle. And what I want to do is see if I can figure out the measure of this angle right here. And we'll call that measure theta. And they tell us a few other things. You might have seen this symbol before. That means that these are right angles or that they have a measure of 90 degrees. So that's a 90-degree angle, that is a 90-degree angle, and that is a 90-degree angle over there. And they also tell us that this angle over here is 32 degrees. So let's see what we can do. And maybe we can solve this in multiple different ways. That's what's really fun about these is there's multiple ways to solve these problems. So if this angle is theta, we have theta is adjacent to this green angle. And if you add them together, you're going to get this right angle. So this pink angle, theta, plus this green angle must be equal to 90 degrees. When you combine them, you get a right angle. So you could call this one-- its measure is And now we have three angles in the triangle, and we just have to solve for theta. Because we know this angle plus this angle plus this angle are going to be equal to 180 degrees. So you have 90 minus theta plus 90 degrees plus 32 degrees-- so I'm going to do that in a different color-- is going to be equal to 180 degrees. The sum of the measures of the angle inside of a triangle add up to 180 degrees. That's all we're doing over here. And so let's see if we can simplify this a little bit. So these two guys-- 90 plus 90's going to be 180, so you get 180 minus theta plus 32 is equal to 180 degrees. And then what else do we have? We have 180 on both sides. We can subtract that from both sides. So that cancels out. That goes to 0. You can add theta to both sides. And you get 32 degrees is equal to theta, or theta is equal to 32 degrees. So it's going to actually be the same measure as this angle right over here. That's one way to do the problem. There's other ways that we could have done the problem. Actually, there's a ton of ways we could have done this. We could have looked at this big triangle over here. And we could've said, look. If this is 90 degrees over here, this is 32 degrees over here, this angle up here is going to be 180 minus 90 degrees minus 32 degrees. Because they all have to add up to 180 degrees. And I just kind of skipped a step there. Actually, let me not skip a step. Let me call this x. If we call the measure of that angle x, we would have x plus 90. I'm looking at the biggest triangle in this diagram right here. x plus 90 plus 32 is going to be equal to 180 degrees."
    },
    {
        "Q": "At 2:20 he integrated g'(x)=1 to get g(x)=x, but shouldn't the integral of g'(x)=1 be g(x)=x+c?\n",
        "A": "That is correct if that is where you were going to end your problem, but since there will be further integrals down the road, you can just add a +C at the end of the problem to encompass all the +C you would have had to put in.",
        "video_name": "iw5eLJV0Sj4",
        "timestamps": [
            140
        ],
        "3min_transcript": "The goal of this video is to try to figure out the antiderivative of the natural log of x. And it's not completely obvious how to approach this at first, even if I were to tell you to use integration by parts, you'll say, integration by parts, you're looking for the antiderivative of something that can be expressed as the product of two functions. It looks like I only have one function right over here, the natural log of x. But it might become a little bit more obvious if I were to rewrite this as the integral of the natural log of x times 1dx. Now, you do have the product of two functions. One is a function, a function of x. It's not actually dependent on x, it's always going to be 1, but you could have f of x is equal to 1. And now it might become a little bit more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as the product of one function, and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already. f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times. So when you figure out what should be f and what should be g, for f you want to figure out something that it's easy to take the derivative of and it simplifies things, possibly if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x, I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x times g of x, which is x, which is xdx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely. This is going to end up equaling x natural log of x"
    },
    {
        "Q": "at 12:22 and earlier, could we have constructed the triangle to have the right angle at the tip of a? ..and would this have any other effect besides flipping the projection from a\u00e2\u0086\u0092b to b\u00e2\u0086\u0092a?\n",
        "A": "a\u00c2\u00b7b and b\u00c2\u00b7a yield the same result, so switching the vectors is fine as long as you evaluate them consistently. For simplicity, I would call the shorter vector a so I m doing most of my work with smaller values. Sal was also unclear on what happens when the projection of a is shorter than the length of b. I doubt it would be an issue but am having trouble visualizing the area in that situation.",
        "video_name": "tdwFdzVqito",
        "timestamps": [
            742
        ],
        "3min_transcript": "Or if I put a light shining down from above and I'd say what's the shadow of a onto b? You'd get nothing. You'd get 0. This arrow has no width, even though I've drawn it to have It has no width. So you would have a 0 down here. The part of a that goes in the same direction as b. No part of this vector goes in the same direction as this vector. So you're going to have this 0 kind of adjacent side times b, so you're going to get something that's 0. So hopefully that makes a little sense. Now let's think about the cross product. The cross product tells us well, the length of a cross b, I painstakingly showed, you is equal to the length of a times the length of b times the sin of the angle between them. So let me do the same example. Let me draw my two vectors. That's my vector a and this is my vector b. And now sin-- SOH CAH TOA. Sin of theta-- SOH CAH TOA-- is equal to opposite over the hypotenuse. So if I were to draw a little right triangle here, so if I were to draw a perpendicular right there, this is theta. What is the sin of theta equal to in this context? The sin of theta is equal to what? It's equal to this side over here. Let me call that just the opposite. It's equal to the opposite side over the hypotenuse. So the hypotenuse is the length of this vector a right there. It's the length of this vector a. So the hypotenuse is the length over my vector a. So if I multiply both sides of this by my length of vector a, I get the length of vector a times the sin of theta is equal to the opposite side. So if we rearrange this a little bit, I can rewrite this as equal to-- I'm just going to swap them. This is equal to b, the length of vector b, times the length of vector a sin of theta. Well this thing is just the opposite side as I've defined it right here. So this right here is just the opposite side, this side right there. So when we're taking the cross product, we're essentially multiplying the length of vector b times the part of a that's going perpendicular to b. This opposite side is the part of a that's going perpendicular to b. So they're kind of opposite ideas. The dot product, you're multiplying the part of a that's going in the same direction as b with b. While when you're taking the cross product, you're multiplying the part of a that's going in the perpendicular direction to b with the length of b. It's a measure, especially when you take the length of this, it's a measure of how perpendicular these two guys are. And this is, it's a measure of how much do they move in the"
    },
    {
        "Q": "At about 5:10, wouldn't the \"- x\" (minus x) in \"-y * (x - x)\" be changed to \"-x\" (negative x)?\n",
        "A": "No because the expression is -y * (x-x) which in words would read, negative y times x minus x. Since any number subtracted by itself is zero then your final product will equal zero.",
        "video_name": "hmtJV49AWio",
        "timestamps": [
            310
        ],
        "3min_transcript": "we multiply these things. So we could view this as negative S, that's that, let me write it a little bit neater. We could view it as negative S times S, times S times T. Times, oops, let me do that in a different color, times S times T. Times T. And do any of the choices look like that? Well almost. Instead of saying negative S times S, this says S times negative S. And because multiplication, once again, I'm not a big fan of using the word because it sounds complicated, but it's commutative. A times B is the same thing as B times A. So I can rewrite this as, I can rewrite this, I can swap these two and write this as S times, S times negative S times T, Times T. All I did is I swapped these two. This negative S and this S. I just swapped them and I got exactly what I have right over here. Now let's just make sure that this one does not apply and maybe the easiest way is to try to simplify this. And the best way I could think about that is by distributing this S. So if I distribute this S, what I'm going to get, this is going to be equal to S times T, which is ST. Or I could even, I can write it like this. I could write it S times T, like that. And then I have minus S times S, so minus S times S. I could write it that was or I could write minus S squared if I want to. That's the same thing as S times S. But this is very different. This is very different. Here I'm just taking the product of three variables here, I have two different terms, taking the product of two variables here and then the product I guess you could say So this is not, this is not the same thing. Which of the following expressions are equal to negative X times, and then in parentheses negative Y times X? And I forgot to mention it, but like always, pause the video, try to work them out by yourself before I do them. Alright, select all that apply. So let's just try to manipulate this a little bit. So once again, multiplication, it's associative. I could, so it's negative X times, times negative Y times X. So, the way it's written here, I could do these first, that's essentially what's written over here. Or it's associative. Instead, I could do these first. And the reason why I find this interesting is a negative times a negative is going to be a positive. So this is going to be the same thing, this thing over here is going to be the same thing as positive X times positive Y. Negative times a negative is a positive. So you're gonna get positive X times Y and then you're multiplying by an X again."
    },
    {
        "Q": "In 3:45 (or so) he says that the derivative of ln2 = ln2. How is this possible, if the derivative of lnx=1/x? shouldn't then the derivative of ln2 = 1/2?\n",
        "A": "Gotcha, so you re saying there s a big difference in whether or not an x is involved with the ln. derivative of lnx=1/x therefore derivative of lncx (where c is a constant) = 1/cx. And then derivative of lnc=lnc, because its a number, without a variable. Is that right? (This is a response to Tysaki, not another answer)",
        "video_name": "Mci8Cuik_Gw",
        "timestamps": [
            225
        ],
        "3min_transcript": "And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect So we took the derivative of e to the something with respect to that something-- that's this right here, it's just e to that something. And then we're going to multiply that by, this is just an application of the chain rule, of the derivative of that something with respect to x. So the derivative of natural log of 2 times x with respect to x is just going to be natural log of 2. This is just going to be natural log of 2. The derivative of a times x is just going to be equal to a. This is just the coefficient on the x. And just to be clear, this is the derivative of natural log of 2 times x with respect to x. So we're essentially done. But we can simplify this even further. This thing right over here can be rewritten. And let me draw a line here just to make it clear that this equals sign is a continuation from what we did up there. But this e to the natural log of 2x, we can rewrite that, using this exact same exponent property, as e to the natural log of 2, and then And of course, we're multiplying it times the natural log of 2, so times the natural log of 2. Well, what is e to the natural log of 2? Well, we already figured that out. That is exactly equal to 2. This right over here is equal to 2. And so now we can simplify. This whole thing, the derivative of 2 to the x, is equal to-- and I'll switch the order a little bit-- it is the natural log of 2, that's this part right over here, times 2 to the x. Or we could write it as 2 to the x times the natural log of 2."
    },
    {
        "Q": "\nAt 3:12, when we get our final answer, why did we integrate just sinx and not 1? Shouldn't it be in the end \"xsinx + xcosx + c\" or x (sinx +cosx) + c?",
        "A": "Yes yes! Thankyou :) I was terribly confused.",
        "video_name": "bZ8YAHDTFJ8",
        "timestamps": [
            192
        ],
        "3min_transcript": "If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x of this, which is just negative cosine of x. And then we could throw in a plus c right at the end of it. And doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant which could even be negative. And so this is all going to be equal to-- we get our drum roll now-- it's going to be x times sine of x, subtract a negative, that becomes a positive, plus cosine of x plus c. And we are done. We were able to take the antiderivative of something that we didn't know how to take the antiderivative of before. That was pretty interesting."
    },
    {
        "Q": "At 1:48, why does g(x) = sinx. Why is it not g(x) = sinx + c?\n",
        "A": "He is just noting what g (x) and g(x) are. The actual integration happens in the problem. When he does the integration the integral sign and dx disappear and that is your signal to add the +c to your answer before you put the little box around it. If the problem had more than one integral you would not put +c every time you did an integration. Only when the last integral disappears would you add a +c which represents all the arbitrary constants from all the integrations that happened.",
        "video_name": "bZ8YAHDTFJ8",
        "timestamps": [
            108
        ],
        "3min_transcript": "In the last video, I claimed that this formula would come handy for solving or for figuring out the antiderivative of a class of functions. Let's see if that really is the case. So let's say I want to take the antiderivative of x times cosine of x dx. Now if you look at this formula right over here, you want to assign part of this to f of x and some part of it to g prime of x. And the question is, well do I assign f of x to x and g prime of x to cosine of x or the other way around? Do I make f of x cosine of x and g prime of x, x? And that thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here. And here where we have the derivative of f of x times g of x. So what you want to do is assign f of x so that the derivative of f of x is actually simpler than f of x. And assign g prime of x that, if you were to take its antiderivative, it doesn't really become any more complicated. So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler, If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x"
    },
    {
        "Q": "\nat 2:33 I don't understand, why does he write expected value as m rather than E(X). Is it the same thing or not?",
        "A": "he wrote mu, it just looked liked M",
        "video_name": "ry81_iSHt6E",
        "timestamps": [
            153
        ],
        "3min_transcript": "If this is 60% chance of success there has to be a 40% 70% chance of success, 30% chance of failure. Now with this definition of this-- and this is the most general definition of a Bernoulli Distribution. It's really exactly what we did in the last video, I now want to calculate the expected value, which is the same thing as the mean of this distribution, and I also want to calculate the variance, which is the same thing as the expected squared distance of a value from the mean. So let's do that. So what is the mean over here? What is going to be the mean? Well that's just the probability weighted sum of the values that this could take on. So there is a 1 minus p probability that we get So there's 1 minus p probability of getting 0, so times 0. And then there is a p probability of getting 1, plus p times 1. Well this is pretty easy to calculate. 0 times anything is 0. So that cancels out. And then p times 1 is just going to be p. So pretty straightforward. The mean, the expected value of this distribution, is p. And p might be here or something. So once again it's a value that you cannot actually take on in this distribution, which is interesting. But it is the expected value. Now what is going to be the variance? What is the variance of this distribution? Remember, that is the weighted sum of the squared distances from the mean. Now what's the probability that we get a 0? There's a 1 minus p probability that we get a 0. So that is the probability part. And what is the squared distance from 0 to our mean? Well the squared distance from 0 to our mean-- let me write it over here-- it's going to be 0, that's the value we're taking on-- let me do that in blue since I already wrote the 0-- 0 minus our mean-- let me do this in a new color-- minus our mean. That's too similar to that orange. Let me do the mean in white. 0 minus our mean, which is p plus the probability that we get a 1, which is just p-- this is the squared distance, It's the probability weighted sum of the squared distances from the mean. Now what's the distance-- now we've got a 1-- and what's the difference between 1 and the mean? It's 1 minus our mean, which is going to be p over here."
    },
    {
        "Q": "how did Sal get 3/3 when there is only one three present? @2:05 in the video.\n",
        "A": "The average change is calculated using (y1-y1)/(x2-x1), or the slope, right? Here y is a function of f, that means we have (f(x2) - f(x1))/(x2 - x1) We are told that x1=2 and x2=5 We are told that f(x) = x\u00c2\u00b2 - 6x + 8. That means f(x2) = f(5) = 5\u00c2\u00b2 - (6)(5) + 8 = 25 - 30 + 8 = 3 And that f(x1) = f(2) = 2\u00c2\u00b2 - (6)(2) + 8 = 4 - 12 + 8 = 0. THEREFORE (y1-y1)/(x2-x1) = (f(x2) - f(x1))/(x2 - x1) = (3 - 0)/(5 - 2) = 3/3.",
        "video_name": "S_YIUXy-WFM",
        "timestamps": [
            125
        ],
        "3min_transcript": "Let's say I have some function f of x that is defined as being equal to x squared minus 6x plus 8 for all x. And what I want to do is show that for this function we can definitely find a c in an interval where the derivative at the point c is equal to the average rate of change over that interval. So let's give ourselves an interval right over here. Let's say we care about the interval between 2 and 5. And this function is definitely continuous over this closed interval, and it's also differentiable over it. And it just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, that's in the open interval such that the derivative at c Or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see f of 5 minus f of 2, f of 5 is, let's see, f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8. So this is equal to 3/3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be in this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7/2, which is the same thing as 3 and 1/2. So it's definitely in this interval right over here. So we've just found our c is equal to 7/2. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis."
    },
    {
        "Q": "At 3:01 shouldn't it be:\n2a^2-4a\n----------\n(a+2)(a+2)\n",
        "A": "No, if you multiply it out using FOIL, you ll see (a+2)(a-2) = a^2 - 2a + 2a - 4. The a terms cancel out and you are left with a^2 - 4.",
        "video_name": "IKsi-DQU2zo",
        "timestamps": [
            181
        ],
        "3min_transcript": "is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there. be very careful here-- you're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3, so what does this simplify to? You have a squared minus a plus-- let's see, negative 4 plus 3 is negative 1, all of that over a plus 2 times a plus 2. We could write that as a plus 2 squared. Now, we might want to factor this numerator out more, to just make sure it doesn't contain a common factor with the denominator. The denominator is just 2a plus 2 is multiplied by themselves. And you can see from inspection a plus 2 will not"
    },
    {
        "Q": "At 1:17, isn't it supposed to be 150/1000 instead of 150/100? Or is it just for the decimals?\n",
        "A": "Sal is starting with 150%. A percent by definition is a ratio that has a denominator of 100. So, it is 150/100. Hope this helps.",
        "video_name": "xEDnwEOOf7Y",
        "timestamps": [
            77
        ],
        "3min_transcript": "So we have 0.79 minus 4/3 minus 1/2 plus 150%. So we have four different numbers written in different formats. Here it's a decimal, here we have two fractions, and then here we have a percentage. So the easiest thing to do would be to write all of these in the same format. And for me, the easiest format to do this computation in would be to write them all as fractions. And the reason why I want to do that, in particular, is because 4/3, when you divide by 3, when you divide 1/3, 2/3, 4/3, you're going to have a repeating decimal. So to avoid that, I want to put all of these-- I want to rewrite all of these as fractions. So let's do them one at a time. So 0.79, this is the same thing as 79/100, so I'll just write it that way. So this is the same thing as 79 over 100. Then of course, we have minus 4/3. Then we have minus 1/2. And then finally, we have-- I don't Finally we have 150%. Well, 150%, percent literally means per cent, per hundred. So this is plus 150 per 100. So now we've written them all as fractions. And in order to do all the subtraction and addition, we have to find a common denominator. So what's the least common multiple of 100, 3, 2, and 100? Well, 100 is divisible by 2, so 100 is actually the least common multiple of 102. So we really have to just find the least common multiple between 100 and 300. And that's just going to be 300. There's no other common factors between 100 and 3. So let's write all of them with 300 as the common denominator. So let me do this in this reddish color. So 79 over 100 is the same thing. If I were to write it over 300, to go from 100 to 300 in the denominator, I'm multiplying by 3, so I have to multiply the numerator by 3 as well. Let's see, 80 times 3 would be 240. So it's going to be 3 less than that. So 240 minus 3 is 237. Now 4/3. Well, to get the denominator to be 300, we have to multiply the denominator by 100, so we have to multiply the numerator by 100 as well. 1/2, if our denominator is 300, we multiplied the denominator by 150 to go from 200 to 300, so we have to multiply the numerator by 150. And then finally, 150 over 100, well, we're multiplying the denominator by 3 to get to 300, to go from 100 to 300. So we have to do the same thing in the numerator. So 3 times 150 is 450."
    },
    {
        "Q": "\nAt 0:57, what is the difference between a zig and a zag?",
        "A": "The direction.",
        "video_name": "EdyociU35u8",
        "timestamps": [
            57
        ],
        "3min_transcript": "So you're me and you're in math class, and you're supposed to be learning about logarithms, which you figure is probably something to do with avant-garde percussion performances. But every time you try to pay attention, you find that your teacher's explanations of logarithms inspires nothing but a bit of performance art involving drumming on wood. Anyway, your teacher gives you a dirty look. So you stop drumming and start doodling. You feel a need for motion, excitement, so you decide to do a flip book on the corner of your notebook. It's pretty easy because the paper is thin. You can trace the previous frame, modifying it just a little. And if you really want to get into the Zen zone of doodle bliss, you can make each new frame be modified according to some simple rule like, keep adding one new petal around a spiral, or add another dot in the next place in the spiral. Or how about, keep making the squiggle squigglier. But what does squigglier really mean? Like you could just increase the curvature of the squiggle, or you could make it squiggle more between the squiggles. You want to figure out an exact squiggle rule so that you can really get into the squiggle zone. So you discretize the squiggle into a zigzag line. On the next layer, you could just deepen the zag, Or maybe zig's too good to zig zag, and zag's too good to zag zig. So the next would be, zigzag zag zig, zigza-- wait, no. That was a zag, and those get turned into zags zigs. So it should be zigzag, zag zig zag zig zig zag. And next, zig zag zag zigzag, zig zig zigzag, zag zig-- wait, was this a zig? Maybe if we put it all onto some sort of reference diagram, that generates each new pattern based on-- No! This was supposed to be about mindless zigging in the squiggle zone. This is unacceptable. So maybe just pretend each time you get to a new stage, the old path is all just zigzag zigzag zigzag zigzag. And to keep it all neat and orderly, you decide to try to make all the lines the same length, always with right angles between them. Here we go. [SINGING] What if we tried starting with three lines? Zig zag zug. OK. So then each zig should get a zig zag zug, and each zag a zug zag zig? And then the zug gets, well, maybe it just goes back and forth, whether it goes on the inside or the outside. OK, wait now where did it start? And everything's running into itself, and getting bigger, and doesn't fit on the paper. OK, maybe if it were a little more open it wouldn't run into itself. Say like, half a hexagon, so you can keep all the angles perfect. And then trapezoids in and out. Yeah, this is totally going to work. Lookit. In fact, like this, you could even make it go inside first, and back and forth, and it wouldn't run into itself. Next, you could make it go even closer together here, and still not run into itself. Except that means you'd have to start on the outside this time. But that's OK, because next time you can go on the inside again. And now it's an easy pattern."
    },
    {
        "Q": "In 6:52, why does the paper's edge stay the same instead of shrinking just like the infinitely small paper is?\n",
        "A": "because of how she folded it. if she folded it alternating sides the edge would shrink",
        "video_name": "EdyociU35u8",
        "timestamps": [
            412
        ],
        "3min_transcript": "OK, but there's probably some rule. Suddenly, a note lands on your desk from your friend Sam. Who writes, looks like you're concentrating pretty hard. Don't tell me you're actually doing math. As, if. You write back, no way. I'm just doodling this. And just to make extra clear it's not math, you turn it into a dragon, and name it the dragon curve. Yes. You don't want to crumple up your awesome dragon doodle, but you do have to throw it two rows over. So you neatly fold it into a note spear. Which just means you're folding it in half again and again, until it's easy to javelin across the room the moment the teachers back is turned-- Bam! Yes! Perfect landing. You watch as Sam unfolds it. And suddenly you feel like you see something familiar. Some sort of similarity between the paper and-- is that possible? You take your diagram, and fold it in half. And half again. And again. And wow, not only does it look like it's doing the same thing, probably, but it's also showing a new way to do it. you can just copy the old one, and add it 90 degrees from the other, which is totally traceable. Bam. Well, as long as you can keep track of what end to start from. And you don't even have to keep track of what order to draw the lines in. You just need to keep things roughly on a square grid so things line up. Until it gets too big for your paper and you have to dragon-ize it. It's funny, because one way it gets bigger and bigger. If you go on forever, it'll be infinitely big. But with the first way, it stays basically the same size. You just draw more details. Which means if you do it forever, the line will still get infinitely long, but the total size will stay the same. Will that even work? An infinitely long line all squiggled up into a finite area? And then with folding paper, the whole thing gets smaller and smaller until maybe it disappears entirely. Which you suppose makes sense because the edge of the paper stays the same length, no matter how you fold it. You can't make it longer and longer like copying it, where the length doubles each time. Maybe this grid really would fill up until that infinitely long line is squiggled up into an actual solid two dimensional triangle with no empty space left in it. Maybe that would make sense. Or be crazy. You know lines are infinitely thin, but if you had infinitely much of it, maybe the infinities would cancel out or something. Like the line gets closer and closer to itself until it actually touches, but doesn't overlap, but with no space between. Which would make no sense, unless you do it everywhere at once, so who can tell the difference. Yeah, that totally sounds legit. Although, in this one, there's holes that never get filled at any point, so you'd still have an infinite line, but it doesn't fill up space, it's just all holes forever, but it also never overlaps, So where is all that infinite line going? Anyway, class is over, so you pack up and save the question. After all, you've got math class again tomorrow."
    },
    {
        "Q": "\nAt 2:24 she mentions e. What is e?",
        "A": "e is a mathematical constant that show up in many areas of mathematics.",
        "video_name": "5iUh_CSjaSw",
        "timestamps": [
            144
        ],
        "3min_transcript": "Voiceover:Hello and welcome to that one day of the year when, well, everyone else is building up how great Pi is. I'm here to tear it down, because you deserve the truth. Forget about the part where Pi isn't the correct circle concept. This Pi day, I'm not about how people worship Pi for being infinite for going on forever. First of all, Pi is not infinite. It is more three, but you know, less than four. There are cultures where three is the biggest number, so I don't want to be insensitive, but trust me on this four is not infinite and neither is Pi. I know it's not about it's magnitude, it's about all those digits, infinite digits going on forever, but first of all it doesn't go anywhere. It just is. There's no time element. If you had a number line, Pi would be exactly one point on that number line sitting perfect still right now. It's not going to start wondering off on an infinite journey that takes forever, or even on a finite journey that takes forever, or an infinite journey that takes finite time. Secondly, yeah, so it's got infinite digits. So what, one-third has infinite digits. There's exactly as exactly as many digits in one-third and in Pi as in 99.9999 repeating. Oh, and there's also as many digits as in numbers like, say, five, I know, big number. It's even more than four, so, it's piratically like double infinity. Which, it actuality kind of is, because in decimal innovation there's secretly infinite zeros in all of these places. Zero's going out to forever. Ooh! So mysterious, and then zero's going the other way too. Which is actually not any more zeros than if they only went one way. No. Pi is not especially infinite in any way, it's more like in-between-finite. There's an infinite number of rational numbers. for any two factions you can find another fraction that's between them again and again and again. There's never any fractions that are right next to each other on the number line. But, despite there's a infinite amount of rational numbers, Pi isn't one of them. and you can find an infinite number of rational numbers that are closer to Pi on either side. Pi is between all of them in one of the gaps. It isn't infinite. It's in-between-finite. So what, you think that's special, as if there's just one hole in the rational number line exactly where Pi is, and once you plug that in with a super special number, you're good to go? Maybe, a few more for E and [towel] and square root too. No! Super nope! The in-between-[finiteness] of Pi, its irrationality is an incredibility un-special property. Turns out, most real numbers are irrational. It's the nicely packaged rational numbers that are weird. In fact, if you threw a dart and picked a random number off the number line, the chance of getting a rational number is exactly zero. I'll get into kinds of infinities some other time, but [unintelligible] to say the number of rational numbers, like the number of digits in Pi, is the small and unimpressive countable infinity. While the number of irrational numbers is so much bigger than countable infinity,"
    },
    {
        "Q": "\nin 2:18, can you make it until 'z' with using all the alphabets in a row?",
        "A": "Sure you can! Neat stuff!",
        "video_name": "a5z-OEIfw3s",
        "timestamps": [
            138
        ],
        "3min_transcript": "OK, so fractal fractions. 5 equals 5. Bear with me now. Let's explode this 5 into fifths. 25 fifths to be precise. Now I want to split it into two parts. Say, 17/5 plus 8/5. Could have been 1 plus 24, or whatever, I don't care. OK, now I'm ready for the fun part. Since this 5 down here is just as much a 5 as any 5 is 5, such as this whole thing, which is equivalent to 5, let's go ahead and replace that 5 with this messier looking but still very fively 5. Oh, and now we've got more fives, so we can do it again. And again. And again. And then you can give someone this whole thing and be like, whoa, it's 5. Making things look more confusing than they actually are is a delicate art, but it speaks to the true heart of algebra, which is that you can shuffle numbers around all day and as long as you follow the rules, it all works out. OK, here's another one. Say you want to do something with 7. Maybe you could use 7/7, which is 1. So you need six more. Why not just add it? 7 equals 7/7 plus 6. 7 equals 7/7, plus 6/7, over 7 plus 6, plus 6. Instead of writing it all again, why not just extend this way? There we go. This equals 7. And you can actually take this all the way to infinity. The 7's kind of disappear. But then again, it didn't really matter what they were the first place, as long as they're the same. All these 7's could have been 3, or a billion, or pi to the i, and this would still equal 7. As long as this numerator equals this denominator this fraction equals 1. And whatever else you may think about algebra, at least it has the courtesy to make 1 plus 6 be 7 every time. The fractal structure of this first fraction was like a binary tree. Each layer with twice as many terms as the one above it, growing exponentially. And this one does too, but sideways. But awesomely enough, this is obviously an ABA CABA DABA CABA pattern. That's a fractal pattern that's actually found lots of places, but I'm not going to get into that right now. Point is, if you name this innermost layer and then try to read it from top to bottom, you get ABA CABA DABA CABA. And if your fraction was infinite, you'd get ABA CABA DABA CABA EABA CABA DABA CABA FABA CABA DABA CABA EABA CABA DABA CABA GABA CABA-- and so on. Anyway, a foolish algebra teacher would teach you that algebra is about solving equations. As if the goal of life were to get x on one side, and everything else on the other. As if every fiber of your being should cry out in protest when you see x on the left side, and yet more x on the right side. But you could replace that x with what it equals, and then you could do it again, and again, and each time your equation is still true. How's that for getting rid of the x on this side? And you can make equations even more confusing by remembering special numbers and identities. Write whatever you want, as long as you can sneak in a multiplied by 0, you don't even have to bother knowing what the rest is. Or, knowing that all you need is the top and bottom of the equation to be the same to get 1."
    },
    {
        "Q": "At 1:44, why do we use binary trees for situations like this in math? We can use some other, like just figuring out what kinds of parts and adding them.\n",
        "A": "Binary trees are powerful tools for solving search space problems, but in the video at 1:44, Vi is pointing out the resemblence of the expressions to binary trees.",
        "video_name": "a5z-OEIfw3s",
        "timestamps": [
            104
        ],
        "3min_transcript": "OK, so fractal fractions. 5 equals 5. Bear with me now. Let's explode this 5 into fifths. 25 fifths to be precise. Now I want to split it into two parts. Say, 17/5 plus 8/5. Could have been 1 plus 24, or whatever, I don't care. OK, now I'm ready for the fun part. Since this 5 down here is just as much a 5 as any 5 is 5, such as this whole thing, which is equivalent to 5, let's go ahead and replace that 5 with this messier looking but still very fively 5. Oh, and now we've got more fives, so we can do it again. And again. And again. And then you can give someone this whole thing and be like, whoa, it's 5. Making things look more confusing than they actually are is a delicate art, but it speaks to the true heart of algebra, which is that you can shuffle numbers around all day and as long as you follow the rules, it all works out. OK, here's another one. Say you want to do something with 7. Maybe you could use 7/7, which is 1. So you need six more. Why not just add it? 7 equals 7/7 plus 6. 7 equals 7/7, plus 6/7, over 7 plus 6, plus 6. Instead of writing it all again, why not just extend this way? There we go. This equals 7. And you can actually take this all the way to infinity. The 7's kind of disappear. But then again, it didn't really matter what they were the first place, as long as they're the same. All these 7's could have been 3, or a billion, or pi to the i, and this would still equal 7. As long as this numerator equals this denominator this fraction equals 1. And whatever else you may think about algebra, at least it has the courtesy to make 1 plus 6 be 7 every time. The fractal structure of this first fraction was like a binary tree. Each layer with twice as many terms as the one above it, growing exponentially. And this one does too, but sideways. But awesomely enough, this is obviously an ABA CABA DABA CABA pattern. That's a fractal pattern that's actually found lots of places, but I'm not going to get into that right now. Point is, if you name this innermost layer and then try to read it from top to bottom, you get ABA CABA DABA CABA. And if your fraction was infinite, you'd get ABA CABA DABA CABA EABA CABA DABA CABA FABA CABA DABA CABA EABA CABA DABA CABA GABA CABA-- and so on. Anyway, a foolish algebra teacher would teach you that algebra is about solving equations. As if the goal of life were to get x on one side, and everything else on the other. As if every fiber of your being should cry out in protest when you see x on the left side, and yet more x on the right side. But you could replace that x with what it equals, and then you could do it again, and again, and each time your equation is still true. How's that for getting rid of the x on this side? And you can make equations even more confusing by remembering special numbers and identities. Write whatever you want, as long as you can sneak in a multiplied by 0, you don't even have to bother knowing what the rest is. Or, knowing that all you need is the top and bottom of the equation to be the same to get 1."
    },
    {
        "Q": "\nIn 2:29 Sal made delta= f(epsilon) but I think it should be epsilon=f(delta), shouldn't it ?",
        "A": "In this kind of proof delta is a function of epsilon, even though epsilon refers to the y-axis. In his game, Sal has someone pick an epsilon first. Then he finds the delta to go with it. Thus, epsilon is the independent variable, not delta.",
        "video_name": "0sCttufU-jQ",
        "timestamps": [
            149
        ],
        "3min_transcript": "In the last video, we took our first look at the epsilon-delta definition of limits, which essentially says if you claim that the limit of f of x as x approaches C is equal to L, then that must mean by the definition that if you were given any positive epsilon that it essentially tells us how close we want f of x to be to L. We can always find a delta that's greater than 0, which is essentially telling us our distance from C such that if x is within delta of C, then f of x is going to be within epsilon of L. If we can find a delta for any epsilon, then we can say that this is indeed the limit of f of x as x approaches C. Now, I know what you're thinking. This seems all very abstract. I want to somehow use this thing. And what we will do in this video is use it and to rigorously prove that a limit actually exists. So, right over here, I've defined a function, f of x. It's equal to 2x everywhere except for x equals 5. but when x is equal to 5, it's just equal to x. So I could have really just written 5. It's equal to 5 when x is equal to 5. It's equal to itself. And so we've drawn the graph here. Everywhere else, it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. And if I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. The closer I get to 5, the closer f of x seems to be getting to 10. And so, you might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. the epsilon-delta definition to actually prove it. And the way that most of these proofs typically go is we define delta in the abstract. And then we essentially try to come up with a way that given any epsilon, we can always come up with a delta. Or another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. But maybe I shouldn't use f again. But delta equals function of epsilon that is defined for any positive epsilon. So you give me an epsilon, I just put into our little formula or little function box. And I will always get you a delta. And if I can do that for any epsilon that'll always give you a delta, where this is true, that if x is within delta of C, then the corresponding f of x is going to be within epsilon of L. Then the limit definitely exists. So let's try to do that. So let's think about being within delta of our C."
    },
    {
        "Q": "At 6:46, why isn't delta divided by 2? He says that it is delta/2 but also multiplied by 2 so the 2s will cancel, but I don't see where the 2 that is multiplied is coming from.\n",
        "A": "This is taken from the function. We want to show that |x-5|<\u00e2\u0088\u0082 in the form of |f(x)-L|<\u00ce\u00b5. That is to show that \u00e2\u0088\u0082 is a function of \u00ce\u00b5 .",
        "video_name": "0sCttufU-jQ",
        "timestamps": [
            406
        ],
        "3min_transcript": "this is just going to be the absolute value of 2x minus 10. And it's going to be less than on the right-hand side, you just end up with a 2 delta. Now, what do we have here on the left-hand side? Well, this is f of x as long as x does not equal 5. And this is our limit. So we can rewrite this as f of x minus L is less than 2 delta. And this is for x does not equal 5. This is f of x, this literally is our limit. Now this is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. In terms of epsilon, this has it in terms of delta. So, how can we define delta so that 2 delta is essentially going to be epsilon? Well, this is our chance. We're going to make 2 delta equal epsilon. Or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. Let me switch colors just to ease the monotony. If you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2. It's just going to be less than epsilon. So this is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going And remember, it has to be true for any positive epsilon. But you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. So our limit is up here, so our epsilon is 0.5. So it would literally be the range I want to be between 10 plus epsilon would be 10.5. And then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25. So that'll give us a range between 4.75 and 5.25. And as long as we pick an x between 4.75 and 5.25, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. And so, you give me any epsilon, I can just apply this formula right over here"
    },
    {
        "Q": "\nAt 4:17, when the calculator is shown in the video, is it correct to say that the symbol n! is like the one that can facilitate the factorial multiplication? Do all scientific calculators have it?",
        "A": "Yes, that s correct. n! represents the factorial as long as n is non-negative integer. It is more than just like it. It IS it. I would assume that all scientific calculators have a factorial function, but I wouldn t rule out the possibility of the existence of a calculator deemed scientific to exclude it.",
        "video_name": "SbpoyXTpC84",
        "timestamps": [
            257
        ],
        "3min_transcript": "Or maybe another hand is I have the eight cards, 1, 2, 3, 4, 5, 6, 7, 8, and then I have the 9 of spades. If we were thinking of these as two different hands, because we have the exact same cards, but they're in different order, then what I just calculated would make a lot of sense, because we did it based on order. But they're telling us that the cards can be sorted however the player chooses, so order doesn't matter. So we're overcounting. We're counting all of the different ways that the same number of cards can be arranged. So in order to not overcount, we have to divide this by the ways in which nine cards can be rearranged. So we have to divide this by the way nine cards can be rearranged. So how many ways can nine cards be rearranged? If I have nine cards and I'm going to pick one of nine to be in the first slot, well, that means I have 9 ways to put something in the first slot. Then in the second slot, I have 8 ways of putting a card first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is. times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280."
    },
    {
        "Q": "Why is there a blank space on 2:12\n",
        "A": "the pie was eaten 5 + 3 slices. and the pie initially have 9 slices. so, 8/9. (left 1 blank space). hope that help",
        "video_name": "u2hLYcmI5y4",
        "timestamps": [
            132
        ],
        "3min_transcript": "Brandon ate 5 slices of apple-- of pie. I'm just assuming it's apple pie. They didn't tell me that. Gabriela ate 3 slices. If there were originally 9 slices, what fraction of the pie was eaten? So let me see if I can draw this thing out. So let me draw the pie. I will draw the pie in a yellowish color. So let me try my best. So let's see how good I am at drawing a pie. So I'll just draw it from the top view as a circle. And there're 9 slices. I think it's a reasonable assumption to say that they're 9 equal slices. So we have 9 equal slices of pie. And I'll just make sure they're initially 9 equal slices. What fraction of the pie was eaten? So let's first divide this into 9 sections. So one way I could do that, I could divide it into 3 sections first, so it looks like a peace symbol. It actually looks more like the Mercedes emblem. So I'll draw it into 3 sections first. So let's see, I'll draw like that and like that. Keep in mind, I'm trying to make these as equal as possible. So bear with me if they don't look 100% equal, but I'm trying. I am trying my best. So 9 equal slices-- so that looks pretty respectable. So here's our pie that initially had 9 equal slices. Now, they tell us that Brandon ate 5 slices of pie. So Brandon eats-- he seems like a hungry young man-- so he eats 1, 2, 3, 4, 5. You could say that he ate 5/9 of the pie. But that's not it. That's not what they're asking. It's saying total, not just how much did Brandon eat, but how much was eaten total between Brandon and Gabriela. And they tell us Gabriela ate 3 slices. So she ate 1-- sounds like they didn't really So now let's answer their question. What fraction of the pie was eaten? Well, we know that there was a total of 9 equal slices of pie. What fraction was eaten? Well, as we see, 1, 2, 3, 4, 5, 6, 7, 8. 8/9 of the pie was eaten. So let's actually type that in. And then we will do it right over here. We'd say 8, and we use that little slash symbol on your computer like this. 8 over 9 or 8 divided by 9 or 8/9 of the pie was eaten. Let's do a couple more of these. Ishaan ate 2 slices of pizza. Omar ate 3 slices. If there were originally 8 slices, what fraction of the pie is remaining? So this is interesting. Actually, let me copy and paste this"
    },
    {
        "Q": "\nat 2:08, when he was rewriting the expressions, how come he did 6(2^2) instead of 6(2)^2?",
        "A": "They mean the same thing. So, either form is acceptable.",
        "video_name": "I9eLKDbc8og",
        "timestamps": [
            128
        ],
        "3min_transcript": "The surface area of a cube is equal to the sum of the areas of its six sides. Let's just visualize that. I like to visualize things. So if that's the cube, we can see three sides. Three sides are facing us. But then if it was transparent, we see that there are actually six sides of a cube. So there's this one-- one, two, three in front-- and then one-- this is the bottom. This is in the back, and this is also in the back. So you have three sides of the cube. So I believe what they're saying. The surface area of a cube with side length x-- so if this is x, if this is x, if this is x-- is given by the expression 6x squared. That also makes sense. The area of each side is going to be x times x is x squared, and there's six of them. So it's going to be 6x squared. Jolene has two cube-shaped containers that she wants to paint. One cube has side length 2. So this is one cube right over here. I'll do my best to draw it. So this right over here has side length 2, The other cube has side length 1.5. So the other cube is going to be a little bit smaller. It has side length 1.5. So it's 1.5 by 1.5 by 1.5. What is the total surface area that she has to paint? Well, we know that the surface area of each cube is going to be 6x squared, where x is the dimensions of that cube. So the surface area of this cube right over here is going to be 6. And now-- let me do it in that color of that cube-- it's going to be 6 times x, where x is the dimension of the cube. And then the cube all has the same dimensions, so its length, width, and depth is all the same. So for this cube, the surface area is going to be 6 times 2 squared. And then the surface area of this cube is going to be 6 times 1.5 squared. it's going to be the sum of the two cubes. So we're just going to add these two things. And so if we were to compute this first one right over here, this is going to be 6 times 4. This is 24. And this one right over here, this is going to be a little bit hairier. Let's see. 15 times 15 is 225. So 1.5 times 1.5 is 2.25. So 1.5 squared is 2.25. And 2.25 times 6-- so let me just multiply that out. 2.25 times 6. Let's see. We're going to have 6 times 5 is 30. 6 times 2 is 12, plus 3 is 15. 6 times 2 is 12, plus 1 is 13. I have two numbers behind the decimal-- 13.5. So it's going to be 13.5."
    },
    {
        "Q": "At 1:27 am pacific time, what does it mean to evaluate an expression?\n",
        "A": "make all the appropriate calculations to obtain the final result of the expression",
        "video_name": "I9eLKDbc8og",
        "timestamps": [
            87
        ],
        "3min_transcript": "The surface area of a cube is equal to the sum of the areas of its six sides. Let's just visualize that. I like to visualize things. So if that's the cube, we can see three sides. Three sides are facing us. But then if it was transparent, we see that there are actually six sides of a cube. So there's this one-- one, two, three in front-- and then one-- this is the bottom. This is in the back, and this is also in the back. So you have three sides of the cube. So I believe what they're saying. The surface area of a cube with side length x-- so if this is x, if this is x, if this is x-- is given by the expression 6x squared. That also makes sense. The area of each side is going to be x times x is x squared, and there's six of them. So it's going to be 6x squared. Jolene has two cube-shaped containers that she wants to paint. One cube has side length 2. So this is one cube right over here. I'll do my best to draw it. So this right over here has side length 2, The other cube has side length 1.5. So the other cube is going to be a little bit smaller. It has side length 1.5. So it's 1.5 by 1.5 by 1.5. What is the total surface area that she has to paint? Well, we know that the surface area of each cube is going to be 6x squared, where x is the dimensions of that cube. So the surface area of this cube right over here is going to be 6. And now-- let me do it in that color of that cube-- it's going to be 6 times x, where x is the dimension of the cube. And then the cube all has the same dimensions, so its length, width, and depth is all the same. So for this cube, the surface area is going to be 6 times 2 squared. And then the surface area of this cube is going to be 6 times 1.5 squared. it's going to be the sum of the two cubes. So we're just going to add these two things. And so if we were to compute this first one right over here, this is going to be 6 times 4. This is 24. And this one right over here, this is going to be a little bit hairier. Let's see. 15 times 15 is 225. So 1.5 times 1.5 is 2.25. So 1.5 squared is 2.25. And 2.25 times 6-- so let me just multiply that out. 2.25 times 6. Let's see. We're going to have 6 times 5 is 30. 6 times 2 is 12, plus 3 is 15. 6 times 2 is 12, plus 1 is 13. I have two numbers behind the decimal-- 13.5. So it's going to be 13.5."
    },
    {
        "Q": "\nat 7:57, why do the sun and moon travel across the sky?",
        "A": "because it needs its orbit 24 hours and the moon goes down the sun comes up and when the sun goes down the moon comes up",
        "video_name": "I9eLKDbc8og",
        "timestamps": [
            477
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nat 0:30, Sal said the first one is irrational. but 8/2 is four and square of 4 is 2 which is rational.",
        "A": "it was sqrt of 8 by 2 not 8/2",
        "video_name": "d9pO2z2qvXU",
        "timestamps": [
            30
        ],
        "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here."
    },
    {
        "Q": "At 0:18 Sal states that if you take the square root of a non-perfect square you will always end up with an irrational number. My question is this... 'What is a non-perfect square?'\n",
        "A": "A perfect square is created when you multiply a number with itself. For example: 3 * 3 = 9. So, 9 is a perfect square. A number that does not have matching factors would be a non-perfect square. Hope this helps.",
        "video_name": "d9pO2z2qvXU",
        "timestamps": [
            18
        ],
        "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here."
    },
    {
        "Q": "\nAt 8:17, how is cos 1/3 x zero? Is it because the y-value is zero which makes the cos zero? And when you multiply -2.5 you get zero?\n\nThanks for the help!",
        "A": "Yes, cos[(1/3)x] is 0 at that point because the y-value is 0. Also, yes, when you multiply that value, 0, by -2.5, you still get zero.",
        "video_name": "uBVhtGL9y88",
        "timestamps": [
            497
        ],
        "3min_transcript": "So given this change, we're now multiplying by negative 2.5, what is going to be-- well actually, let's think about a few things. What was the amplitude in the first two graphs right over here? Well there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these case, the minimum is negative 1, maximum is 1. The difference is 2, half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is, once again, 1. What's going to be the amplitude for this thing right over here? Well the amplitude is going to be the absolute value of what's multiplying the cosine function. the amplitude is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point it would go up by a factor of 2 and 1/2. But it's a negative 2.5, so at each point, you're going to stretch it out and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5. That'd be negative-- let me make it clear. this would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there. It could serve to be useful. Now when cosine of 1/3 x is 0, it doesn't matter what you multiply it by, you're still going to get 0 right over here. Now, when cosine of 1/3 x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well cosine of 1/3 x, we see, is negative 1. Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right-- let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there. And then when cosine of 1/3 x is equal to 0, doesn't matter what we multiply it by, we get to 0."
    },
    {
        "Q": "in 5:31 you said the M was negitive 800 why\n",
        "A": "The negative 800 is paired with the W, not the M.",
        "video_name": "VuJEidLhY1E",
        "timestamps": [
            331
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:43, wouldn't the logicians know if their forehead was painted? They would obviously feel that somebody painted/was painting their forehead.\n",
        "A": "Yes, I agree. They would feel it, wouldn t they? Maybe the paint is special or something :)",
        "video_name": "rBaCDC52NOY",
        "timestamps": [
            103
        ],
        "3min_transcript": "There's a new reality television program, and it's called the Blue-- I should probably write it in blue-- but it's called the Blue Forehead Room. And what they do in this reality television program-- and you'll have to bear with me, because the show probably wouldn't be that interesting to watch-- but it's interesting to predict what happens. What they do is, they take a room. They'll call it the blue forehead room. And let's see, that's kind of a top view of the room. And let's say there's a door here. None of this is relevant to the actual problem. This is the door, right there. And what they do is they get 100 perfect logicians to sit in this room, in a circle. So they're all sitting in a circle in this room. Now, before the game even starts, before they even enter the room the first time, the logicians are told two things. forehead painted blue. At least one of you has your forehead-- And they all get their foreheads painted, so that obviously if you're the only guy who has your forehead painted. But you just don't know what color it is. So all of them have different color foreheads. Or, we don't know. But all they're told is, obviously I've painted your forehead. At least one of the people in the room that you will enter will have their forehead painted blue. And then they're also told that as soon as you deduce that your forehead is blue, you need to leave the room. that I set this up properly. They're all outside of the room. No-one's inside the room. And let's say they're blindfolded. And while they're blindfolded, they essentially have the thing painted onto their forehead. So they can't see the paintbrush or anything. So they really don't know what's on their forehead. And then after that, they all enter the room. And they all sit in a circle like this, so that they can all look at each other. And let's say when they enter, the lights are off. So the lights are off, and then the protocol is that the lights will be turned on, and then they can all look around at each other. There's no reflective surfaces. They can't look into each other's eyes and try to see No tricks like that. There are no mirrors in this room."
    },
    {
        "Q": "\n3:45 - 3:58 Can you think of a formula to help me with all this information you just gave me? I think it would really help me during school.",
        "A": "What he is describing is the variance of a population, which he goes over in a video with the same title.",
        "video_name": "iHXdzfF7UEs",
        "timestamps": [
            225,
            238
        ],
        "3min_transcript": "Well, the sample mean, which we would denote by lowercase x with a bar over it, is just the sum of all of these divided by the number of data points we have. So let's see we have 1.5 plus 2.5 plus 4 plus 2 plus 1 plus 1. And all of that divided by 6, which gives-- let's see, the numerator 1.5 plus 2.5 is 4, plus 4 is 8, plus 2 is 10, plus 2 more is 12. So it's going to be 12 over 6, which is equal to 2 hours of television. So at least for your sample, you say, my sample mean is two hours of television. It's a statistic that is trying to estimate this parameter, this thing that's very hard to know. But it's our best shot. Maybe we get a better answer if we get more data points. But this is we have so far. well, I don't want to just estimate my population mean. I also want to estimate another parameter. I also am interested in estimating my population variance. So once again, since we can't survey every one in the population, this is pretty much impossible to know. But we're going to attempt to estimate of this parameter. We attempted to estimate the mean. Now we will also attempt to estimate this parameter, this variance parameter. So how would you do it? Well, reasonable logic would say, well, we maybe we'll do the same thing with a sample as we would have done with the population. When you're doing the population variance, you would take each data point in the population, find the distance between that and the normal population mean, take the square of that difference, and then add up all the squares of those differences, and then divide by the number of data points you have. So let's try that over here. So let's try to find-- take each of these data points, in a different color-- each of these data points, and find the difference between that data point and our sample mean-- not the population mean, we don't know what the population mean-- the sample mean. So that's that first data point plus the second data point-- so it's 4 minus 2 squared plus 1 minus 2 squared. And this is what you would have done if you were taking a population variance. If this was your entire population, this is how you would you find a population mean here, if this was your entire population. And you find the squared distances from each of those data points and then divide by the number of data points. So let's just think about this a little bit. 1 minus 2 squared. Then you have 2.5 minus 2-- 2 being the sample mean-- squared. Let me see, this green color. Plus 2 minus 2 squared."
    },
    {
        "Q": "\nat 9:32, Why didn't Sal write it as sqrt((a+b+c/2)*(a-b+c/2))",
        "A": "because the second (a+b+c)/2 is a part of (a+b+c)/2 - 2b/2. for multiplying how you said he would have had to multiply (a+b+c)/2 with -2b/2 .",
        "video_name": "nZu7IZLhJRI",
        "timestamps": [
            572
        ],
        "3min_transcript": "This is the same thing as b minus c minus a. Right? And all of that over 4. Now, I can rewrite this whole expression. I don't want to run out of space. I can rewrite this whole expression as, well 4 is the product of 2 times 2. So our whole area expression has been, arguably, simplified to it equals the square root-- and this is really the home stretch-- of this right here, which I can just write as a plus b plus c over 2. That's that term right there. Times this term. Times that term. minus b, that's the same thing as a plus b plus c minus 2b. These two things are equivalent. You have an a, you have a c, and then b minus 2b is going to be equal to minus b. Right? b minus 2b, that's minus b. So this next term is going to be a plus b plus c minus 2b, over 2. Or instead of writing it like that, let me write this over 2 minus this over 2. And then our next term right here. Same exact logic. That's the same thing as a plus b plus c minus 2a, all of that over 2. If we add the minus 2a to the a we get minus a. So we get b plus c minus a. So all this over 2, or we can split the denominators just like that over 2. And then one last term. And you might already recognize the rule of Heron's formula popping up. I was thinking not the rule of Heron-- Heron's formula. That term right there is the exact same thing as a plus b plus c minus 2c. You take 2c away from the c you get a minus c, and then you still have the a and the b. And then all of that over 2. You could write that over 2 minus that over 2. And, of course, we're taking the square root of all of this stuff. Now, if we define an S to be equal to a plus b plus c over 2, then this equation simplifies a good bit. This right here is S. That right there is S. That right there is S."
    },
    {
        "Q": "I have noticed that when talking about Vectors Sal generally denotes them as a column matrix like he did at 3:00 when describing vector x?\nIS there a reason to do so or could your represent them as a row vector also?\n",
        "A": "You could use a row vector without changing any meaning. It only becomes significant when mixing vectors with matrices, where a row and column are very different and in that context one or the other is used as appropriate.",
        "video_name": "gAbadNuQEjI",
        "timestamps": [
            180
        ],
        "3min_transcript": "Where you said, oh, a function is, you just give me some number and I'll give you another number. Or I'll do something to that number. While it can be much more general than that. It's an association between any member of one set and some other members of another set. Now, we know that vectors are members of sets. Right? In particular, if we say that some vector x is a member of some set -- let me just say it's a member of rn, because that's what we deal with -- all that means is that this is just a particular representation of an n-tuple. Remember what rn was. rn we defined way back at the beginning of the linear algebra playlist. We defined it as the set of all n-tuples -- x1, x2, xn, where your x1's, x2's, all the way to So your rn is most definitely a set. This could be rn. And obviously the use of the letter n is arbitrary. It could be rm, it could be rs. n is just kind of a placeholder for how many tuples we have. It could be r5. It could be 5 tuples. And when we say that a vector x is a member of rn, we're just saying that it's another way of writing one of these n-tuples. And all of our vectors so far are column vectors -- that's the only type that we've defined so far -- and we say it's this ordered list where each of the members are a member of r's. It's an ordered list of n -- it's an ordered list of n-components -- x1, x2, all the way to xn -- where each of those guys, or each of those x1's, x2's all the way to xn's, are a member of the real numbers. That's, by definition, what we mean when we say that x is a member of rn. here -- let's say that this set right here is rn and then let me just change, just to be general, let me create another set right there and call that set right there rm. Just a different number. It it could be the same as n, it could be different. This is m-tuples, that's n-tuples. We've defined that vectors can be members of rn. So you could have some vector here and then, if you associate with that vector in rn -- if you associate it with some vector in are rm -- if you associate it with, let's call that vector y, if you make this association, that too is a function. And that might have already been obvious to you and this would be a function that's mapping from rn to rm."
    },
    {
        "Q": "\nAt 1:16 Sal says that in order to have a vertical asymptote \"it cannot be defined there. \" What does he mean by this? What is he referring to as it? Why can \"it\" not be defined?",
        "A": "It means that at that particular x value, the equation doesn t work so the y value is undefined. For example if the function is a fraction and at a particular x value the bottom of the fraction is equal to 0, then the output would be undefined since we can t divide by 0, so there would be an asymptote on the graph.",
        "video_name": "onmNaDrxwmo",
        "timestamps": [
            76
        ],
        "3min_transcript": "Voiceover:So, we've got three functions graphed here. The function f in magenta. We have the function g in this green color. And we have the function h in this dotted purple color. And then we have three potential expressions, or three expressions that could be potential definitions, for f, g, and h. And what I want to do in this video, is try to match them, try to match the function to a definition. I encourage you to pause this video and try to think about it on your own before I work through it. There's a couple of ways that we could approach this. One, we could think about what the graphs of each of these look like, and then think about which of these function's graphs look like that. Or, we could we could look at the function graph and think about the vertical and horizontal asymptotes, and think about which of these expressions would have a vertical or horizontal asymptotes at that point. So, I'm actually going to do it the second way. Let's look at the graphs. I tend to be a little bit more visual, or I like to look at graphs. So it looks like - f, I'll start with f. f looks like it has a vertical asymptote at x equals 5. Or, sorry, at x equals negative 5. at x is equal to negative five. Let's think about which of these would have a vertical asymptote at x equals negative 5. In order to have a vertical asymptote, it can't be defined there. So that'sthe first way I would think about it. And then even if isn't defined there, we need to make sure that it's an actual vertical asymptote, and not just a hole at that point, not just a point discontinuity. So, let's think about it. So, this first expression is actually defined at x equals negative 5. The only reason why it wouldn't be defined, if somehow we got a 0 in the denominator. But if you have negative 5 minus 5, that's negative 10. So this one's defined at that point, so that's not f. This one's also defined at x equals negative 5. The denominator does not become 0, so that's not f. This one, when x is equal to negative 5, this denominator does become 0. So this seems, I just used purely deductive reasoning, this looks like my best candidate for f of x. But let's confirm that it's consistent with the other things that we're seeing over here. If I look at the graph, it looks like there's a horizontal asymptote. Especially as x gets to larger and larger values, it looks like f of x is approaching 1. f of x is approaching 1. Now, is that the same case over here? As x gets larger and larger and larger, as x approaches infinity, well, then the negative 2 and the plus 5 don't matter. As x approaches infinity, this is going to approximate, for very large x's, it's going to be x over x. We look at the highest degree terms, and so that is going to approach 1, as x gets very very very large. Then the negative two, subtracting the 2 in the numerator and adding 5 in the denominator are going to matter less and less and less, because the x gets so big. So, this will approach 1. So, that seems consistent from that point of view. And let's see, is there anything else interesting? Well, when does this equal 0? Well, the numerator equals 0 when x is equal to 2. And we see that's indeed the case for f right over there. So, I'm feeling really good that this is our f of x."
    },
    {
        "Q": "\nAround 3:53, why does 10 to the power of both sides cancel out the base 10 and the log_10 exponent?",
        "A": "Since that s how logarithms work; exponentiation and logarithms are inverse operations to each other. Since log(10) a is whatever 10 needs to be raised in to get a, 10^(log(10) a) = a.",
        "video_name": "Kv2iHde7Xgw",
        "timestamps": [
            233
        ],
        "3min_transcript": "log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3."
    },
    {
        "Q": "\nI'm getting confused at 3:42, when he offers an alternate way of solving the equation. Can someone theoretically explain to me how raising both sides of the equation to the log's base solves the problem?\nI understand the lesson up until that part.",
        "A": "logarithms are just inverse functions of exponential functions so that the base and the exponents cancel and equal 1 .try this logany base (withthat number)=1 as well exponets leading coeffitient with raised with any logsame numbe =1 let say 10^x(power)=100 by logarithm rules it inverse it intern of x log(10_base)(100)=x so that x=2 log( 10^x(power))=log(100) this simplifies to x=log 100 or 2",
        "video_name": "Kv2iHde7Xgw",
        "timestamps": [
            222
        ],
        "3min_transcript": "log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3."
    },
    {
        "Q": "\nat 1:24 is 3-4 the same as 4-3",
        "A": "3-4 is not the same as 4-3 because in subtracting, there is no Commutative Property. 3-4 would be -1, while 4-3 would be positive 1.",
        "video_name": "AO9bHbUdg-M",
        "timestamps": [
            84
        ],
        "3min_transcript": "Voiceover:Let's explore what it means to subtract numbers. So let's say that I want to figure out what 4, what 4 minus 3 is. 4 minus 3. So one way to think about this, is you start with 4 objects. And so let me just draw 4 objects. So there I have 1, 2, 3, and 4. And when I ... So this is the 4 objects right over here. And when I say minus 3, or if I'm going to subtract 3 from the 4, one way to think about it is, I'm going to take 3 of these 4 objects away. So let's do that. So I'm going to take away 1, I'm going to take away 2, and I'm going to take away 3. Notice, I took away 1, 2, 3 objects. Well, if I start with 4 and I take away 3, I subtract 3, 4 minus 3, I am left with, I am left with, 1 object, right over here. So 4 minus 3 is equal to, is equal to 1. Fascinating. Let's do another one of these. Let's figure out what 5, what 5, what 5 minus 2 is. And let's write it this way. Let's say we want to figure out something, some question mark, some question mark ... Actually, let me just clear this out. I'll just do it right over here, actually. So let's say we have some question mark. So there's some unknown number right over here. So I'll just put a question mark over here. And we say that unknown number is equal to, is equal to 5, 5 minus 2. Minus 2. So what is this going to be? Well let's visualize it. This means I have 5 things and I'm going to take away 2 of them and this is going to be what I have leftover after I start with 5 and take away 2. So I have 1, 2, 3, 4, 5 things. Now I am going to take away 2 of them. So I'm going to take away 1 and 2. So I took away 1, 2 objects. So how many do I have leftover? Well, I have leftover these, these ... Let me do this in a different color. I have leftover these purple things right over here. So how many is that? How many do I have leftover?"
    },
    {
        "Q": "At 12:46, how did he factor a -x out of +17x^2??\n",
        "A": "Sal divided 17x^2 by -x. Another way to think about this is to say what does -x have to be multiplied by in order to get +17x^2.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            766
        ],
        "3min_transcript": "h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So a plus b will be equal to the middle coefficient. And a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now, I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is to teach you to factor things a little And this is a little bit of a add-on. I was going to make a whole video on this. But I think, on some level, it might be a little obvious for you. So let's say we had-- let me get a good one here. squared, minus 70x. Immediately, you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to third power. And the first thing you should realize is that every term here is divisible by x. So let's factor out an x. Or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times-- negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. The x's cancel out. And now, you have something that might look a little bit familiar. We have just a standard quadratic where the leading We just have to find two numbers whose product is 70, and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70. You add them up, you get negative 17. So this part right here is going to be x minus 10, times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out. And that'll get it into a form that you might recognize. Hopefully, you found this helpful. I want to reiterate what I showed you at the beginning of I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient. But to some degree, you're going to find out easier ways to do this, especially with the quadratic formula, in not too long."
    },
    {
        "Q": "\nAt 9:36, you put fhx instead of fhx^2 when you multiplied fx and hx, why?",
        "A": "Sal just made a mistake. It has been corrected already if you watch the video again.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            576
        ],
        "3min_transcript": "to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm I think that'll confuse you, because I use a's and b's here. They won't be the same thing. So let me use completely different letters. Let's say I have fx plus g, times hx plus, I'll use j instead of i. You'll learn in the future why don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx which is fhx. And then, fx times j. So plus fjx. And then, we're going to have g times hx. So plus ghx. And then g times j. Plus gj. Or, if we add these two middle terms, you have fh times x, Plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative 1 coefficient here, we look for two numbers that add up to this, whose product is equal to the product of that times that. Well, here we have two numbers that add up-- let's say that a is equal to fj. That is a. And b is equal to gh. So a plus b is going to be equal to that middle coefficient. And then what is a times b? a times b is going to be equal to fj times gh. We could just reorder these terms. We're just multiplying"
    },
    {
        "Q": "I am confused why at about 7:40, the 6x needed to come before the 1x. Why couldn't it have been the other way around??\n",
        "A": "The 6x did not need to come first. You can absolutely write out the new polynomial as: 6x^2 + 1x + 6x + 1. If you finish the factoring process, you will get the same 2 binomial factors as Sal has in the video. Your intermediate steps will look a little different, but the end result will be the same. Hope this helps.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            460
        ],
        "3min_transcript": "Let's factor out the x plus 7. We get x plus 7, times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping. And we factored it into two binomials. Let's do another example of that, because it's a little But once you get the hang of it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, which is equal to 6. And we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. What are the-- well, the obvious one is 1 and 6, right? 1 plus 6 is 7. So we have a is equal to 1. Or let me not even assign them. The numbers here are 1 and 6. Now, we want to split this into a 1x and a 6x. But we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squar ed here, plus-- and so I'm going to put the 6x first because 6 and 6 share a factor. And then, we're going to have plus 1x, right? 6x plus 1x equals 7x . That was the whole point. They had to add up to 7 . And then we have the final plus 1 there. Now, in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times-- 6x squar ed divided by 6x is just an x. 6x divided by 6x is just a 1. to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm"
    },
    {
        "Q": "At, 1:55, if it is a*b, won't it be 4*25, instead of 4*-21?\n",
        "A": "In this case a and b are not from the classical form a x\u00c2\u00b2 + b x + c, but rather two independent variables to use the method. You could call them p and q if thats easier for you. Or in short: For a x\u00c2\u00b2 + b x + c you search p and q such as p * q = a * c and p + q = b As Sal didn t use the a, b, c in the equation, he just chose them as variables.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            115
        ],
        "3min_transcript": "In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as the leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or negative 1 where this 4 is sitting. All of a sudden now, we have this 4 here. So what I'm going to teach you is a technique called, factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. To some degree, it'll become obsolete once you learn the quadratic formula, because, frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique. And then at the end of this video, I'll actually show you why it works. So what we need to do here, is we need to think of two numbers, a and b, where a times b is equal 4 times So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a plus b, need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is. So we go, 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference. Because that's essentially what you're going to do, if one is negative and one is positive. Too far apart. Let's see you could do 3-- I'm jumping the gun. 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 42 is negative 40-- too far apart. 3 and-- Let's see, 3 goes into 84-- 3 goes into 8 2 times. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. Goes exactly 8 times. So 3 and 28. This seems interesting. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25."
    },
    {
        "Q": "\nSal mentions the Z-Score table at 12:56. How were the values on this table calculated - where did they come from?",
        "A": "What could have been made more clear is that Sal is looking for the probability for X to be between 0 and up to 2+(2.02*0.099)=2.19998 liters of water. Since we know that it is a normal distribution we can write: X is N(2 , 0.099) and we look for P(0<X<2.19998) Then, if we don t have a z-table, only a texas instrument we can write: normalcdf(0 , 2.19998 , 2 , 0.099) = 0.9783..... Which is the number that Sal finds in the z- table Hope it helps :)",
        "video_name": "0ZstEh_8bYc",
        "timestamps": [
            776
        ],
        "3min_transcript": "standard deviations-- or maybe I should write it this way-- more than-- Let me write it down here where I have more space. So this all boils down to the probability of running out of water is the probability that the sample mean will be more than-- just the 50 that we happened to select-- remember, if we take a bunch of samples of 50 and plot all of them we'll get this whole distribution. But the one 50, the group of 50 that we happened to select, the probability of running out of water is the same thing as the probability of the mean of those people, will be more than 2.020 standard deviations above the mean of this distribution. And here we just have to look up our Z-table. Remember, this 2.02 is just this value right here. 0.2 divided by 0.09. I just had to pause the video because there's some type of fighter jet outside or something. So anyway, hopefully they won't come back. But anyway, so we need to figure out the probability that the sample mean will be more than 2.02 standard deviations above the mean. And to figure that out we go to a Z-table, and you could find this pretty much anywhere. Usually it's in any stat book or on the internet, wherever. And so essentially we want to know the probability-- the Z-table will tell you how much area is below this value. So if you go to z of 2.02-- that was the value that we You have 2.02, it was-- so you go for the first digit. We go to 2.0, and it was 2.02. 2.02 is right over there. So 2.02 is right over there. So this 0.9783-- let me write it down over here-- this 0.9783-- I want to be very careful. 0.9783, that Z-table, that's not this value over here. This 0.9783 on the Z-table, that is giving us this whole area over here. It's giving us the probability that we are below that value. That we are less than 2.02 standard deviations above the mean. So it's giving us that value over here. So to answer our question, to answer this probability, we just have to subtract this from 1 because these will all add up to 1. So we just take our calculator back out and we just take 1 minus 0.9783 is equal to 0.0217. So this right here is 0.0217."
    },
    {
        "Q": "At 4:56 he wrote strelches but meant stretches, right?\n",
        "A": "He wrote stretches all right. You can barely see the first stroke of the t but it s there.",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            296
        ],
        "3min_transcript": "negative 10/3. Then, I'm sorry, this is positive 10/9, right? Negative 1/3 times negative 10/3, negatives cancel out. 10/9. Don't want to make a mistake here. These are sequences. You might also see the word a series. And you might even see a geometric series. A series, the most conventional use of the word series, means a sum of a sequence. So for example, this is a geometric sequence. A geometric series would be 90 plus negative 30, plus 10, plus negative 10/3, plus 10/9. So a general way to view it is that a series is the sum of a sequence. I just want to make that clear because that used to confuse But anyway, let's go back to the notion of a geometric sequence, and actually do a word problem that deals with one of these. So they're telling us that Anne goes bungee jumping off of a bridge above water. On the initial jump, the cord stretches by 120 feet. So on a1, our initial jump, the cord stretches by 120 feet. We could write it this way. We could write, jump, and then how much the cord stretches. So on the initial jump, on jump one, the cord stretches 120 feet. the original jump, and then each additional bounce stretches the rope 60% of the previous stretch. So here, the common ratio, where each successive term in our sequence is going to be 60% of the previous term. Or it's going to be 0.6 times the previous term. So on the second jump, we're going to start 60% of that, or 0.6 times 120. Which is equal to what? That's equal to 72. Then on the third jump, we're going to stretch 0.6 of 72, or 0.6 times this. So it would be 0.6 times 0.6 times 120. Notice, over here, so on the fourth jump we're going to have 0.6 times 0.6 times 0.6 times 120."
    },
    {
        "Q": "\nat 9:00 he says that you multipy 0.6 to the nth power by 120. But a sub 1 is suppose to be the first term right? so therefore, wouldnt the equation be 0.6 to the nth power multiplied by 72?",
        "A": "No, our original amount of stretch when Anne first jumped was 120, so the second term is 0.6(120), and that second term is also the first bounce.",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            540
        ],
        "3min_transcript": "So you have 0.6 to the 0th power, and you've just got a 1 here. And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1, and notice 2 minus 1 is the first power, and we have exactly one 0.6 here. So I figured it was n minus 1 because when n is 2, we have one 0.6, when n is 3, we have two 0.6's multiplied by themselves. When n is 4, we have 0.6 to the third power. So whatever n is, we're taking 0.6 to the n minus 1 power, and of course we're multiplying that times 120. Now and the question they also ask us, what will be the rope stretch on the 12th bounce? And over here I'm going to use the calculator. and actually let me correct this a little bit. bounce, and we could call the jump the zeroth bounce. Let me change that. This isn't wrong, but I think this is where they're going with the problem. So you can view the initial stretch as the zeroth bounce. So instead of labeling it jump, let me label it bounce. So the initial stretch is the zeroth bounce, then this would be the first bounce, the second bounce, the third bounce. And then our formula becomes a lot simpler. Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120, right? On the zeroth bounce, that was our original stretch, you get 0.6 to the 0, that's 1 times 120. 0.6 times the previous stretch, or the previous bounce. So this has it in terms of bounces, which I think is what the questioner wants us to do. So what about the 12th bounce? Using this convention right there. So if we do the 12th bounce, let's just get our calculator out. We're going to have 120 times 0.6 to the 12th power. And hopefully we'll get order of operations right, because exponents take precedence over multiplication, so it'll just take the 0.6 to the 12th power only. And so this is equal to 0.26 feet. So after your 12th bounce, she's going to be barely moving. She's going to be moving about 3 inches on that 12th bounce."
    },
    {
        "Q": "\nAt 7:21, Sal says the first jump is 1.6(120). He meant 0.6(120), right? I just want to make sure so I don't get confused.",
        "A": "When Sal did the first jump the 1 was for the 0.6 in the first jump. Hope it helps.",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            441
        ],
        "3min_transcript": "the original jump, and then each additional bounce stretches the rope 60% of the previous stretch. So here, the common ratio, where each successive term in our sequence is going to be 60% of the previous term. Or it's going to be 0.6 times the previous term. So on the second jump, we're going to start 60% of that, or 0.6 times 120. Which is equal to what? That's equal to 72. Then on the third jump, we're going to stretch 0.6 of 72, or 0.6 times this. So it would be 0.6 times 0.6 times 120. Notice, over here, so on the fourth jump we're going to have 0.6 times 0.6 times 0.6 times 120. the previous jump. So if we wanted to make a general formula for this, just based on the way we've defined it right here. So the stretch on the nth jump, what would it be? So let's see, we start at 120 times 0.6 to the what? To the n minus 1. How did I get this? Let me write this a little bit here. So this is equal to 0.6, actually let me write the 120 first. This is equal to 120 times 0.6 to the n minus 1. How did I get that? Well we're defining the first jump as stretching 120 feet. So you have 0.6 to the 0th power, and you've just got a 1 here. And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1, and notice 2 minus 1 is the first power, and we have exactly one 0.6 here. So I figured it was n minus 1 because when n is 2, we have one 0.6, when n is 3, we have two 0.6's multiplied by themselves. When n is 4, we have 0.6 to the third power. So whatever n is, we're taking 0.6 to the n minus 1 power, and of course we're multiplying that times 120. Now and the question they also ask us, what will be the rope stretch on the 12th bounce? And over here I'm going to use the calculator. and actually let me correct this a little bit."
    },
    {
        "Q": "\nWhy did sal only consider multiplying the identity matrix with v on the lamda side but not the v with A at 3:04?",
        "A": "Because he wants to put the vector v in matrix form using the Identity matrix and A is already in that form",
        "video_name": "rfm0wQObxjk",
        "timestamps": [
            184
        ],
        "3min_transcript": "eigenvector. You know, hopefully from that last video, we have a little bit of appreciation of why they're useful. But now in this video let's at least try to determine what some of them are. You know, based on what we know so far, if you show me an eigenvector I can verify that it definitely is the case, or an eigenvalue. I could verify the case. But I don't know a systematic way of solving for either of them. So let's see if we can come up with something. So in general, we're looking for solutions to the equation A times v is equal to lambda v. It's equal to lambda times the vector. Now one solution might immediately pop out at you, and that's just v is equal to the 0 vector. And that definitely is a solution, although it's not normally considered to be an eigenvector just because one, it's not a useful basis vector. It doesn't add anything to a basis. It doesn't add really the amount of vectors that you can span when you throw the basis vector in there. associated with it. Because if v is equal to 0, any eigenvalue will work for that. So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors. So we're looking for vectors that are not equal to the 0 vector. So given that, let's see if we can play around with this equation a little bit and see if we can at least come up with eigenvalues maybe in this video. So we subtract Av from both sides, we get the 0 vector is equal to lambda v minus A times v. Now, we can rewrite v as-- v is just the same thing as the identity matrix times v, right? v is a member of Rn. The identity matrix n by n. You just multiply and we're just going to get v again. So if I rewrite v this way, at least on this part of the expression-- and let me swap sides-- so then I'll get lambda times-- instead of v I'll write the identity is equal to the 0 vector. Now I have one matrix times v minus another matrix times v. Matrix vector products, they have the distributive property. So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v. And that's going to be equal to 0, right? This is just some matrix right here. And the whole reason why I made this substitution is so I can write this as a matrix vector product instead of just a scalar vector product. And that way I was able to essentially factor out the v and just write this whole equation as essentially, some matrix vector product is equal to 0. Now, in order-- if we assume that this is the case, and we're assuming-- remember, we're assuming that v"
    },
    {
        "Q": "I don't quite understand how you can tell whether or not to write and/or. In 5:15 it shows two different equations with an or in between them. But in the example above they were written together and used with an and. Could you explain this a little bit more in depth?\n",
        "A": "When you graph for an and problem, the line shouldn t go outside of the graph, staying in a closed area like the first example. In an or problem, there should be two different lines, both going off the graph in opposite directions.",
        "video_name": "iI_2Piwn_og",
        "timestamps": [
            315
        ],
        "3min_transcript": "Let's do positive 21, and let's do a negative 21 here. So we want all of the numbers, so whatever this thing is, that are greater than or equal to 21. They're more than 21 away from 0. Their absolute value is more than 21. Well, all of these negative numbers that are less than negative 21, when you take their absolute value, when you get rid of the negative sign, or when you find their distance from 0, they're all going to be greater than 21. If you take the absolute value of negative 30, it's going to be greater than 21. Likewise, up here, anything greater than positive 21 will also have an absolute value greater than 21. So what we could say is 7x needs to be equal to one of these numbers, or 7x needs to be equal to one of these numbers out here. So we could write 7x needs to be one of these numbers. These are all of the numbers that are less than or equal to negative 21, or 7x-- let me do a different color here-- or 7x has to be one of these numbers. And that means that 7x has to be greater than or equal to positive 21. I really want you to kind of internalize what's going on here. If our absolute value is greater than or equal to 21, that means that what's inside the absolute value has to be either just straight up greater than the positive 21, or less than negative 21. Because if it's less than negative 21, when you take its absolute value, it's going to be more than 21 away from 0. Hopefully that make sense. We'll do several of these practice problems, so it really gets ingrained in your brain. But once you have this set up, and this just becomes a compound inequality, divide both sides of this equation by Or you divide both sides of this by 7, you get x is greater than or equal to 3. So I want to be very clear. This, what I drew here, was not the solution set. This is what 7x had to be equal to. I just wanted you to visualize what it means to have the absolute value be greater than 21, to be more than 21 away from 0. This is the solution set. x has to be greater than or equal to 3, or less than or equal to negative 3. So the actual solution set to this equation-- let me draw a number line-- let's say that's 0, that's 3, that is negative 3. x has to be either greater than or equal to 3. That's the equal sign. Or less than or equal to negative 3."
    },
    {
        "Q": "\n@ 11:13 I don't get why you multiply everything by 7. Why not just multiply the two fractions and be done with it... And could you just subtract by 9 from both sides and come up with the right answer?",
        "A": "If you multiply by seven, it erases the denominator on both sides, making integers, no longer fractions. Plus, you can t multiply the fractions because there is a variable as a numerator, along with an integer. Subtracting 9 wouldn t give you the answer right away as you still would have had to isolate x.",
        "video_name": "iI_2Piwn_og",
        "timestamps": [
            673
        ],
        "3min_transcript": "But it comes from the same logic. This has to evaluate to something that is less than a away from 0. Now, if we go to the other side, if you have something of the form f of x is greater than a. That means that this thing has to evaluate to something that is further than a away from 0. So that means that f of x is either just straight up greater than positive a, or f of x is less than negative a. If it's less than negative a, maybe it's negative a minus another 1, or negative 5 plus negative a. Then, when you take its absolute value, it'll become a plus 5. So its absolute value is going to be greater than a. So I just want to-- you could memorize this if you want, but I really want you to think about this is just saying, OK, to be more than a away from 0. Let's do one more, because I know this can be a little bit confusing. And I encourage you to watch this video over and over and over again, if it helps. Let's say we have the absolute value of 2x-- let me do another one over here. Let's do a harder one. Let's say the absolute value of 2x over 7 plus 9 is greater than 5/7. So this thing has to evaluate to something that's more than 5/7 away from 0. So this thing, 2x over 7 plus 9, it could just be straight up greater than 5/7. Or it could be less than negative 5/7, because if it's less than negative 5/7, its absolute value is going to be greater than 5/7. Or 2x over 7 plus 9 will be less than negative 5/7. And then we just solve both of these equations. See if we subtract-- let's just multiply everything by 7, just to get these denominators out of the way. So if you multiply both sides by 7, you get 2x plus 9 times 7 is 63, is greater than 5. Let's do it over here, too. You'll get 2x plus 63 is less than negative 5. Let's subtract 63 from both sides of this equation, and you get 2x-- let's see. 5 minus 63 is 58, 2x is greater than 58. If you subtract 63 from both sides of this equation, you get 2x is less than negative 68. Oh, I just realized I made a mistake here. You subtract 63 from both sides of this, 5 minus 63 is"
    },
    {
        "Q": "At 3:06, can't Sal just find a integer between 23 and 25(which is 24)??\n",
        "A": "Yes, because the difference between 23 and 25 is 2, it s easy to see that 24 is the answer. But he wants to show what to do in case the numbers were not so nice. Imagine, instead of 23 and 25, the two middle numbers were 231473 and 258872 (in some huge science experiment). Then it would not be so easy to find the number halfway in between without any work. So he wants to show that you should add them together and divide by 2, in case they are ugly numbers.",
        "video_name": "k3aKKasOmIw",
        "timestamps": [
            186
        ],
        "3min_transcript": "So the sum of all the numbers is 206. And then we want to divide 206 by 8. So if I say 206 divided by 8 gets us 25.75. So the mean is equal to 25.75. So this is one way to kind of measure the center, the central tendency. Another way is with the median. And this is to pick out the middle number, the median. And to figure out the median, what we want to do is order these numbers from least to greatest. So it looks like the smallest number here is 20. Then, the next one is 21. There's no 22 here. 23 and a 23. So 23 and a 23. And no 24's. There's a 25. 25. There's no 26, 27, 28. There is a 29. 29. Then you have your 32. 32. And then you have your 33. 33. So what's the middle number now that we've ordered it? So we have 1, 2, 3, 4, 5, 6, 7, 8 numbers. We already knew that. And so there's actually going to be two middles. If you have an even number, there's actually two numbers that qualify for close to the middle. And to actually get the median, we're going to average them. So 23 will be one of them. That, by itself, can't be the median because there's three less than it and there's four greater than it. And 25 by itself can't be the median because there's three larger than it and four less than it. So what we do is we take the mean of these two numbers and we pick that as the median. So if you take 23 plus 25 divided by 2, that's 48 over 2, So even though 24 isn't one of these numbers, the median is 24. So this is the middle number. So once again, this is one way of thinking about central tendency. If you wanted a number that could somehow represent the middle. And I want to be clear, there's no one way of doing it. This is one way of measuring the middle. Let me put that in quotes. The middle. If you had to represent this data with one number. And this is another way of representing the middle. Then finally, we can think about the mode. And the mode is just the number that shows up the most in this data set. And all of these numbers show up once, except we have the 23, it shows up twice. And since because 23 shows up the most, it shows up twice. Every other number shows up once, 23 is our mode."
    },
    {
        "Q": "at 4:28, what are the formulae of mean media and mode?\n",
        "A": "mean is the sum of the numbers divided by the total number of numbers in the sum. mode is the most frequent number median is the halfway point in the sorted list of numbers.",
        "video_name": "k3aKKasOmIw",
        "timestamps": [
            268
        ],
        "3min_transcript": "So the sum of all the numbers is 206. And then we want to divide 206 by 8. So if I say 206 divided by 8 gets us 25.75. So the mean is equal to 25.75. So this is one way to kind of measure the center, the central tendency. Another way is with the median. And this is to pick out the middle number, the median. And to figure out the median, what we want to do is order these numbers from least to greatest. So it looks like the smallest number here is 20. Then, the next one is 21. There's no 22 here. 23 and a 23. So 23 and a 23. And no 24's. There's a 25. 25. There's no 26, 27, 28. There is a 29. 29. Then you have your 32. 32. And then you have your 33. 33. So what's the middle number now that we've ordered it? So we have 1, 2, 3, 4, 5, 6, 7, 8 numbers. We already knew that. And so there's actually going to be two middles. If you have an even number, there's actually two numbers that qualify for close to the middle. And to actually get the median, we're going to average them. So 23 will be one of them. That, by itself, can't be the median because there's three less than it and there's four greater than it. And 25 by itself can't be the median because there's three larger than it and four less than it. So what we do is we take the mean of these two numbers and we pick that as the median. So if you take 23 plus 25 divided by 2, that's 48 over 2, So even though 24 isn't one of these numbers, the median is 24. So this is the middle number. So once again, this is one way of thinking about central tendency. If you wanted a number that could somehow represent the middle. And I want to be clear, there's no one way of doing it. This is one way of measuring the middle. Let me put that in quotes. The middle. If you had to represent this data with one number. And this is another way of representing the middle. Then finally, we can think about the mode. And the mode is just the number that shows up the most in this data set. And all of these numbers show up once, except we have the 23, it shows up twice. And since because 23 shows up the most, it shows up twice. Every other number shows up once, 23 is our mode."
    },
    {
        "Q": "at 1:21, shouldn't`t the answer be 6^1/15? why is it 6^3/5?\n",
        "A": "Because you re multiplying 1/5 by 3, or 3/1. 3*1=3, which is the numerator, and 5*1=5, which is the denominator. 3/5.",
        "video_name": "Ht-YXje4R2g",
        "timestamps": [
            81
        ],
        "3min_transcript": "Let's see if we can simplify 6 to the 1/2 power times the fifth root of 6 and all of that to the third power. And I encourage you to pause this video and try it on your own. So let me actually color code these exponents, just so we can keep track of them a little better. So that's the 1/2 power in blue. This is the fifth root here in magenta. And let's see. In green, let's think about this third power. So one way to think about this fifth root is that this is the exact same thing as raising this 6 to the 1/5 power, so let's write it like that. So this part right over here, we could rewrite as 6 to the 1/5 power, and then that whole thing gets raised to the third power. And of course, we have this 6 to the 1/2 power out here, 6 to the 1/2 power times all of this business right over here. and then raise that whole thing to another exponent? Well we've already seen in our exponent properties, that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the-- 3 times 1/5 is 3/5-- 6 to the 3/5 power. And of course, we're multiplying that times 6 to the 1/2 power. 6 to the 1/2 power times 6 to the 3/5 power. And now, if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way, because these all equal each other. This is the same thing as 6 being raised to the 1/2 plus 3/5 power, 1/2 plus 3 over 5. Well, we could find a common denominator. It would be 10, so that's the same thing as-- actually let me just write it this way-- this is the same thing as 6 to the-- instead of 1/2, we can write it as 5/10. Plus 3/5 is the same thing as 6/10 power, which is the same thing-- and we deserve a little bit of a drum roll here, this wasn't that long of a problem-- 6 to the 11/10 power. I'll just write it all, 11/10 power. And so, that looks pretty simplified to me. I guess we're done."
    },
    {
        "Q": "If in the first example at 1:55 the (1/5) is not the starting point or part of the common ratio, what happens to it ? Because if it is part of the common ratio then 4.5*(1/5) is less than 1 making the common ratio converge. Can someone please explain ? Any help would be appreciated. Thanks !\n",
        "A": "(1/5) is the starting point when n=0. You are correct saying that (1/5) is not part of the common ratio (which is 4.5). He obtained the (1/5) by simplifying the initial problem to the ar^n form. Since it s (1/5)*(4.5^n), or as he writes it (1/5)*(45/10)^n, as n reaches infinity, the multiplication by (1/5) does not change it diverging to infinity. Someone please correct me if I m wrong, but since the upper limit is infinity, it doesn t really matter what the starting point is with respect to convergence and divergence.",
        "video_name": "KNrEAJeRIdo",
        "timestamps": [
            115
        ],
        "3min_transcript": "We have three different geometric series here. And what I want you to think about is, which of these converge and which of these diverge? Or another way of thinking about it, it converges if the sum comes up to a finite value. It diverges if the sum goes to infinity, or even possibly negative infinity. So pause the video and think about that. Well, the general way to think about it-- and we've seen it in a previous video-- is that you're going to have a converging geometric series. Infinite geometric series is if the absolute value of your common ratio is greater than zero and less than one. So we just have to think about, what is the absolute value of the common ratios over here? And it's not as obvious, because they didn't just write one term or one number to an exponent. They wrote several numbers to an exponents. So we'll have to do a little bit of algebraic manipulation. So 5 n to the negative 1, that's the same thing. Let me do that over here. 5 n to the negative 1 is the same thing as 5 to the n over 5. to the negative 1, which is the same thing as 5 to the n over 5. And so 5 to the n over 5, 5 to the n over 5 times 9 10 to the n. 9 over 10 to the n. This is the same thing. This is equal to 1 over 5 times 5 times 9 over 10 to the nth power. And now the common ratio becomes a little bit more clear. This right over here would be our-- well, it won't be our first term anymore, because we're starting at n equals 2. So we won't view it that way. But if we look at our common ratio, this right over here is going to be our common ratio. And this is absolute value greater than or less than 1. Well, 5 times 9 is 45, divided by 10. So this is greater than 1. The absolute value is greater than 1. So this one is going to diverge. This sum is going to go to infinity. Now let's think about this one over here. I encourage you to pause it if that one helped you a little bit. Let's try to rewrite this in a way that the common ratio is a little bit more obvious. So 3/2 to the n times 1 over 9 to the n plus 2. Well, we could think about this as 1-- let me write it this way. 3/2 to the n times 1 over 9 to the n plus 2 is the same thing as-- let me write it this way-- 9 to the n times 9 squared. And if we write it that way, then this is going to be the same thing as-- let's write it this way-- 1 over 9 squared, which is 81, times 3/2 to the n times 1 over 9 to the n is the same thing as 1 over 9 to the n."
    },
    {
        "Q": "\nAround 3:13, what was the mental math Sal did with the fractions? I understand that he was actually multiplying 1*3 and 2*9 and simplifying, but he muttered something about, \"3 over 9... so we're gonna get 1.\" If he was using some kind of shorthand trick for multiplying fractions, that's a trick I'd love to know.",
        "A": "I can t read his mind, but my guess would be: 3/2 * 1/9 = 3/9 * 1/2 = 1/3 * 1/2 = 1/6. In other words, he immediately saw that 3 and 9 are related : 9 is divisible by 3, so if you have those on opposite sides of the fraction line (one above, one below) then you can simplify the fraction by dividing both by 3.",
        "video_name": "KNrEAJeRIdo",
        "timestamps": [
            193
        ],
        "3min_transcript": "to the negative 1, which is the same thing as 5 to the n over 5. And so 5 to the n over 5, 5 to the n over 5 times 9 10 to the n. 9 over 10 to the n. This is the same thing. This is equal to 1 over 5 times 5 times 9 over 10 to the nth power. And now the common ratio becomes a little bit more clear. This right over here would be our-- well, it won't be our first term anymore, because we're starting at n equals 2. So we won't view it that way. But if we look at our common ratio, this right over here is going to be our common ratio. And this is absolute value greater than or less than 1. Well, 5 times 9 is 45, divided by 10. So this is greater than 1. The absolute value is greater than 1. So this one is going to diverge. This sum is going to go to infinity. Now let's think about this one over here. I encourage you to pause it if that one helped you a little bit. Let's try to rewrite this in a way that the common ratio is a little bit more obvious. So 3/2 to the n times 1 over 9 to the n plus 2. Well, we could think about this as 1-- let me write it this way. 3/2 to the n times 1 over 9 to the n plus 2 is the same thing as-- let me write it this way-- 9 to the n times 9 squared. And if we write it that way, then this is going to be the same thing as-- let's write it this way-- 1 over 9 squared, which is 81, times 3/2 to the n times 1 over 9 to the n is the same thing as 1 over 9 to the n. it's not going to change the value of that 1. And so that's going to be the same thing as 1/81 times-- let's see. 3 divided by 9, we're going to get 1 over 6 to the n. And so here it's a little bit clearer that our common ratio is 1/6. Its absolute value is clearly less than 1. So this is going to converge. This is actually going to give you a finite value for the sum. Finally, let's go on to this one right over here. So let's see. 1 over 3 to the n minus 1. So let me rewrite the whole thing. This is 2 to the n. And I'm going to write it over 3 to the n times 3 Same exact thing, which is the same thing as 3 to the negative 1 in the denominator. I could write it in the numerator. So it's going to be 3 times 2 to the n over 3 to the n, which is equal to 3 times 2/3 to the n. So all this business is the same thing as 3 times 2/3 to the n."
    },
    {
        "Q": "At 1:20 you verify that you're in radian mode. Why do you need to be in radian mode versus degrees?\n",
        "A": "In calculus the angles are generally taken to be in radians measure rather than in degrees. The value of the trigonometric ratios will not be the same in radians as the numerically equivalent degree measure. Such as sin(90 radians) = 0.894 while sin(90 degrees) = 1",
        "video_name": "t7NvlTgMsO8",
        "timestamps": [
            80
        ],
        "3min_transcript": "Consider the table with function values for f of x is equal to x squared over 1 minus cosine x at positive x-values near 0. Notice that there is one missing value in the table. This is the missing one right here. Use a calculator to evaluate f of x at x equals 0.1 and enter this number in the table rounded to the nearest thousandth. From the table, what does the one-sided limit, the limit as x approaches 0 from the positive direction of x squared over 1 minus cosine of x, appear to be? So let's see what they did. They evaluated when x equals 1, f of x is 2.175. When x gets even a little bit closer to 0-- and once again we're approaching 0 from values larger than 0, that's what this little superscript positive tells us-- we're at 0.5 and we're at 2.042. Then when we even closer to 0, 0.2 f of x is 2.007. I'm guessing when I'm getting even closer it's going to be even closer to 2 right over here. Get my calculator out. So I want to evaluate x squared over 1 minus cosine of x when x is equal to 0.1. Let me actually verify that I'm in radian mode, because otherwise, I might get a strange answer. So I am in radian mode. Let me evaluate it. So I'm going to have 0.1 squared divided by 1 minus cosine of 0.1, and this gets me 2.0016. And let's see, they want us to round to the nearest thousandth. So that would be 2.002. Type that in, 2.002. And so it looks like the limit is approaching 2. crossed 2.005 from 2.007 to 2.002. So let's check our answer, and we got it right. I always find it fun to visualize these things. And that's what a graphing calculator is good for. It can actually graph things. So let's graph this right over here. So go in to graph mode. Let me redefine my function here. So let's see, it's going to be x squared divided by 1 minus cosine of x. And then let me make sure that the range of my graph is right. So I'm zoomed in at the right the part that I care about. So let me go to the range. And let's see, I care about approaching x from the-- or approaching 0 from the positive direction, but as long as I see values around 0, I should be fine."
    },
    {
        "Q": "\nAt 8:56, he begins to factor 108. Why does he factor? Is he putting his answer into its lowest terms.",
        "A": "Yes, he is. Because the square root of 108 is not in its lowest terms, he will go ahead and simplify it. In general, if a number is not fully simplified; then just put in simplest terms... usually the simplified version is also easier to work with",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            536
        ],
        "3min_transcript": "Let's say A is equal to 6. And then we say B-- this colored B-- is equal to question mark. And now we can apply the Pythagorean theorem. A squared, which is 6 squared, plus the unknown B squared is equal to the hypotenuse squared-- is equal to C squared. Is equal to 12 squared. And now we can solve for B. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example we solved for the hypotenuse. We solved for C. So that's why it's always important to recognize that A squared plus B squared plus C squared, C is the length So let's just solve for B here. So we get 6 squared is 36, plus B squared, is equal to 12 squared-- this 12 times 12-- is 144. Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there."
    },
    {
        "Q": "\nI can follow with no issue up to 8:35 - but then you lose me completely. Why do you suddenly try to find the square root of the answer? Why isn't 108 the answer? Can't I put 108 as the answer?\n\nAbsolutely nothing that comes after that point makes any sense to me. Did I miss a video?",
        "A": "You have b^2 = 108, and you do not want to know what the squared is, you want to know what the length of b is. Just like other opposites you should already know (add-subtract, divide-multiply), this is introducing a new opposite square-square root. When I take the \u00e2\u0088\u009ab^2, they cancel to just get b, but the opposites rule is whatever I do to one side, I have to do to the other, so b = \u00e2\u0088\u009a108. Hope this helps.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            515
        ],
        "3min_transcript": "Let's say A is equal to 6. And then we say B-- this colored B-- is equal to question mark. And now we can apply the Pythagorean theorem. A squared, which is 6 squared, plus the unknown B squared is equal to the hypotenuse squared-- is equal to C squared. Is equal to 12 squared. And now we can solve for B. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example we solved for the hypotenuse. We solved for C. So that's why it's always important to recognize that A squared plus B squared plus C squared, C is the length So let's just solve for B here. So we get 6 squared is 36, plus B squared, is equal to 12 squared-- this 12 times 12-- is 144. Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there."
    },
    {
        "Q": "\nAt 7:36, why did Sal square the hypotenuse? Couldn't he have just left out the squares and done 6+B=12?",
        "A": "No because the eqution is a squared+b squared=c squared",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            456
        ],
        "3min_transcript": "Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B. Let's say A is equal to 6. And then we say B-- this colored B-- is equal to question mark. And now we can apply the Pythagorean theorem. A squared, which is 6 squared, plus the unknown B squared is equal to the hypotenuse squared-- is equal to C squared. Is equal to 12 squared. And now we can solve for B. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example we solved for the hypotenuse. We solved for C. So that's why it's always important to recognize that A squared plus B squared plus C squared, C is the length So let's just solve for B here. So we get 6 squared is 36, plus B squared, is equal to 12 squared-- this 12 times 12-- is 144. Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same"
    },
    {
        "Q": "\nAt 10:00, why did you just take out one of the threes and make it its own square root?",
        "A": "He broke it apart into what is perfect squares and what is not - I can take out 2 2s as one and 2 3s as 3 to get 6, but there is nothing to go with the other 3, so it will stay in. He does this to illustrate the difference, not as a practice of simplifying a root.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            600
        ],
        "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6."
    },
    {
        "Q": "Can anyone tell me why he separated the 3 at 10:00? I'd be most grateful! :3\n",
        "A": "He separated the three so that he got a perfect square on the other side (36), so that he could write 6 \u00e2\u0088\u009a3 instead of \u00e2\u0088\u009a108. He could have done like this: \u00e2\u0088\u009a( 2 * 2 * 3 * 3 * 3 ) \u00e2\u0088\u009a( 2 * 2 ) \u00e2\u0088\u009a( 3 * 3 ) \u00e2\u0088\u009a( 3 ) <-- As you can see, he separates them in groups of two equal numbers, and then the three is left. 2 * 3 * \u00e2\u0088\u009a3 6 \u00e2\u0088\u009a3",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            600
        ],
        "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6."
    },
    {
        "Q": "At 5:28, why do we need to take the positive square root of the number equal C Squared?\n",
        "A": "Good question. Typically, we take the principal square root of a number (the positive square root) because there is no such thing as negative length (yet). c^2 cannot have a negative value or else the laws of mathematics have been violated.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            328
        ],
        "3min_transcript": "theorem tells us. So let's say that C is equal to the length of the hypotenuse. So let's call this C-- that side is C. Let's call this side right over here A. And let's call this side over here B. So the Pythagorean theorem tells us that A squared-- so the length of one of the shorter sides squared-- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared. Now let's do that with an actual problem, and you'll see that it's actually not so bad. So let's say that I have a triangle that looks like this. Let me draw it. Let's say this is my triangle. It looks something like this. And let's say that they tell us that this is the right angle. That this length right here-- let me do this in different length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this."
    },
    {
        "Q": "\nAt 1:28, does the pythagorean theorm only apply to right triangles? If so, why?",
        "A": "The Pythagorean Theorem only applies to right triangles because that is the only triangle with a hypotenuse. Remember, the hypotenuse is the side opposite of the right angle. Since acute and obtuse triangles do not have a right angle, there is no hypotenuse. The Pythagorean Theorem cannot be applied because you cannot find or use a hypotenuse length as the c value. I hope that helps!",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            88
        ],
        "3min_transcript": "In this video we're going to get introduced to the Pythagorean theorem, which is fun on its own. But you'll see as you learn more and more mathematics it's one of those cornerstone theorems of really all of math. It's useful in geometry, it's kind of the backbone of trigonometry. You're also going to use it to calculate distances between points. So it's a good thing to really make sure we know well. So enough talk on my end. Let me tell you what the Pythagorean theorem is. So if we have a triangle, and the triangle has to be a right triangle, which means that one of the three angles in the triangle have to be 90 degrees. And you specify that it's 90 degrees by drawing that little box right there. So that right there is-- let me do this in a different color-- a 90 degree angle. And a triangle that has a right angle in it is called a right triangle. So this is called a right triangle. Now, with the Pythagorean theorem, if we know two sides of a right triangle we can always figure out the third side. And before I show you how to do that, let me give you one more piece of terminology. The longest side of a right triangle is the side opposite the 90 degree angle-- or opposite the right angle. So in this case it is this side right here. This is the longest side. And the way to figure out where that right triangle is, and kind of it opens into that longest side. That longest side is called the hypotenuse. And just so we always are good at identifying the hypotenuse, let me draw a couple of more right triangles. So let's say I have a triangle that looks like that. Let me draw it a little bit nicer. So let's say I have a triangle that looks like that. And I were to tell you that this angle right here is 90 degrees. In this situation this is the hypotenuse, because it is opposite the 90 degree angle. It is the longest side. Let me do one more, just so that we're good at recognizing the hypotenuse. So let's say that that is my triangle, and this is the 90 degree angle right there. And I think you know how to do this already. You go right what it opens into. That is the hypotenuse. That is the longest side. So once you have identified the hypotenuse-- and let's say that that has length C."
    },
    {
        "Q": "\nOK so at 2:33 he subtracts 2x from 2x and I don't understand why. Can somebody help me? I'm so confused.....",
        "A": "lol what you need help with ?",
        "video_name": "f15zA0PhSek",
        "timestamps": [
            153
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt the 6:10 in the video instead of dividing 5 by 3 he kept it as 5/3, why is that? Would it work either way? Thank you :)",
        "A": "It does work either way but it s just simpler to keep fractions through the end so your more precise",
        "video_name": "f15zA0PhSek",
        "timestamps": [
            370
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\n3:02 y do u subtract 2x from both sides why dont u subtract 3 from both sides instead?",
        "A": "You could do that, but this way is more simple and easier to solve.",
        "video_name": "f15zA0PhSek",
        "timestamps": [
            182
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At roughly 2:22, why does Sal plug in x for -2 if x is just less than -2 and not equal to it?\n",
        "A": "-2 is still the limit you are trying to reach, so you find what the limit is and put a circle around it. Think that it could get as close to -2 as you can imagine (-2.00000000000000000001), but it would never quite reach it. Guess what would happen it you put this close number in, it would round to the same number as -2. Hope this helps.",
        "video_name": "PQiXRrT_14o",
        "timestamps": [
            142
        ],
        "3min_transcript": "- [Voiceover] So, I have this somewhat hairy function definition here, and I want to see if we can graph it. And this is a piecewise function. It's defined as a different, essentially different lines. You see this right over here, even with all the decimals and the negative signs, this is essentially a line. It's defined by this line over this interval for x, this line over this interval of x, and this line over this interval of x. Well, let's see if we can graph it. I encourage you, especially if you have some graph paper, to see if you can graph this on your own first before I work through it. So, let's think about this first interval. When negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75. So this is going to be a line, a downward sloping line, and the easiest way I can think about graphing it is let's just plot the endpoints here, and then draw the line. So, when x is equal to 10, sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way. Let me do it over here where I do the, times negative 10 plus 4.75. That is going to be equal to, let's see, the negative times a negative is a positive, and then 10 times this is going to be 1.25 plus 4.75. That is going to be equal to six. So, we're going to have the point negative 10 comma six. And that point, and it includes, so x is defined there, it's less than or equal to, and then we go all the way to negative two. So, when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see negative times negative is positive, two times this is going to be point, is going to be positive 0.25 plus 4.75. It's going to be equal to positive five. Now, we might be tempted, we might be tempted, but remember, this interval does not include negative two. It's up to and including, it's up to negative two, not including. So, I'm gonna put a little open circle there, and then I'm gonna draw the line. And then I'm gonna draw the line. I am going to draw my best attempt, my best attempt, at the line. Now, let's do the next interval. The next interval, this one's a lot more straightforward. We started x equals negative two, when x equals negative two negative two plus seven is, negative two plus seven is five. So, negative two, so negative two comma five, so it actually includes that point right over there. So we're actually able to fill it in, and then when x is negative one, negative one plus seven is going to be positive six. Positive six, but we're not including x equals negative one"
    },
    {
        "Q": "\nAt 2:27, do you always move the decimal place two to the left? Thanks:)",
        "A": "Only if its divided by 100",
        "video_name": "_SpE4hQ8D_o",
        "timestamps": [
            147
        ],
        "3min_transcript": "Let's get started with some problems. Let's see. First problem: what is 15% of 40? The way I do percent problems is I just convert the percentage to a decimal and then I multiply it times the number that I'm trying to get the percentage of. So 15% as a decimal is 0.15. You learned that from the percent to decimal conversion video, hopefully. And we just multiply this times 40. So let's say 40 times 0.15. 5 times 0 is 0. 5 times 4 is 20. Put a 0 there. And then 1 times 0 is 0. 1 times 4 is 4. And you get 6 0 0. 1, 2. No decimals up there, so you go 1, 2 and you put the decimal there. So 15% of 40 is equal to 0.15 times 40, which equals 6.00. Well, that's just the same thing as 6. Let's do another problem. Hopefully, that didn't confuse you too much. And I'm going to try to confuse you this time just in case you weren't properly confused the last time. What is 0.2% of-- let me think of a number-- of 7. So a lot of people's inclinations would just say, oh, 0.2%, that's the same thing as 0.2. And if that was your inclination you would be wrong. Because remember, this isn't 0.2. This is 0.2%. You could say that this is 0.2/100, which is, if you multiply the numerator and denominator by 10, is the same thing as 2/1,000. Or you can just do the technique where you move the decimal space over 2 to the left. In which case, if you're starting with 0.2 and you move the decimal space 2 to the left, you go bam. Whoops! Bam, bam. That's where the decimal goes. So it's 0.002. This is key. 0.2% is the same thing as 0.002. This can always trip you up and I've made this careless mistake all the time, so don't feel bad if you ever do it. But just always pay careful attention if you see a decimal and a percentage at the same time. So now that we've figured out how to write this percentage as a decimal we just have to multiply it times the number that we want to take the percentage of. So we say 0.002 times 7."
    },
    {
        "Q": "Does it matter if you put the decimal on the top at 0:46.\n",
        "A": "If you are talking about the multiplication, then no not at all. It s just more common for the smaller number to be on the bottom.",
        "video_name": "_SpE4hQ8D_o",
        "timestamps": [
            46
        ],
        "3min_transcript": "Let's get started with some problems. Let's see. First problem: what is 15% of 40? The way I do percent problems is I just convert the percentage to a decimal and then I multiply it times the number that I'm trying to get the percentage of. So 15% as a decimal is 0.15. You learned that from the percent to decimal conversion video, hopefully. And we just multiply this times 40. So let's say 40 times 0.15. 5 times 0 is 0. 5 times 4 is 20. Put a 0 there. And then 1 times 0 is 0. 1 times 4 is 4. And you get 6 0 0. 1, 2. No decimals up there, so you go 1, 2 and you put the decimal there. So 15% of 40 is equal to 0.15 times 40, which equals 6.00. Well, that's just the same thing as 6. Let's do another problem. Hopefully, that didn't confuse you too much. And I'm going to try to confuse you this time just in case you weren't properly confused the last time. What is 0.2% of-- let me think of a number-- of 7. So a lot of people's inclinations would just say, oh, 0.2%, that's the same thing as 0.2. And if that was your inclination you would be wrong. Because remember, this isn't 0.2. This is 0.2%. You could say that this is 0.2/100, which is, if you multiply the numerator and denominator by 10, is the same thing as 2/1,000. Or you can just do the technique where you move the decimal space over 2 to the left. In which case, if you're starting with 0.2 and you move the decimal space 2 to the left, you go bam. Whoops! Bam, bam. That's where the decimal goes. So it's 0.002. This is key. 0.2% is the same thing as 0.002. This can always trip you up and I've made this careless mistake all the time, so don't feel bad if you ever do it. But just always pay careful attention if you see a decimal and a percentage at the same time. So now that we've figured out how to write this percentage as a decimal we just have to multiply it times the number that we want to take the percentage of. So we say 0.002 times 7."
    },
    {
        "Q": "\ni just noticed that around 1:25, sal has left 2^2=4 in the table of powers of 2 given in the right side of this video. is that a mistake?",
        "A": "how do we convert a floating point number like 1.21 into binary?",
        "video_name": "bvcXEJbEzSs",
        "timestamps": [
            85
        ],
        "3min_transcript": "- [Voiceover] Let's now see if we can convert a larger decimal representation to binary. So let's say that we have the number 114 and this is its decimal representation. See if you can pause the video and rewrite this in its binary representation. So I'm assuming you have at least tried. Now we can work on this together. So as always, we just want to decompose this into the sum of powers of two. You can always decompose this and any number into a sum of powers of two. We can once again just remind ourselves the powers of two. Two to the zero is one, two to the first power is two, two to the third power is eight, two to the fourth power is 16, two to the fifth power is 32, two to the sixth power is 64, two to the seventh power is 128, and that gets us large enough. We've already gotten larger than the number here. So let's see, 114 can be rewritten as, or equal to that is 64. So we can rewrite it as 64 plus, what's going to be left over, 64 plus 50, now we're going to have to rewrite 50 as the sum of the powers of two. And let's see, 50 can be rewritten as the largest power of two that is less than or equal to 50 is 32. So we can rewrite it as 32 plus, 32 plus 18 and now we have to rewrite 18 as the sum of some powers of two. Well 18, the largest power of two that is less than or equal to 18 is 16. So this is going to be 16 and then 16 plus, 16 plus two, and lucky for us, two, well I guess not that lucky, we had to do this a good bit, 114 is equal to, lemme give myself enough real estate here, is equal to 64 plus 32 plus 16 plus 2. I've just written 114 as the sum of powers of two. And once again we can read this as one 64 plus one 32 plus one 16 plus one two. Now we're ready to really rewrite this in binary. Let's just write the different place values. So remember, this is the ones, this right over here is the one's place value or the one's place, I should say. Lemme just do this in a different color. So this is going to be the ones, then we're gonna have the twos, then we're gonna have the fours,"
    },
    {
        "Q": "At around 2:00-2:10, Sal said divide by 2/3 on both sides. How would that look like in an equation?\n",
        "A": "suppose you have 3x = 5y - 1 Dividing both sides by 2/3 means: (2/3) * 3x = 2/3 * (5y - 1) (2*3x)/3 = (2*5y)/3 - 2/3 2x = 10y - 2/3",
        "video_name": "Ye13MIPv6n0",
        "timestamps": [
            120,
            130
        ],
        "3min_transcript": "So we have our scale again. And we've got some masses on the left hand side and some masses on the right hand side. And we see that our scale is balanced. We have the same total mass on the left hand side that we have on the right hand side. Instead of labeling the mystery masses as question mark, I've labeled them all x. And since they all have an x on it, we know that each of these have the same mass. But what I'm curious about is, what is that mass? What is the mass of each of these mystery masses, I guess we could say? And so I'll let think about that for a second. How would you figure out what this x value actually is? How many kilograms is the mass of each of these things? What could you do to either one or both sides of this scale? I'll give you a few seconds to think about that. So you might be tempted to say, well if I could end up with just one mystery mass on the left hand side, and if I keep my scale balanced, then that thing's going to be equal to whatever I have on the right hand side. And that part would actually be a true statement. on the left hand side, you might say, well why don't I just remove two of them? You might just say, well why don't I just remove-- let me do it a good color for removing-- why don't I just remove that one and that one? And then I'll just be left with that right over there. But if you just removed these two, then the left hand side is going to become lighter or it's going to have a lower mass than the right hand side. So it's going to move up and the right hand side is going to move down. And then you might say, OK, I understand. Whatever I have to do to the left hand side, I have to do to the right hand side in order to keep my scale balanced. So you might say, well why don't I remove two of these mystery masses from the right hand side? But that's a problem too because you don't know what this mystery mass is. You could try to remove two from this, but how many of these blocks represent a mystery mass? We actually don't know. But you might then say, well let's see, I've got three of these things here. If I essentially multiply what I have here by 1/3 and if I only leave a 1/3 of the stuff here, then the scale should be balanced. If this has the total mass as this, then 1/3 of this total mass is going to be the same thing as 1/3 of that total mass. So let's just keep only 1/3 of this here. So that's the equivalent to multiplying by 1/3. So if we're only going to keep 1/3 there, we're going to be left with only one of the masses. And if we only keep 1/3 here, let's see, we have one, two, three, four, five, six, seven, eight, nine masses. If we multiply this by 1/3, or if we only keep 1/3 of it there, 1/3 times 9 is 3. So we're going to remove these . And so we have 1/3 of what we originally had on the right hand side and 1/3 of what we originally had on the left hand side. And they will be balanced because we took 1/3 of the same total masses. And so what you're left with is just one of these mystery masses, this x thing right over here, whatever x might be."
    },
    {
        "Q": "\nAt 5:57 where did Sal get 4 and 2 from, in the equation 3(4)-12(2)+9?",
        "A": "He just substituted 2 into the v(t) formula, and in case you ask why he did that, he did it to get a value for 2 to draw it in the graph (since 2 is the value between 1 and 3)",
        "video_name": "SE1ltVuE5yM",
        "timestamps": [
            357
        ],
        "3min_transcript": "Let's see. What two numbers, when you take a product, get 3, and when you add them, you get negative 4? Well, that's going to be t minus 3 times t minus 1 is equal to 0. How can this expression be equal to 0? Well if either of these are equal to 0, if either t minus 3 is 0 or t minus 1 is 0, it's going to be equal to 0. So t could be equal to 3, or t could be equal to 1. If t is 3 or t is 1, either of these are equal to 0, or this entire expression up here is going to be equal to 0. And since our coefficient on the t squared term is positive, we know this is going to be an upward opening parabola. So let's see if we can plot velocity as a function of time. So that is my velocity axis. This right over here is my time axis. And let's say this is 1 times 1 second, or I'm assuming this is in seconds-- 2, 3, 4. just because 1 and 3 are significant-- 1, 2, and 3. And they're not going to be-- I'm going to squash to the vertical scale a little bit. But this right over here, let's say that is 9, a velocity of 9. And so when t equals 0, our velocity is 9. When t equals 1, then our velocity is going to be 0. We get that right over here. 3 minus 12 plus 9, that's 0. And when t is equal to 3 our velocity is 0 again. Our vertex is going to be right in between those, when t is equal to 2-- right in between these two 0's. And we could figure out what that velocity is if we like. It's going to be 3 times 4 minus 12 times 2 plus 9. So what is that? That's 12 minus 24 plus 9. So that is negative 12 plus 9. Did I do that-- 12, yep, negative 3. So you're going to be-- negative 3 might be-- that's 9, so that's positive. So it might be something like this. So the graph of our velocity as a function of time is going to look something like this. And we only care about positive time. It's going to look something like this. So let's think. Remember, this is velocity. This is our velocity as a function of time. Now let's think about when is the velocity less than 0 and the acceleration is less than 0? So let's think about this case right over here? When is this the case? Both of them are going to be less than 0. Well, velocity is the less than 0 over this entire interval, this entire magenta interval. But the acceleration isn't less than 0 that entire time. Remember, the acceleration is the rate of change of velocity."
    },
    {
        "Q": "From 0:48 to 0:53, why does he times 7 by 1, and 7 by 2?\n",
        "A": "21 goes into 160 7 times, so you multiply 7x21, he just broke it down so it would be easier and said 7 times 1 =7 and 7x2=14 (147). Bringing down the 147 to minus it from 160. :)",
        "video_name": "NM8qTo361ic",
        "timestamps": [
            48,
            53
        ],
        "3min_transcript": "Let's see if we can express 16/21 as a decimal. Or we could call this 16 twenty-firsts. This is also 16 divided by 21. So we can literally just divide 21 into 16. And because 21 is larger than 16, we're going to get something less than 1. So let's just literally divide 21 into 16. And we're going to have something less than 1. So let's add some decimal places here. We're going to round to the nearest thousandths in case our digits keep going on, and on, and on. And let's start dividing. 21 goes into 1 zero times. 21 goes into 16 zero times. 21 goes into 160-- well, 20 would go into 160 eight times. So let's try 7. Let's see if 7 is the right thing. So 7 times 1 is 7. 7 times 2 is 14. And then when we subtract it, we should get a remainder less than 21. If we pick the largest number here where, if I multiply it by 21, I get close to 160 And so if we subtract, we do get 13. So that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there. But you could regroup. You could say this is a 10. And then this would be a 5. 10 minus 7 is 3. 5 minus 4 is 1. 1 minus 1 is 0. Now let's bring down a 0. 21 goes into 130. Would 6 work? It looks like 6 would work. 6 times 21 is 126. So that looks like it works. So let's put a 6 there. 6 times 1 is 6. 6 times 2 is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30. So now this becomes a 2. 10 minus 6 is 4. 2 minus 2 is 0. 1 minus 1 is 0. Now let's bring down another 0. 21 goes into 40, well, almost two times, 1 times 21 is 21. And now let's subtract. This is a 10. This becomes a 3. 10 minus 1 is 9. 3 minus 2 is 1. And we're going have to get this digit. Because we want to round to the nearest thousandth. So if this is 5 or over, we're going round up. If this is less than 5, we're going to round down. So let's bring another 0 down here. And 21 goes into 190. Let's see, I think 9 will work. 9 times 1 is 9. 9 times 2 is 18. When you subtract, 190 minus 189 is 1. And we could keep going on, and on, and on. But we already have enough digits to round to the nearest thousandth. This digit right over here is greater than or equal to 5. So we will round up in the thousandths place."
    },
    {
        "Q": "He mentioned it at 2:54-2:57\n",
        "A": "mentioned what?",
        "video_name": "NM8qTo361ic",
        "timestamps": [
            174,
            177
        ],
        "3min_transcript": "And so if we subtract, we do get 13. So that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there. But you could regroup. You could say this is a 10. And then this would be a 5. 10 minus 7 is 3. 5 minus 4 is 1. 1 minus 1 is 0. Now let's bring down a 0. 21 goes into 130. Would 6 work? It looks like 6 would work. 6 times 21 is 126. So that looks like it works. So let's put a 6 there. 6 times 1 is 6. 6 times 2 is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30. So now this becomes a 2. 10 minus 6 is 4. 2 minus 2 is 0. 1 minus 1 is 0. Now let's bring down another 0. 21 goes into 40, well, almost two times, 1 times 21 is 21. And now let's subtract. This is a 10. This becomes a 3. 10 minus 1 is 9. 3 minus 2 is 1. And we're going have to get this digit. Because we want to round to the nearest thousandth. So if this is 5 or over, we're going round up. If this is less than 5, we're going to round down. So let's bring another 0 down here. And 21 goes into 190. Let's see, I think 9 will work. 9 times 1 is 9. 9 times 2 is 18. When you subtract, 190 minus 189 is 1. And we could keep going on, and on, and on. But we already have enough digits to round to the nearest thousandth. This digit right over here is greater than or equal to 5. So we will round up in the thousandths place. we can say that this is 0.76. And then this is where we're going around up-- 762."
    },
    {
        "Q": "\nWait, why do we divide 21 into 16 at 0:13? I've been troubled by this for a while. If you can tell me i'll love you forever <3",
        "A": "otherwise it would be more than 1",
        "video_name": "NM8qTo361ic",
        "timestamps": [
            13
        ],
        "3min_transcript": "Let's see if we can express 16/21 as a decimal. Or we could call this 16 twenty-firsts. This is also 16 divided by 21. So we can literally just divide 21 into 16. And because 21 is larger than 16, we're going to get something less than 1. So let's just literally divide 21 into 16. And we're going to have something less than 1. So let's add some decimal places here. We're going to round to the nearest thousandths in case our digits keep going on, and on, and on. And let's start dividing. 21 goes into 1 zero times. 21 goes into 16 zero times. 21 goes into 160-- well, 20 would go into 160 eight times. So let's try 7. Let's see if 7 is the right thing. So 7 times 1 is 7. 7 times 2 is 14. And then when we subtract it, we should get a remainder less than 21. If we pick the largest number here where, if I multiply it by 21, I get close to 160 And so if we subtract, we do get 13. So that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there. But you could regroup. You could say this is a 10. And then this would be a 5. 10 minus 7 is 3. 5 minus 4 is 1. 1 minus 1 is 0. Now let's bring down a 0. 21 goes into 130. Would 6 work? It looks like 6 would work. 6 times 21 is 126. So that looks like it works. So let's put a 6 there. 6 times 1 is 6. 6 times 2 is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30. So now this becomes a 2. 10 minus 6 is 4. 2 minus 2 is 0. 1 minus 1 is 0. Now let's bring down another 0. 21 goes into 40, well, almost two times, 1 times 21 is 21. And now let's subtract. This is a 10. This becomes a 3. 10 minus 1 is 9. 3 minus 2 is 1. And we're going have to get this digit. Because we want to round to the nearest thousandth. So if this is 5 or over, we're going round up. If this is less than 5, we're going to round down. So let's bring another 0 down here. And 21 goes into 190. Let's see, I think 9 will work. 9 times 1 is 9. 9 times 2 is 18. When you subtract, 190 minus 189 is 1. And we could keep going on, and on, and on. But we already have enough digits to round to the nearest thousandth. This digit right over here is greater than or equal to 5. So we will round up in the thousandths place."
    },
    {
        "Q": "At about 8:00 or 8:10, does \u00e2\u0088\u0086y/\u00e2\u0088\u0086x equal dy/dx?\n",
        "A": "Yes. delta y / delta x essentially means the change in y divided by the change in x. This is exactly what dy/dx means.",
        "video_name": "EKvHQc3QEow",
        "timestamps": [
            480,
            490
        ],
        "3min_transcript": "And then maybe near the end he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. What we could see at any given moment the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates over here, it seems like his rate of change of distance, which would be roughly-- or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. When you average it out, it gets to 23 and 1/2 miles per hour. And I looked it up, Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds is closer to 30 miles per hour. You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x, and figure out what the change of y is around it, or as we go past that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x get smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this, and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. at that point in the curve. Or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. And the notation we use for the derivative is a dy over dx. And that's why I reserved the letter y. And then you say, well, how does this relate to the word differential? Well, the word differential is relating-- this dy is a differential, dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope. Or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is zero, you're going to get something that's undefined. You can't divide by 0. So we take the limit as it approaches 0."
    },
    {
        "Q": "at 8:23, did he mean \"why I reserved the letter 'd'\"?\n",
        "A": "Yes, I think so. He circled the d, and it would make more sense - he didn t use d more distance because it would have been confusing to write dd/dt",
        "video_name": "EKvHQc3QEow",
        "timestamps": [
            503
        ],
        "3min_transcript": "And then maybe near the end he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. What we could see at any given moment the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates over here, it seems like his rate of change of distance, which would be roughly-- or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. When you average it out, it gets to 23 and 1/2 miles per hour. And I looked it up, Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds is closer to 30 miles per hour. You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x, and figure out what the change of y is around it, or as we go past that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x get smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this, and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. at that point in the curve. Or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. And the notation we use for the derivative is a dy over dx. And that's why I reserved the letter y. And then you say, well, how does this relate to the word differential? Well, the word differential is relating-- this dy is a differential, dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope. Or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is zero, you're going to get something that's undefined. You can't divide by 0. So we take the limit as it approaches 0."
    },
    {
        "Q": "At 1:57 wouldn't he have to distribute to the square root 5 also?\n",
        "A": "The -1 has to be distributed to each term within the parentheses. Terms are things that are added or subtracted. There are only two terms within the parentheses, 2x\u00e2\u0088\u009a5 and -4. So the answer to your question is no. If there had been 3 terms, say, 2x + \u00e2\u0088\u009a5 - 4, then you would have been correct.",
        "video_name": "KvMyZY9upuA",
        "timestamps": [
            117
        ],
        "3min_transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three."
    },
    {
        "Q": "At 2:02, how is -1 times -4 equal to +5? Why is it not +4?\n",
        "A": "This is a known mistake that s already been corrected in the video.",
        "video_name": "KvMyZY9upuA",
        "timestamps": [
            122
        ],
        "3min_transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three."
    },
    {
        "Q": "\nAt 2:03, shouldn't that be positive 4?",
        "A": "This is a known error in the video. A box pops up and tells you this is an error and tells you the correct value is 4.",
        "video_name": "KvMyZY9upuA",
        "timestamps": [
            123
        ],
        "3min_transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three."
    },
    {
        "Q": "At 0:03 and various other times in the video, Sal says \"principal square root.\" I understand the the principal square is the positive square root, but how does Sal know to read it as principal square root and not just square root? The signs are the same. Is it because we can assume that it is a function so it has to be the principal square root because otherwise it would just be a relation?\n",
        "A": "The symbol in the video, \u00e2\u0088\u009a , means principle square root . It does not mean square root. Square root is written with a plus or minus symbol in front, so it looks like this: \u00c2\u00b1\u00e2\u0088\u009a So, for instance, if we used the number 25 , they would look like this \u00e2\u0088\u009a25 = 5 (principle square root) \u00c2\u00b1\u00e2\u0088\u009a25 = \u00c2\u00b15 (square root) Unless you write the plus or minus symbol, it is always just the principle square root, not the square root. So that s how you can tell the difference. I hope this helps.",
        "video_name": "KvMyZY9upuA",
        "timestamps": [
            3
        ],
        "3min_transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three."
    },
    {
        "Q": "\nAt 2:26, it said that you have to multiply 20x5=-100, so what would happen if like for example 20x5 equals more than 100, how will we make it out of 100? Please respond quickly!\nThanks!! :)",
        "A": "you don t always have to multiply by 5, its just an example. For example if the demonater was 9 isted of 20 we would multiply by 11.111111 instead of 5",
        "video_name": "AjYil74WrVo",
        "timestamps": [
            146
        ],
        "3min_transcript": "In the United States, 13 out of every 20 cans are recycled. What percent of cans are recycled? So 13 out of every 20 are recycled. So 13/20, or 13 over 20, could also be viewed as 13 divided 20, or 13 divided by 20. And if we do this, we'll get a decimal, and it's fairly straightforward to convert that decimal into a percentage. So 13 divided 20. We have the smaller number in this case being divided by the larger number. So we're going to get a value less than 1. Since we're going to get a value less than 1, let's put a decimal right over here. And let's add a couple of zeroes, as many zeroes as we would need. And we could say, hey, look, 20 goes into 13 zero times. 0 times 20 is 0. And then 13 minus 0 is 13. Now you bring down a 0. 20 goes into 130. So 6 times 20 is 120. So it's going to six times. 6 times 20 is 120. You subtract. You get a 10. Let's bring down another 0. 20 goes into 100 five times. 5 times 20 is 100. And we are done. So this, written as a decimal, is 0.65. So as a decimal, it's 0.65. And if you want to write it as a percentage you essentially multiply this number by 100. Or another way you could say is you shift the decimal point over two spots to the right. So this is going to be equal to 65%. Now, there's another way you could have done it. You could have said, look, percent literally means per hundred. So 13 out of 20 is going to be equal to what over 100? To go from 20 to 100, you would multiply by 5. So let's multiply the numerator by 5 as well. And 13 times 5, let's see, that's 15 plus 50, which is 65. So this would have been a faster way to do it, especially if you recognize it's pretty easy to go from 20 to 100. You multiply it by 5. So we would do the same thing with the 13. And so you would get 65/100, which is the same thing as 65 per-- let me write this percent symbol-- 65%. And just a reminder, percent literally means per hundred, 65 per hundred, 65%."
    },
    {
        "Q": "At 3:25 isn't that method also called the cover-up method?\n",
        "A": "Yes, it is. I think of it as eliminating a term (by multiplying by zero), but that s the way I learned it. A rose by any other name would smell as sweet :-)",
        "video_name": "6CFE60iP2Ug",
        "timestamps": [
            205
        ],
        "3min_transcript": "And in point-slope form, if you know that some, if you know that there's an equation where the line that represents the solutions of that equation has a slope M. Slope is equal to M. And if you know that X equals, X equals A, Y equals B, satisfies that equation, then in point-slope form you can express the equation as Y minus B is equal to M times X minus A. This is point-slope form and we do videos on that. But what I really want to get into in this video is another form. And it's a form that you might have already seen. And that is standard form. Standard. Standard form. And standard form takes the shape of AX plus BY is equal to C, And what I want to do in this video, like we've done in the ones on point-slope and slope-intercept is get an appreciation for what is standard form good at and what is standard form less good at? So let's give a tangible example here. So let's say I have the linear equation, it's in standard form, 9X plus 16Y is equal to 72. And we wanted to graph this. So the thing that standard form is really good for is figuring out, not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can find out the y-intercept pretty clearly from standard form and the x-intercept. The x-intercept isn't so easy to figure out from these other forms right over here. So how do we do that? Well to figure out the x and y-intercepts, let's just set up a little table here, X comma Y, and so the x-intercept is going to happen when Y is equal to zero. when X is equal to zero. So when Y is zero, what is X? So when Y is zero, 16 times zero is zero, that term disappears, and you're left with 9X is equal to 72. So if nine times X is 72, 72 divided by nine is eight. So X would be equal to eight. So once again, that was pretty easy to figure out. This term goes away and you just have to say hey, nine times X is 72, X would be eight. When Y is equal to zero, X is eight. So the point, let's see, Y is zero, X is one, two, three, four, five, six, seven, eight. That's this point, that right over here. This point right over here is the x-intercept. When we talk about x-intercepts we're referring to the point where the line actually intersects the x-axis. Now what about the y-intercept? Well, we said X equals zero, this disappears. And we're left with 16Y is equal to 72. And so we could solve,"
    },
    {
        "Q": "At 5:56, why is 8 becoming a negative? Why isn't 4.5 a negative?\n",
        "A": "The reason 4.5 isn t a negative is because Sal had said From 8 to 0 we would have to change x (8) to -8. So 8 - 8 = 0. Simple. Same for 0 to 4.5. 0 + 4.5 = 4.5. Hoped this helped, I am also trying to learn this by myself for the first time, so all the details might not be crystal clear for me and for you too.",
        "video_name": "6CFE60iP2Ug",
        "timestamps": [
            356
        ],
        "3min_transcript": "So we could say, alright 16Y is equal to 72. And then divide both sides by 16. We get Y is equal to 72 over 16. And let's see, what is that equal to? That is equal to, let's see, they're both divisible by eight, so that's nine over two. Or we could say it's 4.5. So when X is zero, Y is 4.5. And so, we could plot that point as well. X is zero, Y is one, two, three, 4.5. And just with these two points, two points are enough to graph a line, we can now graph it. So let's do that. So let me, oops, though I was using the tool that would draw a straight line. Let me see if I can... So the line will look something like that. I've just graphed, I've just graphed, this is the line that represents all the X and Y pairs 9X plus 16Y is equal to 72. Now, I mentioned standard form's good at certain things and the good thing that standard form is, where it's maybe somewhat unique relative to the other forms we looked at, is it's very easy to figure out the x-intercept. It was very easy to figure out the x-intercept from standard form. And it wasn't too hard to figure out the y-intercept either. If we looked at slope-intercept form, the y-intercept just kinda jumps out at you. At point-slope form, neither the x nor the y-intercept kind of jump out at you. The place where slope-intercept or point-slope form are frankly better is that it's pretty easy to pick out the slope here, while in standard form you would have to do a little bit of work. You could use these two points, you could use the x and y-intercepts as two points and figure out the slope from there. So you can literally say, \"Okay, if I'm going from \"this point to this point, my change in X \"to go from eight to zero is negative eight. \"And to go from zero to 4.5,\" Let me. So when you go from eight to zero, your change in X is equal to negative eight. And to go from zero to 4.5, your change in Y is going to be 4.5. So your slope, once you've figured this out, you could say, \"Okay, this is going to be \"change in Y, 4.5, over change in X, \"over negative 8.\" And since I, at least I don't like a decimal up here, let's multiply the numerator and the denominator by two. You get negative nine over 16. Now once again, we had to do a little bit of work here. We either use these two points, it didn't just jump immediately out of this, although you might see a little bit of a pattern of what's going on here. But you still have to think about is it negative? Is it positive? You have to do a little bit of algebraic manipulation. Or, what I typically do if I'm looking for the slope, I actually might put this into, into one of the other forms. Especially slope-intercept form. But standard form by itself, great for figuring out both the x and y-intercepts"
    },
    {
        "Q": "\nAt around 2:45, Sal says that the input would become negative faster but by what I see on the graphs it just looks like the y values are increasing faster rather than the inputs/x values. It actually seems as if the x value growth has become slower after it compressed. Am I wrong? Please help.",
        "A": "It depends on where you are in the function. Whatever happens to y = f(x) happens twice as fast, as x changes, So at x=a, f(x) is f(a), but f(2x) is already up to f(2a).",
        "video_name": "4H5JZnytOfE",
        "timestamps": [
            165
        ],
        "3min_transcript": "If you apply the transformation at the point f equals, at the point negative six comma f of negative six. You get to the point negative three, g of negative three right over there. Let's do a couple of more. If you look at f is, so f of two, looks like it corresponds to g of one, f of two corresponds to g of one. So let's write that down. F of two looks like it corresponds to g of one. And once again, I'm looking at the where the functions hit the same value and also optically I'm just looking at, well, looks like it's the same part of the function if we assumed g of x is a squeezed version of f of x. And so in general, it looks like for given x. So we could say f of x is going to be equal to g of, well whatever you have it here, it seems like we have half the value over here. So g of x over two. if you want to think of the other way, if you want to say g of x is going to be f of... Well, whatever we have here, it's f of twice that. So f of two x. And we see that that is one of the choices. The g of x is equal to f of two x. Whatever the x that you input into g of x, you get that same value out of the function when you input two times that into f of x. These seem to validate that. It looks optically like that and we shrunk it down. One way to think about it when you take, when you multiply the input into a function by a number larger than one, it's going to compress. It's going to make things happen faster. The input to the function is going to increase or become negative faster so it thins it up. And if that doesn't make intuitive sense, you can also just try some of these values. And I encourage you to try more. Find the corresponding points And you'll see over and over again that to get the same value, you have to put two times as much into f as you have to put into g. Hopefully that helps."
    },
    {
        "Q": "Could someone explain to me why at 1:40 he puts his old Y values as his new X values? I don't understand the reasoning behind that. Please and thank you. :)\n",
        "A": "He took the old y-values and used them as his new x-values because the function: log_b (x) is an inverse of the function: y=b^x. Because they are inverse functions, they are able to be reflected across the line y=x. Hope this helps!",
        "video_name": "DhW9pz5Vfwo",
        "timestamps": [
            100
        ],
        "3min_transcript": "Voiceover:The three points plotted below are on the graph of y is equal to b to the x power. Based only on these three points, plot the three corresponding points that must be on the graph of y is equal to log base b of x by clicking on the graph. I've actually copy and pasted this problem on my little scratch pad so I can mark it up a little bit. What is this first function? This first function is telling us so x, and this is y is equal to b to the x power. When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the one power or b to the first power is equal to four. y is equal to four. Another way of thinking of this y or four is equal to b to the first power and actually we can do [said] must be four. That's this point right over there. This point is telling us that b to the second power When x is equal to two, b to the second power, y is equal to 16. Now we want to plot the three corresponding points on this function. Let me draw another table here. Now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. What are the possibilities here? What I want to do is think ... Let's take these values because these are essentially inverse functions log is the inverse of exponents. If we take the points one, four, and 16. What is y going to be here? y is going to be log base b of one. This is saying what power I need to raise b to If we assume that b is non zero and that's a reasonable assumption because b to different powers are non zero, this is going to be zero for any non zero b. This is going to be zero right there, over here. We have the point one comma zero, so it's that point over there. Notice this point corresponds to this point, we have essentially swapped the x's and y's. In general when you're taking an inverse you're going to reflect over the line, y is equal to x and this is clearly reflection over that line. Now let's look over here, when x is equal to four what is log base b of four. What is the power I need to raise b to to get to four. We see right over here, b to the first power is equal to four. We already figured that out, when I take b to the first power is equal to four. This right over here is going to be equal to one. When x is equal to four, y is equal to one. Notice once again, it is a reflection over the line"
    },
    {
        "Q": "\nAt 5:43: Why is it: x^2 - 3x + 2x - 6 instead of: x^2 - 5x - 6 ?\nBecause isn't it \"x^2 minus POSITIVE 3x plus 2x...\"?\nIf it is \"x^2 minus NEGATIVE 3x plus 2x...\" why aren't there two - signs?",
        "A": "nice question. the -3x + 2x is simple -3x + 2x = ? negative 3x plus 2x it -1x see? -3x + 2x = -2x + 1x = -1x = -x your welcome",
        "video_name": "fGThIRpWEE4",
        "timestamps": [
            343
        ],
        "3min_transcript": "This would be equal to a times x plus a times 2. So over here, you can see when x minus 3 is the same thing as a, we're just distributing it. And now we would do the distributive property again. In this case, we're distributing the x now onto the x minus 3. We're going to distribute the 2 onto the x minus 3. You might be used to seeing the x on the other side, but either way, we're just multiplying it. So this is going to be-- I'll stay color coded. This is going to be x times x, minus 3 times x, plus x times 2-- I'm going through great pains to keep it color coded for you. I think it's helping-- minus 3 times 2. And soon you're going to get used to this. We can do it in one step. You're actually multiplying every term in this one by every term in that one, and we'll figure out faster ways But I really want to show you the idea here. So what's this going to equal? This is going to equal x squared. This right here is going to be minus 3x. This is going to be plus 2x. And then this right here is going to be minus 6. And so this is going to be x squared minus 3 of something, plus 2 of something, that's minus 1 of that something. Minus x, minus 6. We've multiplied those two. Now before we move on and do another problem, I want to show you that you can kind of do this in your head as well. You don't have to go through all of these steps. I just want to show you really that this is just the distributive property. The fast way of doing it, if you had x minus 3, times x plus 2, you literally just want to multiply every term So you'd say, this x times that x, so you'd have x squared. Then you'd have this x times that 2, so plus 2x. Then you'd have this minus 3 times that x, minus 3x. And then you have the minus 3, or the negative 3, times 2, which is negative 6. And so when you simplify, once again you get x squared minus x minus 6. And it takes a little bit of practice to really get used to it. Now the next thing I want to do-- and the principal is really the exact same way-- but I'm going to multiply a binomial times a trinomial, which many people find daunting. But we're going to see, if you just stay calm, it's not too bad. 3x plus 2, times 9x squared, minus 6x plus 4. Now you could do it the exact same way that we did the previous video. We could literally take this 3x plus 2, distribute it onto"
    },
    {
        "Q": "at 1:25 why didn't you combine like terms with the 2x,4x, and 2x\n",
        "A": "You have to multiply these terms first.",
        "video_name": "fGThIRpWEE4",
        "timestamps": [
            85
        ],
        "3min_transcript": "In this video I'm just going to multiply a ton of polynomials, and hopefully that'll give you enough exposure to feel confident when you have to multiply any for yourselves. Let's start with a fairly simple problem. Let's say we just want to multiply 2x times 4x minus 5. Well, we just straight up use the distributive property here. And really, when we do all of these polynomial multiplications, all we're doing is the distributive property repeatedly. But let's just do the distributive property here. This is 2x times 4x, plus 2x times negative 5. Or we could say negative 5 times 2x. So you'd say, minus 5 times 2x. All I did is distribute the 2x. This first term is going to be equal to-- we can multiply the coefficients. Remember, 2x times 4x is the same thing as-- you can rearrange the order of multiplication. This is the same thing as 2 times 4, times x times x. Remember, x to the 1, times x to the 1, add the exponents. I mean, you know x times x is x squared. So this first term is going to be 8x squared. And the second term, negative 5 times 2 is negative 10x. Not too bad. Let's do a slightly more involved one. Let's say we had 9x to the third power, times 3x squared, minus 2x, plus 7. So once again, we're just going to do the distributive property here. So we're going to multiply the 9x to the third times each of these terms. So 9x to the third times 3x squared. I'll write it out this time. In the next few, we'll start doing it a little bit in our heads. So this is going to be 9x to the third times 3x squared. way-- minus 2x times 9x to the third, and then plus 7 times 9x to the third. So sometimes I wrote the 9x to the third first, sometimes we wrote it later because I wanted this negative sign here. But it doesn't make a difference on the order that you're multiplying. So this first term here is going to be what? 9 times 3 is 27 times x to the-- we can add the exponents, we learned that in our exponent properties. This is x to the fifth power, minus 2 times 9 is 18x to the-- we have x to the 1, x to the third-- x to the fourth power. Plus 7 times 9 is 63x to the third. So we end up with this nice little fifth degree polynomial. Now let's do one where we are multiplying two binomials."
    },
    {
        "Q": "\nAt 4:00, why do we do the cosine relative to angle E and not angle D?(if we do this the adjacent angle becomes DC)",
        "A": "Your question isn t clear enough to answer. Angles are written with three letters, and sides are written with two letters. There are 4 angles at point E, and two angles at point D. DC is the side connecting points D and C. Using that information, what is your question?",
        "video_name": "EBKNtjZAjXg",
        "timestamps": [
            240
        ],
        "3min_transcript": "So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally, it's one of the sides of the angle that is not the hypotenuse, so let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down sohcahtoa just to remind ourselves. So soh cah toa. Sine of an angle is opposite over hypotenuse, cosine of an angle is adjacent over hypotenuse. is going to be equal to the length of the adjacent side. So that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC? So DEC-- D, E, C. So that's this angle right over here. I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well once again, cosine is adjacent over hypotenuse. So cosine of angle DEC, the adjacent side to this, well that's this right over here. Well that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is, I could call it EC. It's the length of segment EC. And then the hypotenuse is this right over here. The length of the hypotenuse-- the hypotenuse is side DE, or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is one of-- we do get one of the angles They give us this 41 degrees. And the ratio of this green side over-- the length of this green side over this orange side, what would that be in terms of, if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side, and the orange side is still the hypotenuse. So relative to 41 degrees-- so let's"
    },
    {
        "Q": "\nAround 2:06, he says that the area of all the smaller triangles are 18, and we know that the longer leg is also 18. Is just a coincidence?\nAt 5:45, he writes that AF, and HE, both equal 6, but AE equals 12, leaving no room for FH, at 8:26, he says that FH is equal to 2. But Af(6), HE(6), and FH(2), does not equal AE(12).\nWhy is that?",
        "A": "Yeah, it is just a coincidence. He actually does not say that HE = 6, he says that FE = 6 and FH = 2 and 6-2 = 4 so really HE = 4. The total is 6 + 2 + 4 = 12. Hope that helps!",
        "video_name": "k45QTFCHSVs",
        "timestamps": [
            126,
            345,
            506
        ],
        "3min_transcript": "So we're told that AE is equal to 12. That's this side right over here. And EC is equal to 18. And then they've drawn a bunch of the medians here for us. So we know that they are medians because, when they intersect the opposite side, they're telling us that this length is equal to this length, or that ED is equal to DC, CB is equal to BA, AF is equal to FE, or that F, B, and D are the midpoints and that G, then, would be the centroid where the medians intersect. And so the first thing they ask us is, what is the area of BGC? So BGC right here. That is this triangle right over there. And to figure out that area, we just have to remind ourselves that the three medians of a triangle divide a triangle into six triangles that have equal area. So if we know the area of the entire triangle-- and I think we can figure this out. This is a right triangle. They're telling us that. AE-- this entire distance right over here-- is going to be 12. Let me make sure I have enough space. This entire distance right over here is 18. They tell us that. So the area of AEC is going to be equal to 1/2 times the base-- which is 18-- times the height-- which is 12-- which is equal to 9 times 12, which is 108. That's the area of this entire right triangle, triangle AEC. If we want the area of BGC or any of these smaller of the six triangles-- if we ignore this little altitude right over here, the ones that are bounded by the medians-- then we just have to divide this by 6. Because they all have equal area. We've proven that in a previous video. So the area of BGC is equal to the area of AEC, the entire triangle, divided by 6, which is 108 divided by 6. You get 10 and then 48. Looks like it would be 18. And that's right because it would be-- 108 is the same thing as 18 times 6. So we did our first part. The area of that right over there is 18. And if we wanted, we could say, hey, the area of any of these triangles-- the ones that are bounded by the medians-- this is going to be 18. This is going to be 18. This entire FGE triangle is going to be 18, but we did this first part right over there. Now they ask us, what is the length of AG? So AG is the distance. It's the longer part of this median right over here. And to figure out what AG is, we just have to remind ourselves that the centroid is always 2/3 along the way of the medians, or it divides the median into two segments that have a ratio of 2 to 1. So if we know the entire length of this median, we could just take 2/3 of that. And that'll give us the length of AG."
    },
    {
        "Q": "I think you did math wrong at 12:00.. should be +8/7 not -8/7 since - a negative 8/7..\n",
        "A": "It is -8/7, which was created by multiply -4/7 * 2. A negative * a positive = a negative. Hope this helps.",
        "video_name": "iUQR0enP7RQ",
        "timestamps": [
            720
        ],
        "3min_transcript": "And then minus 3/7. It's a little hairy. We ended up with fractions here and things. But let's confirm that this really is the inverse of the matrix B. Let's multiply them out. So before I do that I have to create some space. I don't even need this anymore. OK. So let's confirm that that times this, or this times that, is really equal to the identity matrix. So let's do that. So let me switch colors. So B inverse is 5/7, if I haven't made any careless mistakes. Minus 4/7. 2/7. And minus 3/7. 3 minus 4. 2 minus 5. And this is going to be the product matrix. I need some space to do my calculations. Let me switch colors. I'm going to take this row times this column. So 5/7 times 3 is what? 15/7. Plus minus 4/7 times 2. So minus 4/7 times 2 is minus-- let me make sure that's right-- 5 times 3 is 15/7. Minus 4-- oh right, right-- 4 times 2, so minus 8/7. So 5 times minus 4 is minus 20/7. Plus minus 4/7 times minus 5. That is plus 20/7. My brain is starting to slow down, having to do matrix multiplications with fractions with negative numbers. But this is a good exercise for multiple parts of the brain. But anyway. So let's go down and do this term. So now we're going to multiply this row times this column. So 2/7 times 3 is 6/7. Plus minus 3/7 times 2. So that's minus 6/7. One term left."
    },
    {
        "Q": "At 7:40 is the absolute value sign used for the determinant because it is analogous to the absolute value of a number?\n",
        "A": "I don t think so. I think that the absolute value signs have different meanings when they are used in different places. In this case, it denotes the determinant. Corresponding to numbers, it is the distance from zero. This is just my speculation, since I m just learning about matrices, so you don t have to take my word for it.",
        "video_name": "iUQR0enP7RQ",
        "timestamps": [
            460
        ],
        "3min_transcript": "So if I have a matrix A, and that is a, b, c, d. And I want to calculate its inverse. Its inverse is actually-- and this is going to seem like voodoo. In future videos, I will give you a little bit more intuition for why this works, or I'll actually show you how this came about. But for now it's almost better just to memorize the steps, just so you have the confidence that you know that you can calculate an inverse. It's equal to 1 over this number times this. a times d minus b times c. ad minus bc. And this quantity down here, ad minus bc, that's called the determinant of the matrix A. And we're going to multiply that. This is just a number. This is just a scalar quantity. And we're going to multiply that by-- you switch the a and the d. So you're left with d and a. And you make these two, you make the bottom left and the top right, you make them negative. So minus c minus b. And the determinant-- once again, this is something that you're just going to take a little bit on faith right now. In future videos, I promise to give you more tuition. But it's actually kind of sophisticated to learn what the determinant is. And if you're doing this in your high school class, you kind of just have to know how to calculate it. Although I don't like telling you that. So what is this? This is also call the determinant of A. So you might see on an exam, figure out the determinant of A. So let me just tell you that. And that's denoted by A in absolute value signs. And that's equal to ad minus bc. So another way of saying this, this could be 1 over the determinant. So you could write A inverse is equal to 1 over the determinant of A times d minus b minus c, a. But let's apply this to a real problem, and you'll see that it's actually not so bad. So let's change letters, just so you know it doesn't always have to be an A. Let's say I have a matrix B. And the matrix B is 3-- I'm just going to pick random numbers-- minus 4, 2 minus 5. Let's calculate B inverse. So B inverse is going to be equal to 1 over the determinant of B. What's the determinant? It's 3 times minus 5 minus 2 times minus 4. So 3 times minus 5 is minus 15, minus 2 times minus 4. 2 times minus 4 is minus 8. We're going to subtract that. So it's plus 8. And we're going to multiply that times what?"
    },
    {
        "Q": "\ncouldnt sal have just multiplied the 500a+300c=3100 by -1 at 6:09",
        "A": "Well there are man ways of solving equations, I used an entirely different method to solve it. It really depends on your style.",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            369
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 4:48 how would you know you can bring the other number (-300c) into the first equation? i don't understand how anyone would guess that you just pick that number instead of using the numbers you have (eg. 200c) can someone please explain? Thank you Edit: i know this question was answered in another comment but i still don't understand what it means",
        "A": "It is like you are subtracting the 500a+300c=3100 from the 500a+200c=2900, so the 300c is subtracted from 200c, and the equation is 200c-300c, and -300c is in the equation",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            288
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:39 couldn't you multiply the bottom equation by -1. Making it read -500a-300c= -3100\n\nIsn't it the same thing? Just making sure. Thanks.\n",
        "A": "It s already in that form at 6:39 so multiplying the bottom by -1 will force you to either divide the bottom equation by -1 or multiply the top equation by -1. So in total, it wouldn t do anything.",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            399
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nWhat did Sal mean at 0:28?",
        "A": "He means that the right angle equals 90\u00c2\u00b0.",
        "video_name": "2mzuFKCuDg4",
        "timestamps": [
            28
        ],
        "3min_transcript": "Match these angles with their measures. There should be one angle in each category. We have 30 degrees, 60 degrees, 90 degrees, and 180 degrees. So there's a few that are very easy to spot out. This angle right over here, it makes a complete line. So this is clearly 180 degrees. So this one's 180 degrees. The other one that might jump out at you is this one right over here. This is clearly a right angle, or a 90-degree angle right over here. So this is 90 degrees. And so we have to think which of these is 30 and which is 60 degrees? Well, just by comparing it that way, you know that the one that opens out more is going to be 60 degrees and the one that opens out less is going to be 30 degrees. And you can see that this, the opening here, is half the size of the opening right over here. So it makes sense that this is 30, this is 60, and it also makes sense that this angle is 1/3 the measure of this angle right over here, that this is 90 degrees. And then if you were to double this, then this part of the angle would go all the way out, So let's check our answer and make sure we actually got this one right. Let's do one more, just for fun. Which of these angles has a measure of 60 degrees? So there's a couple that we can immediately rule out. This angle here is clearly more than 90 degrees. It's an obtuse angle. This angle here is 180 degrees, so we're going to have to pick between these two. And we can assume that these are all at scale. So if we know that a right angle would have this line straight up and down, this looks like it's 2/3 of the way there. So I would go with this one right over there. Now, you might say, well, what about this one? Well, this one looks like it's about 1/3 of a right angle. If we were to do 2/3, then 3/3, it looks like we would get to a right angle. So this one looks much closer to 30 degrees. So which of these angles has a measure of 60? That one right over there. You could immediately rule out that one and that one, and then these take a little bit closer inspection."
    },
    {
        "Q": "\nAt 2:17, when he takes the square root of the equation, should he not put in a plus or minus (+-), since squaring a positive or a negative will give you the same result?",
        "A": "It would ve gave the same answer in the end.",
        "video_name": "n4EK92CSuBE",
        "timestamps": [
            137
        ],
        "3min_transcript": "Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it"
    },
    {
        "Q": "at 2:15 Sal substitutes sin^2 for 2/3 x^2. So can we substitute any value for sin?\n",
        "A": "Yes. You can substitute anything to see if it works. If it doesn t work you will find it out very quickly. However, for this particular identity, the coefficient of the x^2 can be any constant. But the 1, -x^2, and sqrt still need to be there for this identity to work. I hope this helps!",
        "video_name": "n4EK92CSuBE",
        "timestamps": [
            135
        ],
        "3min_transcript": "Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it"
    },
    {
        "Q": "\nok so at 1:05 in the vid how do u know that? also HOW RU SO SMART BOY",
        "A": "He s a maths genius he knows everything about maths",
        "video_name": "GiSpzFKI5_w",
        "timestamps": [
            65
        ],
        "3min_transcript": "We're asked to simplify 8 plus 5 times 4 minus, and then in parentheses, 6 plus 10 divided by 2 plus 44. Whenever you see some type of crazy expression like this where you have parentheses and addition and subtraction and division, you always want to keep the order of operations in mind. Let me write them down over here. So when you're doing order of operations, or really when you're evaluating any expression, you should have this in the front of your brain that the top priority goes to parentheses. And those are these little brackets over here, or however Those are the parentheses right there. That gets top priority. Then after that, you want to worry about exponents. There are no exponents in this expression, but I'll just write it down just for future reference: exponents. One way I like to think about it is parentheses always takes top priority, but then after that, we go in descending order, or I guess we should say in-- well, yeah, in When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want"
    },
    {
        "Q": "Why at 2:40 does Sal draw an empty bracket?\n",
        "A": "cuse he uses it later",
        "video_name": "GiSpzFKI5_w",
        "timestamps": [
            160
        ],
        "3min_transcript": "When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want here like this. You could imagine putting some more parentheses. Let me do it in that same purple. You could imagine putting some more parentheses right here to really emphasize the fact that you're going to do the division first. So 10 divided by 2 is 5, so this will result in 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate this parentheses, so this results-- what's 6 plus 5? Well, that's 11. So we're left with the 20-- let me write it all down again. We're left with 8 plus 20 minus 6 plus 5, which is 11, plus 44. And now that we have everything at this level of operations, we can just go left to right. So 8 plus 20 is 28, so you can view this as 28 minus 11 plus 44."
    },
    {
        "Q": "6:08\nHow can we visualize it as a a scalar field ? I didn't understand.\n",
        "A": "Consider it this way, anywhere in the volume is associated with a value. You can find a numeric value for all points in it given x, y, and z values. These are scalars (not vectors, no direction) lying in a field of three dimensions.",
        "video_name": "vxQvL_WhBGU",
        "timestamps": [
            368
        ],
        "3min_transcript": "now you have the other two integrals left. So I didn't write the other two integrals down. So then you're left with two integrals. You're left with dy and dx. And y goes from 0 to 4 and x goes from 0 to 3. I'm definitely going to run out of space. And now you take the antiderivative of this with respect to y. So what's the antiderivative of this with respect to y? Let me erase some stuff just so I don't get too messy. I was given the very good suggestion of making it scroll, but, unfortunately, I didn't make it scroll enough this time. So I can delete this stuff, I think. Oops, I deleted some of that. But you know what I deleted. OK, so let's take the antiderivative with respect to y. I'll start it up here where I have space. squared over 2, 2's cancel out. So you get xy squared. And y goes from 0 to 4. And then we still have the outer integral to do. x goes from 0 to 3 dx. And when y is equal to 4 you get 16x. And then when y is 0 the whole thing is 0. So you have 16x integrated from 0 to 3 dx. And that is equal to what? 8x squared. And you evaluate it from 0 to 3. When it's 3, 8 times 9 is 72. And 0 times 8 is 0. So the mass of our figure-- the volume we figured out last time was 24 meters cubed. I erased it, but if you watched the last video that's what we learned. But it's mass is 72 kilograms. And we did that by integrating this 3 variable density Or in three-dimensions you can view it as a scalar field, right? At any given point, there is a value, but not really a direction. And that value is a density. But we integrated the scalar field in this volume. So that's kind of the new skill we learned with the triple integral. And in the next video I'll show you how to set up more complicated triple integrals. But the real difficulty with triple integrals is-- and I think you'll see that your calculus teacher will often do this-- when you're doing triple integrals, unless you have a very easy figure like this, the evaluation-- if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated for example, a density function. The integral gets very hairy, very fast. And it's often very difficult or very time consuming to evaluate it analytically just using your traditional calculus skills. So you'll see that on a lot of calculus exams when they start doing the triple integral, they just want you to set it up."
    },
    {
        "Q": "\nVi,\nIf you want five thousand in Roman numerals (see 0:52), you put a V with a bar over it. Just letting you know.",
        "A": "That s actually a medieval change. During Pythagoras s time you would actually count out the M s, because the Romans never needed to measure anything larger than years, which were only a few thousand.",
        "video_name": "X1E7I7_r3Cw",
        "timestamps": [
            52
        ],
        "3min_transcript": "Ok, so I've been learning about Pythagoras and the dirt on him is just too good. You've probably heard of the Pythagorean Theorem but not the part where Pythagoras was a crazy cult leader who thought he'd made a deal with a god thousands of years ago and could remember all of his past lives. Oh, and he killed a guy. I mean maybe it was a long time ago and he was afraid of beans. As in beans, they just like freaked him out or something, I don't know. But mostly I want to talk about the murdery part. See, Pythagoras and his cult of Pythagoreans had this cool-kids club where they'd talk about proportions all day. They'd be like, \"Hey, I drew a two by three rectangle using a straight-edge and compass. Isn't that awesome?\" And then someone would be like, \"Hey guys, I have a box that's two by three and a half?\" And the cool kids would be like, \"Three and a half? That's not a number! Get out of our club!\" And then they'd make the units half the length and call it four by seven, and everything was okay. Even if your box is 2.718 by 6.28 you can just divide your units into thousandths and you'd get a box that's a nice, even 2,718 by 6,280. It's not a simple proportion but hey, the box still has a whole number proportion, so Pythagoras is happy. Unless it's a box of beans, then he freaks out. I'd like to imagine what it would be like to Maybe you think of numbers as being on a line. Numbers one way, zero, and negative numbers the other, and there are numbers between them: fractions, rationals, filling in the gaps. But Pythagoras didn't think about numbers like this at all. They weren't points in a continuum they were each their own, separate being, which was still pretty modern because before that people only thought of numbers as adjectives, numbers of. In Pythagoras's world, there is no number between seven and eight, and there is no number three over two so much as a relationship between three and two, a proportion. Six to four has the same relationship because the numbers share this evenness, which when accounted for makes it three to two. The universe to Pythagoras was made up of these relationships. Mathematics wasn't numbers, mathematics was between the numbers. Though while people admire how much Pythagoras loved proportions, there's a dark flip side to that obsession. How far was he willing to go to protect the proportions he loved? Would he kill for them? Would he die for them? And the answer was he'd go pretty far until beans got involved. It's time for Time Line Time. I want the context, because in school today if you bring out the ruler and compass and are like, \"Let's do some geometry! Let's draw two lines at 90 degree angles using a straight-edge and compass! Here's a happy square!\" Then you've probably had years of math class already and think of geometry as being harder than adding big numbers together. You probably think that zero is a simple, easy concept and have heard of decimals too. Well, here's now, 2012. Here's Einstein, Euler, Newton and Da Vinci - - that sure was a while ago! Now let's go all the way back to when Arabic numerals were invented and brought to the West by Fibbonacci. Before that, arithmetic was nightmarishly hard, so if you can multiply multi-digit numbers together you can go back in time and impress the beans out of Pythagoras. And before that there was no concept of zero, except in India where zero was discovered around here. And if you keep going back you get to the year one, (there's no year zero, of course, because zero hadn't been invented) and back a bit more you get to folk like Aristotle, Euclid, Archimedes and then finally Pythagoras, all the way back in 6th century BCE. Point is, you can do some pretty cool mathematics without having a good handle on arithmetic and people did for a long time."
    },
    {
        "Q": "Umm, at 4:28, the big \"E\" symbol pops up, and I have no idea what it means. Where should I go to learn the proper prerequisite information before I watch the rest of this video?\n",
        "A": "That big E is actually the Greek letter sigma. It means the sum of . To understand how it s used, type sigma into the Search box in the upper left-hand corner. The video on sigma notation will be listed there.",
        "video_name": "iPwrDWQ7hPc",
        "timestamps": [
            268
        ],
        "3min_transcript": "so 2ab squared, and then b times a squared is ba squared, or a squared b, a squared b. I'll multiply b times all of this stuff. Now let's multiply a times all this stuff. a times b squared is ab squared, ab squared. a times 2ab is 2a squared b, 2a squared b, and then a times a squared is a to the 3rd power. Now when we add all of these things together, we get, we get a to the 3rd power plus, let's see, we have 1 a squared b plus another, plus 2 more a squared b's. That's going to be 3a squared b plus 3ab squared. 2ab squared plus another ab squared plus b to the 3rd power. Just taking some of the 3rd power, this already took us a little reasonable amount of time, and so you can imagine how painful it might get to do something like a plus b to the 4th power, or even worse, if you're trying to find a plus b to the 10th power, or to the 20th power. This would take you all day or maybe even longer than that. It would be incredibly, incredibly painful. That's where the binomial theorem becomes useful. What is the binomial theorem? The binomial theorem tells us, let me write this down, binomial theorem. Binomial theorem, it tells us that if we have a binomial, and I'll just stick with the a plus b for now, if I have, and I'm going to try to color code this a little bit, if I have the binomial a plus b, and I'm going to raise it the nth power, I'm going to raise this to the nth power, the binomial theorem tells us that this is going to be equal to, and the notation is going to look a little bit complicated at first, but then we'll work through an actual example, is going to be equal to the sum from k equals 0, k equals 0 to n, this n and this n are the same number, of ... I don't want to ... that's kind of a garish color ... of n choose k, n choose k, and we'll review that in a second; this comes straight out of combinatorics; n choose k times a to the n minus k, n minus k, times b, times b to the k,"
    },
    {
        "Q": "\nAt 4:06, Mr. Sanderson said \"[the gradient vector] points in the direction of steepest descent\" but actually meant \"[the gradient vector] points in the direction of steepest ascent\"",
        "A": "He s actually saying ascent, but it s a poor trancription. In the previous video he writes ascent out on the screen",
        "video_name": "ZTbTYEMvo10",
        "timestamps": [
            246
        ],
        "3min_transcript": "X looks like a constant. And the derivative is just that constant, x. And this can be visualized as a vector field in the xy plane as well. You know, at every given point, xy, so you kind of go like x equals two, y equals one, let's say. So that would be x equals two, y equals one. You would plug in the vector and see what should be output. And at this point, the point is two, one. The desired output kind of swaps those. So we're looking somehow to draw the vector one, two. So you would expect to see the vector that has an x component of one and a y component of two. Something like that. But it's probably gonna be scaled down because of the way we usually draw vector fields. And the entire field looks like this. So I'll go ahead and erase what I had going on. Since this is a little bit clearer. And remember, we scaled down all the vectors. The color represents length. So red here is super-long. Blue is gonna be kind of short. And one thing worth noticing. if the vector is crossing a contour line, it's perpendicular to that contour line. Wherever you go. this vector's perpendicular to the contour line. Over here, perpendicular to the contour line. And this happens everywhere. And it's for a very good reason. And it's also super-useful. So let's just think about what that reason should be. Let's zoom in on a point. So I'm gonna clear up our function here. Clear up all of the information about it. And just zoom in on one of those points. So let's say like right here. We'll take that guy and kind of imagine zooming in and saying what's going on in that region? So you've got some kind of contour line. And it's swooping down like this. And that represents some kind of value. Let's say that represents the value f equals two. And, you know, it might not be a perfect straight line. But the more you zoom in, the more it looks like a straight line. And when you want to interpret the gradient vector. the gradient in the context of a graph, I said it points in the direction of steepest descent. So if you imagine all the possible vectors kind of pointing away from this point, the question is, which direction should you move to increase the value of f the fastest? And there's two ways of thinking about that. One is to look at all of these different directions and say which one increases x the most? But another way of doing it would be to get rid of them all and just take a look at another contour line that represents a slight increase. All right, so let's say you're taking a look at a contour line, another contour line. Something like this. And maybe that represents something that's right next to it. Like 2.1. That represents, you know, another value that's very close. And if I were a better artist, and this was more representative, it would look like a line that's parallel to the original one. Because if you change the output the set of in points that look like it is pretty much the same but just shifted over a bit. So another way we can think about the gradient here is to say of all of the vectors that move"
    },
    {
        "Q": "I don't understand why at 12:14 Sal says that g dot f maps to the identity matrix of f. Why shouldn't it be the identity matrix of y? If g maps from Y-->X and f from X-->Y shouldn't the composition of these functions be a mapping from Y-->Y?\n",
        "A": "gof = g(f) maps from the domain of f (= X) to the codomain of f (= Y), and g maps from Y back to X (first you do f, then g - right to left ).",
        "video_name": "-eAzhBZgq28",
        "timestamps": [
            734
        ],
        "3min_transcript": "we're going to start dealing with these notions with transformations and matrices in the very near future. So it's good to be exposed to it in this more precise form. Now the first thing you might ask is let's say that I have a function f, and there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question is, is f inverse unique? Actually probably the obvious question is how do you know when something's invertible. We're going to talk a lot about that in the very near future. But let's say we know that f is invertible. How do we know, or do we know whether f inverse is unique? To answer that question, let's assume it's not unique. that satisfy our two constraints that can act as inverse functions of f. Let's say that g is one of them. So let's say g is a mapping. Remember f is a mapping from X to Y. Let's say that g is a mapping from Y to X such that if I apply f to something and then apply g to it-- so this gets me from X to Y. Then when I do the composition with g, that gets me back into X. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. I'm assuming that g is an inverse of f. Now let's say that h is another inverse of f. By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from Y to X. Then if I take the composition of h with f, I have to get the identity matrix on the set X. Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The composition of the inverse with the function has to become the identity matrix on x. Then the composition of the function with the inverse has to be the identity function on Y. Let's write that. So g is an inverse of f. It implies this. It also implies-- I'll do it in yellow-- that the"
    },
    {
        "Q": "At 6:33 what is a codomain?\n",
        "A": "If the range of f (the set of all the y such that y = f(x)) is a subset of the real numbers, then the real numbers might be a codomain.",
        "video_name": "-eAzhBZgq28",
        "timestamps": [
            393
        ],
        "3min_transcript": "function, f inverse, that's a mapping from Y to X such that if I take the composition of f inverse with f, this is equal to the identity function over X. So let's think about what's happening. This is just part of it actually. Let me just complete the whole definition. This is true, this has to be true, and the composition of f with the inverse function has to be equal to the identity function over Y. So let's just think about what's this saying. There's some function-- I'll call it right now, this called the inverse of f-- and it's a mapping from Y to X. So f is a mapping from X to Y. We showed that. This is the mapping of f right there. It goes in that direction. We're saying there has be some other function, f inverse, that's a mapping from Y to X. So let's write it here. So f inverse is a mapping from Y to X. f inverse, if you give me some value in set Y, I go to set X. So this guy's domain is this guy's codomain, and this guy's codomain is this guy's domain. Fair enough. But let's see what it's saying. It's saying that the composition of f inverse with f, has to be equal to the identity matrix. So essentially it's saying if I apply f to some value in X-- right, if you think about what's this composition doing-- this guy's going from X to Y. This guy goes from Y to X. f is going from X to Y. Then f inverse is going from Y to X. So this composition is going to be a mapping from X to X, which the identity function needs to do. It needs to go from X to X. They're saying this equals the identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this is that f-- let me do it in another color-- the composition of f inverse with f of some member of the set X is equal to the identity"
    },
    {
        "Q": "At 2:40 he gives the number 0.714141414\nI fully understand the answer that he gives to solve this problem.\nBut it seems to me that the repeating section is 14, not 41.\nWhy doesn't 100x - x work?\n",
        "A": "it depends whether you want to use the repeating section 14 or 41 because the repeating period or cycle is just every two digits. Some numbers like 1/23, 1/4093 or 1/8779 have repeating cycles of 22 digits",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            160
        ],
        "3min_transcript": "So 100x is going to be equal to-- the decimal is going to be there now, so it's going to be 36.363636 on and on and on forever. And then let me rewrite x over here. We're going to subtract that from the 100x. x is equal to 0.363636 repeating on and on forever. And notice when we multiplied by 100x, the 3's and the 6's still line up with each other when we align the decimals. And you want to make sure you get the decimals lined up appropriately. And the reason why this is valuable is now that when we subtract x from 100x, the repeating parts will cancel out. So let's subtract. Let us subtract these two things. So on the left-hand side, we have 100x minus x. So that gives us 99x. And then we get, on the right-hand side, this part cancels out with that part. And we're just left with 36. are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're"
    },
    {
        "Q": "At 5:05 can you just multiply by 10 every time your numerator is a decimal?\n",
        "A": "Yes, but only if the decimal stops at the tenths place...if it goes to the hundredths place, multiply by 100. If it goes to the thousandths place, multiply by 1000, and so on...",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            305
        ],
        "3min_transcript": "the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix. to get rid of this decimal. So let's multiply the numerator by 10 and the denominator by 10. And so we get 707/990. Let's do one more example over here. So let's say we had something like-- let me write this way-- 3.257 repeating, and we want to convert this into a fraction. So once again, we set this equal to x. And notice, this is going to be 3.257257257. The 257 is going to repeat on and on and on. Since we have three digits that are repeating, we want to think about 1,000x, 10 to the third power times x. And that'll let us shift it just right so that the repeating parts can cancel out. So 1,000x is going to be equal to what?"
    },
    {
        "Q": "\nAt 4:25 I was confused on why it was changed to p^2-17p+4 why is it plus 4?",
        "A": "This is Algebra II. The process of going from equation 1 to equation involves multiplying by 4. This is one of the most common operations in Algebra. If the process confuses you, consider reviewing the courses of Algebra that involve solving quadratic equations.",
        "video_name": "GDppV18XDCs",
        "timestamps": [
            265
        ],
        "3min_transcript": "in a row, of getting anything else twice in a row? Well, you're just going to multiply this probability times itself. It's going to be 1 minus p times 1 minus p, or we could just write that as 1 minus p squared. So this right over here is the probability of getting a different candy, any other candy, twice in a row. So prob of any other non-\"Honey Bunny\" candy, any other candy, twice in a row. Now, they tell us that this probability needs to be greater than 2 and 1/4 times the probability of getting \"Honey Bunny\" in one try. So is greater than 2 and 1/4 times the probability of getting \"Honey Bunny\" in one So we have just set up the first part. We have written an inequality that models the situation. Now let's actually solve this inequality. And so to do that, I will just expand 1 minus p squared out. 1 minus p squared is the same thing as-- well, I'll just multiply it out. So this is going to be 1 squared minus 2p plus p squared. And that's going to be greater than 2 and 1/4 p. Now let's see. If we subtract 2 and 1/4 p from both sides, we're going to be left with-- and I'm going to reorder this. We're going to get p squared. So you have minus 2p minus 2 and 1/4 p, so that's going to get us minus 4 and 1/4 p. greater than 0. And so let's think about solving this quadratic right over here. And under which circumstances is this greater than 0? To think about it, let's factor it. And actually, before we factor it, let's simplify it a little bit. I don't like having this 17/4 right over here, so let's multiply both sides times 4. And since 4 is a positive number, it's not going to change the sign, the direction of this inequality. So we could rewrite this as 4p squared minus 17p plus 4 is greater than 0. What are the roots of this? And we could use the quadratic formula if we wanted to do it really quick. We could probably do it other ways."
    },
    {
        "Q": "\nI don't understand the braiding thing Vi does at 1:33. Someone please help!",
        "A": "All she does is put 2 like an 8, and weaves the other, whoch she cut in half, into the others.",
        "video_name": "4tsjCND2ZfM",
        "timestamps": [
            93
        ],
        "3min_transcript": "So say your vector field green bean casserole is in the oven, and now it's time to think about a nice, crispy onion topping. Normal people might just use, for instance, French's French fried onions in a can, put super awesome people use a real French person, and real fresh onions, to make their own fresh onion toroids. And they fry free linked with the Brunnian property to get Borromean onion rings. The Borromean rings show up in many forms, they come flat and in 3D, round, rectangly, triangly. But, the important thing is not the way the rings appear, but the way they are connected to each other. The thing about the Borromean rings is that no two of the rings are actually linked together. Ignore the pink and look at just the green and brown. They're sitting on top of each other, not linked. And if you just look at the green and pink, or pink and brown, it's the same thing. And yet, all three together are linked inseparably. So to make your Borromean rings out of onion rings, you will have to cut one of your rings and then fasten it back together with a toothpick or something, which can be removed after frying. Or you can use the fourth dimension. And luckily I have a four-dimensional guest to help me out. If you're stuck in three dimensions, you can think of it like this. Now, the third ring which I have cut, is going to go outside of the outside ring, but inside of the inside ring. Each ring is wholly out of, and wholly inside of the other two rings so that no two are linked, but all three are. You can also think of laying two on each other flat, one on top of the other. And then having the third weave through them, so that it goes over the one on top, and under the one on bottom. The result can be made to be flatter or more spherical, in some you can see the relationship that Borromean rings have with braids. Sure the orange, yellow, and red ribbons are all twisted together, but no two strands are twisted together. If I pull out the orange one, the other two fall apart. Some people and cultures and stuff think of this togetherness property as a metaphor for unity. So when you eat Borromean onion rings, you get to feel all deep and symbolic. But don't forget to save enough to put on top of your green bean matherole. And there we go. At this point I've got a gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes, a vector field green bean matherole with Borromean onion rings, All I need is a double helix cut ham, and of course, the crowning glory of this feast which I will tell you about next time."
    },
    {
        "Q": "who's she cooking with at 0:11?\n",
        "A": "Marc ten Bosch, the 4 dimensional Frenchman who invented the Borromean Onion Rings.",
        "video_name": "4tsjCND2ZfM",
        "timestamps": [
            11
        ],
        "3min_transcript": "So say your vector field green bean casserole is in the oven, and now it's time to think about a nice, crispy onion topping. Normal people might just use, for instance, French's French fried onions in a can, put super awesome people use a real French person, and real fresh onions, to make their own fresh onion toroids. And they fry free linked with the Brunnian property to get Borromean onion rings. The Borromean rings show up in many forms, they come flat and in 3D, round, rectangly, triangly. But, the important thing is not the way the rings appear, but the way they are connected to each other. The thing about the Borromean rings is that no two of the rings are actually linked together. Ignore the pink and look at just the green and brown. They're sitting on top of each other, not linked. And if you just look at the green and pink, or pink and brown, it's the same thing. And yet, all three together are linked inseparably. So to make your Borromean rings out of onion rings, you will have to cut one of your rings and then fasten it back together with a toothpick or something, which can be removed after frying. Or you can use the fourth dimension. And luckily I have a four-dimensional guest to help me out. If you're stuck in three dimensions, you can think of it like this. Now, the third ring which I have cut, is going to go outside of the outside ring, but inside of the inside ring. Each ring is wholly out of, and wholly inside of the other two rings so that no two are linked, but all three are. You can also think of laying two on each other flat, one on top of the other. And then having the third weave through them, so that it goes over the one on top, and under the one on bottom. The result can be made to be flatter or more spherical, in some you can see the relationship that Borromean rings have with braids. Sure the orange, yellow, and red ribbons are all twisted together, but no two strands are twisted together. If I pull out the orange one, the other two fall apart. Some people and cultures and stuff think of this togetherness property as a metaphor for unity. So when you eat Borromean onion rings, you get to feel all deep and symbolic. But don't forget to save enough to put on top of your green bean matherole. And there we go. At this point I've got a gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes, a vector field green bean matherole with Borromean onion rings, All I need is a double helix cut ham, and of course, the crowning glory of this feast which I will tell you about next time."
    },
    {
        "Q": "\nat 1:01 , i do not get why it is 20-6, when it is 200-60",
        "A": "because you can take away a 0 from the end of the numbers, the you can add a zero to the answer i.e. 200-6 20-6 = 14 which becomes 140",
        "video_name": "iivtjjdSu9I",
        "timestamps": [
            61
        ],
        "3min_transcript": "What I want to do in this video is show you a way of subtracting numbers that is different than the regrouping technique. And this is closer to what I actually do in my head. And this might not be what you see in school, so be careful while you're looking at this. Some people might not fully approve of what I'm about to show you. But the idea, the reason why I'm showing you this is to understand that there's not just one way to do things. As long as you understand the underlying principles, what these numbers represent, and you do things that make sense, you should be OK. And what's neat about this technique is that we don't have to regroup. We're just going to start with the hundreds place and keep on moving. So, for example, if I think of 301, I first want to subtract 100. So 300 minus 100 is going to get me to 200. Now I need to subtract 60. And so I can think about what is 20-- two, zero-- minus 6. So this is essentially saying what is 200 minus 60? Well, 20 minus 6 is 14. So I just subtract, and now I'm left with 14. So I really just have to think about what-- well, or I could do what 141 minus 9 is. Well, 141, it's a little bit more mental computation than you might be used to. But 141 minus 9 is going to be 132. So we are left with 132. Let's do this one, same technique. And I encourage you to pause the video and try it yourself, this same technique. Well, 9 minus 2, it's really 900 minus 200. You're going to be left with 700. Then 71 minus 8, let's see, 11 minus 8 is 13. So you're going to be left with 63. And now we have to subtract 633 minus 6. 13 minus 6 is 7. So it's going be 627. And once again, try to pause the video and do it on your own. go to 72 minus 8, which is really 720 minus 80. But let's just think in terms of 72 minus 8. Well, 12 minus 8 is 4, so 72 minus 8 is going to be 64. So now this is the same thing as 641 minus 8. Well, 41 minus 8 is going to be 33. I have to do a little bit of mental computation here. So this will result in 633."
    },
    {
        "Q": "\nat 2:37, why did Sal use the phrase \"Not to beat a dead horse\"? Is it a simile, metaphor, or WHAT? Also, i like horses. I am very much offended.",
        "A": "It means not to belabor the point. Yes, it is a figurative phrase.",
        "video_name": "AuD2TX-90Cc",
        "timestamps": [
            157
        ],
        "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths."
    },
    {
        "Q": "What does beat a dead horse mean? Because at 2:36, Sal said, \"Not to beat a dead horse...\"\n",
        "A": "Beat a dead horse means, Keep doing what we have done enough of. , or Waste energy doing something that useless.",
        "video_name": "AuD2TX-90Cc",
        "timestamps": [
            156
        ],
        "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths."
    },
    {
        "Q": "what does Sal mean at 2:35 \"not to beat a dead horse\"?\n",
        "A": "The phrase means to keep doing or saying something over and over even after it has become pointless .",
        "video_name": "AuD2TX-90Cc",
        "timestamps": [
            155
        ],
        "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths."
    },
    {
        "Q": "I might be asking a question explained in another video, but at 0:47, he said that, \"We have two angles (angle CDE and angle ABC) that are congruent.\" Why is that?\n\n(I am young in learning this)\n",
        "A": "In geometry, the alternate interior angle rule says that those angles are congruent. Try searching for that rule.",
        "video_name": "R-6CAr_zEEk",
        "timestamps": [
            47
        ],
        "3min_transcript": "In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then, we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles. So they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles, and they are going to be congruent. Or you could say that, if you continue this transversal, you would have a corresponding angle with CDE right up here and that this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles and two of the corresponding angles are the same. And that by itself is enough to establish similarity. We actually could show that this angle and this angle but we don't have to. So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices. And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. So we know triangle ABC is similar to triangle-- so this vertex A corresponds to vertex E over here. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. EDC. Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA."
    },
    {
        "Q": "\nat 4:26, wouldn't Aij be the determinate of the submatrix, not the submatrix itself?",
        "A": "No, Sal defined Aij to be the submatrix itself.",
        "video_name": "H9BWRYJNIv4",
        "timestamps": [
            266
        ],
        "3min_transcript": "So this is going to be an n minus 1 by n minus 1 matrix. So if this is 7 by 7, the submatrix is going to be 6 by 6, one less in each direction. So this is going to be the n minus 1 by n minus 1 matrix you get if you essentially ignore or if you take away-- maybe I should say take away. Let's say ignore, like the word ignore. If you ignore the i-th row, this right here is the row, the i-th row and the j-th column of a. So, for example, let's go back to our 3 by 3 right here. This thing could be denoted based on that definition I-- this term right here. We could denote the matrix when you get rid of the first column and the first row or the first row and the first column, we could call this thing right here, we could call that big matrix A11. So this was big matrix A11. This is big matrix A21, or actually, this matrix was called C, so this would be C11 right there. We could call this one, this would be matrix C12. Because if you get rid of the first row, let me get rid of the first row, right? The first term is your row. If you get rid of the first row and the second column, this is the matrix that's left over: 2, 3, 4, 1. So this is this guy and this guy. 2, 3, 4, 1. C, But this one is C12. I know it's a little bit messy there. So that's all we mean by the submatrix. Very similar to what we did in the 3 by 3 case. You essentially get rid of-- so if you want to find out this guy's submatrix, you would call that a11, and you would literally cross out the first row and the first column, and everything left over here would be that submatrix. Now, with that out of the way, we can create a definition, and it might seem a little circular to you at first, and on some level it is. We're going to define the determinant of A to be equal to-- this is interesting. It's actually a recursive definition. I'll talk about that in a second. It's equal to-- we start with a plus. It's equal to a11 times the submatrix if you remove this guy's row and column. So that, by definition, is just A, big capital A11's"
    },
    {
        "Q": "At 6:26, we get the same answer by treating the question as ratios. That is, 0.6:0.7 as 0.5:x. Is the reasoning sound?\n",
        "A": "Yup, because ratios are just fractions, so in the end everything will work out; your reasoning is sound.",
        "video_name": "6xPkG2pA-TU",
        "timestamps": [
            386,
            360
        ],
        "3min_transcript": "Well, we know a lot of this information. We know the probability of A given B is 0.7. So let me write that, I'll scroll down a little bit. This is 0.7. We know that the probability of B is 0.5. So this is 0.5. So we know that the probability of A and B is the product of these two things. That's going to be 0.35. Seven times five is 35 or, I guess you could say, half of .7 is 0.35. .5 of .7. And that is going to be equal to what we need to figure out. Probability of B given A times probability of A. But we know probability of A. We know that that is 0.6. We know that this is 0.6. So just like that, we've set up a situation, an equation, where we can solve for the probability of B given A. The probability of B given A. Notice, let me just rewrite it right over here. 0.6, 0.6 times the probability of B given A. Times that, right over there. And I'll just copy and paste it so I don't have to keep changing colors. That, over there, is equal to 0.35. Is equal to 0.35. And so to solve for the probability of B given A, we can just divide both sides by 0.6. 0.6, 0.6 and we get the probability of B given A is equal to ... Let me get our calculator out. So 0.35 divided by, divided by 0.6 and we deserve a little bit of a drum roll here, is .5833 ... It keeps going. They tell us to round to the nearest hundredth. So it's 0.58. So notice, this is equal to 0 ... or I'll say approximately equal to 0.58. Once again, verifying that these are dependent. The probability that B happens given A is true, is higher than just the probability that B by itself, or without knowing anything else. Just the probability of B is lower than the probability of B given that you know, given that you know A has happened, or event A is true. And we're done."
    },
    {
        "Q": "\nAt 1:30 , Sal wrote down cos 58 with no parentheses. Usually he uses parentheses when he has a degree measure. What is the correct way to do it?",
        "A": "It s better to write it using parentheses,",
        "video_name": "yiH6GoscimY",
        "timestamps": [
            90
        ],
        "3min_transcript": "We are told that the cosine of 58 degrees is roughly equal to 0.53. And that's roughly equal to, because it just keeps going on and on. I just rounded it to the nearest hundredth. And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle. One of the angles is already labeled 32 degrees. Figure out what all of the angles are, and then use the fundamental definitions, your sohcahtoa definitions, to see if you can figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90 plus another 90 is going to be 180 degrees. Or another way to think about is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down sohcahtoa. Soh, sine, is opposite over hypotenuse. Cah, cosine, is adjacent over hypotenuse. Toa, tangent, is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is a 58 degree angle. The side that is adjacent to it is-- let me do it in this color-- is side BC right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is a hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Now let's think about what the sine of 32 degrees would be. Well, sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite it? Well, it opens up onto BC. And what's the length of the hypotenuse? It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine-- I want to do that in that pink color-- the sine of 32 degrees"
    },
    {
        "Q": "at 5:04 it confuses me that the two sides of a seemingly isosceles triangle to have different angles at both acute sides is that impossible or am i mistaken\n",
        "A": "it may look like an isosceles triangle, but it is not. If the 2 seemingly equal sides were in fact of equal length, the 2 angles would have each been 30\u00c2\u00b0, instead of 25\u00c2\u00b0 and 35\u00c2\u00b0.",
        "video_name": "D5lZ3thuEeA",
        "timestamps": [
            304
        ],
        "3min_transcript": "I want to make it a little bit more obvious. So let's say a triangle like this. If this angle is 60 degrees, maybe this one right over here is 59 degrees. And then this angle right over here is 61 degrees. Notice they all add up to 180 degrees. This would be an acute triangle. Notice all of the angles are less than 90 degrees. A right triangle is a triangle that has one angle that is exactly 90 degrees. So for example, this right over here would be a right triangle. Maybe this angle or this angle is one that's 90 degrees. And the normal way that this is specified, people wouldn't just do the traditional angle measure and write 90 degrees here. They would draw the angle like this. And that tells you that this angle right over here is 90 degrees. And because this triangle has a 90 degree angle, and it could only have one 90 degree angle, this is a right triangle. So that is equal to 90 degrees. Now you could imagine an obtuse triangle, based on the idea that an obtuse angle is larger than 90 degrees, an obtuse triangle is a triangle that has one angle that is larger than 90 degrees. So let's say that you have a triangle that looks like this. Maybe this is 120 degrees. And then let's see, let me make sure that this would make sense. Maybe this is 25 degrees. Or maybe that is 35 degrees. And this is 25 degrees. Notice, they still add up to 180, or at least they should. But the important point here is that we have an angle that is a larger, that is greater, than 90 degrees. Now, you might be asking yourself, hey Sal, can a triangle be multiple of these things. Can it be a right scalene triangle? Absolutely, you could have a right scalene triangle. In this situation right over here, actually a 3, 4, 5 triangle, a triangle that has lengths of 3, 4, and 5 actually is a right triangle. And this right over here would be a 90 degree angle. You could have an equilateral acute triangle. In fact, all equilateral triangles, because all of the angles are exactly 60 degrees, all equilateral triangles are actually acute. So there's multiple combinations that you could have between these situations and these situations right over here."
    },
    {
        "Q": "At 3:13 you are multiplying the results by -1.....where did the \"-1\" come from?\n",
        "A": "Sal multiplied both sides of the equation by -1 because it was in the form -p = -17. We are trying to solve for p not for negative p. The fastest way to transform one side of an equation from negative to positive or vice versa is to multiply both sides by -1 because only the sign changes and not the value.",
        "video_name": "bRwJ-QCz9XU",
        "timestamps": [
            193
        ],
        "3min_transcript": "And I'm going to rewrite it over here. So we have 4 over p minus 1 is equal to 5 over p plus 3. So the first thing we could do, especially because we can assume now that neither of these expressions And this is going to be defined, since we've excluded these values of p, is to get the p minus 1 out of the denominator. We can multiply the left hand side by p minus 1. But remember, this is an equation. If you want them to continue to be equal, anything you do left hand side, you have to do to the right hand side. So I'm multiplying by p minus 1. Now I also want to get this p plus 3 out of the denominator here on the right hand side. So the best way to do that is multiply the right hand side by p plus 3. But if I do that to the right hand side, I also have to do that to the left hand side. p plus 3. And so what happens? We have a p minus 1 in the numerator, p minus 1 at the denominator. These cancel out. So you have just a 1 of the denominator, or you have no denominator anymore. And the left hand side simplifies to distribute the 4, 4 times p plus 3. So that is 4p plus 12. And then the right hand side, you have plus 3 canceling with a p plus 3. This is p plus 3 divided by p plus 3. And all you're left with is 5 times p minus 1. If you distribute the 5 you get 5p minus 5. And now this is a pretty straightforward linear equation to solve. We just want to isolate the p's on one side and the constants on the other. So let's subtract 5p from both sides. I'll switch colors. So let's subtract 5p from both sides. And we get on the left hand side, 4p minus 5p is negative p plus 12. Is equal to, these cancel out, is equal to negative 5. And then we could subtract 12 from both sides. negative 5 minus 12 is negative 17. And we're almost done. We can multiply both sides by negative 1 or divide both sides by negative 1 depending on how you want to view it. And we get, negative one times negative p is. So let me just scroll down a little bit so I have a little bit of real estate. That's positive p is equal to 17. p is equal to 17. And let's verify that this really works. Well, it wasn't one of our excluded values, but just in case, let's verify that it really works. If we go, if we have p is 17. We get 4 over 17 minus 1, needs to be equal to 5 over 17 plus 3. I'm just putting 17 in for p, because that's our solution. So this is the same thing as 4 over 16, needs to be the same thing as 5 over 20."
    },
    {
        "Q": "At 1:15, why does Sal write a 16 instead of a 6?\n",
        "A": "That s bracket start (6 months) .",
        "video_name": "BKGx8GMVu88",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's say that you are desperate for a dollar. So you come to me the local loan shark, and you say hey I need to borrow a dollar for a year. I tell you I'm in a good mood, I willing to lend you that dollar that you need for a year. I will lend it to you for the low interest of 100% per year. 100% per year. How much would you have to pay me in a year? You're going to have to pay the original principal what I lent you plus 100% of that. Plus one other dollar. Which is clearly going to be equal to $2. You say oh gee, that's a lot to have to pay to pay back twice what I borrowed. There's a possibility that I might have the money in 6 months. What kind of a deal could you get me for that Mr. Loanshark? I say oh gee, if your willing to pay back in 6 months, then I'll just charge you half the You borrow one dollar, so in 6 months, I will charge you 50% interest. 50% interest over 6 months. This, of course, was 1 year. How much would you have to pay? Well, you would have to pay the original principal what you borrowed. The one dollar plus 50% of that one dollar. Plus 0.50, and that of course is equal to 1.5. That is equal to $1.50. I'll just write it like this. $1.50. Now you say well gee that's I guess better. What happens if I don't have the money then? If I still actually need a year. We actually have a system for that. What I'll do is just say that okay, you don't have the money for me yet. I'll essentially ... we could think about it. I will just lend that amount that you need for you for another 6 months. We'll lend that out. same interest rate at 50% for the next 6 months. Then you'll owe me the principal a $1.50 plus 50% of the principal, plus 75 cents. Plus 75 cents, and that gets us to $2.25. That equals $2.25. Another way of thinking about it is to go from $1 over the first period, you just multiply that times 1.5. If your going to grow something by 50%, you just multiply it times 1.5. If your going to grow it by another 50%, you can multiply by 1.5 again. One way of thinking about it that 50% interest is the same thing as multiplying by 1.5. Multiplying by 1.5. If you start with 1 and multiply by 1.5 twice, this is going to be the same thing. $2.25 is going to be 1 multiplied by 1.5 twice."
    },
    {
        "Q": "\nat 3:00 ,why does the square root of 2 times the square root of 6 equals the square root of 12 ?",
        "A": "Because if you multiply 2 and 6, you get 12. You wouldn t be able to combine/simplify them if you were adding, but since you are multiply, it s ok.",
        "video_name": "yAH3722GrP8",
        "timestamps": [
            180
        ],
        "3min_transcript": "So let's do that. So we get x squared minus the principal square root of 6 times this term-- I'll do it in yellow-- times x squared. And then we have plus this thing again. We're just distributing it. It's just like they say. It's sometimes not that intuitive because this is a big expression, but you can treat it just like you would treat a variable over You're distributing it over this expression over here. And so then we have x squared minus the principal square root of 6 times the principal square root of 2. And now we can do the distributive property again, but what we'll do is we'll distribute this x squared onto each of these terms and distribute the square root of 2 onto each of these terms. It's the exact same thing as here, it's just you could imagine writing it like this. x plus y times a is still going to be ax plus ay. as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3."
    },
    {
        "Q": "2:10, if |r| > 1, the number will become massively huge. How can we come up with the formula \u00ce\u00a3 = a / (1 - r) if |r| > 1? Please notice that I fully understand \u00ce\u00a3 = a / (1 - r) if 0 < |r| < 1. Did I miss something there?\n",
        "A": "You re right--the formula \u00ce\u00a3 = a / (1 - r) does not apply if |r| > 1. So our formula for the sum of a geometric series only applies if |r| < 1, which Sal begins to address around 2:30.",
        "video_name": "b-7kCymoUpg",
        "timestamps": [
            130
        ],
        "3min_transcript": "In a previous video, we derived the formula for the sum of a finite geometric series where a is the first term and r is our common ratio. What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind blowing because, or actually more than mildly mind blowing, because you're taking the sum of an infinite things but as we see, you can actually get a finite value depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just a limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals zero to n of a times r to the k. as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't have to keep switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than one, for r is equal to one, and actually let me make it clear-- let's think about it for the absolute values of r is greater than one, the absolute values of r equal to one, and then the absolute value of r less than one. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than one, as this exponent explodes, as it approaches infinity, this number is just going to become massively, And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to one, then the denominator is going to become zero. And we're going to be dividing by that denominator, and this formula just breaks down. But where this formula can be helpful, and where we can get this to actually give us a sensical result, is when the absolute value of r is between zero and one. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to zero. So let's think about the case where the absolute value of r is greater than zero, and it is less than one. What's going to happen in that case? Well, the denominator is going to make sense, right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than one, and you take it to higher and higher and higher"
    },
    {
        "Q": "At 0:13 Sal talks about a vertical cut. What is a vertical cut?\n",
        "A": "It means you re cutting up and down. A horizontal cut would be cutting from side to side.",
        "video_name": "hoa1RBk4dTo",
        "timestamps": [
            13
        ],
        "3min_transcript": "So I have a three-dimensional solid right over here. And I want to imagine what type of a shape I would get if I were to make a vertical cut. And just to refresh ourselves, or give us a sense of what a vertical cut is, imagine if this was made out of jello or something kind of fairly soft. But it's still a three-dimensional solid. And I were to make a cut-- let's say I were to make a cut right over here. So let's say I had this big, sharp metal thing-- let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to cut-- I'm going to make it go straight you want to call it, this rectangular pyramid of jello. And what would be the resulting shape of the intersection between the jello and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions. Because this purple surface is a two-dimensional-- you could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it-- it will look like this. This is a vertical cut. So I've brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there. It would get all the way to the bottom right over there. And along this side, it would cut right there. And along that side, it would cut right over there. So what's the resulting shape in two dimensions of essentially the intersection between the slicer"
    },
    {
        "Q": "\n9:00 Does that mean, for instance, that the range of ground movement in 2011 Japan was 100 times greater than in 1989 Loma Prieta?",
        "A": "Yes; you re moving from 7.0 to 9.0, a difference of 2. On the logarithmic scale we use, that s 10^2 or 100.",
        "video_name": "RFn-IGlayAg",
        "timestamps": [
            540
        ],
        "3min_transcript": "But remember, this is a logarithmic scale. And I encourage you to watch the videos we made on the logarithmic scale. On a logarithmic scale, a fixed distance is not a fixed amount of movement on that scale. Or a change on that scale is not a fixed linear distance. It is actually a scaling factor. And you're not scaling by 1.2 over here. You're scaling by 10 to the 1.2 power. So this is times 10 to the 1.2 power. So I'll get my calculator out right over here. And let's figure what that is. So you could imagine what it's going to be. 10 to the first power is 10. And then you have 0.2. So it's going to be, let's just do it, 10 to the 1.2 power. It's 15.8. So it's roughly 16 times stronger. So whatever shaking that was just felt on the east coast, and maybe some of you all watching this might have felt it, Loma Prieta earthquake was 16 times stronger than the earthquake-- let that we just had on the east coast. So that's a dramatic difference. Even though this caused some damage, and this is kind of shaking on a pretty good scale. Imagine 16 times as much shaking. And how much damage that would cause. I actually just met a reporter who told me that she was in her backyard during the Loma Prieta earthquake, not too far from where I live now. And she says all the cars were like jumping up and down. So it was a massive, massive earthquake. Now let's think about the Japanese earthquake. We could think about how much stronger it was than Loma Prieta. So remember, this isn't, you don't just think of this as just 2 times stronger. It is 10 to the second times stronger. And we know how to figure that out. 10 to the second power is 100. So this right over here. So cars were jumping up and down at the Loma Prieta earthquake. The Japanese earthquake was 100 times stronger. 100 times stronger than Loma Prieta. And if you compare it to the east coast earthquake, that occurred in August of 2011. So massive, massive, massive earthquake. And just to get a sense of how much stronger the Chilean earthquake was in 1960-- and just to-- there's some fascinating outcomes of the Japanese earthquake. It was estimated that Japan, just over the course of the earthquake got 13 feet wider. So this is doing something to the actual shape of a huge island. And on top of that, it's estimated that because of the shaking, and the distortions in earth caused by that shaking, that the day on earth got one millionth of a second shorter. A little over a millionth of a second shorter. So you might say, hey, that's only a millionth of a second. But I'd say, hey, look it actually changed the day of the earth. A very fundamental thing. It actually matters when people send things into space, and probes into Mars, is that they are to be able to know that our day just got a millionth of a second shorter."
    },
    {
        "Q": "At 2:41 Sal says 1 -3= -1\nIs that a mistake or am I just not understanding something?\n",
        "A": "It said in the corner that Sal meant -2, but accidentally said -1. He just made a mistake.",
        "video_name": "s4cLM0l1gd4",
        "timestamps": [
            161
        ],
        "3min_transcript": "Well, it gets to y equals negative 2. So what's halfway between 4 and negative 2? Well, you could eyeball it, or you could count, or you could, literally, just take the average between 4 and negative 2. So 4-- so the midline is going to be the horizontal line-- y is equal to 4 plus negative 2 over 2. Just literally the mean, the arithmetic mean, between 4 and negative 2. The average of 4 and negative 2, which is just going to be equal to one. So the line y equals 1 is the midline. So that's the midline right over here. And you see that it's kind of cutting the function where you have half of the function is above it, and half of the function is below it. So that's the midline. Now, let's think about the amplitude. Well, the amplitude is how much this function And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there."
    },
    {
        "Q": "\nat 4:13 in the video when he talks about periods, does the point have to reach zero in order to determine the period, or can you start from anywhere, like start from 4, 6 or anywhere on the graph?",
        "A": "Period can be derived from any two points in the domain of the function that have the exact same y value and have identical slopes.",
        "video_name": "s4cLM0l1gd4",
        "timestamps": [
            253
        ],
        "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2."
    },
    {
        "Q": "At 4:21, why is the period not 2pi over two (simplifies to pi)? Why is it two?\n",
        "A": "Why would it be 2\u00f0\u009d\u009c\u008b/2 = \u00f0\u009d\u009c\u008b? Nothing in the problem suggests that it is! The period of a function is the smallest \u00f0\u009d\u0091\u009d > 0 such that \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5 + \u00f0\u009d\u0091\u009d) for all \u00f0\u009d\u0091\u00a5 \u00e2\u0088\u0088 Dom \u00f0\u009d\u0091\u0093. In this case, the function repeats every 2 units away from the point being considered, hence by definition, the period of the function is 2.",
        "video_name": "s4cLM0l1gd4",
        "timestamps": [
            261
        ],
        "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2."
    },
    {
        "Q": "How can you derive an already derived function (0:13) ?\nI mean, given that f(x)=2x^2, it's derivative would be f '(x)=4x.\nSo, if i understood correctly, to get the antiderivative, i would have to derive f ' (x), which is F(x)=4\nbut... thats not equal to f(x).\nI'm confused.\n",
        "A": "So, if i understood correctly You didnt. If you do a deriavative of a derivative you get the sencond derivative. f(x)=2x^2 f (x)=4x f (x)=4 To get the antiderivative, you have to find a function such that its derivative is your original function. F(x) = 2/3 x^3 Do the derivative and you get back to: F (x) = f(x) = 2 x^2",
        "video_name": "61ecnr8m04U",
        "timestamps": [
            13
        ],
        "3min_transcript": "In the last video we looked at a function and tried to draw its derivative. Now in this video, we're going to look at a function and try to draw its antiderivative. Which sounds like a very fancy word, but it's just saying the antiderivative of a function is a function whose derivative is that function. So for example, if we have f of x, and let's say that the antiderivative of f of x is capital F of x. And this tends to be the notation, when you're talking about an antiderivative. This just means that the derivative of capital F of x, which is equal to, you could say capital f prime of x, is equal to f of x. So we're going to try to do here is, we have our f of x. And we're going to try to think about what's a possible function that this could be the derivative of? when you start looking at integral calculus. But there's actually many possible functions that this could be the derivative of. And our goal in this video is just to draw a reasonable possibility. So let's think about it a little bit. So let's, over here on the top, draw y is equal to capital F of x. So what we're going to try to draw is a function where its derivative could look like this. So what we're essentially doing is, when we go from what we're draw up here to this, we're taking the derivative. So let's think about what this function could look like. So when we look at this derivative, it says over this interval over this first interval right over here, let me do this in purple, it says over this interval from x is equal to 0 all the way to whatever value of x, this right over here, it says that the slope is a constant positive 1. So let me draw a line with a slope of a constant positive 1. And I could shift that line up and down. I will just pick a reasonable one. So I could have a line that looks something like this. I want to draw a slope of positive 1 as best as I can. So let's say it looks something like this. And I could make the function defined here or undefined here. The derivative is undefined at this point. I could make the function defined or undefined as I see fit. This will probably be a point of discontinuity on the original function. It doesn't have to be, but I'm just trying to draw a possible function. So let's actually just say it actually is defined at that point right over there. But since this is going to be discontinuous, the derivative is going to be undefined at that point. So that's that first interval. Now let's look at the second interval. The second interval, from where that first interval ended, all the way to right over here. The derivative is a constant negative 2."
    },
    {
        "Q": "at 0:43 sal says there is a positive trend between them, i do not see any?\n",
        "A": "The scores appear to be getting generally higher as study time is increased, a trend that teachers expect to happen. (positive trend is as independent increases, dependent also increases). Lowest scores are on the left and highest scores are on the right (even though the max score is not with max study time). What do you see that thinks it is not positive?",
        "video_name": "Jpbm5YgciqI",
        "timestamps": [
            43
        ],
        "3min_transcript": "The graphs below show the test grades of the students in Dexter's class. The first graph shows the relationship between test grades and the amount of time the students spent studying. So this is study time on this axis and this is the test grade on this axis. And the second graph shows the relationship between test grades and shoes size. So shoe size on this axis and then test grade. Choose the best description of the relationship between the graphs. So first, before looking at the explanations, let's look at the actual graphs. So this one on the left right over here, it looks like there is a positive linear relationship right over here. I could almost fit a line that would go just like that. And it makes sense that there would be, that the more time that you spend studying, the better score that you would get. Now for a certain amount of time studying, some people might do better than others, but it does seem like there's this relationship. really much of a relationship. You see the shoe sizes, for a given shoe size, some people do not so well and some people do very well. Someone with a size 10 and 1/2, it looks like, someone it looks like they flunked the exam. Someone else, looks like they got A minus or a B plus And it really would be hard to somehow fit a line here. No matter how you draw a line, these dots don't seem to form a trend. So let's see which of these choices apply. There's a negative linear relationship between study time and score. No, that's not true. It looks like there's a positive linear relationship. The more you study, the better your score would be. A negative linear relationship would trend downwards like that. There is a non-linear relationship between study time and score and a negative linear relationship between shoe size and score. Well that doesn't seem right either. A non-linear relationship, it would not be easy to fit a line to it. And this one seems like a line would be very reasonable. between shoe size and score. So I wouldn't pick this one either. There's a positive linear relationship between study time and score. That's right. And no relationship between shoe size and score. Well, I'm going to go with that one. Both graphs show positive linear trends of approximately equal strength. No, not at all. This one doesn't show a linear relationship of really any strength."
    },
    {
        "Q": "\nat 0:36 what is a parabola?",
        "A": "the graph of the quadratic function. This ------> y = x^2 is just an example of the parent quadratic function. it could be something as broad as y = 456x^2+567x. The parent parabolic graph is an even graph because it is symmetric over the y-axis.",
        "video_name": "0A7RR0oy2ho",
        "timestamps": [
            36
        ],
        "3min_transcript": "Let's see if we can learn a thing or two about conic sections. So first of all, what are they and why are they called conic sections? Actually, you probably recognize a few of them already, and I'll write them out. They're the circle, the ellipse, the parabola, and the hyperbola. Hyperbola. And you know what these are already. When I first learned conic sections, I was like, oh, I know what a circle is. I know what a parabola is. And I even know a little bit about ellipses and hyperbolas. Why on earth are they called conic sections? So to put things simply because they're the intersection of a plane and a cone. And I draw you that in a second. But just before I do that it probably makes sense to just draw them by themselves. And I'll switch colors. Circle, we all know what that is. Actually let me see if I can pick a thicker line for my circles. It's all the points that are equidistant from some center, and that distance that they all are that's the radius. So if this is r, and this is the center, the circle is all the points that are exactly r away from this center. We learned that early in our education what a circle is; it makes the world go round, literally. Ellipse in layman's terms is kind of a squished circle. It could look something like this. Let me do an ellipse in another color. So an ellipse could be like that. Could be like that. It's harder to draw using the tool I'm drawing, but it could also be tilted and rotated around. But this is a general sense. And actually, circles are a special case of an ellipse. It's an ellipse where it's not stretched in one dimension It's kind of perfectly symmetric in every way. Parabola. You've learned that if you've taken algebra two and you probably have if you care about conic sections. A parabola looks something like this, kind of a U shape and you know, the classic parabola. I won't go into the equations right now. Well, I will because you're probably familiar with it. y is equal to x squared. And then, you could shift it around and then you can even have a parabola that goes like this. That would be x is equal to y squared. You could rotate these things around, but I think you know the general shape of a parabola. We'll talk more about how do you graph it or how do you know what the interesting points on a parabola actually are. And then the last one, you might have seen this before, is a hyperbola. It almost looks like two parabolas, but not quite, because the curves look a little less U-ish and a little more open. But I'll explain what I mean by that. So a hyperbola usually looks something like this. So if these are the axes, then if I were to draw-- let"
    },
    {
        "Q": "At 12:10 Sal says that if he has time then he would make a video so I was wondering If he can make a video seeing as to that there are no exercises to make sure that I have learned the material. Where would I go to ask the site staff or Sal to ask him to make a video as a continuation to Parametric equations?\n",
        "A": "If you click on ask a question... look on the right side. there is an option to request a feature, rather than post a question.",
        "video_name": "IReD6c_njOY",
        "timestamps": [
            730
        ],
        "3min_transcript": "So if you have the shape, you don't know what parametric equation you can go back to. You could make up one, but you don't know which one it'll go back to. And just to kind of hit the point home of kind of making up a parametric equation. Sometimes that's asked of you. So let's just do a very simple example. Let's start with a normal function of x. So let's say that we have the equation y is equal to x squared plus x. I just made it up. And let's say that you were asked to turn this into a parametric equation. And this is often a very hard thing for people because there is not one right answer. You can turn into an arbitrary number of parametric equations. I just may do something really crazy and arbitrary. I could say x is equal-- When I have y defined explicitly in terms of x like this, I can make x into anything I could say it's cosine of t minus the natural log of t. That's just a random thing. But then if x is this, then y is going to be equal to-- I just substitute this back in --cosine of t minus ln of t squared plus cosine of t minus ln of t. We just converted this into a parametric equation or a set of parametric equations. I could've also just written x is equal to t. And then what would y equal? y would be equal to t squared plus t. You might say well what's the difference between this parametric equation and this parametric equation? Well, they're both going to have the same shape of their paths. It's going to be something like a parabola. But the rate and the direction with which they progress on those paths will be very very different. It's actually a very interesting thing There are some paths that you could take where-- Let's say Let's say that the shape is-- You know everything we've done so far, we've always been going in one direction, but you could have a-- There are scenarios-- And maybe if I have time, I'll make a video. And this isn't what-- Let's say the shape of the path is some type of, I dunno, let's say it's a circle of some kind. I'm just going to-- That's not this. I'm doing a completely different example right now. I'm just kind of-- Sometimes, with the ellipses we had paths that went counterclockwise and then paths that went clockwise. You can also have paths that kind of isolate in between, move around, move back and forth along the thing. So there's all sorts of parametric equations that you can define. And you can say if t equals from this to this, you'll use this set of parametric equations. If it's another set of t's, use another one. So there's all sorts of crazy things you can do to say what happens as you move along the path. So the difference-- Not that this is the path of that. This is actually more of a parabola. But the difference between this and this is how you move along the shape."
    },
    {
        "Q": "\nAt 9:00 Sal says that making both negative will make the path reverse. Wouldn't it only reverse if one was negative and one positive?",
        "A": "cos -t = cos t so the only thing that really matters is changing the sign of sin t",
        "video_name": "IReD6c_njOY",
        "timestamps": [
            540
        ],
        "3min_transcript": "We get all the way-- Oh. That's the same color I used before. Let me see this. Let me do this color. We're here. Now notice: in the first one, when we went from t equals 0 to t equals pi over 2, we went from here to there. We went kind of a quarter of the way around the ellipse. But now when we went from t equals zero to pi over 2, where did we go? Went halfway around the ellipse. We went all the way from there, all the way over there. And likewise, when we went from t equals pi over 2 to t equals pi with this set of parametric equations, we went another quarter of the ellipse. We went from there to there. But here, when we go from t equals pi ever 2 to t equals pi, we go all of this way. We go back to the beginning part of our ellipse. has the exact same shape of its path as this set of parametric equations. Except it's going around it at twice as fast of a rate. For every time when t increases by pi over 2 here, we go by-- we kind of go a quarter way around the ellipse. But when t increases by pi over 2 here, we go halfway around the ellipse. So the thing to realize-- and I know I've touched on this before --is that even though both of these sets of parametric equations, when you do the algebra, they can kind of be converted into this shape. You lose the information about where our particle is as it's rotating around the ellipse or how fast it's rotating And that's why you need these parametric equations. We can even set up a parametric equation that goes in the other direction. Instead of having these-- and I encourage you to play with that --but if you instead of this, if you just put a minus sign right here. Instead of going in that direction, it would go in this direction. It would go in a clockwise direction. So one thing that you've probably been thinking from the beginning is OK, I was able to go from my parametric equations to this equation of ellipse in terms of just x and y. Can you go back the other way? Could you go from this to this? And, I think you might realize now, that the answer is no. Because there's no way, just with the information that you're given here, to know that you should go to this parametric equation or this parametric equation or any of an infinite number of parametric equations. I mean anything of the form x is equal to 3 cosine of really anything times t and y is equal to 3 times cosine of-- As long as it's the same anything-- I drew the two squiggly marks the same. --as long as these two things are the same, then you"
    },
    {
        "Q": "At 0:12 are the numbers on the line tenths or wholes\n",
        "A": "The blue lines are whole numbers and the little black lines are tenths.",
        "video_name": "qb0QSP7Sfz4",
        "timestamps": [
            12
        ],
        "3min_transcript": "- [Voiceover] Where is the point on the number line? Well, here it is, here is the point. But I'm guessing that they're asking not literally just to find it and look at it but what number is this point graphed at. Where is this on the number line? So, one thing we know pretty quickly is the number is between three and four. It's greater than three but it's not quite four. But to figure out how much greater than three we need to know what these black tick marks represent. So, between three and four there is one, two, three, four, five, six, seven, eight, nine, ten equal spaces. So, each of these distances, each of these equal spaces, is one tenth or one tenth of the distance between three and four. It's one out of ten equal spaces. So, if that's one tenth and this next space is another one tenth. And then we have to travel one more tenth So, we went three, we know it's three. Plus, one, two, three tenths. Three and three tenths. Or, let's write this as a decimal, let's look at it as a decimal. If we wanted, we could have our ones place value and then after the ones, the decimal and the tenths. So, for the ones, there's three ones. And how many tenths did we see here? There were three tenths. So, either way we can say three and three tenths or three and three tenths. Our decimal, our point is 3.3 on the number line."
    },
    {
        "Q": "\nAt 3:24, Sal is just dividing both the sides by tan... Am I right? Thanks in advance :)",
        "A": "Yes...you could think about it that way.",
        "video_name": "aHzd-u35LuA",
        "timestamps": [
            204
        ],
        "3min_transcript": "And they've given us two pieces of information. They gave us the side that is opposite the angle. And they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write, like I always like to do, soh, cah, toa. And these are really by definition. So you just have to know this, and soh cah toa helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. We can write that the tangent of theta is equal to the length of the opposite side-- 324 meters-- over the length of the adjacent side-- over 54 meters. What angle, when I take its tangent, gives me 324/54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the Inverse Tan Function. So we could rewrite this as we're going to take the inverse tangent-- and sometimes it's written as tangent with this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324/54. And just to be clear, what is this inverse tangent? This just literally says, this will return what is the angle that, when I take the tangent of it, gives me 324/54. This says, what is the angle that, when I take the tangent of it, gives me tangent of theta? Theta is the angle that when you get the tangent of it gets you tangent of theta. And so we get theta is equal to inverse tangent of 324/54. Once again, this inverse tangent thing you might find confusing. But all this is saying is, over here, we're saying tangent of some angle is 324/54. This is just saying my angle is whatever angle I need so that when I take the tangent of it, I get 324/54. It's how we will solve for theta. So let's get our calculator out. And let's say that we want our answer in degrees. Well, I'm just going to assume that they want our answers in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the 2nd mode right over here. And actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees."
    },
    {
        "Q": "At 3:56, you change the calculator to degree mode. For CC Geometry, should you keep the calculator at degree mode?\n",
        "A": "It depends on the question, if its asked to write the answer in Radians, there is no other way!",
        "video_name": "aHzd-u35LuA",
        "timestamps": [
            236
        ],
        "3min_transcript": "What angle, when I take its tangent, gives me 324/54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the Inverse Tan Function. So we could rewrite this as we're going to take the inverse tangent-- and sometimes it's written as tangent with this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324/54. And just to be clear, what is this inverse tangent? This just literally says, this will return what is the angle that, when I take the tangent of it, gives me 324/54. This says, what is the angle that, when I take the tangent of it, gives me tangent of theta? Theta is the angle that when you get the tangent of it gets you tangent of theta. And so we get theta is equal to inverse tangent of 324/54. Once again, this inverse tangent thing you might find confusing. But all this is saying is, over here, we're saying tangent of some angle is 324/54. This is just saying my angle is whatever angle I need so that when I take the tangent of it, I get 324/54. It's how we will solve for theta. So let's get our calculator out. And let's say that we want our answer in degrees. Well, I'm just going to assume that they want our answers in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the 2nd mode right over here. And actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. And let me just type in the inverse tangent-- so it's in this yellow color right here-- inverse tangent of 324 divided by 54 is going to be-- and they told us to round to two decimal places-- 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien."
    },
    {
        "Q": "At 1:41 what does he mean by transitive, communicative and commutative?\n",
        "A": "The Commutative Laws say we can swap numbers over and still get the same answer ... a+b = b+a Hope it helps :D Have a great day/night :) :)",
        "video_name": "d8lP5tR2R3Q",
        "timestamps": [
            101
        ],
        "3min_transcript": "Welcome to the presentation on multiplying and dividing negative numbers. Let's get started. I think you're going to find that multiplying and dividing negative numbers are a lot easier than it might look initially. You just have to remember a couple rules, and I'm going to teach probably in the future like I'm actually going to give you more intuition on why these rules work. But first let me just teach you the basic rules. So the basic rules are when you multiply two negative numbers, so let's say I had negative 2 times negative 2. First you just look at each of the numbers as if there was no negative sign. Well you say well, 2 times 2 that equals 4. And it turns out that if you have a negative times a negative, that that equals a positive. So let's write that first rule down. A negative times a negative equals a positive. What if it was negative 2 times positive 2? two numbers without signs. We know that 2 times 2 is 4. But here we have a negative times a positive 2, and it turns out that when you multiply a negative times a positive you get a negative. So that's another rule. Negative times positive is equal to negative. What happens if you have a positive 2 times a negative 2? I think you'll probably guess this one right, as you can tell that these two are pretty much the same thing by, I believe it's the transitive property -- no, no I think it's the communicative property. I have to remember that. But 2 times negative 2, this also equals negative 4. So we have the final rule that a positive times a negative also equals the negative. of the same thing. A negative times a positive is a negative, or a positive times a negative is negative. You could also say that as when the signs are different and you multiply the two numbers, you get a negative number. And of course, you already know what happens when you have a positive times a positive. Well that's just a positive. So let's review. Negative times a negative is a positive. A negative times a positive is a negative. A positive times a negative is a negative. And positive times each other equals positive. I think that last little bit completely confused you. Maybe I can simplify it for you. What if I just told you if when you're multiplying and they're the same signs that gets you a positive result. And different signs gets you a negative result."
    },
    {
        "Q": "\nWho is saying this? 0:00",
        "A": "sal is doing this.",
        "video_name": "d8lP5tR2R3Q",
        "timestamps": [
            0
        ],
        "3min_transcript": "Welcome to the presentation on multiplying and dividing negative numbers. Let's get started. I think you're going to find that multiplying and dividing negative numbers are a lot easier than it might look initially. You just have to remember a couple rules, and I'm going to teach probably in the future like I'm actually going to give you more intuition on why these rules work. But first let me just teach you the basic rules. So the basic rules are when you multiply two negative numbers, so let's say I had negative 2 times negative 2. First you just look at each of the numbers as if there was no negative sign. Well you say well, 2 times 2 that equals 4. And it turns out that if you have a negative times a negative, that that equals a positive. So let's write that first rule down. A negative times a negative equals a positive. What if it was negative 2 times positive 2? two numbers without signs. We know that 2 times 2 is 4. But here we have a negative times a positive 2, and it turns out that when you multiply a negative times a positive you get a negative. So that's another rule. Negative times positive is equal to negative. What happens if you have a positive 2 times a negative 2? I think you'll probably guess this one right, as you can tell that these two are pretty much the same thing by, I believe it's the transitive property -- no, no I think it's the communicative property. I have to remember that. But 2 times negative 2, this also equals negative 4. So we have the final rule that a positive times a negative also equals the negative. of the same thing. A negative times a positive is a negative, or a positive times a negative is negative. You could also say that as when the signs are different and you multiply the two numbers, you get a negative number. And of course, you already know what happens when you have a positive times a positive. Well that's just a positive. So let's review. Negative times a negative is a positive. A negative times a positive is a negative. A positive times a negative is a negative. And positive times each other equals positive. I think that last little bit completely confused you. Maybe I can simplify it for you. What if I just told you if when you're multiplying and they're the same signs that gets you a positive result. And different signs gets you a negative result."
    },
    {
        "Q": "At around 6:30, Sal starts to talk about how the overlap is only supposed to be counted once. But why would it be counted even once if the question is the P (yellow OR cube?) anyways? Doesn't this question not want the area where we can get both outcomes? If that is the case, then is the addition rule really necessary? Couldn't you just simply add together the P(only Yellow), which would be 7, and P(only Cube), which would be 8?\n",
        "A": "P(yellow or cube) includes yellow cubes because you re looking for the probability of either. Since yellow cubes have both traits, they automatically have either and are therefore part of the outcome we re looking for. The rule exists so that they re counted, but not double-counted.",
        "video_name": "QE2uR6Z-NcU",
        "timestamps": [
            390
        ],
        "3min_transcript": "that I've drawn? This Venn diagram is just a way to visualize the different probabilities. And they become interesting when you start thinking about where sets overlap, or even where they don't overlap. So here we are thinking about things that are members of the set yellow. So they're in this set, and they are cubes. So this area right over here-- that's the overlap of these two sets. So this area right over here-- this represents things that are both yellow and cubes, because they are inside both circles. So this right over here-- let me rewrite it right over here. So there's five objects that are both yellow and cubes. Now let's ask-- and this is probably the most interesting thing to ask-- what is the probability of getting something that is yellow or or a cube, a cube of any color? or a cube of any color-- well, we still know that the denominator here is going to be 29. These are all of the equally likely possibilities that might jump out of the bag. But what are the possibilities that meet our conditions? Well, one way to think about it is, well, the probability-- there's 12 things that would meet the yellow condition. So that would be this entire circle right over here-- 12 things that meet the yellow condition. So this right over here is 12. This is the number of yellow. That is 12. And then to that, we can't just add the number of cubes, because if we add the number of cubes, we've already counted these 5. These 5 are counted as part of this 12. One way to think about it is there are 7 yellow objects that are not cubes. Those are the spheres. There are 5 yellow objects that are cubes. And then there are 8 cubes that are not yellow. That's one way to think about. we counted all of this. So we can't just add the number of cubes to it, because then we would count this middle part again. So then we have to essentially count cubes, the number of cubes, which is 13. So the number of cubes, and we'll have to subtract out this middle section right over here. Let me do this. So subtract out the middle section right over here. So minus 5. So this is the number of yellow cubes. It feels weird to write the word yellow in green. The number of yellow cubes-- or another way to think about it-- and you could just do this math right here. 12 plus 13 minus 5 is 20. Did I do that right? 12 minus, yup, it's 20. So that's one way. You just get this is equal to 20 over 29. But the more interesting thing than even the answer of the probability of getting that,"
    },
    {
        "Q": "\nAt 1:37, why do we take half of -6?",
        "A": "We are working with completing the square. It starts from the concept that (x + b) ^2 = x(x+b)+b(x+b) = x^2 + bx + bx + b^2 or x^2 + 2bx + b^2. Since the middle has to come from two middle terms, we divide by two. In the specific problem, we have to say that 2b = - 6, thus when we divide both sides, we get b = -3 and b^2 = (-3)^2 = 9.",
        "video_name": "4Bx06GFyhUA",
        "timestamps": [
            97
        ],
        "3min_transcript": "Which quadratic has the lowest maximum value? So let's figure out the maximum value for each of these-- and they're defined in different ways-- and then see which one is the lowest. And I'll start with the easiest. So h of x. We can just graphically look at it, visually look at it, and say-- what's the maximum point? And the maximum point looks like it's right over here when x is equal to 4. And when x is equal to 4, y or h of x is equal to negative 1. So the maximum for h of x looks like it is negative 1. Now, what's the maximum for g of x? And they've given us some points here and here. Once again, we can just eyeball it, and say-- well, what's the maximum value they gave us? Well, 5 is the largest value. It happens when x is equal to 0. g of 0 is 5. So the maximum value here is 5. Now, f of x. They just give us an expression to define it. And so it's going take a little bit of work to figure out what the maximum value is. The easiest way to do that for a quadratic is to complete the square. And so let's do it. plus 6x minus 1. I never like having this negative here. So I'm going to factor it out. This is the same thing as negative times x squared minus 6x and plus 1. And I'm going to write the plus 1 out here because I'm fixing to complete the square. Now, just as a review of completing the square, we essentially want to add and subtract the same number so that part of this expression is a perfect square. And to figure out what number we want to add and subtract, we look at the coefficient on the x term. It's a negative 6. You take half of that. That's negative 3. And you square it. Negative 3 squared is 9. Now, we can't just add a 9. That would change the actual value of the expression. We have to add a 9 and subtract a 9. And you might say-- well, why are we adding and subtracting the same thing if it doesn't change the value of the expression? And the whole point is so that we can get this first part of the expression to represent a perfect square. So I can rewrite that part as x minus 3 squared and then minus 9-- or negative 9-- plus 1 is negative 8. Let me do that in a different color so we can keep track of things. So this part right over here is negative 8. And we still have the negative out front. And so we can rewrite this as-- if we distribute the negative sign-- negative x minus 3 squared plus 8. Now, let's think about what the maximum value is. And to understand the maximum value, we have to interpret this negative x minus 3 squared. Well, x minus 3 squared-- before we think about the negative-- that is always going to be a positive value. Or it's always going to be non-negative. But then, when we make it negative, it's always going to be non-positive."
    },
    {
        "Q": "\nOkay I'm having troubles at 2:30.\nI don't understand where you got the 4 from, when I multiply 3x -2^2 I get 12 not 4?",
        "A": "He got twelve, but then subtracted it.",
        "video_name": "za0QJRZ-yQ4",
        "timestamps": [
            150
        ],
        "3min_transcript": "Determine the domain and range of the function f of x is equal to 3x squared plus 6x minus 2. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. So essentially any number if we're talking about reals when we talk about any number. So, the domain, the set of valid inputs, the set of inputs over which this function is defined, is all real numbers. So, the domain here is all real numbers. And, for those of you who might say, well, you know, aren't all numbers real? You may or may not know that there is a class of numbers, that are a little bit bizarre when you first learn them, called imaginary numbers and complex numbers. But, I won't go into that right now. But, most of the traditional numbers that you know of, they are part of the set of real numbers. So, you take any real number and you put it here, you can square it, multiply it by 3, then add 6 times it and subtract 2. Now, the range, at least the way we've been thinking about it in this series of videos-- The range is set of possible, outputs of this function. Or if we said y equals f of x on a graph, it's a set of all the possible y values. And, to get a flavor for this, I'm going to try to graph this function right over here. And, if you're familiar with quadratics-- and that's what this function is right over here, it is a quadratic-- you might already know that it has a parabolic shape. And, so its shape might look something like this. And, actually this one will look like this, it's upward opening. But other parabolas have shapes like that. And, you see when a parabola has a shape like this, it won't take on any values below its vertex when it's upward opening, and it won't take on any values above its vertex when it is downward opening. So, let's see if we can graph this and maybe get a sense of its vertex. but let's see how we can think about this problem. So, I'm gonna try some x and y values. There's other ways to directly compute the vertex. Negative b over 2a is the formula for it. It comes straight out of the quadratic formula, which you get from completing the square. Lets try some x values and lets see what f of x is equal to. So, let's try, well this the values we've been trying the last two videos. What happens when x is equal to negative two? Then f of x is 3 times negative 2 squared, which is 4, plus 6 times negative 2, which is 6 times negative 2, so it's minus 12 minus 2. So, this is 12 minus 12 minus 2. So, it's equal to negative 2. Now, what happens when x is equal to negative 1? So, this is going to be 3 times negative 1 squared, which is just 1, minus, or I should say plus 6 times negative 1 which is"
    },
    {
        "Q": "At 2:33, did Sal just choose -2 randomly? If it wasn't random, why did he choose -2?\n",
        "A": "He picked -2 because that is a place that he normally starts his tables and graphs on, it is just habit, there is no specific reason.",
        "video_name": "za0QJRZ-yQ4",
        "timestamps": [
            153
        ],
        "3min_transcript": "So, you take any real number and you put it here, you can square it, multiply it by 3, then add 6 times it and subtract 2. Now, the range, at least the way we've been thinking about it in this series of videos-- The range is set of possible, outputs of this function. Or if we said y equals f of x on a graph, it's a set of all the possible y values. And, to get a flavor for this, I'm going to try to graph this function right over here. And, if you're familiar with quadratics-- and that's what this function is right over here, it is a quadratic-- you might already know that it has a parabolic shape. And, so its shape might look something like this. And, actually this one will look like this, it's upward opening. But other parabolas have shapes like that. And, you see when a parabola has a shape like this, it won't take on any values below its vertex when it's upward opening, and it won't take on any values above its vertex when it is downward opening. So, let's see if we can graph this and maybe get a sense of its vertex. but let's see how we can think about this problem. So, I'm gonna try some x and y values. There's other ways to directly compute the vertex. Negative b over 2a is the formula for it. It comes straight out of the quadratic formula, which you get from completing the square. Lets try some x values and lets see what f of x is equal to. So, let's try, well this the values we've been trying the last two videos. What happens when x is equal to negative two? Then f of x is 3 times negative 2 squared, which is 4, plus 6 times negative 2, which is 6 times negative 2, so it's minus 12 minus 2. So, this is 12 minus 12 minus 2. So, it's equal to negative 2. Now, what happens when x is equal to negative 1? So, this is going to be 3 times negative 1 squared, which is just 1, minus, or I should say plus 6 times negative 1 which is So, this is 3 minus 6 is negative 3 minus 2 is equal negative 5, and that actually is the vertex. And, you know the formula for the vertex, once again, is negative b over 2 a. So, negative b. That's the coefficient on this term right over here. It's negative 6 over 2 times this one right over here, 2 times 3. 2 times 3, this is equal to negative 1. So, that is the vertex, but let's just keep on going right over here. So, what happens when x is equal to 0? These first two terms are 0, you're just left with a negative 2. When x is equal to positive 1. And, this is where you can see that this is the vertex, and you start seeing the symmetry. If you go one above the vertex, f of x is equal to negative 2. If you go one x value below the vertex, or below the x value of the vertex, f of x is equal to negative 2 again. But, let's just keep going. We could try, let's do one more point over here."
    },
    {
        "Q": "I Am SOOOO Lost?\nCan Anyone Help Me\n0:00 - 4:17\n",
        "A": "ask a question and i think someone will come along.in the meantime watch the math videos and play them more than once if you miss what the person is saying",
        "video_name": "tVDslyeLefU",
        "timestamps": [
            0,
            257
        ],
        "3min_transcript": "So I have this rectangular prism here. It's kind of the shape of a brick or a fish tank, and it's made up of these unit cubes. And each of these unit cubes we're saying is 1/4 of a foot by 1/4 of a foot by 1/4 of a foot. So you could almost imagine that this is-- so let me write it this way-- a 1/4 of a foot by 1/4 of a foot by 1/4 of a foot. Those are its length, height, and width, or depth, whatever you want to call it. So given that, what is the volume of this entire rectangular prism going to be? So I'm assuming you've given a go at it. So there's a couple of ways to think about it. You could first think about the volume of each unit cube, and then think about how many units cubes there are. So let's do that. The unit cube, its volume is going to be 1/4 of a foot times 1/4 of a foot times 1/4 of a foot. 1/4 times 1/4 cubic feet, which is often written as feet to the third power, cubic feet. So 1/4 times 1/4 is 1/16, times 1/4 is 1/64. So this is going to be 1 over 64 cubic feet, or 1/64 of a cubic foot. That's the volume of each of these. That's the volume of each of these unit cubes. Now, how many of them are there? Well, you could view them as kind of these two layers. The first layer has 1, 2, 3, 4, 5, 6, 7, 8. That's this first layer right over here. And then we have the second layer down here, which would be another 8. So it's going to be 8 plus 8, or 16. So the total volume here is going is going to be equal to 16/64 cubic feet, which is the same thing. 16/64 is the same thing as 1/4. Divide the numerator and the denominator by 16. This is the same thing as 1/4 of a cubic foot. And that's our volume. Now, there's other ways that you could have done this. You could have just thought about the dimensions of the length, the width, and the height. The width right over here is going to be 2 times 1/4 feet, which is equal to 1/2 of a foot. The height here is the same thing. So it's going to be 2 times 1/4 of a foot, which is equal to 2/4, or 1/2 of a foot."
    },
    {
        "Q": "why would you divide 8:20 to 2:5 when 20 can divide by more then just 5?\n",
        "A": "8 : 20 is the ratio and when you simplify a ratio you need to take BOTH values into consideration. If you watch some Algebra/ Pre-Algebra concepts you will see that what you do to one side needs to be done to the other to balance it out. 8 : 20 can be simplified no lower than 2 : 5 because the greatest divisor of 8 and 20 is 4 so it ends up as 2 :5. I hope this helped.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            500,
            125
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nHow are fractions and ratios the same? How are they different? Is 2:5 the same as 2/5?",
        "A": "Yes, they are the same exact thing.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            125
        ],
        "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to."
    },
    {
        "Q": "So the first ratio number goes on top of the fraction so 4:6 is 4 over 6?\n",
        "A": "Yes, that is correct.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            246
        ],
        "3min_transcript": "8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to. Or we could say it's 2/5, the fraction 2/5, which would sometimes be read as 2 to 5. This is also, when it's written this way, you could also read that as a ratio, depending on the context. In a sentence like this I would read this as 2/5 of the fruit are apples."
    },
    {
        "Q": "\nSo can 2:3 become 2/3? I'm taking notes...",
        "A": "The bottom number does not (nor should it ever) refer to the whole group. The first and second items (example 2:3) added together will sum to the whole group. Instructions for mixing chemicals are given in this notation.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            123
        ],
        "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to."
    },
    {
        "Q": "So for example does 8 to 8 equal to 8:8 and 8/8?\n",
        "A": "Yes... and if you reduce it, the ratio is 1 to 1",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            488
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nIs 2:3, 4:6, and 8:12 equivalent",
        "A": "Yes, they are equivalent.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            123,
            246,
            492
        ],
        "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to."
    },
    {
        "Q": "\nHow can you get 30:39 equal?",
        "A": "If you mean how do you simplify 30:39, then you just divide by three. 30 divided by 3 is 10 and 39 divided by 3 is 13. Your final simplified ratio is 10 to 13. Both 10 to 13 and 30 to 39 are equal though. Hope this helps! :)",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            1839
        ],
        "3min_transcript": ""
    },
    {
        "Q": "\nAt 7:43 in the video, isn't A really A transpose? And what Sal calls r1T, r2T aren't they really the transpose of Column 1 Column2, so maybe they should be labelled c1T, c2T? Because the r1T suggests the transpose of a row to a column.",
        "A": "@Jonh I believe you right. r1T is in reality c1T, but as siddhantsabo said, the notation used was to point you re dealing now with rows instead of columns.",
        "video_name": "QOTjdgmNqlg",
        "timestamps": [
            463
        ],
        "3min_transcript": "And the way that we can write the row space of A, this thing right here, the row space of A, is the same thing as the column space of A transpose. So one way you can rewrite this sentence right here, is that the null space of A is the orthogonal complement of the row space. The row space is the column space of the transpose matrix. And the claim, which I have not proven to you, is that this is the orthogonal complement of this. This is equal to that, the little perpendicular superscript. That's the claim, and at least in the particular example that actually showed you that 2 by 3 matrix. But let's see if this applies generally. So let me write my matrix A like this. So my matrix A, I can write it as just a bunch of row vectors. But just to be consistent with our notation, with vectors we tend to associate as column vectors, so to represent the row vectors here I'm just going to write them as transpose vectors. Because in our reality, vectors will always be column vectors, and row vectors are just transposes of those. r1 transpose, r2 transpose and you go all the way down. We have m rows. So you're going to get rm transpose. Don't let the transpose part confuse you. I'm just saying that these are row vectors. I'm writing transposes there just to say that, look these represent these rows. But if it's helpful for you to imagine them, just imagine this is the first row of the matrix, this is the second row of that matrix, so on and so forth. Now, what is the null space of A? Well that's all of the vectors here. Let me do it like this. The null space of A is all of the vectors x that satisfy the equation that this is going to be equal to the zero vector. Now to solve this equation, what can we do? We've seen this multiple times. This matrix-vector product is essentially the same thing as saying-- let me write it like this-- it's going to be equal to the zero vector in rm. You're going to have m 0's all the way down to the m'th 0. So another way to write this equation, you've seen it before, is when you take the matrix-vector product, you"
    },
    {
        "Q": "At 1:36, how did you get the side length of 10?\n",
        "A": "It s indicated on the diagram along the bottom edge.",
        "video_name": "mtMNvnm71Z0",
        "timestamps": [
            96
        ],
        "3min_transcript": "- What I want to do in this video is get some practice finding surface areas of figures by opening them up into what's called nets. And one way to think about it is if you had a figure like this, and if it was made out of cardboard, and if you were to cut it, if you were to cut it right where I'm drawing this red, and also right over here and right over there, and right over there and also in the back where you can't see just now, it would open up into something like this. So if you were to open it up, it would open up into something like this. And when you open it up, it's much easier to figure out the surface area. So the surface area of this figure, when we open that up, we can just figure out the surface area of each of these regions. So let's think about it. So what's first of all the surface area, what's the surface area of this, right over here? Well in the net, that corresponds to this area, it's a triangle, it has a base of 12 and height of eight. So this area right over here is going to be one half times the base, so times 12, So this is the same thing as six times eight, which is equal to 48 whatever units, or square units. This is going to be units of area. So that's going to be 48 square units, and up here is the exact same thing. That's the exact same thing. You can't see it in this figure, but if it was transparent, if it was transparent, it would be this backside right over here, but that's also going to be 48. 48 square units. Now we can think about the areas of I guess you can consider them to be the side panels. So that's a side panel right over there. It's 14 high and 10 wide, this is the other side panel. It's also this length over here is the same as this length. It's also 14 high and 10 wide. So this side panel is this one right over here. And then you have one on the other side. And so the area of each of these 14 times 10, they are 140 square units. This one is also 140 square units. the area of I guess you can say the base of the figure, so this whole region right over here, which is this area, which is that area right over there. And that's going to be 12 by 14. So this area is 12 times 14, which is equal to let's see. 12 times 12 is 144 plus another 24, so it's 168. So the total area is going to be, let's see. If you add this one and that one, you get 96. 96 square units. The two magenta, I guess you can say, side panels, 140 plus 140, that's 280. 280. And then you have this base that comes in at 168. We want it to be that same color. 168. One, 68. Add them all together, and we get the surface area for the entire figure. And it was super valuable to open it up into this net"
    },
    {
        "Q": "\nAt 3:50, is (350/50)=(350/10)/(50/10)=35/5=7 correct?",
        "A": "That is right. When you get rid of the zeroes, it is the same as 35/7.",
        "video_name": "ccS5Fy5yLjk",
        "timestamps": [
            230
        ],
        "3min_transcript": "well this fraction bar right here is the same as the division sign up here. 350 divided by 50 is the same as 350 over 50. And, when we have a fraction like this fraction down here, we can simplify it. In this case, because there's zeroes on the end, we know they're both multiples of 10, so we can divide them both by 10. We can divide our numerator and our denominator by 10. And, when we divide whole numbers by 10, we have a trick we can use, a pattern, really, which is that the whole number, in this case 350, when it's divided by 10 we drop a zero from the end. So, 350 divided by 10 is 35. 350 could be divided into 10 groups of 35. When we divided 50 by 10, we drop that zero off the end, or another way to think about it is 50 divided into groups of 10 would make five groups. And, then we end up with our simplified fraction of 35 fifths or 35 divided by five, which is the same thing here. So, in both of these cases, we can see that 350 divided by 50 is the same as 35 divided by 5. So, when both whole numbers, when we're dividing whole numbers and they both end in zeroes, we can cancel those zeroes. Basically, we're factoring out a 10. We're taking the 10, the divided by 10 out of both of them, out of both numbers. So, we can cancel those zeroes which leaves us with smaller numbers and at least for me, I find division a lot simpler when I'm working with smaller numbers. Let's try one like 420 divided by 70. So, we can see we have two whole numbers, both end in zero. So, we're gonna cancel those zeroes, basically we're dividing a ten out of both numbers and end up with a simpler division problem of 42 divided by seven. 42 divided by seven equals six, therefore 420 divided by 70 also equals six. And, here's one last one. What is we had 5600 divided by 80? So, the first thing I notice is I have zeroes at the end"
    },
    {
        "Q": "\nSal, at 0:42, how do you know that a-sub2 is equal to -8/5, why is it not a-sub1? Why did you start with 2 instead of 1?",
        "A": "Because i = 2 in the Sigma statement.",
        "video_name": "yvddTWa9ptU",
        "timestamps": [
            42
        ],
        "3min_transcript": "Let's say that we're told that this sum right over here, where our index starts at 2 and we go all the way to infinity, that this infinite series is negative 8/5 plus 16/7 minus 32/9 plus-- and we just keep going on and on forever. And so what I want to do is to explicitly define what a sub n is here. So right now we just say, hey, if you take the sum of a sub n from n equals 2 to infinity, it turns out you get this sum right over here. But let's think about what a sub n-- how we can actually define it in terms of n. And I encourage you to pause the video right now and try it on your own. So the first thing that you might realize is, well, this is the number that we're going to get. Let me write it this way. a sub 2 is equal to negative 8/5. a sub 3 is equal to 16/7. a sub 4 is equal to negative 32/9. Negative 8/5 is the same thing as negative 8 over 5. Let me make that a little bit clearer. So I'll make that a little bit clearer. So this is negative 8/5. Obviously, this is positive, so I don't have to really worry about it too much. And then here, I'm just saying negative 32/9, so it's the same thing as negative 32 over 9. So let's see if we can first find a pattern in the numerator. So when we go from negative 8 to 16, what's happening? Well, we're multiplying by negative 2. Now, to go from 16 to negative 32, we're multiplying by negative 2 again. So you might say, OK, well, whatever we have in the numerator must be a power of negative 2. And, all right, if you say, well, maybe this is negative 2 squared, well, you know that negative 8 isn't negative 2 squared. Negative 2 squared is equal to positive 4. Negative 8, that is equal to negative 2 to the third power. 16 is equal to negative 2 to the fourth power. Negative 32 is equal to negative 2 to the fifth power. So notice, our exponent on the negative 2 is always going to be one more than our index. Our index is 2, our exponent is 3. Our index is 3, our exponent is 4. Our index is 4, our exponent is 5. So that gives a sense that at least the numerator is going to be-- whatever our index is, it's going to be-- so let me write this down. So a sub n is equal to-- well, it's going to be negative 2 to whatever index we're at, to that index plus 1 power. So that's a reasonable way to think about our numerator."
    },
    {
        "Q": "\nSee 1:47 into the video. Referring to the denominator, why is it 0-(-a) and not (-a)-0. Thanks ;)",
        "A": "You could do that, but the answer would be the same. For example, if the 2 points were (9,2) and (7,5), 5-2/7-9 would equal 3/-2, which equals -(3/2.) If you switched the order, it would be 2-5/9-7, which is -3/2. The answers are the same. It does not matter which you put first, as long as when you put the y value of it first, then that point s x value has to be in front too.",
        "video_name": "H6ZNLD1AeM8",
        "timestamps": [
            107
        ],
        "3min_transcript": "Consider the graph of the function f of x that passes through three points as shown. So these are the three points, and this curve in blue is f of x. Identify which of the statements are true. So they give us these statements. This first one says f of negative a is less than 1 minus f of negative a over a. So this seems like some type of a bizarre statement. How are we able to figure out whether this is true from this right over here? So let's just go piece by piece and see if something starts to make sense. So f of negative a, where do we see that here? Well, this is f of negative a. This is the point x equals negative a. So this is negative a, and this is y is equal to f of negative a. So this is f of negative a right over here. And what we know about f of negative a, based on looking at this graph, is that f of negative a is between 0 and 1. So we can write that. 0 is less than f of negative a, which is less than 1. So that's all I can deduce about f of negative a right Now let's look at this crazy statement, 1 minus f of negative a over a. What is this? Well, let's think about what happens if we take the secant line, if we're trying to find the slope of the secant line, between this point and this point, if we wanted to find the average rate of change between the point negative a, f of negative a, and the point 0, 1. If this is our endpoint, our change in y is going to be 1 minus f of negative a. So 1 minus f of negative a is equal to our change in y. And our change in x, going from negative a to 0, so change in x is going to be equal to 0 minus negative a, which is equal to positive a. So this right over here is essentially our change in x over our change in y from this point to this point. It is our average rate of change from this point to this point. or you could say it's the slope of the secant line. So the secant line would look something like this. Slope of the secant line, so this right over here is slope of secant line between from f of negative a to 0 comma 1. So just looking at this diagram right over here, what do we know about this slope? And in particular, can we make any statements about that slope relative to, say, 0 or 1 or anything like that? Well, let's think about what a line of slope 1 would look like. Well, a line of slope 1, especially one that went through this point right over here, would look something like this. A line of slope 1 would look something like this. So this line right over here that I've just"
    },
    {
        "Q": "\nIn the example at 5:50 wouldn't the 2a affect the actual slope of the line because it is not a quadratic?",
        "A": "I m not quite sure what you mean, but we are only comparing the slope of the secant lines, not the actual slope of the curve. The slope of the secant line is \u00ce\u0094y/\u00ce\u0094x",
        "video_name": "H6ZNLD1AeM8",
        "timestamps": [
            350
        ],
        "3min_transcript": "that we just looked at. So this is the same value right over here. So we're comparing this slope right over here, to what's this? f of a minus 1 over a. Well, this is the slope of this secant line, this is the slope of this secant line that I'm drawing in this-- let me do it in more contrast. Let me do it in orange. That is the slope of this secant line. So which one has a higher slope? Well, it's pretty clear that the blue secant line has a higher slope than this orange secant line. But here, it's saying that the blue's slope is lower than the orange. So this is not going to be true. So this is not true. Then finally, let's look at this over here-- f of a minus f of negative a over 2a. So this is the slope. Let me draw this. So this right over here, this is the slope of the secant line between this point Our change in y is f of a minus f of negative a. Our change in x is a minus negative a, which is 2a. So this is this secant line right over here. So let me draw it. So this secant line right over here, so they're comparing that slope to this slope. f of a minus 1 is our change in y, over a is our change in x. So we're comparing it to that one right over there. And you could immediately eyeball this kind of brownish, maroon-- I guess it's kind of a brown color-- this secant line that goes all the way from here to here is clearly steeper than this one right over here. And we know that, that the average rate of change from here to here is going to be higher than the average rate of change from here to here, because at least from negative a to 0, we were increasing at a much faster rate. And then we slowed down to this rate. So the average over the entire interval is definitely going to be more than what we get from 0 to a. This has a higher-- we actually know that this is false. Both of these would have been true if we swapped these signs around, if this was a greater than sign, if this was a greater than sign. So this is the only one that applies."
    },
    {
        "Q": "At 2:59 min, why is the \"divided by 2\" switched with \"times 1/2\"?\n",
        "A": "1/2 and divided by 2 are both 0.5 A Half can be written... As a fraction: 1/2 As a decimal: 0.5 As a percentage: 50%",
        "video_name": "GwycEivqYYI",
        "timestamps": [
            179
        ],
        "3min_transcript": "The figure is 13.09 centimeters. So let's say that this is the figure right over here. So this is the figure that she wants to place in the middle of the page. So this is 13.09 centimeters. So the question is, how much margin should she leave on the left so that the figure is centered? So they're really asking, what should this distance be here so that the figure is centered? And the key thing is, if the figure is centered, this distance is going to be the same as this distance, so that the figure is right in the middle. Whatever this is, it's going to be the same as this. So one way we could think about it-- we could say, OK, let's take 13.09 centimeters from 21.59 And if we do that, then we'll know how much leftover space this plus this combined is. We want half of it to go on the right side, and we want half of it to go on the left side. So let's see how much this left margin and the right margin combined need to be. And that's just 21.59. That's just going to be 21.59 centimeters minus 13.09 centimeters. And this, we get to-- let's see. This is 8.50 centimeters, which is the combined margins. Not just the left or not just the right-- this is the left plus the right margin, is 8.50 centimeters. This is this distance plus this distance. Now, if we want to figure out what the left margin should be, if we want to center it. Well, we just want to split this evenly between the left and the right margin. So we just want to divide this by 2. So if you divide 18.50 by 2-- or multiply it by 1/2, I guess. So if we multiply this by 1/2, you are left with 1/2 of 8.50, is 4.25 centimeters. So if we want it centered, this is going to be 4.25 centimeters. And of course, this is going to be 4.25 centimeters, as well. And you see, when you add 4.25 to 4.25, you get 8.5 centimeters. And then you add that to 13.09. You get the entire width of the piece of paper. So how much margin should she leave on the left? 4.25 centimeters."
    },
    {
        "Q": "\nAt 0:36 Sal starts to extend the number. Do I need to do that all the time?",
        "A": "No, that s only to help you understand the concept. If you get it, that s great. =) You may have to do this when asked to expand the number.",
        "video_name": "jxA8MffVmPs",
        "timestamps": [
            36
        ],
        "3min_transcript": "Find the place value of 3 in 4,356. Now, whenever I think about place value, and the more you do practice problems on this it'll become a little bit of second nature, but whenever I see a problem like this, I like to expand out what 4,356 really is, so let me rewrite So if I were to write it-- and I'll write it in different colors. So 4,356 is equal to-- and just think about how I just said it. It is equal to 4,000 plus 300 plus 50 plus 6. And you could come up with that just based on how we said it: four thousand, three hundred, and fifty-six. Now another way to think about this is this is just like saying this is 4 thousands plus-- or you could even think as 5 tens plus 6. And instead of 6, we could say plus 6 ones. And so if we go back to the original number 4,356, this is the same thing as 4-- I'll write it down. Let me see how well I can-- I'll write it up like this. This is the same thing is 4 thousands, 3 hundreds, 5 tens and then 6 ones. So when they ask what is the place value of 3 into 4,356, we're concerned with this 3 right here, It's in the hundreds place. If there was a 4 here, that would mean we're dealing with 4 hundreds. If there's a 5, 5 hundreds. It's the third from the right. This is the ones place. That's 6 ones, 5 tens, 3 hundreds. So the answer here is it is in the hundreds place."
    },
    {
        "Q": "At 3:40 , i didn't understand why sal used that formula ? May i know which video can i look upon for the formula ?\n",
        "A": "Finding the sum of n squares part 1/ part 2",
        "video_name": "LwhJVURumAA",
        "timestamps": [
            220
        ],
        "3min_transcript": "when this is 7, all the way to 14. You could factor out a 2. And so this is going to become 2 times 1 plus 2 plus 3 all the way to 7. And so you can rewrite this piece right over here as 2 times the sum-- so we're essentially just factoring out the 2-- 2 times the sum, which is the sum from n equals 1 to 7 of n. So this is this piece. We still have this 28 that we have to add. So we have this 28. And we draw the parentheses so you don't think that the 28 is part of this right over here. And now we can do the same thing with this. 3 times n-- we're taking from n equals 1 to 7 of 3 n squared. Doing the same exact thing as we just did in magenta, this is going to be equal to 3 times the sum from n We're essentially factoring out the 3. We're factoring out the 2. n squared. And once again, we can put parentheses just to clarify things. Now, at this point, there are formulas to evaluate each of these things. There's a formula to evaluate this thing right over here. There's a formula to evaluate this thing over here. And you can look them up. And actually, I'll give you the formulas, in case you're curious. This formula, one expression of this formula is that this is going to be n to the third over 3 plus n squared over 2 plus n over 6. That's one formula for that. And one formula for this piece right over here, going from n equals 1 to 7-- sorry. Let me make it clear. This n is actually what your terminal value should be. So this should be 7 to the third power over 3-- I was just mindlessly using the formula-- 7 to the third over 3 plus 7 squared over 2 plus 7/6. So that's this sum. And this sum, you could view it as the average of the first and the last terms. So the first term is 1. The last term is 7. So take their average and then multiply it times the number of terms you have. So times-- you have 7 terms. So what is this middle one going to evaluate to? Well, 1 times-- and of course, we have this 2 out front. This green is just this part right over here. So you have 2 times this. And over here, you have 3 times this business right over here. So if we evaluate this one, 2 times-- let's see. 1 plus 7 is 8, divided by 2 is 4. 4 times 2 is 8. Times 7, it's 56. So that becomes 56. Now, this-- let's see."
    },
    {
        "Q": "at 1:15 you say that the domain is 6. but when you substitute x as 6 in the numerator as well you get\n6^2 - 36/6 - 6\n=36 - 36/6 - 6\n=0/0 = 0\n",
        "A": "The domain is not 6. In fact, the domain is all real numbers EXCEPT 6, because 6 doesn t work in the expression (leads to division by zero). In the simplification you just did, at the end, 0/0 = undefined or indeterminate, 0/0 does not equal 0.",
        "video_name": "ey_b3aPsRl8",
        "timestamps": [
            75
        ],
        "3min_transcript": "Simplify the rational expression and state the domain. Let's see if we can start with the domain part of the question, if we can start with stating the domain. Now, the domain is the set of all of the x values that you can legitimately input into this if you view this as a function, if you said this is f of x is equal to that. The domain is a set of all x values that you could input into this function and get something that is well-defined. The one x value that would make this undefined is the x value that would make the denominator equal 0-- the x value that would make that equal 0. So when does that happen? Six minus x is equal to 0. Let's add x to both sides. We get 6 is equal to x, so the domain of this function is equal to the set of all real numbers except 6. is 6 then you're dividing by 0, and then this expression is undefined. We've stated the domain, now let's do the simplifying the rational expression. Let me rewrite it over here. We have x squared minus 36 over 6 minus x. Now, this might jump out at you immediately, as it's that special type of binomial. It's of the form a squared minus b squared, and we've seen this multiple times. This is equivalent to a plus b times a minus b. And in this case, a is x and b is 6. This top expression right here can be factored as x plus 6 times x minus 6, all of that over 6 minus x. Now, at first you might say, I have a x minus 6 and a 6 minus x. Those aren't quite equal, but what maybe will jump out at Try it out. Let's multiply by negative 1 and then by negative 1 again. Think of it that way. If I multiply by negative 1 times negative 1, obviously, I'm just multiplying the numerator by 1, so I'm not in any way changing the numerator. What happens if we just multiply the x minus 6 by that first negative 1? What happens to that x minus 6? Let me rewrite the whole expression. We have x plus 6, and I'm going to distribute this negative 1. If I distribute the negative 1, I have negative 1 times x is negative x. Negative 1 times negative 6 is plus 6. And then I have a negative 1 out here. I have a negative 1 times negative 1, and all of that is over 6 minus x. Now, negative plus 6. This is the exact same thing as 6 minus x if you just rearrange the two terms. Negative x plus 6 is the same"
    },
    {
        "Q": "At 1:52 how would you solve the equation if you left in the 75?\n",
        "A": "You would factor the expression. You need to find two numbers that add up to 10 and multiply to equal -75. Those two numbers are 15 and -5. Thus, x^2+10x -75 = (x+15)(x-5). Now, solving either bracket for x gives you x = -15 and x = 5.",
        "video_name": "TV5kDqiJ1Os",
        "timestamps": [
            112
        ],
        "3min_transcript": "We're asked to complete the square to solve 4x squared plus 40x minus 300 is equal to 0. So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that, because as long as whatever I do to the left hand side, I also do the right hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x. And then 300 divided by 4 is what? That is 75. Let me verify that. 7 times 4 is 28. You subtract, you get a remainder of 2. Bring down the 0. 4 goes into 20 five times. 5 times 4 is 20. Subtract zero. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way. But it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5. And 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left hand side into a perfect square. And I'm going to start out by kind of getting this 75 out You'll sometimes see it where people leave the 75 on the left hand side. I'm going to put on the right hand side just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left hand side of the equation. plus 75. Those guys cancel out. And I'm going to leave some space here, because we're going to add something here to complete the square that is equal to 75. So all I did is add 75 to both sides of this equation. Now, in this step, this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left hand side becomes a perfect square. And the way we can do that, and saw this in the last video where we constructed a perfect square trinomial, is that this last term-- or I should say, what we see on the left hand side, not the last term, this expression on the left hand side, it will be a perfect square if we have a constant term that is the square of half of the coefficient on the first degree So the coefficient here is 10. Half of 10 is 5."
    },
    {
        "Q": "\nAt 3:40, why did he color in 8/15 instead of 10/15 because there is still 2/15 remaining on the side.?",
        "A": "He wasn t dealing with the whole 15/15 but rather 4/5 of it = 12/15 2/3 x 4/5 = 8/15",
        "video_name": "hr_mTd-oJ-M",
        "timestamps": [
            220
        ],
        "3min_transcript": "to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5. So how can we think about that? Well, we could take this section and divide it into thirds. So let's do that. Divide it into thirds. So we're going divide it into 3 equal sections. So that's 1/3, and then 2/3. So we took each of the 5 equal sections, and we divided them into 3 equal sections. Now what's going to be 2/3 of the 4/5? Well, that's going to be this part right over here. So let me make this clear. This is 1/3 of the 4/5. And then this would be 2/3 of the 4/5. So this right over here, would be 2/3 of the 4/5, or 2/3 times 4/5. But what fraction of the whole does that represent? Well, how many total, how many total equal sections do we now have? Well, we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, So we have 15 equal sections. I'm using a new color. We have 15 equal sections, and that make sense. We started with 5 equal sections, but then we divided each of those into 3 equal sections. So now we have 5 times 3 total equal sections. And then how many of those are now colored in? Well, we see it's 2 times 4. 1, 2, 3, 4, 5, 6, 7, 8. How many of them are in the 2/3 of the 4/5, I should say. And there's 8 of them, 8 of the 15 equals sections. And so there you have it. It should hopefully now make visual sense, or it makes conceptual sense, that 2/3 times 4/5-- you can obviously compute it by just multiplying the numerators, 2 times 4 is 8. And then multiplying the denominators, 3 times 5 is 15-- but hopefully this now makes conceptual sense as 2/3 of 4/5."
    }
]